lab a and lab b computer science work.You can use visual studio 2019 first then copy paste to word doc
CSCI110
– Fundamentals of Computer Science
Homework Assignments
Winter 2022
Instructions for your lab report.
1. You need to create a cover page formatted as following:
(Notes: The cover page needs to be typed and printed)
CSCI110 – Fundamentals of Computer Science
MT SAC College
CSCI110
Lab #: __________________
Description: ______________________
______________________
______________________
Due Date: ___________________
Name: _______________________
Grade: _______________________
Notes: ______________________
CSCI110 – Fundamentals of Computer Science
Requirements for your lab reports
Your submission should include:
1. The cover page
2. The listing of the source code (See details shown below)
3. A screenshot of your program execution
The source code should be organized and presented as:
1. Prolog
a. Program Description
b. Author
c. Input variables
d. Process Flow
e. Output variables
2. A listing of source code with internal comments
Programming requirements:
1. Your program needs to be user-friendly and easy to understand.
2. You need to follow the book’s and my instructions to code your program – no deviation. If you have any questions, please let me know.
CSCI110 – Fundamentals of Computer Science
Homework Assignments
Winter 2022
Lab 1A
Requirements:
Please work on p1.6 on page 26.
Test cases:
1. Please produce the report based on the book’s instructions.
1. Initial principal = 10000 (Please prompt the user for this amount)
1. Print ‘”New balance = “ for 3 years.
Due date: January 11, 2022
Submission includes 1) cover page 2) source listing 3) screen shot of program execution.
CSCI110 – Fundamentals of Computer Science
Lab 1B
Requirements:
Please work on p2.21 on page 69. You need to work on both conditions – the first time is earlier than the second time and the first time is later than the second time. There is no extra credit given for this lab.
Test cases:
1. You need to test both conditions mentioned in the requirements.
1. You need to print a user-friendly message showing the difference between these two times. Do not follow the book by simply printing “8 hours and 30 minutes”.
Due date: January 11, 2022
Submission includes 1) cover page 2) source listing 3) screen shot of program execution.
5
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C + + f o r
E v E r y o n E
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C + + f o r
E v E r y o n E
S e c o n d e d i t i o n
cay Horstmann
San Jose State University
John Wiley & Sons, Inc.
cfe2_fm_pi_xxvi.indd 3 10/28/10 5:08 PM
VICE PRESIDENT AND EXECUTIVE PUBLISHER Don Fowley
EXECUTIVE EDITOR Beth Lang Golub
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COVER PHOTO © Ricardo Azoury/iStockphoto
This book was set in Stempel Garamond by Publishing Services, and printed and bound by RRD Jefferson City.
The cover was printed by RRD Jefferson City.
This book is printed on acid-free paper. ∞
Founded in 1807, John Wiley & Sons, Inc. has been a valued source of knowledge and understanding for more
than 200 years, helping people around the world meet their needs and fulfill their aspirations. Our company is
built on a foundation of principles that include responsibility to the communities we serve and where we live
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Copyright © 2012, 2009 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be re-
produced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, pho-
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Library of Congress Cataloging in Publication Data:
Horstmann, Cay S., 1959-
C++ for everyone / Cay S. Horstmann. — 2nd ed.
p. cm.
Includes index.
ISBN 978-0-470-92713-7 (pbk.)
1. C++ (Computer program language) I. Title.
QA76.73.C153H6685 2010
005.13’3–dc22
2010039907
ISBN 978-0-470-92713-7 (Main Book)
ISBN 978-0-470-92092-3 (Binder-Ready Version)
Printed in the United States of America
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P r E fa C E
v
This book is an introduction to C++ and computer programming that focuses on the
essentials—and on effective learning. The book is designed to serve a wide range of
student interests and abilities and is suitable for a first course in programming for
computer scientists, engineers, and students in other disciplines. No prior program-
ming experience is required, and only a modest amount of high school algebra is
needed. Here are the key features of this book:
Guidance and worked examples help students succeed.
Beginning programmers often ask “How do I start? Now what do I do?” Of course,
an activity as complex as programming cannot be reduced to cookbook-style instruc-
tions. However, step-by-step guidance is immensely helpful for building confidence
and providing an outline for the task at hand. “Problem Solving” sections stress the
importance of design and planning. “How To” guides help students with common
programming tasks. Additional Worked Examples are available online.
Practice makes perfect.
Of course, programming students need to be able to implement nontrivial programs,
but they first need to have the confidence that they can succeed. This book contains a
substantial number of self-check questions at the end of each section. “Practice It”
pointers suggest exercises to try after each section. At the end of each chapter, you
will find a great variety of programming assignments, ranging from simple practice
problems to realistic applications.
teach computer science principles, not just c++ or object-orientation.
This book uses the C++ programming language as a vehicle for introducing com-
puter sci ence concepts. A substantial subset of the C++ language is covered, focusing
on the modern features of standard C++ that make students productive. The book
takes a traditional route, stressing control structures, procedural decomposition, and
array algorithms, before turning to the design of classes in the final chapters.
A visual approach motivates the reader and eases navigation.
Photographs present visual analogies that explain the
nature and behavior of computer concepts. Step-by-
step figures illustrate complex program operations.
Syntax boxes and example tables present a variety
of typical and special cases in a compact format. It
is easy to get the “lay of the land” by browsing the
visuals, before focusing on the textual material.
Focus on the essentials while being
technically accurate.
An encyclopedic coverage is not helpful for a begin-
ning programmer, but neither is the opposite—
reducing the material to a list of simplistic bullet points. In this book, the essentials
are presented in digestible chunks, with separate notes that go deeper into good prac-
tices or language features when the reader is ready for the additional information.
You will not find artificial over-simplifications that give an illusion of knowledge.
Visual features help the reader
with navigation.
cfe2_fm_pi_xxvi.indd 5 10/28/10 5:08 PM
vi Preface
new to This Edition
Problem Solving Strategies
This edition adds practical, step-by-step illustrations of techniques that can help stu-
dents devise and evaluate solutions to programming problems. Introduced where
they are most relevant, these strategies address barriers to success for many students.
Strategies included are:
• Algorithm Design (with pseudocode)
• First Do It By Hand (doing sample
calculations by hand)
• Flowcharts
• Test Cases
• Hand-Tracing
• Storyboards
• Reusable Functions
• Stepwise Refinement
• Adapting Algorithms
• Discover Algorithms by Manipulat-
ing Physical Objects
• Draw a Picture (pointer diagrams)
• Tracing Objects (identifying state and
behavior)
• Discovering Classes
optional Engineering Exercises
End-of-chapter exercises have been enhanced with problems from scientific and
engineering domains. Geared to students learning C++ for a technical major, the
exercises are designed to illustrate the value of programming in those fields. Addi-
tional exercises are available on the book’s companion web site.
new and reorganized Topics
All chapters were revised and enhanced to respond to user feedback and improve the
flow of topics. Loop algorithms are now introduced explicitly in Chapter 4. Debug-
ging is now introduced in a lengthy Worked Example in Chapter 5. Arrays are cov-
ered before vectors are introduced in Chapter 6, and a new section on vector algo-
rithms builds on the array algorithms presented earlier in the chapter. A new optional
section on structure types is now in Chapter 7. New example tables, photos, and
exercises appear throughout the book.
a Tour of the Book
The core material of the book is:
Chapter 1. Introduction
Chapter 2. Fundamental Data Types
Chapter 3. Decisions
Chapter 4. Loops
Chapter 5. Functions
Chapter 6. Arrays and Vectors
In a course for engineers with a need for systems and embedded programming, you
will want to cover Chapter 7 on pointers. Sections 7.1 and 7.4 are sufficient for using
pointers with polymorphism in Chapter 10.
cfe2_fm_pi_xxvi.indd 6 10/28/10 5:08 PM
Preface vii
File processing is the subject of Chapter 8. Section 8.1 can be covered sooner for
an intro duction to reading and writing text files. The remainder of the chapter gives
addi tional material for practical applications.
Chapters 9 and 10 introduce the object-oriented features of C++. Chapter 9
introduces class design and implementation. Chapter 10 covers inheritance and
polymorphism.
Four additional chapters are available on the Web. They can be used individu-
ally for a capstone chapter, or they can be combined for teaching a two-semester
course. (They can also be incorporated into a custom print version of the text; ask
your Wiley sales representative for details.)
Chapter 11. Recursion
Chapter 12. Sorting and Searching
Chapter 13. Lists, Stacks, and Queues
Chapter 14. Sets, Maps, and Priority Queues
Figure 1 shows the dependencies between the chapters.
Figure 1
Chapter Dependencies
Section 8.1
contains the core
material
Sections
7.1 and 7.4 are
required
A gentle
introduction to recursion
is optional
1. Introduction
2. Fundamental
Data Types
3. Decisions
4. Loops
5. Functions
6. Arrays
and Vectors
9. Classes 11. Recursion 12. Sorting and
Searching
8. Streams
13. Lists, Stacks,
and Queues
14. Sets, Maps,
Priority Queues
7. Pointers
10. Inheritance
Fundamentals
Online
cfe2_fm_pi_xxvi.indd 7 10/28/10 5:08 PM
viii Walkthrough
a Walkthrough of the Learning aids
The pedagogical elements in this book work together to focus on and reinforce key
concepts and fundamental principles of programming, with additional tips and detail
organized to support and deepen these fundamentals. In addition to traditional fea-
tures, such as chapter objectives and a wealth of exercises, each chapter contains ele-
ments geared to today’s visual learner.
4.3 The for Loop 143
4.4 The for Loop
It often happens that you want to execute a sequence of statements a given number of
times. You can use a while loop that is controlled by a counter, as in the following
example:
counter = 1; // Initialize the counter
while (counter <= 10) // Check the counter
{
cout << counter << endl;
counter++; // Update the counter
}
Because this loop type is so common, there is a special form for it, called the for loop
(see Syntax 4.2).
for (counter = 1; counter <= 10; counter++)
{
cout << counter << endl;
}
Some people call this loop count-controlled. In contrast,
the while loop of the preceding section can be called an
event-controlled loop because it executes until an event
occurs (for example, when the balance reaches the target).
Another commonly-used term for a count-controlled
loop is definite. You know from the outset that the loop
body will be executed a definite number of times––ten
times in our example. In contrast, you do not know how
many iterations it takes to accumulate a target balance.
Such a loop is called indefinite.
The for loop neatly groups the initialization, condi-
tion, and update expressions together. However, it is
important to realize that these expressions are not exe-
cuted together (see Figure 3).
The for loop is
used when a
value runs from a
starting point to an
ending point with a
constant increment
or decrement.
You can visualize the
for loop as an orderly
sequence of steps.
Syntax 4.2 for Statement
for (int i = 5; i <= 10; i++)
{
sum = sum + i;
}
This loop executes 6 times.
See page 147.
These three
expressions should be related.
See page 147.
This initialization
happens once
before the loop starts.
The loop is
executed while
this condition is true.
This update is
executed after
each iteration.
The variable i is
defined only in this for loop.
See page 144.
Annotations explain
required components
and point to more information
on common errors or best practices
associated with the syntax.
Throughout each chapter,
margin notes show where
new concepts are introduced
and provide an outline of key ideas.
Annotated syntax boxes
provide a quick, visual overview
of new language constructs.
Like a variable in a computer
program, a parking space has
an identifier and a contents.
Analogies to everyday objects are
used to explain the nature and behavior
of concepts such as variables, data
types, loops, and more.
cfe2_fm_pi_xxvi.indd 8 10/28/10 5:08 PM
Walkthrough ix
6.5 Problem Solving: Discovering Algorithms by Manipulating Physical Objects 277
Now how does that help us with our problem, switching the first and the second
half of the array?
Let’s put the first coin into place, by swapping it with the fifth coin. However, as
C++ programmers, we will say that we swap the coins in positions 0 and 4:
Table 3 Variable Names in C++
Variable Name Comment
can_volume1 Variable names consist of letters, numbers, and the underscore
character.
x In mathematics, you use short variable names such as x or y. This is
legal in C++, but not very common, because it can make programs
harder to understand (see Programming Tip 2.1 on page 38).
! Can_volume Caution: Variable names are case-sensitive.
This variable name is different from can_volume.
6pack Error: Variable names cannot start with a number.
can volume Error: Variable names cannot contain spaces.
double Error: You cannot use a reserved word as a variable name.
ltr/fl.oz Error: You cannot use symbols such as / or.
Step 1 Determine the inputs and outputs.
In our sample problem, we have these inputs:
• purchase price1 and fuel efficiency1, the price and fuel efficiency (in mpg) of the first car
• purchase price2 and fuel efficiency2, the price and fuel efficiency of the second car
We simply want to know which car is the better buy. That is the desired output.
H O W T O 1 . 1 Describing an Algorithm with Pseudocode
Before you are ready to write a program in C++, you need to develop an algorithm—a method
for arriving at a solution for a particular problem. Describe the algorithm in pseudocode: a
sequence of precise steps formulated in English.
For example, consider this problem: You have the choice of
buying two cars. One is more fuel efficient than the other, but also
more expensive. You know the price and fuel efficiency (in miles
per gallon, mpg) of both cars. You plan to keep the car for ten
years. Assume a price of $4 per gallon of gas and usage of 15,000
miles per year. You will pay cash for the car and not worry about
financing costs. Which car is the better deal?
WO R K E D E X A M P L E 1 . 1 Writing an Algorithm for Tiling a Floor
This Worked Example shows how to develop an algorithm for laying
tile in an alternating pattern of colors.
Memorable photos reinforce
analogies and help students
remember the concepts.
Example tables support beginners
with multiple, concrete examples.
These tables point out common
errors and present another quick
reference to the section’s topic.
Problem Solving sections teach
techniques for generating ideas and
evaluating proposed solutions, often
using pencil and paper or other
artifacts. These sections emphasize
that most of the planning and problem
solving that makes students successful
happens away from the computer.
How To guides give step-by-step
guidance for common programming
tasks, emphasizing planning and
testing. They answer the beginner’s
question, “Now what do I do?” and
integrate key concepts into a
problem-solving sequence.
Worked Examples apply
the steps in the How To to
a different example, showing
how they can be used to
plan, implement, and test
a solution to another
programming problem.
A recipe for a fruit pie may say to use any kind of fruit.
Here, “fruit” is an example of a parameter variable.
Apples and cherries are examples of arguments.
pie(fruit) pie(fruit)
Next, we swap the coins in positions 1 and 5:
cfe2_fm_pi_xxvi.indd 9 10/28/10 5:08 PM
x Walkthrough
•
Figure 3 Parameter Passing
1 Function call result1 =
side_length =
2 Initializing function parameter variable result1 =
side_length = 2
3 About to return to the caller result1 =
side_length =
volume = 8
2
4 After function call result1 = 8
double result1 = cube_volume(2);
double volume = side_length * side_length * side_length;
return volume;
double result1 = cube_volume(2);
double result1 = cube_volume(2);
The parameter variable side_length of the cube_volume function is created. 1
• The parameter variable is initialized with the value of the argument that was
passed in the call. In our case, side_length is set to 2. 2
• The function computes the expression side_length * side_length * side_length,
which has the value 8. That value is stored in the variable volume. 3
• The function returns. All of its variables are removed. The return value is trans-
ferred to the caller, that is, the function calling the cube_volume function. 4
11. Write the for loop of the invtable.cpp program as a while loop.
12. How many numbers does this loop print?
for (int n = 10; n >= 0; n–)
{
cout << n << endl;
}
13. Write a for loop that prints all even numbers between 10 and 20 (inclusive).
14. Write a for loop that computes the sum of the integers from 1 to n.
15. How would you modify the for loop of the invtable.cpp program to print all bal-
ances until the invest ment has doubled?
Practice It Now you can try these exercises at the end of the chapter: R4.2, R4.7, P4.12.
S E L F C H E C K
ch05/cube.cpp
1 #include
2
3 using namespace std;
4
5 /**
6 Computes the volume of a cube.
7 @param side_length the side length of the cube
8 @return the volume
9 */
10 double cube_volume(double side_length)
11 {
12 double volume = side_length * side_length * side_length;
13 return volume;
14 }
15
16 int main()
17 {
18 double result1 = cube_volume(2);
19 double result2 = cube_volume(10);
20 cout << "A cube with side length 2 has volume " << result1 << endl;
21 cout << "A cube with side length 10 has volume " << result2 << endl;
22
23 return 0;
24 }
Program Run
A cube with side length 2 has volume 8
A cube with side length 10 has volume 1000
Figure 3
Execution of a
for Loop
for (counter = 1; counter <= 10; counter++)
{
cout << counter << endl;
}
Initialize counter1
for (counter = 1; counter <= 10; counter++)
{
cout << counter << endl;
}
Check condition2
for (counter = 1; counter <= 10; counter++)
{
cout << counter << endl;
}
Execute loop body3
for (counter = 1; counter <= 10; counter++)
{
cout << counter << endl;
}
Update counter4
for (counter = 1; counter <= 10; counter++)
{
cout << counter << endl;
}
Check condition again5
counter = 1
counter = 1
counter = 1
counter = 2
counter = 2
Progressive figures trace code
segments to help students visualize
the program flow. Color is used
consistently to make variables and
other elements easily recognizable.
Optional engineering exercises
engage students with applications
from technical fields.
Program listings are carefully
designed for easy reading,
going well beyond simple
color coding. Functions are set
off by a subtle outline.
Self-check exercises at the
end of each section are designed
to make students think through
the new material—and can
spark discussion in lecture.
Engineering P7.20 Write a program that simulates the control
software for a “people mover” system, a set of
driverless trains that move in two concentric
circular tracks. A set of switches allows trains
to switch tracks.
In your program, the outer and inner tracks
should each be divided into ten segments.
Each track segment can contain a train that
moves either clockwise or counterclockwise.
A train moves to an adjacent segment in its track or, if that segment is occupied, to
the adjacent segment in the other track.
Define a Segment structure. Each segment has a pointer to the next and previous
segments in its track, a pointer to the next and previous seg ments in the other track,
cfe2_fm_pi_xxvi.indd 10 10/28/10 5:08 PM
Walkthrough xi
Hand-Tracing
A very useful technique for understanding whether a program
works correctly is called hand-tracing. You simulate the pro-
gram’s activity on a sheet of paper. You can use this method with
pseudocode or C++ code.
Get an index card, a cocktail napkin, or whatever sheet of
paper is within reach. Make a column for each variable. Have the
program code ready. Use a marker, such as a paper clip, to mark
the current statement. In your mind, execute statements one at a
time. Every time the value of a variable changes, cross out the old
value and write the new value below the old one.
For example, let’s trace the tax program with the data from the
program run on page 95. In lines 13 and 14, tax1 and tax2 are initial-
ized to 0.
6 int main()
7 {
8 const double RATE1 = 0.10;
9 const double RATE2 = 0.25;
10 const double RATE1_SINGLE_LIMIT = 32000;
11 const double RATE1_MARRIED_LIMIT = 64000;
12
13 double tax1 = 0;
14 double tax2 = 0;
15
In lines 18 and 22, income and marital_status are
initialized by input statements.
16 double income;
17 cout << "Please enter your income: ";
18 cin >> income;
19
20 cout << "Please enter s for single, m for married: ";
21 string marital_status;
22 cin >> marital_status;
23
Because marital_status is not “s”, we move to the
else branch of the outer if statement (line 36).
Programming Tip 3.6
Hand-tracing helps you
understand whether a
program works correctly.
marital
tax1 tax2 income status
0 0
marital
tax1 tax2 income status
0 0 80000 m
Using Undefined Variables
You must define a variable before you use it for the first time. For example, the following
sequence of statements would not be legal:
double can_volume = 12 * liter_per_ounce;
double liter_per_ounce = 0.0296;
In your program, the statements are compiled in order. When the compiler reaches the first
statement, it does not know that liter_per_ounce will be defined in the next line, and it reports
an error.
Common Error 2.1
Random Facts provide historical and
social information on computing—for
interest and to fulfill the “historical and
social context” requirements of the
ACM/IEEE curriculum guidelines.
According to legend,
the first bug was
found in the Mark II, a huge electrome-
chanical computer at Harvard Univer-
sity. It really was caused by a bug—a
moth was trapped in a relay switch.
Actually, from the note that the
operator left in the log book next to
the moth (see the photo), it appears as
if the term “bug” had already been in
active use at the time.
The First Bug
The pioneering computer scientist
Maurice Wilkes wrote, “Somehow, at
the Moore School and afterwards, one
had always assumed there would be
no particular difficulty in getting pro-
grams right. I can remember the exact
instant in time at which it dawned on
me that a great part of my future life
would be spent finding mistakes in my
own programs.”
Random Fact 4.1 The First Bug
A Sorting Algorithm
A sorting algorithm rearranges the elements of a sequence so that they are stored in sorted
order. Here is a simple sorting algorithm, called selection sort. Consider sorting the following
array values:
[0] [1] [2] [3] [4]
11 9 17 5 12
An obvious first step is to find the smallest element. In this case the smallest element is 5,
stored in values[3]. You should move the 5 to the beginning of the array. Of course, there is
already an element stored in values[0], namely 11. Therefore you cannot simply move val-
ues[3] into values[0] without moving the 11 somewhere else. You don’t yet know where the 11
should end up, but you know for certain that it should not be in values[0]. Simply get it out of
the way by swapping it with values[3]:
5 9 17 11 12
Now the first element is in the correct place. In the foregoing figure, the darker color indicates
the portion of the array that is already sorted.
Next take the minimum of the remaining entries values[1]…values[4]. That minimum
value, 9, is already in the correct place. You don’t need to do anything in this case, simply
extend the sorted area by one to the right:
5 9 17 11 12
Repeat the process. The minimum value of the unsorted region is 11, which needs to be
swapped with the first value of the unsorted region, 17:
5 9 11 17 12
Special Topic 6.2
Special Topics present optional
topics and provide additional
explanation of others.
Common Errors describe the kinds
of errors that students often make,
with an explanation of why the errors
occur, and what to do about them.
Programming Tips explain
good programming practices,
and encourage students to be
more productive with tips and
techniques such as hand-tracing.
cfe2_fm_pi_xxvi.indd 11 10/28/10 5:09 PM
xii Preface
appendices
Appendix A contains a programming style guide. Using a style guide for program-
ming assignments benefits students by directing them toward good habits and reduc-
ing gratuitous choice. The style guide is available in electronic form so that instruc-
tors can modify it to reflect their preferred style.
Appendices B and C summarize C++ reserved words and operators. Appendix D
lists character escape sequences and ASCII character code values. Appendix E docu-
ments all of the library functions and classes used in this book.
Additional appendices available from the book’s companion web site include an
expanded version of Appendix E that includes the functions and classes used in the
four optional chapters, 11–14, plus appendices that cover number systems, bit and
shift operations, and a comparison of C++ and Java.
Student and Instructor resources
The following resources for students and instructors can be obtained by visiting
www.wiley.com/college/horstmann. Two companion web sites accompany the book—
one for students, and a password-protected site for instructors only.
• Additional exercises geared to the scientific and engineering problem domains
• Worked Examples that apply the problem-solving steps in the book to other
realistic examples (identified in the book by an icon, )
• Source code for all examples in the book
• Solutions to all review and programming exercises (for instructors only)
• Lecture presentation slides (in PowerPoint format) that summarize each chapter
and include code listings and figures from the book (for instructors only)
• A test bank that focuses on skills, not just terminology (for instructors only)
• Four additional chapters on recursion, sorting and searching, and data structures
• The programming style guide in electronic form
W O R K E D E X A M P L E 2 . 1 Computing Travel Time
In this Worked Example, we develop a hand calculation to compute
the time that a robot requires to retrieve an item from rocky terrain.
Pointers in the book
describe what students
will find on the Web.
Visit the C++ for Everyone companion web sites at www.wiley.com/college/horstmann.
cfe2_fm_pi_xxvi.indd 12 10/28/10 5:09 PM
www.wiley.com/college/horstmann
www.wiley.com/college/horstmann
acknowledgments xiii
acknowledgments
Many thanks to Beth Golub, Tom Kulesa, Andre Legaspi, Elizabeth Mills, Michael
Berlin, and Lisa Gee at John Wiley & Sons, and to the team at Publishing Services for
their hard work and sup port for this book project. An especially deep acknowledg-
ment and thanks to Cindy Johnson, who, through enormous patience and attention
to detail, made this book a reality. We would also like to thank Jonathan Tolstedt,
North Dakota State University, for his high-quality solutions; Brent Seales, Univer-
sity of Kentucky, for revising and enhancing the test bank; and to Evan Gallagher,
Polytechnic Institute of New York University, for his creative PowerPoint slides.
We are very grateful to the many individuals who reviewed and/or class tested
this and the first edition of the book. We value their many valuable suggestions for
improvement. They include:
Charles D. Allison, Utah Valley State College
Fred Annexstein, University of Cincinnati
Stefano Basagni, Northeastern University
Noah D. Barnette, Virginia Tech
Susan Bickford, Tallahassee Community College
Ronald D. Bowman, University of Alabama, Huntsville
Peter Breznay, University of Wisconsin, Green Bay
Richard Cacace, Pensacola Junior College, Pensacola
Kuang-Nan Chang, Eastern Kentucky University
Joseph DeLibero, Arizona State University
Subramaniam Dharmarajan, Arizona State University
Mary Dorf, University of Michigan
Marty Dulberg, North Carolina State University
William E. Duncan, Louisiana State University
John Estell, Ohio Northern University
Waleed Farag, Indiana University of Pennsylvania
Stephen Gilbert, Orange Coast Community College
Kenneth Gitlitz, New Hampshire Technical Institute
Daniel Grigoletti, DeVry Institute of Technology, Tinley Park
Barbara Guillott, Louisiana State University
Charles Halsey, Richland College
Jon Hanrath, Illinois Institute of Technology
Neil Harrison, Utah Valley University
Jurgen Hecht, University of Ontario
Steve Hodges, Cabrillo College
Jackie Jarboe, Boise State University
Debbie Kaneko, Old Dominion University
Mir Behrad Khamesee, University of Waterloo
Sung-Sik Kwon, North Carolina Central University
cfe2_fm_pi_xxvi.indd 13 10/28/10 5:09 PM
xiv acknowledgments
Lorrie Lehman, University of North Carolina, Charlotte
Cynthia Lester, Tuskegee University
Yanjun Li, Fordham University
W. James MacLean, University of Toronto
LindaLee Massoud, Mott Community College
Charles W. Mellard, DeVry Institute of Technology, Irving
Ethan V. Munson, University of Wisconsin, Milwaukee
Philip Regalbuto, Trident Technical College
Don Retzlaff, University of North Texas
Jeff Ringenberg, University of Michigan, Ann Arbor
John P. Russo, Wentworth Institute of Technology
Kurt Schmidt, Drexel University
Brent Seales, University of Kentucky
William Shay, University of Wisconsin, Green Bay
Michele A. Starkey, Mount Saint Mary College
William Stockwell, University of Central Oklahoma
Jonathan Tolstedt, North Dakota State University
Boyd Trolinger, Butte College
Muharrem Uyar, City College of New York
Mahendra Velauthapillai, Georgetown University
Kerstin Voigt, California State University, San Bernardino
David P. Voorhees, Le Moyne College
Salih Yurttas, Texas A&M University
A special thank you to all of our class testers:
Pani Chakrapani and the students of the University of Redlands
Jim Mackowiak and the students of Long Beach City College, LAC
Suresh Muknahallipatna and the students of the University of Wyoming
Murlidharan Nair and the students of the Indiana University of South Bend
Harriette Roadman and the students of New River Community College
David Topham and the students of Ohlone College
Dennie Van Tassel and the students of Gavilan College
cfe2_fm_pi_xxvi.indd 14 10/28/10 5:09 PM
C o n T E n T S
xv
PrEfaCE v
SPECIaL fEaTurES xx
InTroDuCTIon 1
1.1 What is Programming 2
1.2 The anatomy of a Computer 3
1.3 Machine Code and Programming Languages 6
1.4 Becoming familiar with your Programming Environment 7
1.5 analyzing your first Program 11
1.6 Errors 15
1.7 Problem Solving: algorithm Design 17
funDaMEnTaL DaTa TyPES 29
2.1 variables 30
2.2 arithmetic 40
2.3 Input and output 48
2.4 Problem Solving: first Do It By Hand 52
2.5 Strings 56
DECISIonS 75
3.1 The if Statement 76
3.2 Comparing numbers and Strings 82
3.3 Multiple alternatives 90
3.4 nested Branches 94
3.5 Problem Solving: flowcharts 99
3.6 Problem Solving: Test Cases 102
3.7 Boolean variables and operators 103
3.8 application: Input validation 109
LooPS 131
4.1 The while Loop 132
4.2 Problem Solving: Hand-Tracing 139
4.3 The for Loop 142
4.4 The do Loop 148
4.5 Processing Input 150
cHAPter 1
cHAPter 2
cHAPter 3
cHAPter 4
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xvi Contents
4.6 Problem Solving: Storyboards 154
4.7 Common Loop algorithms 157
4.8 nested Loops 165
4.9 random numbers and Simulations 168
funCTIonS 193
5.1 functions as Black Boxes 194
5.2 Implementing functions 196
5.3 Parameter Passing 199
5.4 return values 202
5.5 functions Without return values 206
5.6 Problem Solving: reusable functions 208
5.7 Problem Solving: Stepwise refinement 210
5.8 variable Scope and Global variables 218
5.9 reference Parameters 220
5.10 recursive functions (optional) 226
arrayS anD vECTorS 249
6.1 arrays 250
6.2 Common array algorithms 256
6.3 arrays and functions 265
6.4 Problem Solving: adapting algorithms 269
6.5 Problem Solving: Discovering algorithms by Manipulating
Physical objects 274
6.6 Two-Dimensional arrays 278
6.7 vectors 284
PoInTErS 307
7.1 Defining and using Pointers 308
7.2 arrays and Pointers 314
7.3 C and C++ Strings 320
7.4 Dynamic Memory allocation 325
7.5 arrays and vectors of Pointers 329
7.6 Problem Solving: Draw a Picture 332
7.7 Structures and Pointers (optional) 336
STrEaMS 351
8.1 reading and Writing Text files 352
8.2 reading Text Input 358
cHAPter 5
cHAPter 6
cHAPter 7
cHAPter 8
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Contents xvii
Available online at www.wiley.com/college/horstmann.
8.3 Writing Text output 361
8.4 String Streams 363
8.5 Command Line arguments 365
8.6 random access and Binary files 372
CLaSSES 389
9.1 object-oriented Programming 390
9.2 Specifying the Public Interface of a Class 392
9.3 Data Members 395
9.4 Member functions 397
9.5 Constructors 403
9.6 Problem Solving: Tracing objects 407
9.7 Problem Solving: Discovering Classes 414
9.8 Separate Compilation 417
9.9 Pointers to objects 422
InHErITanCE 441
10.1 Inheritance Hierarchies 442
10.2 Implementing Derived Classes 446
10.3 overriding Member functions 451
10.4 virtual functions and Polymorphism 455
rECurSIon (WEB onLy)
11.1 Triangle numbers
11.2 Thinking recursively
11.3 recursive Helper functions
11.4 The Efficiency of recursion
11.5 Permutations
11.6 Mutual recursion
SorTInG anD SEarCHInG (WEB onLy)
12.1 Selection Sort
12.2 Profiling the Selection Sort algorithm
12.3 analyzing the Performance of the Selection Sort algorithm
12.4 Merge Sort
12.5 analyzing the Merge Sort algorithm
12.6 Searching
cHAPter 9
cHAPter 10
cHAPter 11
cHAPter 12
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xviii Contents
Available online at www.wiley.com/college/horstmann.
LISTS, STaCkS, anD QuEuES (WEB onLy)
13.1 using Linked Lists
13.2 Implementing Linked Lists
13.3 The Efficiency of List, array, and vector operations
13.4 Stacks and Queues
SETS, MaPS, anD PrIorITy QuEuES (WEB onLy)
14.1 Sets
14.2 Binary Search Trees
14.3 Maps
14.4 Priority Queues
14.5 Heaps
aPPEnDIx a C++ LanGuaGE CoDInG GuIDELInES 481
aPPEnDIx B rESErvED WorD SuMMary 489
aPPEnDIx C oPEraTor SuMMary 491
aPPEnDIx D CHaraCTEr CoDES 493
aPPEnDIx E C++ LIBrary SuMMary 495
aPPEnDIx f nuMBEr SySTEMS (WEB onLy)
aPPEnDIx G BIT anD SHIfT oPEraTIonS (WEB onLy)
aPPEnDIx H a C++ / Java CoMParISon (WEB onLy)
GLoSSary 499
InDEx 507
CrEDITS 529
cHAPter 13
cHAPter 14
APPendiceS
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Contents xix
SynTax BoxES
assignment 34
C++ Program 12
Class Definition 393
Comparisons 83
Constructor with Base-Class Initializer 451
Defining an array 251
Defining a vector 285
Derived-Class Definition 448
Dynamic Memory allocation 325
for Statement 144
function Definition 197
if Statement 78
Input Statement 48
Member function Definition 400
output Statement 13
Pointer Syntax 310
Two-Dimensional array Definition 279
variable Definition 31
while Statement 133
Working with file Streams 354
AlPHAbeticAl liSt oF
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xx Special features
C H a P T E r
Available online at www.wiley.com/college/horstmann.
Programming
Tips
Special Topics random facts
Backup Copies 11 Escape Sequences 14 The EnIaC and the Dawn
of Computing 5
Standards organizations 7
Choose Descriptive variable
names 38
Do not use Magic numbers 39
Spaces in Expressions 46
numeric Types in C++ 38
numeric ranges and
Precisions 39
Casts 46
Combining assignment
and arithmetic 47
The Pentium floating-Point
Bug 47
International alphabets
and unicode 61
Brace Layout 79
always use Braces 80
Tabs 81
avoid Duplication in Branches 82
Compile With Zero Warnings 85
Hand-Tracing 97
Make a Schedule and
Make Time for unexpected
Problems 103
The Selection operator 81
Lexicographic ordering
of Strings 86
The switch Statement 93
Short-Circuit Evaluation
of Boolean operators 108
De Morgan’s Law 108
The Denver airport
Luggage Handling System 89
artificial Intelligence 112
use for Loops for Their
Intended Purpose only 147
Choose Loop Bounds That
Match your Task 147
Count Iterations 147
flowcharts for Loops 149
Clearing the failure State 153
The Loop-and-a-Half Problem
and the break Statement 153
redirection of Input and
output 154
The first Bug 138
Software Piracy 172
function Comments 199
Don’t Modify Parameter
variables 201
keep functions Short 216
Tracing functions 216
Stubs 217
avoid Global variables 220
Prefer return values to
reference Parameters 225
function Declarations 203
Constant references 225
The Explosive Growth of
Personal Computers 230
Common
Errors
How Tos
and
Worked Examples
1 Introduction omitting Semicolons 14
Misspelling Words 16
Describing an algorithm with
Pseudocode 20
Writing an algorithm for Tiling
a floor
2 fundamental
Data Types
using undefined variables 37
using uninitialized variables 37
unintended Integer Division 43
unbalanced Parentheses 44
forgetting Header files 45
roundoff Errors 45
Computing Travel Time
Carrying out Computations 54
Computing the Cost
of Stamps
3 Decisions a Semicolon after the
if Condition 80
Confusing = and == 85
Exact Comparison of floating-
Point numbers 86
The Dangling else Problem 98
Combining Multiple
relational operators 107
Confusing && and ||
Conditions 107
Implementing an if
Statement 87
Extracting the Middle
4 Loops Infinite Loops 136
Don’t Think “are We
There yet?” 137
off-by-one Errors 137
Writing a Loop 162
Credit Card Processing
5 functions Missing return value 203
Implementing a function 204
Matching and replacing Parts
of a String
using a Debugger
Calculating a Course Grade
Thinking recursively 229
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www.wiley.com/college/horstmann
Special Features xxi
Available online at www.wiley.com/college/horstmann.
Programming
Tips
Special Topics Random Facts
Backup Copies 11 Escape Sequences 14 The ENIAC and the Dawn
of Computing 5
Standards Organizations 7
Choose Descriptive Variable
Names 38
Do Not Use Magic Numbers 39
Spaces in Expressions 46
Numeric Types in C++ 38
Numeric Ranges and
Precisions 39
Casts 46
Combining Assignment
and Arithmetic 47
The Pentium Floating-Point
Bug 47
International Alphabets
and Unicode 61
Brace Layout 79
Always Use Braces 80
Tabs 81
Avoid Duplication in Branches 82
Compile With Zero Warnings 85
Hand-Tracing 97
Make a Schedule and
Make Time for Unexpected
Problems 103
The Selection Operator 81
Lexicographic Ordering
of Strings 86
The switch Statement 93
Short-Circuit Evaluation
of Boolean Operators 108
De Morgan’s Law 108
The Denver Airport
Luggage Handling System 89
Artificial Intelligence 112
Use for Loops for Their
Intended Purpose Only 147
Choose Loop Bounds That
Match Your Task 147
Count Iterations 147
Flowcharts for Loops 149
Clearing the Failure State 153
The Loop-and-a-Half Problem
and the break Statement 153
Redirection of Input and
Output 154
The First Bug 138
Software Piracy 172
Function Comments 199
Don’t Modify Parameter
Variables 201
Keep Functions Short 216
Tracing Functions 216
Stubs 217
Avoid Global Variables 220
Prefer Return Values to
Reference Parameters 225
Function Declarations 203
Constant References 225
The Explosive Growth of
Personal Computers 230
Common
Errors
How Tos
and
Worked Examples
1 Introduction Omitting Semicolons 14
Misspelling Words 16
Describing an Algorithm with
Pseudocode 20
Writing an Algorithm for Tiling
a Floor
2 Fundamental
Data Types
Using Undefined Variables 37
Using Uninitialized Variables 37
Unintended Integer Division 43
Unbalanced Parentheses 44
Forgetting Header Files 45
Roundoff Errors 45
Computing Travel Time
Carrying out Computations 54
Computing the Cost
of Stamps
3 Decisions A Semicolon After the
if Condition 80
Confusing = and == 85
Exact Comparison of Floating-
Point Numbers 86
The Dangling else Problem 98
Combining Multiple
Relational Operators 107
Confusing && and ||
Conditions 107
Implementing an if
Statement 87
Extracting the Middle
4 Loops Infinite Loops 136
Don’t Think “Are We
There Yet?” 137
Off-by-One Errors 137
Writing a Loop 162
Credit Card Processing
5 Functions Missing Return Value 203
Implementing a Function 204
Matching and Replacing Parts
of a String
Using a Debugger
Calculating a Course Grade
Thinking Recursively 229
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xxii Special features
C H a P T E r
Available online at www.wiley.com/college/horstmann.
Programming
Tips
Special Topics random facts
use arrays for Sequences
of related values 255
Prefer vectors over arrays 289
Sorting with the C++ Library 263
a Sorting algorithm 263
Binary Search 264
Constant array Parameters 269
an Early Internet Worm 255
The first Programmer 290
use a Separate Definition for
Each Pointer variable 313
Program Clearly,
not Cleverly 319
Pointers and references 314
using a Pointer to Step
Through an array 318
Constant Pointers 320
Working with C Strings 323
Embedded Systems 336
Encryption algorithms 368
Databases and Privacy 377
all Data Members Should
Be Private; Most
Member functions
Should Be Public 402
const Correctness 402
Make Parallel vectors into
vectors of objects 416
Initializer Lists 405
overloading 406
Destructors and resource
Management 424
Electronic voting Machines 412
open Source and
free Software 426
use a Single Class for variation
in values, Inheritance for
variation in Behavior 450
Don’t use Type Tags 462
Calling the Base-Class
Constructor 451
virtual Self-Calls 463
The Limits of Computation 469
Common
Errors
How Tos
and
Worked Examples
6 arrays and vectors Bounds Errors 254
omitting the Column
Size of a Two-
Dimensional array
Parameter 284
Working with arrays 271
rolling the Dice
a World Population Table
7 Pointers Confusing Pointers
with the Data to
Which They Point 313
returning a Pointer to
a Local variable 319
Dangling Pointers 328
Memory Leaks 328
Working with Pointers 334
Producing a Mass Mailing
8 Streams Processing Text files 369
Looking for for
Duplicates
9 Classes forgetting a Semicolon 395
Trying to Call a Constructor 405
Implementing a Class 409
Implementing a
Bank account Class
10 Inheritance Private Inheritance 449
replicating Base-Class
Members 450
forgetting the Base-
Class name 455
Slicing an object 462
Developing an
Inheritance Hierarchy 464
Implementing an
Employee Hierarchy for
Payroll Processing
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www.wiley.com/college/horstmann
Special Features xxiii
Available online at www.wiley.com/college/horstmann.
Programming
Tips
Special Topics Random Facts
Use Arrays for Sequences
of Related Values 255
Prefer Vectors over Arrays 289
Sorting with the C++ Library 263
A Sorting Algorithm 263
Binary Search 264
Constant Array Parameters 269
An Early Internet Worm 255
The First Programmer 290
Use a Separate Definition for
Each Pointer Variable 313
Program Clearly,
Not Cleverly 319
Pointers and References 314
Using a Pointer to Step
Through an Array 318
Constant Pointers 320
Working with C Strings 323
Embedded Systems 336
Encryption Algorithms 368
Databases and Privacy 377
All Data Members Should
Be Private; Most
Member Functions
Should Be Public 402
const Correctness 402
Make Parallel Vectors into
Vectors of Objects 416
Initializer Lists 405
Overloading 406
Destructors and Resource
Management 424
Electronic Voting Machines 412
Open Source and
Free Software 426
Use a Single Class for Variation
in Values, Inheritance for
Variation in Behavior 450
Don’t Use Type Tags 462
Calling the Base-Class
Constructor 451
Virtual Self-Calls 463
The Limits of Computation 469
Common
Errors
How Tos
and
Worked Examples
6 Arrays and Vectors Bounds Errors 254
Omitting the Column
Size of a Two-
Dimensional Array
Parameter 284
Working with Arrays 271
Rolling the Dice
A World Population Table
7 Pointers Confusing Pointers
with the Data to
Which They Point 313
Returning a Pointer to
a Local Variable 319
Dangling Pointers 328
Memory Leaks 328
Working with Pointers 334
Producing a Mass Mailing
8 Streams Processing Text Files 369
Looking for for
Duplicates
9 Classes Forgetting a Semicolon 395
Trying to Call a Constructor 405
Implementing a Class 409
Implementing a
Bank Account Class
10 Inheritance Private Inheritance 449
Replicating Base-Class
Members 450
Forgetting the Base-
Class Name 455
Slicing an Object 462
Developing an
Inheritance Hierarchy 464
Implementing an
Employee Hierarchy for
Payroll Processing
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xxiv Special features
C H a P T E r
Available online at www.wiley.com/college/horstmann.
Programming
Tips
Special Topics random facts
Library functions for Sorting
and Binary Search
The Quicksort algorithm
Defining an ordering for
Sorting objects
Cataloging your necktie
Collection
reverse Polish notation
Defining an ordering for
Container Elements
Constant Iterators
Discrete Event Simulations
Common
Errors
How Tos
and
Worked Examples
11 recursion (WEB onLy) Infinite recursion
Tracing Through recursive
functions
12 Sorting and Searching
(WEB onLy)
13 Lists, Stacks, and
Queues (WEB onLy)
14 Sets, Maps, and Priority
Queues (WEB onLy)
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www.wiley.com/college/horstmann
Special Features xxv
Available online at www.wiley.com/college/horstmann.
Programming
Tips
Special Topics Random Facts
Library Functions for Sorting
and Binary Search
The Quicksort Algorithm
Defining an Ordering for
Sorting Objects
Cataloging Your Necktie
Collection
Reverse Polish Notation
Defining an Ordering for
Container Elements
Constant Iterators
Discrete Event Simulations
Common
Errors
How Tos
and
Worked Examples
11 Recursion (WEB ONLY) Infinite Recursion
Tracing Through Recursive
Functions
12 Sorting and Searching
(WEB ONLY)
13 Lists, Stacks, and
Queues (WEB ONLY)
14 Sets, Maps, and Priority
Queues (WEB ONLY)
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1C h a p t e r
1
I n t r o d u C t I o n
to learn about the architecture of
computers
to learn about machine languages and
higher-level programming languages
to become familiar with your compiler
to compile and run your first C++ program
to recognize compile-time and run-time errors
to describe an algorithm with pseudocode
to understand the activity of programming
C h a p t e r G o a l s
C h a p t e r C o n t e n t s
1.1 What Is ProgrammIng? 2
1.2 the anatomy of a ComPuter 3
Random Fact 1.1: the enIaC and the dawn of
Computing 5
1.3 maChIne Code and ProgrammIng
Languages 6
Random Fact 1.2: standards organizations 7
1.4 BeComIng famILIar WIth your
ProgrammIng envIronment 7
Programming Tip 1.1: Backup Copies 11
1.5 anaLyzIng your fIrst
Program 11
Syntax 1.1: C++ program 12
Syntax 1.2: output statement 13
Common Error 1.1: omitting semicolons 14
Special Topic 1.1: escape sequences 14
1.6 errors 15
Common Error 1.2: Misspelling Words 16
1.7 ProBLem soLvIng: aLgorIthm
desIgn 17
How To 1.1: describing an algorithm with
pseudocode 20
Worked Example 1.1: Writing an algorithm for
tiling a Floor
cfe2_ch01_p1_28.indd 1 10/25/10 1:03 PM
2
Just as you gather tools, study a project, and make a plan for
tackling it, in this chapter you will gather up the basics you
need to start learning to program. after a brief introduction
to computer hardware, software, and programming in
general, you will learn how to write and run your first
C++ program. You will also learn how to diagnose and
fix programming errors, and how to use pseudocode to
describe an algorithm—a step-by-step description of how
to solve a problem—as you plan your programs.
1.1 What Is programming?
You have probably used a computer for work or fun. Many people use computers for
everyday tasks such as electronic banking or writing a term paper. Computers are
good for such tasks. They can handle repetitive chores, such as totaling up numbers
or placing words on a page, without getting bored or exhausted.
The flexibility of a computer is quite an amazing phenomenon. The same machine
can balance your checkbook, print your term paper, and play a game. In contrast,
other machines carry out a much nar rower range of tasks; a car drives and a toaster
toasts. Computers can carry out a wide range of tasks because they execute different
programs, each of which directs the computer to work on a specific task.
The computer itself is a machine that stores data (numbers, words, pictures), inter-
acts with devices (the monitor, the sound system, the printer), and executes programs.
A computer program tells a computer, in minute detail, the sequence of steps that are
needed to fulfill a task. The physical computer and peripheral devices are collectively
called the hardware. The programs the computer executes are called the software.
Today’s computer programs are so sophisticated that it is hard to believe that they
are composed of extremely primitive operations. A typical operation may be one of
the following:
• Put a red dot at this screen position.
• Add up these two numbers.
• If this value is negative, continue the program at a certain instruction.
The computer user has the illusion of smooth interaction because a program contains
a huge number of such operations, and because the computer can execute them at
great speed.
The act of designing and implementing computer programs is called program-
ming. In this book, you will learn how to program a computer—that is, how to direct
the computer to execute tasks.
To write a computer game with motion and sound effects or a word processor
that supports fancy fonts and pictures is a complex task that requires a team of many
highly skilled programmers. Your first programming efforts will be more mundane.
The concepts and skills you learn in this book form an important foundation, and
you should not be disappointed if your first programs do not rival the sophis ticated
software that is familiar to you. Actually, you will find that there is an immense thrill
even in sim ple programming tasks. It is an amazing experience to see the computer
precisely and quickly carry out a task that would take you hours of drudgery, to
Computers execute
very basic
instructions in
rapid succession.
a computer program
is a sequence
of instructions
and decisions.
programming is the
act of designing and
implementing
computer programs.
make small changes in a program that lead to immediate improvements, and to see the
computer become an extension of your mental powers.
1. What is required to play music on a computer?
2. Why is a CD player less flexible than a computer?
3. What does a computer user need to know about programming in order to play a
video game?
Practice It Now you can try these exercises at the end of the chapter: R1.1, R1.4.
1.2 the anatomy of a Computer
To understand the programming process, you need to have a rudimentary under-
standing of the building blocks that make up a computer. We will look at a personal
computer. Larger computers have faster, larger, or more powerful components, but
they have fundamentally the same design.
At the heart of the computer lies the central processing unit (CPU) (see Figure 1).
It consists of a sin gle chip, or a small number of chips. A computer chip (integrated
circuit) is a component with a plastic or metal housing, metal connectors, and inside
wiring made principally from silicon. For a CPU chip, the inside wiring is enor-
mously complicated. For example, the Pentium chip (a popular CPU for personal
computers at the time of this writing) is composed of several million structural ele-
ments, called transistors.
The CPU performs program control and data processing. That is, the CPU locates
and executes the program instructions; it carries out arithmetic operations such as
addition, subtraction, multiplication, and division; it fetches data from external mem-
ory or devices and stores data back.
The computer stores data and programs. There are two kinds of storage. Primary
storage is made from memory chips: electronic circuits that can store data, provided
they are supplied with electric power. Secondary storage, usually a hard disk, pro-
vides less expensive storage that persists without elec tricity. A hard disk consists of
rotating platters, which are coated with a magnetic material, and read/write heads,
which can detect and change the magnetic flux on the platters (see Figure 2).
s e L f C h e C k
the central
processing unit (Cpu)
performs program
control and
data processing.
storage devices
include memory and
secondary storage.
figure 1
Central processing unit
cfe2_ch01_p1_28.indd 2 10/25/10 1:03 PM
1.2 the anatomy of a Computer 3
make small changes in a program that lead to immediate improvements, and to see the
computer become an extension of your mental powers.
1. What is required to play music on a computer?
2. Why is a CD player less flexible than a computer?
3. What does a computer user need to know about programming in order to play a
video game?
Practice It Now you can try these exercises at the end of the chapter: R1.1, R1.4.
1.2 the anatomy of a Computer
To understand the programming process, you need to have a rudimentary under-
standing of the building blocks that make up a computer. We will look at a personal
computer. Larger computers have faster, larger, or more powerful components, but
they have fundamentally the same design.
At the heart of the computer lies the central processing unit (CPU) (see Figure 1).
It consists of a sin gle chip, or a small number of chips. A computer chip (integrated
circuit) is a component with a plastic or metal housing, metal connectors, and inside
wiring made principally from silicon. For a CPU chip, the inside wiring is enor-
mously complicated. For example, the Pentium chip (a popular CPU for personal
computers at the time of this writing) is composed of several million structural ele-
ments, called transistors.
The CPU performs program control and data processing. That is, the CPU locates
and executes the program instructions; it carries out arithmetic operations such as
addition, subtraction, multiplication, and division; it fetches data from external mem-
ory or devices and stores data back.
The computer stores data and programs. There are two kinds of storage. Primary
storage is made from memory chips: electronic circuits that can store data, provided
they are supplied with electric power. Secondary storage, usually a hard disk, pro-
vides less expensive storage that persists without elec tricity. A hard disk consists of
rotating platters, which are coated with a magnetic material, and read/write heads,
which can detect and change the magnetic flux on the platters (see Figure 2).
s e L f C h e C k
the central
processing unit (Cpu)
performs program
control and
data processing.
storage devices
include memory and
secondary storage.
figure 1
Central processing unit
cfe2_ch01_p1_28.indd 3 10/25/10 1:03 PM
4 Chapter 1 Introduction
figure 2
a hard disk
Programs and data are typically stored on the hard disk and loaded into memory
when the program starts. The program then updates the data in memory and writes
the modified data back to the hard disk.
To interact with a human user, a computer requires peripheral devices. The com-
puter transmits infor mation (called output) to the user through a display screen,
speakers, and printers. The user can enter information (called input) by using a key-
board or a pointing device such as a mouse.
Some computers are self-contained units, whereas others are interconnected
through networks. Through the network cabling, the computer can read data and
programs from central storage locations or send data to other computers. For the
user of a networked computer it may not even be obvious which data reside on the
computer itself and which are transmitted through the network.
figure 3 schematic design of a personal Computer
Printer
Mouse
Keyboard
Ports
CPU
Memory
Disk
Controller
Hard disk
CD/DVD drive
Monitor
Speakers
Internet
Graphics
card
Sound
card
Network
card
cfe2_ch01_p1_28.indd 4 10/25/10 1:03 PM
1.2 the anatomy of a Computer 5
Figure 3 gives a schematic overview of the architecture of a personal computer.
Program instructions and data (such as text, numbers, audio, or video) are stored on
the hard disk, on an optical disk such as a DVD, or elsewhere on the network. When
a program is started, it is brought into memory, where the CPU can read it. The CPU
reads the program one instruction at a time. As directed by these instructions, the
CPU reads data, modifies it, and stores it. Some program instructions will cause the
CPU to place dots on the display screen or printer or to vibrate the speaker. As these
actions happen many times over and at great speed, the human user perceives images
and sound. Some program instructions read user input from the keyboard or mouse.
The program analyzes the nature of these inputs and then executes the next appropri-
ate instruction.
4. Where is a program stored when it is not currently running?
5. Which part of the computer carries out arithmetic operations, such as addition
and multiplication?
Practice It Now you can try these exercises at the end of the chapter: R1.2, R1.3.
s e L f C h e C k
the enIaC (electronic
numerical integrator
and computer) was the first usable
electronic computer. It was designed
by J. presper eckert and John Mauchly
at the university of pennsylvania and
was completed in 1946—two years
before transistors were invented. the
computer was housed in a large room
and consisted of many cabinets con-
taining about 18,000 vacuum tubes
(see Figure 2). Vacuum tubes burned
out at the rate of several tubes per day.
an attendant with a shopping cart full
of tubes constantly made the rounds
and replaced defective ones. the com-
puter was programmed by connecting
wires on panels. each wir ing configura-
tion would set up the computer for a
particular problem. to have the com-
puter work on a different problem, the
wires had to be replugged.
Work on the enIaC was supported
by the u.s. army, which was interested
in computations of ballistic tables that
would give the trajectory of a projec-
tile, depending on the wind resis tance,
initial velocity, and atmospheric con-
ditions. to compute the trajecto ries,
one must find the numerical solu-
tions of certain differential equations;
hence the name “numerical integra-
tor”. Before machines like the enIaC
were developed, humans did this kind
of work, and until the 1950s the word
“computer” referred to these people.
the enIaC was later used for peace-
ful purposes, such as the tabulation of
u.s. Census data.
figure 4 the enIaC
Random Fact 1.1 the enIaC and the dawn of Computing
cfe2_ch01_p1_28.indd 5 10/25/10 1:03 PM
6 Chapter 1 Introduction
1.3 Machine Code and programming languages
On the most basic level, computer instructions are extremely primitive. The proces-
sor executes machine instructions. A typical sequence of machine instructions is
1. Move the contents of memory location 40000 into the CPU.
2. If that value is > 100, continue with the instruction that is stored in memory
location 11280.
Actually, machine instructions are encoded as numbers so that they can be stored in
memory. On a Pen tium processor, this sequence of instruction is encoded as the
sequence of numbers
161 40000 45 100 127 11280
On a processor from a different manufacturer, the encoding would be different.
When this kind of pro cessor fetches this sequence of numbers, it decodes them and
executes the associated sequence of com mands.
How can we communicate the command sequence to the computer? The simplest
method is to place the actual numbers into the computer memory. This is, in fact,
how the very earliest computers worked. However, a long program is composed of
thousands of individual commands, and it is a tedious and error-prone affair to look
up the numeric codes for all commands and place the codes manually into memory.
As already mentioned, computers are really good at automating tedious and error-
prone activi ties. It did not take long for computer scientists to realize that the com-
puters themselves could be har nessed to help in the programming process.
Computer scientists devised high level programming
lan guages that allow programmers to describe tasks,
using a syn tax that is more closely related to the prob-
lems to be solved. In this book, we will use the C++
programming language, which was developed by Bjarne
Stroustrup in the 1980s.
Over the years, C++ has grown by the addition of
many features. A standardization process culminated
in the publi cation of the international C++ standard
in 1998. A minor update to the standard was issued in
2003, and a major revi sion is expected to come to fruition
around 2011. At this time, C++ is the most commonly
used language for develop ing system software such as
databases and operating systems. Just as importantly, C++ is increasingly used for
program ming “embedded systems”, small computers that control devices such as
automobile engines or cellular telephones.
Here is a typical statement in C++:
if (int_rate > 100) { cout << "Interest rate error"; }
This means, “If the interest rate is over 100, display an error message”. A special com-
puter program, a compiler, translates this high-level description into machine instruc-
tions for a particular processor.
High-level languages are independent of the underlying hardware. C++ instruc-
tions work equally well on an Intel Pentium and a processor in a cell phone. Of
course, the compiler-generated machine instruc tions are different, but the program-
mer who uses the compiler need not worry about these differences.
Computer programs
are stored as machine
instructions in a code
that depends on the
processor type.
Bjarne Stroustrup
C++ is a general-
purpose language
that is in widespread
use for systems
and embedded
programming.
high-level
programming
languages are
independent of
the processor.
6. Is the compiler a part of the computer hardware or software?
7. Does a person who uses a computer for office work ever run a compiler?
8. What are the most important uses for C++?
Practice It Now you can try these exercises at the end of the chapter: R1.5.
1.4 Becoming Familiar with Your
programming environment
Many students find that the tools they need as programmers are very different from
the software with which they are familiar. You should spend some time making your-
self familiar with your programming environment. Because computer systems vary
widely, this book can only give an outline of the steps you need to follow. It is a good
idea to participate in a hands-on lab, or to ask a knowledgeable friend to give you a
tour.
step 1 Start the C++ development environment.
Computer systems differ greatly in this regard. On many computers there is an inte-
grated development environment in which you can write and test your programs.
On other computers you first launch an editor, a program that functions like a word
processor, in which you can enter your C++ instructions; then open a console win-
dow and type commands to execute your program. You need to find out how to get
started with your environment.
step 2 Write a simple program.
The traditional choice for the very first program in a new programming language is a
program that dis plays a simple greeting: “Hello, World!”. Let us follow that tradi-
tion. Here is the “Hello, World!” pro gram in C++:
s e L f C h e C k
two standards orga-
nizations, the ameri-
can national standards Institute (ansI)
and the International organization for
standardization (Iso), have jointly
developed the definitive standard for
the C++ language.
Why have standards? You encoun ter
the benefits of standardization every
day. When you buy a light bulb, you
can be assured that it fits in the socket
without having to measure the socket
at home and the bulb in the store. In
fact, you may have experi enced how
painful the lack of stan dards can be if
you have ever purchased a flashlight
with nonstand ard bulbs. replacement
bulbs for such a flashlight can be dif-
ficult and expen sive to obtain.
the ansI and Iso standards organi-
zations are associations of industry
professionals who develop standards
for everything from car tires and
credit card shapes to programming
languages. having a standard for a
programming language such as C++
means that you can take a program
that you developed on one system
with one manufacturer’s compiler to a
different system and be assured that it
will continue to work.
Random Fact 1.2 standards organizations
set aside some time
to become familiar
with the programming
environment that you
will use for your
class work.
cfe2_ch01_p1_28.indd 6 10/25/10 1:03 PM
1.4 Becoming Familiar with Your programming environment 7
6. Is the compiler a part of the computer hardware or software?
7. Does a person who uses a computer for office work ever run a compiler?
8. What are the most important uses for C++?
Practice It Now you can try these exercises at the end of the chapter: R1.5.
1.4 Becoming Familiar with Your
programming environment
Many students find that the tools they need as programmers are very different from
the software with which they are familiar. You should spend some time making your-
self familiar with your programming environment. Because computer systems vary
widely, this book can only give an outline of the steps you need to follow. It is a good
idea to participate in a hands-on lab, or to ask a knowledgeable friend to give you a
tour.
step 1 Start the C++ development environment.
Computer systems differ greatly in this regard. On many computers there is an inte-
grated development environment in which you can write and test your programs.
On other computers you first launch an editor, a program that functions like a word
processor, in which you can enter your C++ instructions; then open a console win-
dow and type commands to execute your program. You need to find out how to get
started with your environment.
step 2 Write a simple program.
The traditional choice for the very first program in a new programming language is a
program that dis plays a simple greeting: “Hello, World!”. Let us follow that tradi-
tion. Here is the “Hello, World!” pro gram in C++:
s e L f C h e C k
two standards orga-
nizations, the ameri-
can national standards Institute (ansI)
and the International organization for
standardization (Iso), have jointly
developed the definitive standard for
the C++ language.
Why have standards? You encoun ter
the benefits of standardization every
day. When you buy a light bulb, you
can be assured that it fits in the socket
without having to measure the socket
at home and the bulb in the store. In
fact, you may have experi enced how
painful the lack of stan dards can be if
you have ever purchased a flashlight
with nonstand ard bulbs. replacement
bulbs for such a flashlight can be dif-
ficult and expen sive to obtain.
the ansI and Iso standards organi-
zations are associations of industry
professionals who develop standards
for everything from car tires and
credit card shapes to programming
languages. having a standard for a
programming language such as C++
means that you can take a program
that you developed on one system
with one manufacturer’s compiler to a
different system and be assured that it
will continue to work.
Random Fact 1.2 standards organizations
set aside some time
to become familiar
with the programming
environment that you
will use for your
class work.
cfe2_ch01_p1_28.indd 7 10/25/10 1:03 PM
8 Chapter 1 Introduction
#include
using namespace std;
int main()
{
cout << "Hello, World!" << endl;
return 0;
}
We will examine this program in the next section.
No matter which programming environment you use, you begin your activity by
typing the pro gram statements into an editor window.
Create a new file and call it hello.cpp, using the steps that are appropriate for your
environment. (If your environment requires that you supply a project name in addi-
tion to the file name, use the name hello for the project.) Enter the program instruc-
tions exactly as they are given above. Alternatively, locate an electronic copy in the
source files for the programs in this book and paste it into your editor.
As you write this program, pay careful attention to the various symbols, and
keep in mind that C++ is case sensitive. You must enter upper- and lowercase letters
exactly as they appear in the program listing. You cannot type MAIN or Endl. If you are
not careful, you will run into problems—see Common Error 1.2 on page 16.
step 3 Compile and run the program.
The process for building and running a C++ program depends greatly on your pro-
gramming environ ment. In some integrated development environments, you simply
push a button. In other environments, you may have to type commands. When you
run the test program, the message
Hello, World!
will appear somewhere on the screen (see Figures 5 and 6).
an editor is a program
for entering and
modifying text, such
as a C++ program.
C++ is case sensitive.
You must be careful
about distinguishing
between upper- and
lowercase letters.
the compiler
translates C++
programs into
machine code.
figure 5
running the hello
program in an
Integrated
development
environment
It is useful to know what goes on behind the scenes when your program gets built.
First, the compiler translates the C++ source code (that is, the statements that you
wrote) into machine instructions. The machine code contains only the translation of
the code that you wrote. That is not enough to actually run the program. To display a
string on a window, quite a bit of low-level activity is necessary. The implementors of
your C++ development environment provided a library that includes the definition
of cout and its functionality. A library is a collection of code that has been pro-
grammed and translated by someone else, ready for you to use in your program.
(More complicated pro grams are built from more than one machine code file and
more than one library.) A program called the linker takes your machine code and the
necessary parts from the C++ library and builds an execut able file. (Figure 7 gives an
overview of these steps.) The executable file is usually called hello.exe or hello,
depending on your computer system. You can run the executable program even after
you exit the C++ development environment.
step 4 Organize your work.
As a programmer, you write programs, try them out, and improve them. You store
your programs in files. Files have names, and the rules for legal names differ from one
system to another. Some systems allow spaces in file names; others don’t. Some dis-
tinguish between upper- and lowercase letters; others don’t. Most C++ compilers
require that C++ files end in an extension .cpp, .cxx, .cc, or .C; for example, test.cpp.
Files are stored in folders or directories. A folder can contain files as well as other
folders, which themselves can contain more files and folders (see Figure 8). This hier-
archy can be quite large, and you need not be concerned with all of its branches.
the linker combines
machine code with
library code into an
executable program.
figure 7 From source Code to executable program
CompilerEditor Linker
Executable
ProgramSource File
Library files
Machine code
cfe2_ch01_p1_28.indd 8 10/25/10 1:03 PM
1.4 Becoming Familiar with Your programming environment 9
figure 6
Compiling and
running the hello
program in a
Console Window
It is useful to know what goes on behind the scenes when your program gets built.
First, the compiler translates the C++ source code (that is, the statements that you
wrote) into machine instructions. The machine code contains only the translation of
the code that you wrote. That is not enough to actually run the program. To display a
string on a window, quite a bit of low-level activity is necessary. The implementors of
your C++ development environment provided a library that includes the definition
of cout and its functionality. A library is a collection of code that has been pro-
grammed and translated by someone else, ready for you to use in your program.
(More complicated pro grams are built from more than one machine code file and
more than one library.) A program called the linker takes your machine code and the
necessary parts from the C++ library and builds an execut able file. (Figure 7 gives an
overview of these steps.) The executable file is usually called hello.exe or hello,
depending on your computer system. You can run the executable program even after
you exit the C++ development environment.
step 4 Organize your work.
As a programmer, you write programs, try them out, and improve them. You store
your programs in files. Files have names, and the rules for legal names differ from one
system to another. Some systems allow spaces in file names; others don’t. Some dis-
tinguish between upper- and lowercase letters; others don’t. Most C++ compilers
require that C++ files end in an extension .cpp, .cxx, .cc, or .C; for example, test.cpp.
Files are stored in folders or directories. A folder can contain files as well as other
folders, which themselves can contain more files and folders (see Figure 8). This hier-
archy can be quite large, and you need not be concerned with all of its branches.
the linker combines
machine code with
library code into an
executable program.
figure 7 From source Code to executable program
CompilerEditor Linker
Executable
ProgramSource File
Library files
Machine code
cfe2_ch01_p1_28.indd 9 10/25/10 1:03 PM
10 Chapter 1 Introduction
figure 8 a Folder hierarchy
However, you should create folders for organizing your work. It is a good idea to
make a separate folder for your programming class. Inside that folder, make a sepa-
rate folder for each assignment.
Some programming environments place your programs into a default location if
you don’t specify a folder yourself. In that case, you need to find out where those files
are located.
Be sure that you understand where your files are located in the folder hierarchy.
This information is essential when you submit files for grading, and for making
backup copies.
You will spend many hours creating and improving C++ programs. It is easy to
delete a file by acci dent, and occasionally files are lost because of a computer mal-
function. To avoid the frustration of recre ating lost files, get in the habit of making
backup copies of your work on a memory stick or on another computer.
9. How are programming projects stored on a computer?
10. What do you expect to see when you load an executable file into your text
editor?
11. What do you do to protect yourself from data loss when you work on program-
ming projects?
Practice It Now you can try these exercises at the end of the chapter: R1.6.
develop a strategy
for keeping backup
copies of your work
before disaster
strikes.
s e L f C h e C k
Backup Copies
Backing up files on a memory stick is an easy and convenient stor-
age method for many people. Another increasingly popular form
of backup is Internet file storage. Here are a few pointers to keep
in mind.
• Back up often. Backing up a file takes only a few seconds, and
you will hate yourself if you have to spend many hours
recreating work that you could have saved easily. I rec ommend
that you back up your work once every thirty minutes.
• Rotate backups. Use more than one directory for backups, and rotate them. That is, first
back up onto the first directory. Then back up onto the second directory. Then use the
third, and then go back to the first. That way you always have three recent backups. If
your recent changes made matters worse, you can then go back to the older version.
• Pay attention to the backup direction. Backing up involves copying files from one place to
another. It is important that you do this right—that is, copy from your work location to
the backup location. If you do it the wrong way, you will overwrite a newer file with an
older version.
• Check your backups once in a while. Double-check that your backups are where you think
they are. There is nothing more frustrating than to find out that the backups are not there
when you need them.
• Relax, then restore. When you lose a file and need to restore it from backup, you are likely
to be in an unhappy, nervous state. Take a deep breath and think through the recovery
process before you start. It is not uncommon for an agitated computer user to wipe out the
last backup when trying to restore a damaged file.
1.5 analyzing Your First program
In this section, we will analyze the first C++ program in detail. Here again is the
source code:
ch01/hello.cpp
1 #include
2
3 using namespace std;
4
5 int main()
6 {
7 cout << "Hello, World!" << endl;
8 return 0;
9 }
The first line,
#include
tells the compiler to include a service for “stream input/output”. You will learn in
Chapter 8 what a stream is. For now, you should simply remember to add this line
into all programs that perform input or output.
programming tip 1.1
cfe2_ch01_p1_28.indd 10 10/25/10 1:03 PM
1.5 analyzing Your First program 11
Backup Copies
Backing up files on a memory stick is an easy and convenient stor-
age method for many people. Another increasingly popular form
of backup is Internet file storage. Here are a few pointers to keep
in mind.
• Back up often. Backing up a file takes only a few seconds, and
you will hate yourself if you have to spend many hours
recreating work that you could have saved easily. I rec ommend
that you back up your work once every thirty minutes.
• Rotate backups. Use more than one directory for backups, and rotate them. That is, first
back up onto the first directory. Then back up onto the second directory. Then use the
third, and then go back to the first. That way you always have three recent backups. If
your recent changes made matters worse, you can then go back to the older version.
• Pay attention to the backup direction. Backing up involves copying files from one place to
another. It is important that you do this right—that is, copy from your work location to
the backup location. If you do it the wrong way, you will overwrite a newer file with an
older version.
• Check your backups once in a while. Double-check that your backups are where you think
they are. There is nothing more frustrating than to find out that the backups are not there
when you need them.
• Relax, then restore. When you lose a file and need to restore it from backup, you are likely
to be in an unhappy, nervous state. Take a deep breath and think through the recovery
process before you start. It is not uncommon for an agitated computer user to wipe out the
last backup when trying to restore a damaged file.
1.5 analyzing Your First program
In this section, we will analyze the first C++ program in detail. Here again is the
source code:
ch01/hello.cpp
1 #include
2
3 using namespace std;
4
5 int main()
6 {
7 cout << "Hello, World!" << endl;
8 return 0;
9 }
The first line,
#include
tells the compiler to include a service for “stream input/output”. You will learn in
Chapter 8 what a stream is. For now, you should simply remember to add this line
into all programs that perform input or output.
programming tip 1.1
cfe2_ch01_p1_28.indd 11 10/25/10 1:03 PM
12 Chapter 1 Introduction
syntax 1.1 C++ program
#include
using namespace std;
int main()
{
cout << "Hello, World!" << endl;
return 0;
}
Every program includes one or
more headers for required services
such as input/output.
Every program that uses standard
services requires this directive.
Every program
has a main function.
Replace this
statement when you
write your own
programs.
The statements
of a function
are enclosed
in braces. Each statement ends in a semicolon.
See page 14.
The next line,
using namespace std;
tells the compiler to use the “standard namespace”. Namespaces are a mechanism for
avoiding naming conflicts in large programs. You need not be concerned about
namespaces. For the programs that you will be writing in this book, you will always
use the standard namespace. Simply add using namespace std; at the top of every pro-
gram that you write, just below the #include directives.
The construction
int main()
{
...
return 0;
}
defines a function called main that “returns” an “integer” (that is, a whole number
without a fractional part, called int in C++) with value 0. This value indicates that the
program finished successfully. A function is a collection of programming instruc-
tions that carry out a particular task. Every C++ program must have a main func tion.
Most C++ programs contain other functions besides main, but it will take us until
Chapter 5 to dis cuss functions and return values.
For now, it is a good idea to consider all these parts as the plumbing that is neces-
sary to write a simple program. Simply place the code that you want to execute inside
the braces of the main function. (The basic structure of a C++ program is shown in
Syntax 1.1.)
To display values on the screen, you use an entity called cout and the << operator
(sometimes called the insertion operator). For example, the statement
cout << 39 + 3;
displays the number 42.
every C++ program
contains a function
called main.
use cout and the <<
operator to display
values on the screen.
The statement
cout << "Hello";
displays the string Hello. A string is a sequence of characters. You must enclose the
contents of a string inside quotation marks so that the compiler knows you literally
mean the text "Hello" and not a function with the same name.
You can send more than one item to cout. Use a << before each one of them. For
example,
cout << "The answer is " << 6 * 7;
displays The answer is 42 (in C++, the * denotes multiplication).
The endl symbol denotes an end of line marker. When this marker is sent to cout,
the cursor is moved to the first column in the next screen row. If you don’t use an end
of line marker, then the next displayed item will simply follow the current string on
the same line. In this program we only printed one item, but in general we will want
to print multiple items, and it is a good habit to end all lines of output with an end of
line marker.
Finally, note that each statement in C++ ends in a semicolon, just as every English
sentence ends in a period.
12. How do you modify the hello.cpp program to greet you instead?
13. What is wrong with this program?
#include
using namespace std;
int main()
{
cout << Goodbye, World! << endl;
return 0;
}
14. What does the following sequence of statements print?
cout << "Hello";
cout << "World";
enclose text strings in
quotation marks.
use + to add two
numbers and * to
multiply two
numbers.
send endl to cout to
end a line of
displayed output.
end each statement
with a semicolon.
s e L f C h e C k
cfe2_ch01_p1_28.indd 12 10/25/10 1:03 PM
1.5 analyzing Your First program 13
syntax 1.2 output statement
cout << "The answer is" << 6 * 7 << endl;
Sending endl
to cout starts
a new line.
Data sent to cout
is displayed in a
console window.
You can send
strings and numbers
to cout.
Add a << symbol before
each item to be displayed.
Strings
are enclosed in
quotation marks.
* denotes
multiplication.
The statement
cout << "Hello";
displays the string Hello. A string is a sequence of characters. You must enclose the
contents of a string inside quotation marks so that the compiler knows you literally
mean the text "Hello" and not a function with the same name.
You can send more than one item to cout. Use a << before each one of them. For
example,
cout << "The answer is " << 6 * 7;
displays The answer is 42 (in C++, the * denotes multiplication).
The endl symbol denotes an end of line marker. When this marker is sent to cout,
the cursor is moved to the first column in the next screen row. If you don’t use an end
of line marker, then the next displayed item will simply follow the current string on
the same line. In this program we only printed one item, but in general we will want
to print multiple items, and it is a good habit to end all lines of output with an end of
line marker.
Finally, note that each statement in C++ ends in a semicolon, just as every English
sentence ends in a period.
12. How do you modify the hello.cpp program to greet you instead?
13. What is wrong with this program?
#include
using namespace std;
int main()
{
cout << Goodbye, World! << endl;
return 0;
}
14. What does the following sequence of statements print?
cout << "Hello";
cout << "World";
enclose text strings in
quotation marks.
use + to add two
numbers and * to
multiply two
numbers.
send endl to cout to
end a line of
displayed output.
end each statement
with a semicolon.
s e L f C h e C k
cfe2_ch01_p1_28.indd 13 10/25/10 1:03 PM
14 Chapter 1 Introduction
15. What does the following statement print?
cout << 2 * 2 << 2;
16. What does the following statement print?
cout << "Hello" << endl << endl << "World";
Practice It Now you can try these exercises at the end of the chapter: R1.7, P1.1, P1.2.
omitting semicolons
In C++ every statement must end in a semicolon. Forgetting to type a semicolon is a common
error. It confuses the compiler because the compiler uses the semicolon to find where one
statement ends and the next one starts. The compiler does not use line ends or closing braces to
recognize the ends of statements. For example, the compiler con siders
cout << "Hello, World!" << endl
return 0;
a single statement, as if you had written
cout << "Hello, World!" << endl return 0;
and then it doesn’t understand that statement, because it does not expect the word return in the
middle of an output command. The remedy is simple. Just scan every statement for a terminat-
ing semicolon, just as you would check that every English sentence ends in a period.
escape sequences
How can you display a string containing quotation marks, such as
Hello, "World"
You can’t use
cout << "Hello, "World"";
As soon as the compiler reads "Hello, ", it thinks the string is finished, and then it gets all
confused about World. Compilers have a one-track mind, and if a simple analysis of the input
doesn’t make sense to them, they just refuse to go on, and they report an error. In contrast, a
human would probably realize that the second and third quotation marks were supposed to be
part of the string.
Well, how do we then display quotation marks on the screen? The designers of C++ pro-
vided an escape hatch. Mark each quotation mark with a backslash character (\), like this:
cout << "Hello, \"World\"";
The sequence \" denotes a literal quote, not the end of a string. Such a sequence is called an
escape sequence.
There are a few other escape sequences. If you actually want to show a backslash on the
display, you use the escape sequence \\. The statement
cout << "Hello\\World";
prints
Hello\World
Common error 1.1
special topic 1.1
Finally, the escape sequence \n denotes a newline character that starts a new line on the screen.
The command
cout << "Hello, World!\n";
has exactly the same effect as
cout << "Hello, World!" << endl;
1.6 errors
Programming languages follow very strict conven-
tions. When you talk to another person, and you
scramble or omit a word or two, your conversa tion
partner will usually still understand what you have to
say. But when you make an error in a C++ program,
the compiler will not try to guess what you meant.
(This is actually a good thing. If the compiler were to
guess wrongly, the resulting program would do the
wrong thing—quite possibly with disastrous effects.)
In this section, you will learn how to cope with errors
in your program.
Experiment a little with the hello.cpp program.
What happens if you make a typing error such as
cot << "Hello, World!" << endl;
cout << "Hello, World! << endl;
cout << "Hollo, World!" << endl;
In the first case, the compiler will complain that it has no clue what you mean by cot.
The exact wording of the error message is dependent on the compiler, but it might be
some thing like “Undefined symbol cot”. This is a compile-time error or syntax error.
Something is wrong according to the language rules, and the compiler finds it. When
the compiler finds one or more errors, it will not translate the program to machine
code, and as a consequence there is no program to run. You must fix the error and
compile again. It is common to go through several rounds of fixing compile-time
errors before compilation succeeds for the first time.
If the compiler finds an error, it will not simply stop and give up. It will try to
report as many errors as it can find, so you can fix them all at once. Sometimes, how-
ever, one error throws it off track. This is likely to happen with the error in the second
line. Since the programmer forgot the closing quote, the compiler will keep looking
for the end of the string. In such cases, it is common for the compiler to emit bogus
error reports for neighboring lines. You should fix only those error messages that
make sense to you and then recompile.
The error in the third line is of a different kind. The program will compile and run,
but its output will be wrong. It will print
Hollo, World!
This is a run-time error. The program is syntactically correct and does something,
but it doesn’t do what it is supposed to do. The compiler cannot find the error, and it
must be flushed out when the program runs, by testing it and carefully looking at its
Programmers spend a fair
amount of time fixing compile-
time and run-time errors.
a compile-time error
is a violation of
the programming
language rules that
is detected by
the compiler.
a run-time error
causes a program to
take an action that
the programmer did
not intend.
cfe2_ch01_p1_28.indd 14 10/25/10 1:03 PM
1.6 errors 15
Finally, the escape sequence \n denotes a newline character that starts a new line on the screen.
The command
cout << "Hello, World!\n";
has exactly the same effect as
cout << "Hello, World!" << endl;
1.6 errors
Programming languages follow very strict conven-
tions. When you talk to another person, and you
scramble or omit a word or two, your conversa tion
partner will usually still understand what you have to
say. But when you make an error in a C++ program,
the compiler will not try to guess what you meant.
(This is actually a good thing. If the compiler were to
guess wrongly, the resulting program would do the
wrong thing—quite possibly with disastrous effects.)
In this section, you will learn how to cope with errors
in your program.
Experiment a little with the hello.cpp program.
What happens if you make a typing error such as
cot << "Hello, World!" << endl;
cout << "Hello, World! << endl;
cout << "Hollo, World!" << endl;
In the first case, the compiler will complain that it has no clue what you mean by cot.
The exact wording of the error message is dependent on the compiler, but it might be
some thing like “Undefined symbol cot”. This is a compile-time error or syntax error.
Something is wrong according to the language rules, and the compiler finds it. When
the compiler finds one or more errors, it will not translate the program to machine
code, and as a consequence there is no program to run. You must fix the error and
compile again. It is common to go through several rounds of fixing compile-time
errors before compilation succeeds for the first time.
If the compiler finds an error, it will not simply stop and give up. It will try to
report as many errors as it can find, so you can fix them all at once. Sometimes, how-
ever, one error throws it off track. This is likely to happen with the error in the second
line. Since the programmer forgot the closing quote, the compiler will keep looking
for the end of the string. In such cases, it is common for the compiler to emit bogus
error reports for neighboring lines. You should fix only those error messages that
make sense to you and then recompile.
The error in the third line is of a different kind. The program will compile and run,
but its output will be wrong. It will print
Hollo, World!
This is a run-time error. The program is syntactically correct and does something,
but it doesn’t do what it is supposed to do. The compiler cannot find the error, and it
must be flushed out when the program runs, by testing it and carefully looking at its
Programmers spend a fair
amount of time fixing compile-
time and run-time errors.
a compile-time error
is a violation of
the programming
language rules that
is detected by
the compiler.
a run-time error
causes a program to
take an action that
the programmer did
not intend.
cfe2_ch01_p1_28.indd 15 10/25/10 1:03 PM
16 Chapter 1 Introduction
output. Because run-time errors are caused by logical flaws in the program, they are
often called logic errors. Some kinds of run-time errors are so severe that they gener-
ate an exception: a signal from the processor that aborts the program with an error
message. For example, if your program includes the statement cout << 1 / 0; your
program may terminate with a “divide by zero” exception.
During program development, errors are unavoidable. Once a program is longer
than a few lines, it requires superhuman concentration to enter it correctly without
slipping up once. You will find yourself omitting semicolons or quotes more often
than you would like, but the compiler will track down these problems for you.
Run-time errors are more troublesome. The compiler will not find them—in fact,
the compiler will cheerfully translate any program as long as its syntax is correct—
but the resulting program will do some thing wrong. It is the responsibility of the
program author to test the program and find any run-time errors. Program testing is
an important topic that you will encounter many times in this book.
17. Suppose you omit the () characters after main from the hello.cpp program. Will
you get a compile-time error or a run-time error?
18. When you used your computer, you may have experienced a program that
“crashed” (quit spontane ously) or “hung” (failed to respond to your input). Is
that behavior a compile-time error or a run-time error?
19. Why can’t you test a program for run-time errors when it has compiler errors?
Practice It Now you can try these exercises at the end of the chapter: R1.10, R1.11.
misspelling Words
If you accidentally misspell a word, strange things may happen, and it may not always be
completely obvious from the error messages what went wrong. Here is a good example of how
simple spelling errors can cause trouble:
#include
using namespace std;
int Main()
{
cout << "Hello, World!" << endl;
return 0;
}
This code defines a function called Main. The compiler will not consider this to be the same as
the main function, because Main starts with an uppercase letter and the C++ language is case-
sensitive. Upper- and lowercase letters are considered to be completely different from each
other, and to the compiler Main is no better match for main than rain. The compiler will compile
your Main function, but when the linker is ready to build the executable file, it will com plain
about the missing main function and refuse to link the program. Of course, the message “miss-
ing main function” should give you a clue where to look for the error.
If you get an error message that seems to indicate that the compiler is on the wrong track, it
is a good idea to check for spelling and capitalization. In C++, most names use only lowercase
letters. If you misspell the name of a symbol (for example out instead of cout), the compiler will
complain about an “undefined symbol”. This error mes sage is usually a good clue that you
made a spelling error.
the programmer is
responsible for
inspecting and
testing the program
to guard against
run-time errors.
s e L f C h e C k
Common error 1.2
cfe2_ch01_p1_28.indd 16 10/25/10 1:03 PM
1.7 problem solving: algorithm design 17
1.7 problem solving: algorithm design
You will soon learn how to program calculations and decision making in C++. But
before we look at the mechanics of implementing computations in the next chapter,
let’s consider the planning process that pre cedes implementation.
You may have run across advertisements
that encourage you to pay for a computerized
service that matches you up with a love part-
ner. Think how this might work. You fill out
a form and send it in. Others do the same. The
data are processed by a computer program. Is
it rea sonable to assume that the computer can
perform the task of finding the best match for
you? Suppose your younger brother, not the
com puter, had all the forms on his desk. What
instructions could you give him? You can’t say,
“Find the best-looking person of the opposite
sex who likes inline skating and browsing the
Internet”. There is no objective standard for good looks, and your brother’s opinion
(or that of a computer program analyzing the digitized photo) will likely be different
from yours. If you can’t give written instructions for some one to solve the problem,
there is no way the computer can magically solve the problem. The computer can
only do what you tell it to do. It just does it faster, without getting bored or exhausted.
Now consider the following investment problem:
You put $10,000 into a bank account that earns 5 percent interest per year. How many
years does it take for the account balance to be double the original?
Could you solve this problem by hand? Sure. You figure out the balance as follows:
year interest balance
0 10000
1 10000.00 x 0.05 = 500.00 10000.00 + 500.00 = 10500.00
2 10500.00 x 0.05 = 525.00 10500.00 + 525.00 = 11025.00
3 11025.00 x 0.05 = 551.25 11025.00 + 551.25 = 11576.25
4 11576.25 x 0.05 = 578.81 11576.25 + 578.81 = 12155.06
You keep going until the balance is at least $20,000. Then the last number in the year
column is the answer.
Of course, carrying out this computation is intensely boring to you or your
younger brother. But computers are very good at carrying out repetitive calcula-
tions quickly and flawlessly. What is important to the computer is a description of the
steps for finding the solution. Each step must be clear and unam biguous, requiring no
guesswork. Here is such a description:
Start with a year value of 0, a column for the interest, and a balance of $10,000.
year interest balance
0 10000
Finding the perfect partner is not a
problem that a computer can solve.
cfe2_ch01_p1_28.indd 17 10/25/10 1:03 PM
18 Chapter 1 Introduction
Repeat the following steps while the balance is less than $20,000
Add 1 to the year value.
Compute the interest as balance x 0.05 (i.e., 5 percent interest)
Add the interest to the balance.
year interest balance
0 10000
1 500.00 10500.00
14 942.82 19799.32
15 989.96 20789.28
Report the final year value as the answer.
Of course, these steps are not yet in a language that a computer can understand, but
you will soon learn how to formulate them in C++. This informal description is called
pseudocode.
There are no strict requirements for pseudocode because it is read by human read-
ers, not a computer program. Here are the kinds of pseudocode statements that we
will use in this book:
• Use statements such as the following to describe how a value is set or changed:
total cost = purchase price + operating cost
or
Multiply the balance value by 1.05.
or
Remove the first and last character from the word.
• You can describe decisions and repetitions as follows:
If total cost 1 < total cost 2
While the balance is less than $20,000
For each picture in the sequence
Use indentation to indicate which statements should be selected or repeated:
For each car
operating cost = 10 x annual fuel cost
total cost = purchase price + operating cost
Here, the indentation indicates that both statements should be executed for
each car.
• Indicate results with statements such as:
Choose car1.
Report the final year value as the answer.
The exact wording is not important. What is important is that pseudocode describes a
sequence of steps that is
• Unambiguous
• Executable
• Terminating
pseudocode is an
informal description
of a sequence of
steps for solving a
problem.
figure 9
the software development process
A method is unambiguous when there are precise
instructions for what to do at each step and where
to go next. There is no room for guesswork or cre-
ativity. A method is executable when each step can
be carried out in practice. Had we asked to use the
actual interest rate that will be charged in years to
come, and not a fixed rate of 5 percent per year, our
method would not have been executable, because
there is no way for anyone to know what that
interest rate will be. A method is terminating if it
will eventually come to an end. In our example, it
requires a bit of thought to see that the method will
not go on forever: With every step, the balance
goes up by at least $500, so eventually it must reach
$20,000.
A sequence of steps that is unambiguous, executable, and terminating is called an
algorithm. We have found an algorithm to solve our investment problem, and thus
we can find the solution by programming a computer. The existence of an algorithm
is an essential prerequisite for programming a task. You need to first discover and
describe an algorithm for the task that you want to solve before you start program-
ming (see Figure 9).
20. Suppose the interest rate was 20 percent. How long would it take for the invest-
ment to double?
21. Suppose your cell phone carrier charges you $29.95 for up to 300 minutes of
calls, and $0.45 for each additional minute, plus 12.5 percent taxes and fees. Give
an algorithm to compute the monthly charge from a given number of minutes.
22. Consider the following pseudocode for finding the most attractive photo from a
sequence of photos:
Pick the first photo and call it "the best so far".
For each photo in the sequence
If it is more attractive than the "best so far"
Discard "the best so far".
Call this photo "the best so far".
The photo called "the best so far" is the most attractive photo in the sequence.
Is this an algorithm that will find the most attractive photo?
Understand
the problem
Develop and
describe an
algorithm
Translate
the algorithm
into C++
Test the
algorithm with
simple inputs
Compile and test
your program
An algorithm is a recipe for
finding a solution.
an algorithm for
solving a problem is
a sequence of steps
that is unambiguous,
executable, and
terminating.
s e L f C h e C k
cfe2_ch01_p1_28.indd 18 10/25/10 1:03 PM
1.7 problem solving: algorithm design 19
figure 9
the software development process
A method is unambiguous when there are precise
instructions for what to do at each step and where
to go next. There is no room for guesswork or cre-
ativity. A method is executable when each step can
be carried out in practice. Had we asked to use the
actual interest rate that will be charged in years to
come, and not a fixed rate of 5 percent per year, our
method would not have been executable, because
there is no way for anyone to know what that
interest rate will be. A method is terminating if it
will eventually come to an end. In our example, it
requires a bit of thought to see that the method will
not go on forever: With every step, the balance
goes up by at least $500, so eventually it must reach
$20,000.
A sequence of steps that is unambiguous, executable, and terminating is called an
algorithm. We have found an algorithm to solve our investment problem, and thus
we can find the solution by programming a computer. The existence of an algorithm
is an essential prerequisite for programming a task. You need to first discover and
describe an algorithm for the task that you want to solve before you start program-
ming (see Figure 9).
20. Suppose the interest rate was 20 percent. How long would it take for the invest-
ment to double?
21. Suppose your cell phone carrier charges you $29.95 for up to 300 minutes of
calls, and $0.45 for each additional minute, plus 12.5 percent taxes and fees. Give
an algorithm to compute the monthly charge from a given number of minutes.
22. Consider the following pseudocode for finding the most attractive photo from a
sequence of photos:
Pick the first photo and call it "the best so far".
For each photo in the sequence
If it is more attractive than the "best so far"
Discard "the best so far".
Call this photo "the best so far".
The photo called "the best so far" is the most attractive photo in the sequence.
Is this an algorithm that will find the most attractive photo?
Understand
the problem
Develop and
describe an
algorithm
Translate
the algorithm
into C++
Test the
algorithm with
simple inputs
Compile and test
your program
An algorithm is a recipe for
finding a solution.
an algorithm for
solving a problem is
a sequence of steps
that is unambiguous,
executable, and
terminating.
s e L f C h e C k
cfe2_ch01_p1_28.indd 19 10/25/10 1:03 PM
20 Chapter 1 Introduction
23. Suppose each photo in Self Check 22 had a price tag. Give an algorithm for find-
ing the most expen sive photo.
24. Suppose you have a random sequence of black and white marbles and want to
rearrange it so that the black and white marbles are grouped together. Consider
this algorithm:
Repeat until sorted
Locate the first black marble that is preceded by a white marble, and switch them.
What does the algorithm do with the sequence mlmll? Spell out the steps
until the algorithm stops.
25. Suppose you have a random sequence of colored marbles. Consider this
pseudocode:
Repeat until sorted
Locate the first marble that is preceded by a marble of a different color, and switch them.
Why is this not an algorithm?
Practice It Now you can try these exercises at the end of the chapter: R1.13, R1.14.
step 1 Determine the inputs and outputs.
In our sample problem, we have these inputs:
• purchase price1 and fuel efficiency1, the price and fuel efficiency (in mpg) of the first car
• purchase price2 and fuel efficiency2, the price and fuel efficiency of the second car
We simply want to know which car is the better buy. That is the desired output.
step 2 Break down the problem into smaller tasks.
For each car, we need to know the total cost of driving it. Let’s do this computation separately
for each car. Once we have the total cost for each car, we can decide which car is the better deal.
The total cost for each car is purchase price + operating cost.
We assume a constant usage and gas price for ten years, so the operating cost depends on the
cost of driving the car for one year.
The operating cost is 10 x annual fuel cost.
The annual fuel cost is price per gallon x annual fuel consumed.
The annual fuel consumed is annual miles driven / fuel efficiency. For example, if you drive the car
for 15,000 miles and the fuel efficiency is 15 miles/gallon, the car consumes 1,000 gallons.
h o W t o 1 . 1 describing an algorithm with Pseudocode
Before you are ready to write a program in C++, you need to develop an algorithm—a method
for arriving at a solution for a particular problem. Describe the algorithm in pseudocode: a
sequence of precise steps formulated in English.
For example, consider this problem: You have the choice of
buying two cars. One is more fuel efficient than the other, but also
more expensive. You know the price and fuel efficiency (in miles
per gallon, mpg) of both cars. You plan to keep the car for ten
years. Assume a price of $4 per gallon of gas and usage of 15,000
miles per year. You will pay cash for the car and not worry about
financing costs. Which car is the better deal?
step 3 Describe each subtask in pseudocode.
In your description, arrange the steps so that any intermediate values are computed before
they are needed in other computations. For example, list the step
total cost = purchase price + operating cost
after you have computed operating cost.
Here is the algorithm for deciding which car to buy:
For each car, compute the total cost as follows:
annual fuel consumed = annual miles driven / fuel efficiency
annual fuel cost = price per gallon x annual fuel consumed
operating cost = 10 x annual fuel cost
total cost = purchase price + operating cost
If total cost1 < total cost2
Choose car1.
Else
Choose car2.
step 4 Test your pseudocode by working a problem.
We will use these sample values:
Car 1: $25,000, 50 miles/gallon
Car 2: $20,000, 30 miles/gallon
Here is the calculation for the cost of the first car:
annual fuel consumed = annual miles driven / fuel efficiency = 15000 / 50 = 300
annual fuel cost = price per gallon x annual fuel consumed = 4 x 300 = 1200
operating cost = 10 x annual fuel cost = 10 x 1200 = 12000
total cost = purchase price + operating cost = 25000 + 12000 = 37000
Similarly, the total cost for the second car is $40,000. Therefore, the output of the algorithm is
to choose car 1.
define “computer program” and programming.
• Computers execute very basic instructions in rapid succession.
• A computer program is a sequence of instructions and decisions.
• Programming is the act of designing and implementing computer programs.
describe the components of a computer.
• The central processing unit (CPU) performs program control and data
processing.
• Storage devices include memory and secondary storage.
W o r k e d e x a M p l e 1 . 1 Writing an algorithm for tiling a floor
This Worked Example shows how to develop an algorithm for laying
tile in an alternating pattern of colors.
C h a p t e r s u M M a r Y
cfe2_ch01_p1_28.indd 20 10/25/10 1:03 PM
Chapter summary 21
23. Suppose each photo in Self Check 22 had a price tag. Give an algorithm for find-
ing the most expen sive photo.
24. Suppose you have a random sequence of black and white marbles and want to
rearrange it so that the black and white marbles are grouped together. Consider
this algorithm:
Repeat until sorted
Locate the first black marble that is preceded by a white marble, and switch them.
What does the algorithm do with the sequence mlmll? Spell out the steps
until the algorithm stops.
25. Suppose you have a random sequence of colored marbles. Consider this
pseudocode:
Repeat until sorted
Locate the first marble that is preceded by a marble of a different color, and switch them.
Why is this not an algorithm?
Practice It Now you can try these exercises at the end of the chapter: R1.13, R1.14.
step 1 Determine the inputs and outputs.
In our sample problem, we have these inputs:
• purchase price1 and fuel efficiency1, the price and fuel efficiency (in mpg) of the first car
• purchase price2 and fuel efficiency2, the price and fuel efficiency of the second car
We simply want to know which car is the better buy. That is the desired output.
step 2 Break down the problem into smaller tasks.
For each car, we need to know the total cost of driving it. Let’s do this computation separately
for each car. Once we have the total cost for each car, we can decide which car is the better deal.
The total cost for each car is purchase price + operating cost.
We assume a constant usage and gas price for ten years, so the operating cost depends on the
cost of driving the car for one year.
The operating cost is 10 x annual fuel cost.
The annual fuel cost is price per gallon x annual fuel consumed.
The annual fuel consumed is annual miles driven / fuel efficiency. For example, if you drive the car
for 15,000 miles and the fuel efficiency is 15 miles/gallon, the car consumes 1,000 gallons.
h o W t o 1 . 1 describing an algorithm with Pseudocode
Before you are ready to write a program in C++, you need to develop an algorithm—a method
for arriving at a solution for a particular problem. Describe the algorithm in pseudocode: a
sequence of precise steps formulated in English.
For example, consider this problem: You have the choice of
buying two cars. One is more fuel efficient than the other, but also
more expensive. You know the price and fuel efficiency (in miles
per gallon, mpg) of both cars. You plan to keep the car for ten
years. Assume a price of $4 per gallon of gas and usage of 15,000
miles per year. You will pay cash for the car and not worry about
financing costs. Which car is the better deal?
step 3 Describe each subtask in pseudocode.
In your description, arrange the steps so that any intermediate values are computed before
they are needed in other computations. For example, list the step
total cost = purchase price + operating cost
after you have computed operating cost.
Here is the algorithm for deciding which car to buy:
For each car, compute the total cost as follows:
annual fuel consumed = annual miles driven / fuel efficiency
annual fuel cost = price per gallon x annual fuel consumed
operating cost = 10 x annual fuel cost
total cost = purchase price + operating cost
If total cost1 < total cost2
Choose car1.
Else
Choose car2.
step 4 Test your pseudocode by working a problem.
We will use these sample values:
Car 1: $25,000, 50 miles/gallon
Car 2: $20,000, 30 miles/gallon
Here is the calculation for the cost of the first car:
annual fuel consumed = annual miles driven / fuel efficiency = 15000 / 50 = 300
annual fuel cost = price per gallon x annual fuel consumed = 4 x 300 = 1200
operating cost = 10 x annual fuel cost = 10 x 1200 = 12000
total cost = purchase price + operating cost = 25000 + 12000 = 37000
Similarly, the total cost for the second car is $40,000. Therefore, the output of the algorithm is
to choose car 1.
define “computer program” and programming.
• Computers execute very basic instructions in rapid succession.
• A computer program is a sequence of instructions and decisions.
• Programming is the act of designing and implementing computer programs.
describe the components of a computer.
• The central processing unit (CPU) performs program control and data
processing.
• Storage devices include memory and secondary storage.
W o r k e d e x a M p l e 1 . 1 Writing an algorithm for tiling a floor
This Worked Example shows how to develop an algorithm for laying
tile in an alternating pattern of colors.
C h a p t e r s u M M a r Y
Available online at www.wiley.com/college/horstmann.
cfe2_ch01_p1_28.indd 21 10/25/10 1:03 PM
www.wiley.com/college/horstmann
22 Chapter 1 Introduction
describe the process of translating high-level languages to machine code.
• Computer programs are stored as machine instructions in a code that depends on
the processor type.
• C++ is a general-purpose language that is in widespread use for systems and
embedded programming.
• High-level programming languages are independent of the processor.
Become familiar with your C++ programming environment.
• Set aside some time to become familiar with the programming environment that
you will use for your class work.
• An editor is a program for entering and modifying text, such as a C++ program.
• C++ is case sensitive. You must be careful about distinguishing between upper-
and lowercase letters.
• Develop a strategy for keeping backup copies of your work
before disaster strikes.
• The compiler translates C++ programs into machine code.
• The linker combines machine code with library code into
an executable program.
describe the building blocks of a simple program.
• Every C++ program contains a function called main.
• Use cout and the << operator to display values on the screen.
• Enclose text strings in quotation marks.
• Use + to add two numbers and * to multiply two numbers.
• Send endl to cout to end a line of displayed output.
• End each statement with a semicolon.
Classify program errors as compile-time and run-time errors.
• A compile-time error is a violation of the programming language rules that is
detected by the compiler.
• A run-time error causes a program to take an action that the programmer did not
intend.
• The programmer is responsible for inspecting and testing the program to guard
against run-time errors.
Write pseudocode for simple algorithms.
• Pseudocode is an informal description of a sequence of steps
for solving a problem.
• An algorithm for solving a problem is a sequence of steps that
is unambiguous, executable, and terminating.
r1.1 Explain the difference between using a computer program and programming a
computer.
r1.2 Which parts of a computer can store program code? Which can store user data?
r1.3 Which parts of a computer serve to give information to the user? Which parts take
user input?
r1.4 A toaster is a single-function device, but a computer can be programmed to carry out
different tasks. Is your cell phone a single-function device, or is it a programma ble
computer? (Your answer will depend on your cell phone model.)
r1.5 Explain two benefits of using C++ over machine code.
r1.6 On your own computer or on your lab computer, find the exact location (folder or
directory name) of
a. The sample file hello.cpp (after you saved it in your development environment).
b. The standard header file
r1.7 What does this program print?
#include
using namespace std;
int main()
{
cout << "6 * 7 = " << 6 * 7 << endl;
return 0;
}
r1.8 What does this program print?
#include
using namespace std;
int main()
{
cout << "Hello" << "World" << endl;
return 0;
}
Pay close attention to spaces.
r1.9 What does this program print?
#include
using namespace std;
int main()
{
cout << "Hello" << endl << "World" << endl;
return 0;
}
r1.10 Write three versions of the hello.cpp program that have different compile-time
errors. Write a version that has a run-time error.
r1.11 How do you discover compile-time errors? How do you discover run-time errors?
r e V I e W e x e r C I s e s
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review exercises 23
r1.1 Explain the difference between using a computer program and programming a
computer.
r1.2 Which parts of a computer can store program code? Which can store user data?
r1.3 Which parts of a computer serve to give information to the user? Which parts take
user input?
r1.4 A toaster is a single-function device, but a computer can be programmed to carry out
different tasks. Is your cell phone a single-function device, or is it a programma ble
computer? (Your answer will depend on your cell phone model.)
r1.5 Explain two benefits of using C++ over machine code.
r1.6 On your own computer or on your lab computer, find the exact location (folder or
directory name) of
a. The sample file hello.cpp (after you saved it in your development environment).
b. The standard header file
r1.7 What does this program print?
#include
using namespace std;
int main()
{
cout << "6 * 7 = " << 6 * 7 << endl;
return 0;
}
r1.8 What does this program print?
#include
using namespace std;
int main()
{
cout << "Hello" << "World" << endl;
return 0;
}
Pay close attention to spaces.
r1.9 What does this program print?
#include
using namespace std;
int main()
{
cout << "Hello" << endl << "World" << endl;
return 0;
}
r1.10 Write three versions of the hello.cpp program that have different compile-time
errors. Write a version that has a run-time error.
r1.11 How do you discover compile-time errors? How do you discover run-time errors?
r e V I e W e x e r C I s e s
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24 Chapter 1 Introduction
r1.12 Write an algorithm to settle the following question: A bank account starts out with
$10,000. Interest is compounded monthly at 6 percent per year (0.5 percent per
month). Every month, $500 is withdrawn to meet college expenses. After how many
years is the account depleted?
r1.13 Consider the question in Exercise R1.12. Suppose the numbers ($10,000, 6 percent,
$500) were user selectable. Are there values for which the algorithm you developed
would not terminate? If so, change the algorithm to make sure it always terminates.
r1.14 In order to estimate the cost of painting a house, a painter needs to know the surface
area of the exterior. Develop an algorithm for computing that value. Your inputs are
the width, length, and height of the house, the number of windows and doors, and
their dimensions. (Assume the windows and doors have a uniform size.)
r1.15 You want to decide whether you should drive your car to work or take the train.
You know the one-way distance from your home to your place of work, and the fuel
efficiency of your car (in miles per gallon). You also know the one-way price of a
train ticket. You assume the cost of gas at $4 per gallon, and car maintenance at 5
cents per mile. Write an algorithm to decide which commute is cheaper.
r1.16 You want to find out which fraction of your car use is for commuting to work, and
which is for personal use. You know the one-way distance from your home to your
place of work. For a particular period, you recorded the beginning and ending
mile age on the odometer and the number of work days. Write an algorithm to settle
this question.
r1.17 In the problem described in How To 1.1 on page 20, you made assumptions about the
price of gas and the annual usage. Ideally, you would like to know which car is the
better deal without making these assumptions. Why can’t a computer program solve
that problem?
r1.18 The value of p can be computed according to the following formula:
π
4
1
1
3
1
5
1
7
1
9
= − + − + − �
Write an algorithm to compute p. Because the formula is an infinite series and an
algorithm must stop after a finite number of steps, you should stop when you have
the result determined to six significant dig its.
r1.19 Suppose you put your younger brother in charge of backing up your work. Write a
set of detailed instructions for carrying out his task. Explain how often he should do
it, and what files he needs to copy from which folder to which location. Explain how
he should verify that the backup was carried out correctly.
engineering r1.20 The San Francisco taxi commission set the following rates for 2010:
• First 1̸5th of a mile: $3.10
• Each additional 1̸5th of a mile or fraction thereof: $0.45
• Each minute of waiting or traffic delay: $0.45
The charge for “waiting or traffic delay” applies instead of the mileage charge for
each minute in which the speed is slower than the break-even point. The break-even
point is the speed at which 1̸5th of a mile is traversed in one minute.
Develop an algorithm that yields the fare for traveling a given distance in a given
amount of time, assuming that the taxi moves at a constant speed.
engineering r1.21 Suppose you know how long it takes a car to accelerate from 0 to 60 miles per hour.
Develop an algorithm for computing the time required to travel a given distance (for
example 5 miles), assuming that the car is initially at rest, accelerates to a given speed
(for example 25 miles per hour), and drives at that speed until the distance is covered.
Hint: An object that starts at rest and accelerates at a constant rate a for t seconds
travels a distance of .
P1.1 Write a program that prints a greeting of your choice, perhaps in another language.
P1.2 Write a program that prints the message, “Hello, my name is Hal!” Then, on a new
line, the program should print the message “What would you like me to do?” Then
it’s the user’s turn to type in an input. You haven’t yet learned how to do it—just use
the following lines of code:
string user_input;
getline(cin, user_input);
Finally, the program should ignore the user input and print the message “I am sorry,
I cannot do that.”
This program uses the string data type. To access this feature, you must place the line
#include
before the main function.
Here is a typical program run. The user input is printed in color.
Hello, my name is Hal!
What would you like me to do?
Clean up my room
I am sorry, I cannot do that.
When running the program, remember to press the Enter key after typing the last
word of the input line.
P1.3 Write a program that prints out a message “Hello, my name is Hal!” Then, on a new
line, the program should print the message “What is your name?” As in Exercise
P1.2, just use the following lines of code:
string user_name;
getline(cin, user_name);
Finally, the program should print the message “Hello, user name. I am glad to meet
you!” To print the user name, simply use
cout << user_name;
As in Exercise P1.2, you must place the line
#include
before the main function.
Here is a typical program run. The user input is printed in color.
Hello, my name is Hal!
What is your name?
Dave
Hello, Dave. I am glad to meet you!
p r o G r a M M I n G e x e r C I s e s
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programming exercises 25
engineering r1.21 Suppose you know how long it takes a car to accelerate from 0 to 60 miles per hour.
Develop an algorithm for computing the time required to travel a given distance (for
example 5 miles), assuming that the car is initially at rest, accelerates to a given speed
(for example 25 miles per hour), and drives at that speed until the distance is covered.
Hint: An object that starts at rest and accelerates at a constant rate a for t seconds
travels a distance of s at=
1
2
2.
P1.1 Write a program that prints a greeting of your choice, perhaps in another language.
P1.2 Write a program that prints the message, “Hello, my name is Hal!” Then, on a new
line, the program should print the message “What would you like me to do?” Then
it’s the user’s turn to type in an input. You haven’t yet learned how to do it—just use
the following lines of code:
string user_input;
getline(cin, user_input);
Finally, the program should ignore the user input and print the message “I am sorry,
I cannot do that.”
This program uses the string data type. To access this feature, you must place the line
#include
before the main function.
Here is a typical program run. The user input is printed in color.
Hello, my name is Hal!
What would you like me to do?
Clean up my room
I am sorry, I cannot do that.
When running the program, remember to press the Enter key after typing the last
word of the input line.
P1.3 Write a program that prints out a message “Hello, my name is Hal!” Then, on a new
line, the program should print the message “What is your name?” As in Exercise
P1.2, just use the following lines of code:
string user_name;
getline(cin, user_name);
Finally, the program should print the message “Hello, user name. I am glad to meet
you!” To print the user name, simply use
cout << user_name;
As in Exercise P1.2, you must place the line
#include
before the main function.
Here is a typical program run. The user input is printed in color.
Hello, my name is Hal!
What is your name?
Dave
Hello, Dave. I am glad to meet you!
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26 Chapter 1 Introduction
P1.4 Write a program that prints the sum of the first ten positive integers, 1 + 2 + … + 10.
P1.5 Write a program that prints the product of the first ten positive integers, 1 × 2 × … × 10.
(Use * for multiplication in C++.)
P1.6 Write a program that prints the balance of an account that earns 5 percent interest
per year after the first, second, and third year.
P1.7 Write a program that displays your name inside a box on the terminal screen, like
this:
Dave
Do your best to approximate lines with characters such as | – +.
P1.8 Write a program that prints your name in large letters, such as
* * ** **** **** * *
* * * * * * * * * *
***** * * **** **** * *
* * ****** * * * * *
* * * * * * * * *
P1.9 Write a program that prints a face similar to (but different from) the following:
/////
+—–+
(| o o |)
| ^ |
| ‘-‘ |
+—–+
P1.10 Write a program that prints a house that looks exactly like the following:
/\
/ \
+—-+
| .-.|
| | ||
+-+-++
P1.11 Write a program that prints an animal speaking a greeting, similar to (but different
from) the following:
/\_/\ —–
( ‘ ‘ ) / Hello \
( – ) < Junior |
| | | \ Coder!/
(__|__) -----
P1.12 Write a program that prints three items, such as the names of your three best friends
or favorite movies, on three separate lines.
P1.13 Write a program that prints a poem of your choice. If you don’t have a favorite
poem, search the Internet for “Emily Dickinson” or “e e cummings”.
P1.14 Write a program that prints an imitation of a Piet Mondrian painting. (Search the
Internet if you are not familiar with his paintings.) Use character sequences such as
@@@ or ::: to indicate different colors, and use - and | to form lines.
P1.15 Write a program that prints the United States flag, using * and = characters.
engineering P1.16 The atmospheres of the gas giant planets (Jupiter, Saturn, Uranus, and Neptune) are
mostly comprised of hydrogen (H2) followed by helium (He). The atmospheres of
the terrestrial planets are mostly comprised of carbon dioxide (CO2) followed by
nitrogen (N2) for Venus and Mars, and for Earth, mostly Nitrogen (N2) followed by
Oxygen (O2). Write a program that outputs this information in a chart with four
columns for the type of planet, the name of the planet, its primary atmospheric gas,
and secondary atmospheric gas.
engineering P1.17 Write a program that displays the following image, using characters such as / \ - | +
for the lines. Write Ω as “Ohm”.
1. A program that reads the data on the CD and sends output to the speakers and the
screen.
2. A CD player can do one thing—play music CDs. It cannot execute programs.
3. Nothing.
4. In secondary storage, typically a hard disk.
5. The central processing unit.
6. Software.
7. No—a compiler is intended for programmers, to translate high-level programming
instructions into machine code.
8. System software and embedded systems
9. Programs are stored in files, and files are stored in folders or directories.
10. You will see a seemingly random sequence of characters and symbols.
11. You back up your files and folders.
12. Replace "World" with your name, for example:
cout << "Hello, Harry!" << endl;
13. There are no quotes around Goodbye, World!.
14. It prints HelloWorld, without a space or comma.
15. 42, without a space.
16. Hello
World
with a blank line between the two words.
17. A compile-time error.
18. It is a run-time error. After all, the program had been compiled in order for you to
run it.
19. When a program has compiler errors, no executable file is produced, and there is
nothing to run.
a n s W e r s t o s e l F - C h e C k Q u e s t I o n s
cfe2_ch01_p1_28.indd 26 10/25/10 1:03 PM
answers to self-Check Questions 27
engineering P1.16 The atmospheres of the gas giant planets (Jupiter, Saturn, Uranus, and Neptune) are
mostly comprised of hydrogen (H2) followed by helium (He). The atmospheres of
the terrestrial planets are mostly comprised of carbon dioxide (CO2) followed by
nitrogen (N2) for Venus and Mars, and for Earth, mostly Nitrogen (N2) followed by
Oxygen (O2). Write a program that outputs this information in a chart with four
columns for the type of planet, the name of the planet, its primary atmospheric gas,
and secondary atmospheric gas.
engineering P1.17 Write a program that displays the following image, using characters such as / \ - | +
for the lines. Write Ω as “Ohm”.
+
–12 V
5 kΩ
10 kΩ
6 kΩ
4 kΩ
1. A program that reads the data on the CD and sends output to the speakers and the
screen.
2. A CD player can do one thing—play music CDs. It cannot execute programs.
3. Nothing.
4. In secondary storage, typically a hard disk.
5. The central processing unit.
6. Software.
7. No—a compiler is intended for programmers, to translate high-level programming
instructions into machine code.
8. System software and embedded systems
9. Programs are stored in files, and files are stored in folders or directories.
10. You will see a seemingly random sequence of characters and symbols.
11. You back up your files and folders.
12. Replace "World" with your name, for example:
cout << "Hello, Harry!" << endl;
13. There are no quotes around Goodbye, World!.
14. It prints HelloWorld, without a space or comma.
15. 42, without a space.
16. Hello
World
with a blank line between the two words.
17. A compile-time error.
18. It is a run-time error. After all, the program had been compiled in order for you to
run it.
19. When a program has compiler errors, no executable file is produced, and there is
nothing to run.
a n s W e r s t o s e l F - C h e C k Q u e s t I o n s
cfe2_ch01_p1_28.indd 27 10/25/10 1:03 PM
28 Chapter 1 Introduction
20. 4 years:
0 10,000
1 12,000
2 14,400
3 17,280
4 20,736
21. Is the number of minutes at most 300?
a. If so, the answer is $29.95 × 1.125 = $33.70.
b. If not,
1. Compute the difference: (number of minutes) – 300.
2. Multiply that difference by 0.45.
3. Add $29.95.
4. Multiply the total by 1.125. That is the answer.
22. No. The step If it is more attractive than the "best so far" is not executable because there is
no objective way of deciding which of two photos is more attractive.
23. Pick the first photo and call it "the most expensive so far".
For each photo in the sequence
If it is more expensive than "the most expensive so far"
Discard "the most expensive so far".
Call this photo "the most expensive so far".
The photo called "the most expensive so far" is the most expensive photo in the sequence.
24. The first black marble that is preceded by a white one is marked in blue:
mlmll
Switching the two yields
lmmll
The next black marble to be switched is
lmmll
yielding
lmlml
The next steps are
llmml
llmlm
lllmm
Now the sequence is sorted.
25. The sequence doesn’t terminate. Consider the input mlmlm. The first two mar-
bles keep getting switched.
step 1 Determine the inputs and outputs.
The inputs are the floor dimensions (length × width),
measured in inches. The output is a tiled floor.
step 2 Break down the problem into smaller tasks.
A natural subtask is to lay one row of tiles. If you can
solve that task, then you can solve the problem by lay-
ing one row next to the other, starting from a wall, until
you reach the opposite wall.
How do you lay a row? Start with a tile at one wall.
If it is white, put a black one next to it. If it is black, put
a white one next to it. Keep going until you reach the
opposite wall. The row will contain width / 4 tiles.
step 3 Describe each subtask in pseudocode.
In the pseudocode, you want to be more precise about exactly where the tiles are placed.
Place a black tile in the northwest corner.
While the floor is not yet filled, repeat the following steps:
Repeat this step width / 4 – 1 times:
Place a tile east of the previously placed tile. If the previously placed tile was white,
pick a black one; otherwise, a white one.
Locate the tile at the beginning of the row that you just placed. If there is space to the south, place a tile of
the opposite color below it.
step 4 Test your pseudocode by working a problem.
Suppose you want to tile an area measuring 20 × 12 inches.
The first step is to place a black tile in the northwest corner.
Next, alternate four tiles until reaching the east wall. (width / 4 – 1 = 20 / 4 – 1 = 4)
There is room to the south. Locate the tile at the beginning of the completed row. It is black.
Place a white tile south of it.
W o r k e d e x a M p l e 1 . 1 Writing an algorithm for tiling a floor
Your task is to tile a rectangular bathroom floor with alternating black and white tiles measur-
ing 4 × 4 inches. The floor dimensions, measured in inches, are multiples of 4.
cfe2_ch01_p1_28.indd 28 10/25/10 1:03 PM
2C h a p t e r
29
F u n d a m e n ta l
d ata t y p e s
to be able to define and initialize
variables and constants
to understand the properties and
limitations of integer and floating-point numbers
to write arithmetic expressions and assignment statements in C++
to appreciate the importance of comments and good code layout
to create programs that read and process input, and display the results
to process strings, using the standard C++ string type
C h a p t e r G o a l s
C h a p t e r C o n t e n t s
2.1 Variables 30
Syntax 2.1: Variable definition 31
Syntax 2.2: assignment 34
Common Error 2.1: using undefined Variables 37
Common Error 2.2: using uninitialized
Variables 37
Programming Tip 2.1: Choose descriptive
Variable names 38
Special Topic 2.1: numeric types in C++ 38
Special Topic 2.2: numeric ranges and
precisions 39
Programming Tip 2.2: do not use magic
numbers 39
2.2 arithmetic 40
Common Error 2.3: unintended Integer
division 43
Common Error 2.4: unbalanced parentheses 44
Common Error 2.5: Forgetting header Files 45
Common Error 2.6: roundoff errors 45
Programming Tip 2.3: spaces in expressions 46
Special Topic 2.3: Casts 46
Special Topic 2.4: Combining assignment and
arithmetic 47
Random Fact 2.1: the pentium Floating-
point Bug 47
2.3 input and Output 48
Syntax 2.3: Input statement 48
2.4 prOblem sOlVing: First dO it
by hand 52
Worked Example 2.1: Computing travel time
How To 2.1: Carrying out Computations 54
Worked Example 2.2: Computing the Cost
of stamps
2.5 strings 56
Random Fact 2.2: International alphabets
and unicode 61
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30
numbers and character strings (such as the ones on this
display board) are important data types in any C++ program.
In this chapter, you will learn how to work with numbers
and text, and how to write simple programs that perform
useful tasks with them.
2.1 Variables
When your program carries out computations, you will want to store values so that
you can use them later. In a C++ program, you use variables to store values. In this
section, you will learn how to define and use variables.
To illustrate the use of variables, we
will develop a pro gram that solves the
following problem. Soft drinks are sold
in cans and bottles. A store offers a six-
pack of 12-ounce cans for the same price
as a two-liter bottle. Which should you
buy? (12 fluid ounces equal approxi-
mately 0.355 liters.)
In our program, we will define vari-
ables for the number of cans per pack
and for the volume of each can. Then we
will compute the volume of a six-pack in
liters and print out the answer.
2.1.1 Variable definitions
The following statement defines a variable named cans_per_pack:
int cans_per_pack = 6;
A variable is a storage location in a computer program. Each variable has a name and
holds a value.
A variable is similar to a parking space in a parking garage. The parking space has
an identifier (such as “J 053”), and it can hold a vehicle. A variable has a name (such as
cans_per_pack), and it can hold a value (such as 6).
What contains more soda? A six-pack of
12-ounce cans or a two-liter bottle?
a variable is a
storage location with
a name.
Like a variable in a computer
program, a parking space has
an identifier and a contents.
When defining a variable, you usually want to initialize it. That is, you specify the
value that should be stored in the variable. Consider again this variable definition:
int cans_per_pack = 6;
The variable cans_per_pack is initialized with the value 6.
Like a parking space that is restricted to a certain type of vehicle (such as a compact
car, motorcycle, or electric vehicle), a variable in C++ stores data of a specific type.
C++ supports quite a few data types: numbers, text strings, files, dates, and many
others. You must specify the type whenever you define a variable (see Syntax 2.1).
The cans_per_pack variable is an integer, a whole number without a fractional part.
In C++, this type is called int. (See the next section for more information about num-
ber types in C++.)
Note that the type comes before the variable name:
int cans_per_pack = 6;
Table 1 shows variations of variable definitions.
When defining a
variable, you usually
specify an initial
value.
When defining a
variable, you also
specify the type of
its values.
Each parking space is suitable
for a particular type of vehicle,
just as each variable holds a
value of a particular type.
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2.1 Variables 31
syntax 2.1 Variable definition
int cans_per_pack = 6;
A variable definition ends
with a semicolon.
Types introduced in
this chapter are
the number types
int and double
(page 32)
and the string type
(page 57).
Supplying an initial value is optional,
but it is usually a good idea.
See page 37.
See page 33 for rules and
examples of valid names.
Use a descriptive
variable name.
See page 38.
When defining a variable, you usually want to initialize it. That is, you specify the
value that should be stored in the variable. Consider again this variable definition:
int cans_per_pack = 6;
The variable cans_per_pack is initialized with the value 6.
Like a parking space that is restricted to a certain type of vehicle (such as a compact
car, motorcycle, or electric vehicle), a variable in C++ stores data of a specific type.
C++ supports quite a few data types: numbers, text strings, files, dates, and many
others. You must specify the type whenever you define a variable (see Syntax 2.1).
The cans_per_pack variable is an integer, a whole number without a fractional part.
In C++, this type is called int. (See the next section for more information about num-
ber types in C++.)
Note that the type comes before the variable name:
int cans_per_pack = 6;
Table 1 shows variations of variable definitions.
When defining a
variable, you usually
specify an initial
value.
When defining a
variable, you also
specify the type of
its values.
Each parking space is suitable
for a particular type of vehicle,
just as each variable holds a
value of a particular type.
cfe2_ch02_p29_74.indd 31 10/27/10 2:29 PM
32 Chapter 2 Fundamental data types
When a fractional part is required (such as in the number 0.355), we use floating-
point numbers. The most commonly used type for floating-point numbers in C++ is
called double. (If you want to know the reason, read Special Topic 2.1 on page 38.) Here
is the definition of a floating-point variable:
double can_volume = 0.355;
When a value such as 6 or 0.355 occurs in a C++ program, it is called a number literal.
Table 2 shows how to write integer and floating-point literals in C++.
2.1.3 Variable names
When you define a variable, you should pick a name that explains its purpose. For
example, it is better to use a descriptive name, such as can_volume, than a terse name,
such as cv.
In C++, there are a few simple rules for variable names:
1. Variable names must start with a letter or the underscore (_) character, and the
remaining characters must be letters, numbers, or underscores.
2. You cannot use other symbols such as $ or %. Spaces are not permitted inside
names either. You can use an underscore instead, as in can_volume.
3. Variable names are case-sensitive, that is, Can_volume and can_volume are different
names. For that rea son, it is a good idea to use only lowercase letters in variable
names.
4. You cannot use reserved words such as double or return as names; these words
are reserved exclu sively for their special C++ meanings. (See Appendix B.)
Table 3 shows examples of legal and illegal variable names in C++.
use the double type
for floating-point
numbers.
table 3 Variable names in C++
Variable name Comment
can_volume1 Variable names consist of letters, numbers, and the underscore
character.
x In mathematics, you use short variable names such as x or y. This is
legal in C++, but not very common, because it can make programs
harder to understand (see Programming Tip 2.1 on page 38).
! Can_volume caution: Variable names are case-sensitive.
This variable name is different from can_volume.
6pack error: Variable names cannot start with a number.
can volume error: Variable names cannot contain spaces.
double error: You cannot use a reserved word as a variable name.
ltr/fl.oz error: You cannot use symbols such as / or.
2.1.2
table 1 Variable definitions in C++
Variable name Comment
int cans = 6; Defines an integer variable and initializes it with 6.
int total = cans + bottles; The initial value need not be a constant. (Of course, cans and bottles
must have been previously defined.)
int bottles = "10"; error: You cannot initialize a number with a string.
int bottles; Defines an integer variable without initializing it. This can be a cause
for errors—see Common Error 2.2 on page 37.
int cans, bottles; Defines two integer variables in a single statement. In this book, we
will define each variable in a separate statement.
! bottles = 1; caution: The type is missing. This statement is not a definition but
an assignment of a new value to an existing variable—see Section
2.1.4 on page 34.
number types
In C++, there are several different types of numbers. You use the integer number
type, called int in C++, to denote a whole number without a fractional part. For
example, there must be an integer number of cans in any pack of cans—you cannot
have a fraction of a can.
use the int type
for numbers that
cannot have a
fractional part.
table 2 number literals in C++
number type Comment
6 int An integer has no fractional part.
–6 int Integers can be negative.
0 int Zero is an integer.
0.5 double A number with a fractional part has type double.
1.0 double An integer with a fractional part .0 has type double.
1E6 double A number in exponential notation: 1 × 106 or 1000000.
Numbers in exponential notation always have type double.
2.96E-2 double Negative exponent: 2.96 × 10–2 = 2.96 / 100 = 0.0296
100,000 error: Do not use a comma as a decimal separator.
3 1/2 error: Do not use fractions; use decimal notation: 3.5.
cfe2_ch02_p29_74.indd 32 10/27/10 2:29 PM
2.1 Variables 33
When a fractional part is required (such as in the number 0.355), we use floating-
point numbers. The most commonly used type for floating-point numbers in C++ is
called double. (If you want to know the reason, read Special Topic 2.1 on page 38.) Here
is the definition of a floating-point variable:
double can_volume = 0.355;
When a value such as 6 or 0.355 occurs in a C++ program, it is called a number literal.
Table 2 shows how to write integer and floating-point literals in C++.
2.1.3 Variable names
When you define a variable, you should pick a name that explains its purpose. For
example, it is better to use a descriptive name, such as can_volume, than a terse name,
such as cv.
In C++, there are a few simple rules for variable names:
1. Variable names must start with a letter or the underscore (_) character, and the
remaining characters must be letters, numbers, or underscores.
2. You cannot use other symbols such as $ or %. Spaces are not permitted inside
names either. You can use an underscore instead, as in can_volume.
3. Variable names are case-sensitive, that is, Can_volume and can_volume are different
names. For that rea son, it is a good idea to use only lowercase letters in variable
names.
4. You cannot use reserved words such as double or return as names; these words
are reserved exclu sively for their special C++ meanings. (See Appendix B.)
Table 3 shows examples of legal and illegal variable names in C++.
use the double type
for floating-point
numbers.
table 3 Variable names in C++
Variable name Comment
can_volume1 Variable names consist of letters, numbers, and the underscore
character.
x In mathematics, you use short variable names such as x or y. This is
legal in C++, but not very common, because it can make programs
harder to understand (see Programming Tip 2.1 on page 38).
! Can_volume caution: Variable names are case-sensitive.
This variable name is different from can_volume.
6pack error: Variable names cannot start with a number.
can volume error: Variable names cannot contain spaces.
double error: You cannot use a reserved word as a variable name.
ltr/fl.oz error: You cannot use symbols such as / or.
cfe2_ch02_p29_74.indd 33 10/27/10 2:29 PM
34 Chapter 2 Fundamental data types
2.1.4 the assignment statement
You use the assignment statement to place a new value into a variable. Here is an
example:
cans_per_pack = 8;
The left-hand side of an assignment statement consists of a variable. The right-hand
side is an expression that has a value. That value is stored in the variable, overwriting
its previous contents.
There is an important difference between a variable definition and an assignment
statement:
int cans_per_pack = 6; // Variable definition
...
cans_per_pack = 8; // Assignment statement
The first statement is the definition of cans_per_pack. It is an instruction to create a new
variable of type int, to give it the name cans_per_pack, and to initialize it with 6. The
second statement is an assignment statement: an instruction to replace the contents of
the existing variable cans_per_pack with another value.
The = sign doesn’t mean that the left-hand side is equal to the right-hand side. The
expression on the right is evaluated, and its value is placed into the variable on the left.
Do not confuse this assignment operation with the = used in algebra to denote
equality. The assign ment operator is an instruction to do something, namely place
a value into a variable. The mathematical equality states the fact that two values are
equal.
For example, in C++, it is perfectly legal to write
total_volume = total_volume + 2;
It means to look up the value stored in the variable total_volume, add 2 to it, and place
the result back into total_volume. (See Figure 1.) The net effect of executing this state-
ment is to increment total_volume by 2. For example, if total_volume was 2.13 before
execution of the statement, it is set to 4.13 afterwards. Of course, in mathematics it
would make no sense to write that x = x + 2. No value can equal itself plus 2.
an assignment
statement stores
a new value in a
variable, replacing
the previously
stored value.
the assignment
operator = does not
denote mathematical
equality.
syntax 2.2 assignment
double total = 0;
.
.
total = bottles * BOTTLE_VOLUME;
.
.
.
total = total + cans * CAN_VOLUME;
The name of a previously
defined variable
The same name
can occur on both sides.
See Figure 1.
The expression that replaces the previous value
This is an initialization
of a new variable,
NOT an assignment.
This is an assignment.
2.1.5 Constants
When a variable is defined with the reserved word const, its value can never change.
Constants are com monly written using capital letters to distinguish them visually
from regular variables:
const double BOTTLE_VOLUME = 2;
It is good programming style to use named constants in your program to explain the
meanings of numeric values. For example, compare the statements
double total_volume = bottles * 2;
and
double total_volume = bottles * BOTTLE_VOLUME;
A programmer reading the first statement may not understand the significance of the
number 2. The sec ond statement, with a named constant, makes the computation
much clearer.
2.1.6 Comments
As your programs get more complex, you should add comments, explanations for
human readers of your code. Here is an example:
const double CAN_VOLUME = 0.355; // Liters in a 12-ounce can
This comment explains the significance of the value 0.355 to a human reader. The
compiler does not pro cess comments at all. It ignores everything from a // delimiter
to the end of the line.
you cannot change
the value of a
variable that is
defined as const.
use comments to
add explanations
for humans who
read your code. the
compiler ignores
comments.
Just as a television commentator explains the news,
you use comments in your program to explain its behavior.
cfe2_ch02_p29_74.indd 34 10/27/10 2:29 PM
2.1 Variables 35
2.1.5
Figure 1 executing the assignment total_volume = total_volume + 2
1
total_volume =
total_volume + 2
2.13
2
total_volume =
4.13
4.13
total_volume + 2
Constants
When a variable is defined with the reserved word const, its value can never change.
Constants are com monly written using capital letters to distinguish them visually
from regular variables:
const double BOTTLE_VOLUME = 2;
It is good programming style to use named constants in your program to explain the
meanings of numeric values. For example, compare the statements
double total_volume = bottles * 2;
and
double total_volume = bottles * BOTTLE_VOLUME;
A programmer reading the first statement may not understand the significance of the
number 2. The sec ond statement, with a named constant, makes the computation
much clearer.
2.1.6 Comments
As your programs get more complex, you should add comments, explanations for
human readers of your code. Here is an example:
const double CAN_VOLUME = 0.355; // Liters in a 12-ounce can
This comment explains the significance of the value 0.355 to a human reader. The
compiler does not pro cess comments at all. It ignores everything from a // delimiter
to the end of the line.
you cannot change
the value of a
variable that is
defined as const.
use comments to
add explanations
for humans who
read your code. the
compiler ignores
comments.
Just as a television commentator explains the news,
you use comments in your program to explain its behavior.
cfe2_ch02_p29_74.indd 35 10/27/10 2:29 PM
36 Chapter 2 Fundamental data types
You use the // syntax for single-line comments. If you have a comment that spans
multiple lines, enclose it between /* and */ delimiters. The compiler ignores these
delimiters and everything in between.
Here is a typical example, a long comment at the beginning of a program, to explain
the program’s pur pose:
/*
This program computes the volume (in liters) of a six-pack of soda cans
and the total volume of a six-pack and a two-liter bottle.
*/
We are now ready to finish our program. The following program shows the use of
variables, constants, and the assignment statement. The program displays the volume
of a six-pack of cans and the total volume of the six-pack and a two-liter bottle. We
use constants for the can and bottle volumes. The total_volume variable is initialized
with the volume of the cans. Using an assignment statement, we add the bottle
volume.
ch02/volume1.cpp
1 #include
2
3 using namespace std;
4
5 /*
6 This program computes the volume (in liters) of a six-pack of soda
7 cans and the total volume of a six-pack and a two-liter bottle.
8 */
9 int main()
10 {
11 int cans_per_pack = 6;
12 const double CAN_VOLUME = 0.355; // Liters in a 12-ounce can
13 double total_volume = cans_per_pack * CAN_VOLUME;
14
15 cout << "A six-pack of 12-ounce cans contains "
16 << total_volume << " liters." << endl;
17
18 const double BOTTLE_VOLUME = 2; // Two-liter bottle
19
20 total_volume = total_volume + BOTTLE_VOLUME;
21
22 cout << "A six-pack and a two-liter bottle contain "
23 << total_volume << " liters." << endl;
24
25 return 0;
26 }
program run
A six-pack of 12-ounce cans contains 2.13 liters.
A six-pack and a two-liter bottle contain 4.13 liters.
1. Define a variable suitable for holding the number of bottles in a case.
2. What is wrong with the following variable definition?
int ounces per liter = 28.35
s e l F c h e c k
3. Define and initialize two variables, unit_price and quantity, to contain the unit
price of a single bottle and the number of bottles purchased. Use reasonable
initial values.
4. Use the variables defined in Self Check 3 to display the total purchase price.
5. Some drinks are sold in four-packs instead of six-packs. How would you change
the volume1.cpp pro gram to compute the total volume?
6. What is wrong with this comment?
double can_volume = 0.355; /* Liters in a 12-ounce can //
7. Suppose the type of the cans_per_pack variable in the volume1.cpp program was
changed from int to dou ble. What would be the effect on the program?
8. Why can’t the variable total_volume in the volume1.cpp program be declared as
const?
9. How would you explain assignment using the parking space analogy?
practice it Now you can try these exercises at the end of the chapter: R2.1, R2.2, P2.1.
using undefined Variables
You must define a variable before you use it for the first time. For example, the following
sequence of statements would not be legal:
double can_volume = 12 * liter_per_ounce;
double liter_per_ounce = 0.0296;
In your program, the statements are compiled in order. When the compiler reaches the first
statement, it does not know that liter_per_ounce will be defined in the next line, and it reports
an error.
using uninitialized Variables
If you define a variable but leave it uninitialized, then your program can act unpredictably. To
understand why, con sider what happens when you define a variable. Just enough space is set
aside in memory to hold values of the type you specify. For example, with the definition
int bottles;
a block of memory big enough to hold integers is reserved. There is already some value in that
memory. After all, you don’t get freshly minted transistors—just an area of memory that has
previously been used, filled with flotsam left over from prior computations. (In this regard, a
variable differs from a parking space. A parking space can be empty, containing no vehicle. But
a variable always holds some value.)
If you use the variable without initializing it, then that prior value will be used, yielding
unpredictable results. For example, consider the program segment
int bottles; // Forgot to initialize
int bottle_volume = bottles * 2; // Result is unpredictable
There is no way of knowing what value will be computed. If you are unlucky, a plausible value
will happen to appear when you run the program at home, and an entirely different result will
occur when the program is graded.
Common error 2.1
Common error 2.2
cfe2_ch02_p29_74.indd 36 10/27/10 2:29 PM
2.1 Variables 37
3. Define and initialize two variables, unit_price and quantity, to contain the unit
price of a single bottle and the number of bottles purchased. Use reasonable
initial values.
4. Use the variables defined in Self Check 3 to display the total purchase price.
5. Some drinks are sold in four-packs instead of six-packs. How would you change
the volume1.cpp pro gram to compute the total volume?
6. What is wrong with this comment?
double can_volume = 0.355; /* Liters in a 12-ounce can //
7. Suppose the type of the cans_per_pack variable in the volume1.cpp program was
changed from int to dou ble. What would be the effect on the program?
8. Why can’t the variable total_volume in the volume1.cpp program be declared as
const?
9. How would you explain assignment using the parking space analogy?
practice it Now you can try these exercises at the end of the chapter: R2.1, R2.2, P2.1.
using undefined Variables
You must define a variable before you use it for the first time. For example, the following
sequence of statements would not be legal:
double can_volume = 12 * liter_per_ounce;
double liter_per_ounce = 0.0296;
In your program, the statements are compiled in order. When the compiler reaches the first
statement, it does not know that liter_per_ounce will be defined in the next line, and it reports
an error.
using uninitialized Variables
If you define a variable but leave it uninitialized, then your program can act unpredictably. To
understand why, con sider what happens when you define a variable. Just enough space is set
aside in memory to hold values of the type you specify. For example, with the definition
int bottles;
a block of memory big enough to hold integers is reserved. There is already some value in that
memory. After all, you don’t get freshly minted transistors—just an area of memory that has
previously been used, filled with flotsam left over from prior computations. (In this regard, a
variable differs from a parking space. A parking space can be empty, containing no vehicle. But
a variable always holds some value.)
If you use the variable without initializing it, then that prior value will be used, yielding
unpredictable results. For example, consider the program segment
int bottles; // Forgot to initialize
int bottle_volume = bottles * 2; // Result is unpredictable
There is no way of knowing what value will be computed. If you are unlucky, a plausible value
will happen to appear when you run the program at home, and an entirely different result will
occur when the program is graded.
Common error 2.1
Common error 2.2
cfe2_ch02_p29_74.indd 37 10/27/10 2:29 PM
38 Chapter 2 Fundamental data types
choose descriptive Variable names
We could have saved ourselves a lot of typing by using shorter variable names, as in
double cv = 0.355;
Compare this definition with the one that we actually used, though. Which one is easier to
read? There is no compar ison. Just reading can_volume is a lot less trouble than reading cv and
then figuring out it must mean “can volume”.
In practical programming, this is particularly important when programs are written by
more than one person. It may be obvious to you that cv stands for can volume and not cur-
rent velocity, but will it be obvious to the person who needs to update your code years later?
For that matter, will you remember yourself what cv means when you look at the code three
months from now?
numeric types in c++
In addition to the int and double types, C++ has several other numeric types.
C++ has two floating-point types. The float type uses half the storage of the double type
that we use in this book, but it can only store 6–7 digits. Many years ago, when computers had
far less memory than they have today, float was the standard type for floating-point computa-
tions, and programmers would indulge in the luxury of “double precision” only when they
needed the additional digits. Today, the float type is rarely used.
By the way, these numbers are called “floating-point” because of their internal representa-
tion in the computer. Consider numbers 29600, 2.96, and 0.0296. They can be represented in
a very similar way: namely, as a sequence of the significant digits—296—and an indication of
the position of the decimal point. When the values are multiplied or divided by 10, only the
position of the decimal point changes; it “floats”. Computers use base 2, not base 10, but the
principle is the same.
table 4 number types
type typical range typical size
int –2,147,483,648 … 2,147,483,647 (about 2 billion) 4 bytes
unsigned 0 … 4,294,967,295 4 bytes
short –32,768 … 32,767 2 bytes
unsigned short 0 … 65,535 2 bytes
long long –9,223,372,036,854,775,808 … 9,223,372,036,854,775,807 8 bytes
double The double-precision floating-point type, with a range of
about ±10308 and about 15 significant decimal digits
8 bytes
float The single-precision floating-point type, with a range of
about ±1038 and about 7 significant decimal digits
4 bytes
programming tip 2.1
special topic 2.1
In addition to the int type, C++ has integer types short, long, and long long. For each integer
type, there is an unsigned equivalent. For example, the short type typically has a range from
–32,768 to 32,767, whereas unsigned short has a range from 0 to 65,535. These strange-looking
limits are the result of the use of binary numbers in computers. A short value uses 16 binary
digits, which can encode 216 = 65,536 values. Keep in mind that the ranges for integer types are
not standardized, and they differ among compilers. Table 4 contains typical values.
numeric ranges and precisions
Because numbers are represented in the computer with a limited number of digits, they cannot
represent arbitrary integer or floating-point numbers.
The int type has a limited range: On most platforms, it can represent numbers up to a little
more than two billion. For many applications, this is not a problem, but you cannot use an int
to represent the world population.
If a computation yields a value that is outside the int range, the result overflows. No error
is displayed. Instead, the result is truncated to fit into an int, yielding a useless value. For
example,
int one_billion = 1000000000;
cout << 3 * one_billion << endl;
displays –1294967296.
In situations such as this, you can switch to double values. However, read Common Error
2.6 on page 45 for more information about a related issue: roundoff errors.
do not use magic numbers
A magic number is a numeric constant that appears in your code
without explanation. For example,
total_volume = bottles * 2;
Why 2? Are bottles twice as voluminous as cans? No, the reason is
that every bottle con tains 2 liters. Use a named constant to make the
code self-documenting:
const double BOTTLE_VOLUME = 2;
total_volume = bottles * BOTTLE_VOLUME;
There is another reason for using named constants. Suppose circumstances change, and the
bottle volume is now 1.5 liters. If you used a named constant, you make a single change, and
you are done. Other wise, you have to look at every value of 2 in your program and ponder
whether it means a bottle volume, or some thing else. In a program that is more than a few
pages long, that is incredibly tedious and error-prone.
Even the most reasonable cosmic constant is going to change one day. You think there are
seven days per week? Your customers on Mars are going to be pretty unhappy about your silly
prejudice. Make a constant
const int DAYS_PER_WEEK = 7;
special topic 2.2
programming tip 2.2
cfe2_ch02_p29_74.indd 38 10/27/10 2:29 PM
2.1 Variables 39
In addition to the int type, C++ has integer types short, long, and long long. For each integer
type, there is an unsigned equivalent. For example, the short type typically has a range from
–32,768 to 32,767, whereas unsigned short has a range from 0 to 65,535. These strange-looking
limits are the result of the use of binary numbers in computers. A short value uses 16 binary
digits, which can encode 216 = 65,536 values. Keep in mind that the ranges for integer types are
not standardized, and they differ among compilers. Table 4 contains typical values.
numeric ranges and precisions
Because numbers are represented in the computer with a limited number of digits, they cannot
represent arbitrary integer or floating-point numbers.
The int type has a limited range: On most platforms, it can represent numbers up to a little
more than two billion. For many applications, this is not a problem, but you cannot use an int
to represent the world population.
If a computation yields a value that is outside the int range, the result overflows. No error
is displayed. Instead, the result is truncated to fit into an int, yielding a useless value. For
example,
int one_billion = 1000000000;
cout << 3 * one_billion << endl;
displays –1294967296.
In situations such as this, you can switch to double values. However, read Common Error
2.6 on page 45 for more information about a related issue: roundoff errors.
do not use magic numbers
A magic number is a numeric constant that appears in your code
without explanation. For example,
total_volume = bottles * 2;
Why 2? Are bottles twice as voluminous as cans? No, the reason is
that every bottle con tains 2 liters. Use a named constant to make the
code self-documenting:
const double BOTTLE_VOLUME = 2;
total_volume = bottles * BOTTLE_VOLUME;
There is another reason for using named constants. Suppose circumstances change, and the
bottle volume is now 1.5 liters. If you used a named constant, you make a single change, and
you are done. Other wise, you have to look at every value of 2 in your program and ponder
whether it means a bottle volume, or some thing else. In a program that is more than a few
pages long, that is incredibly tedious and error-prone.
Even the most reasonable cosmic constant is going to change one day. You think there are
seven days per week? Your customers on Mars are going to be pretty unhappy about your silly
prejudice. Make a constant
const int DAYS_PER_WEEK = 7;
special topic 2.2
programming tip 2.2
cfe2_ch02_p29_74.indd 39 10/27/10 2:29 PM
40 Chapter 2 Fundamental data types
2.2 arithmetic
In the following sections, you will learn how to carry out arithmetic and mathemati-
cal calculations in C++.
2.2.1 arithmetic operators
C++ supports the same four basic arithmetic operations as a calculator—addition,
subtraction, multi plication, and division—but it uses different symbols for multipli-
cation and division.
You must write a * b to denote multiplication. Unlike in mathematics, you can not
write a b, a . b or a × b. Similarly, division is always indicated with a /, never a ÷ or a
fraction bar.
For example,
a b+
2
becomes (a + b) / 2.
Parentheses are used just as in algebra: to indicate in which order the subexpres-
sions should be com puted. For example, in the expression (a + b) / 2, the sum a + b is
computed first, and then the sum is divided by 2. In contrast, in the expression
a + b / 2
only b is divided by 2, and then the sum of a and b / 2 is formed. Just as in regular alge-
braic notation, mul tiplication and division have a higher precedence than addition
and subtraction. For example, in the expression a + b / 2, the / is carried out first, even
though the + operation occurs further to the left. If both arguments of an arithmetic
operation are integers, the result is an integer. If one or both arguments are floating-
point numbers, the result is a floating-point number. For example, 4 * 0.5 is 2.0.
2.2.2 Increment and decrement
Changing a variable by adding or subtracting 1 is so common that there is a special
shorthand for it, namely
counter++;
counter--;
The ++ increment operator gave the C++ programming language its name. C++ is the
incremental improvement of the C language.
2.2.3 Integer division and remainder
Division works as you would expect, as long as at least one of the numbers involved is
a floating-point number. That is, 7.0 / 4.0, 7 / 4.0, and 7.0 / 4 all yield 1.75. However,
if both numbers are integers, then the result of the division is always an integer, with
the remainder discarded. That is,
7 / 4
evaluates to 1 because 7 divided by 4 is 1 with a remainder of 3 (which is discarded).
This can be a source of subtle programming errors—see Common Error 2.3 on page 43.
use * for
multiplication and /
for division.
the ++ operator adds
1 to a variable; the --
operator subtracts 1.
If both arguments
of / are integers,
the remainder
is discarded.
If you are interested in the remainder only, use the % operator:
7 % 4
is 3, the remainder of the integer division of 7 by 4. The % symbol has no analog in
algebra. It was chosen because it looks similar to /, and the remainder operation is
related to division. The operator is called modulus. (Some people call it modulo or
mod.) It has no relationship with the percent operation that you find on some
calculators.
Here is a typical use for the integer / and % operations. Suppose you have an amount
of pennies in a piggybank:
int pennies = 1729;
You want to determine the value in dollars and
cents. You obtain the dollars through an integer
division by 100.
int dollars = pennies / 100; // Sets dollars to 17
The integer division discards the remainder. To
obtain the remainder, use the % operator:
int cents = pennies % 100; // Sets cents to 29
Another common use of the % operator is to check
whether a number is even or odd. If a number n is
even, then n % 2 is zero.
2.2.4 Converting Floating-point numbers to Integers
When a floating-point value is assigned to an integer variable, the fractional part is
discarded:
double price = 2.55;
int dollars = price; // Sets dollars to 2
Discarding the fractional part is not always what you want. Often, you want to round
to the nearest inte ger. To round a positive floating-point value to the nearest integer,
add 0.5 and then convert to an integer:
int dollars = price + 0.5; // Rounds to the nearest integer
In our example, adding 0.5 turns all values above 2.5 into values above 3. In particular,
2.55 is turned into 3.05, which is then truncated to 3. (For a negative floating-point
value, you subtract 0.5.)
Because truncation is a potential cause for errors, your compiler may issue a warn-
ing that assigning a floating-point value to an integer variable is unsafe. See Special
Topic 2.3 on page 46 on how to avoid this warning.
2.2.5 powers and roots
In C++, there are no symbols for powers and roots. To compute them, you must call
functions. To take the square root of a number, you use the sqrt function. For exam-
ple, is written as sqrt(x). To com pute xn, you write pow(x, n).
the % operator
computes the
remainder of an
integer division.
Integer division and the %
operator yield the dollar and
cent values of a piggybank
full of pennies.
assigning a floating-
point variable to an
integer drops the
fractional part.
cfe2_ch02_p29_74.indd 40 10/27/10 2:29 PM
2.2 arithmetic 41
If you are interested in the remainder only, use the % operator:
7 % 4
is 3, the remainder of the integer division of 7 by 4. The % symbol has no analog in
algebra. It was chosen because it looks similar to /, and the remainder operation is
related to division. The operator is called modulus. (Some people call it modulo or
mod.) It has no relationship with the percent operation that you find on some
calculators.
Here is a typical use for the integer / and % operations. Suppose you have an amount
of pennies in a piggybank:
int pennies = 1729;
You want to determine the value in dollars and
cents. You obtain the dollars through an integer
division by 100.
int dollars = pennies / 100; // Sets dollars to 17
The integer division discards the remainder. To
obtain the remainder, use the % operator:
int cents = pennies % 100; // Sets cents to 29
Another common use of the % operator is to check
whether a number is even or odd. If a number n is
even, then n % 2 is zero.
2.2.4 Converting Floating-point numbers to Integers
When a floating-point value is assigned to an integer variable, the fractional part is
discarded:
double price = 2.55;
int dollars = price; // Sets dollars to 2
Discarding the fractional part is not always what you want. Often, you want to round
to the nearest inte ger. To round a positive floating-point value to the nearest integer,
add 0.5 and then convert to an integer:
int dollars = price + 0.5; // Rounds to the nearest integer
In our example, adding 0.5 turns all values above 2.5 into values above 3. In particular,
2.55 is turned into 3.05, which is then truncated to 3. (For a negative floating-point
value, you subtract 0.5.)
Because truncation is a potential cause for errors, your compiler may issue a warn-
ing that assigning a floating-point value to an integer variable is unsafe. See Special
Topic 2.3 on page 46 on how to avoid this warning.
2.2.5 powers and roots
In C++, there are no symbols for powers and roots. To compute them, you must call
functions. To take the square root of a number, you use the sqrt function. For exam-
ple, x is written as sqrt(x). To com pute xn, you write pow(x, n).
the % operator
computes the
remainder of an
integer division.
Integer division and the %
operator yield the dollar and
cent values of a piggybank
full of pennies.
assigning a floating-
point variable to an
integer drops the
fractional part.
cfe2_ch02_p29_74.indd 41 10/27/10 2:29 PM
42 Chapter 2 Fundamental data types
Figure 2
analyzing an expression
b * pow(1 + r / 100, n)
r
100
r
1 +
100
r
n
1 +
100
b
r
n
× +
1
100
To use the sqrt and pow functions, you must place the line #include
of your program file. The header file
with all C++ systems, as is
As you can see, the effect of the /, sqrt, and pow operations is to flatten out math-
ematical terms. In alge bra, you use fractions, exponents, and roots to arrange expres-
sions in a compact two-dimensional form. In C++, you have to write all expressions
in a linear arrangement. For example, the mathematical expres sion
b
r
n
× +
1
100
becomes
b * pow(1 + r / 100, n)
Figure 2 shows how to analyze such an expression.
the C++ library
defines many
mathematical
functions such as
sqrt (square root)
and pow (raising to a
power).
table 5 arithmetic expressions
mathematical
expression
C++
expression
Comments
x y+
2 (x + y) / 2
The parentheses are required;
x + y / 2 computes x
y
+
2
.
xy
2 x * y / 2
Parentheses are not required; operators with
the same precedence are evaluated left to right.
1
100
+
r
n
pow(1 + r / 100, n)
Remember to add #include
of your program.
a b2 2+ sqrt(a * a + b * b) a * a is simpler than pow(a, 2).
i j k+ +
3 (i + j + k) / 3.0
If i, j, and k are integers, using a denominator
of 3.0 forces floating-point division.
Table 6 shows additional functions that are declared in the
and outputs are float ing-point numbers.
10. A bank account earns interest of p percent per year. In C++, how do you com-
pute the interest earned in one year? Assume variables p and balance of type
double have already been defined.
11. In C++, how do you compute the side length of a square whose area is stored in
the variable area?
12. The volume of a sphere is given by .
If the radius is given by a variable radius of type double, write a C++ expression
for the volume. You may assume that p is defined by a constant PI.
13. What is the value of 1729 / 10 and 1729 % 10?
14. Suppose a punch recipe calls for a given amount of
orange soda, measured in ounces.
int amount = 32;
We can compute the number of 12-ounce cans
needed, assuming that the amount does not evenly
divide into 12:
int cans_needed = amount / 12 + 1;
Use the % operator to determine how many ounces will be left over. For example,
if 32 ounces are required, we need 3 cans and have 4 ounces left over.
practice it Now you can try these exercises at the end of the chapter: R2.3, R2.5, P2.2.
unintended integer division
It is unfortunate that C++ uses the same symbol, namely /, for both integer and floating-point
division. These are really quite different operations. It is a common error to use integer divi-
sion by accident. Consider this segment that computes the average of three integers:
cout << "Please enter your last three test scores: ";
int s1;
int s2;
s e l F c h e c k
Common error 2.3
cfe2_ch02_p29_74.indd 42 10/27/10 2:29 PM
2.2 arithmetic 43
table 6 other mathematical Functions
Function description
sin(x) sine of x (x in radians)
cos(x) cosine of x
tan(x) tangent of x
log10(x) decimal log( ) ( ) >log ,10 0x x
abs(x) absolute value x
Table 6 shows additional functions that are declared in the
and outputs are float ing-point numbers.
10. A bank account earns interest of p percent per year. In C++, how do you com-
pute the interest earned in one year? Assume variables p and balance of type
double have already been defined.
11. In C++, how do you compute the side length of a square whose area is stored in
the variable area?
12. The volume of a sphere is given by V r=
4
3
3π .
If the radius is given by a variable radius of type double, write a C++ expression
for the volume. You may assume that p is defined by a constant PI.
13. What is the value of 1729 / 10 and 1729 % 10?
14. Suppose a punch recipe calls for a given amount of
orange soda, measured in ounces.
int amount = 32;
We can compute the number of 12-ounce cans
needed, assuming that the amount does not evenly
divide into 12:
int cans_needed = amount / 12 + 1;
Use the % operator to determine how many ounces will be left over. For example,
if 32 ounces are required, we need 3 cans and have 4 ounces left over.
practice it Now you can try these exercises at the end of the chapter: R2.3, R2.5, P2.2.
unintended integer division
It is unfortunate that C++ uses the same symbol, namely /, for both integer and floating-point
division. These are really quite different operations. It is a common error to use integer divi-
sion by accident. Consider this segment that computes the average of three integers:
cout << "Please enter your last three test scores: ";
int s1;
int s2;
s e l F c h e c k
Common error 2.3
cfe2_ch02_p29_74.indd 43 10/27/10 2:29 PM
44 Chapter 2 Fundamental data types
int s3;
cin >> s1 >> s2 >> s3;
double average = (s1 + s2 + s3) / 3; // Error
cout << "Your average score is " << average << endl;
What could be wrong with that? Of course, the average of s1, s2, and s3 is
s1 s2 s3+ +
3
Here, however, the / does not mean division in the mathematical sense. It denotes integer
division because both s1 + s2 + s3 and 3 are integers. For example, if the scores add up to 14,
the average is computed to be 4, the result of the integer division of 14 by 3. That integer 4 is
then moved into the floating-point variable average. The remedy is to make the numerator or
denominator into a floating-point number:
double total = s1 + s2 + s3;
double average = total / 3;
or
double average = (s1 + s2 + s3) / 3.0;
unbalanced parentheses
Consider the expression
(-(b * b - 4 * a * c) / (2 * a)
What is wrong with it? Count the parentheses. There are three ( and two ). The parenthe-
ses are unbalanced. This kind of typing error is very common with complicated expressions.
Now consider this expression.
-(b * b - (4 * a * c))) / (2 * a
This expression has three ( and three ), but it still is not correct. In the middle of the expres-
sion,
-(b * b - (4 * a * c))) / (2 * a
↑
there are only two ( but three ), which is an error. In the middle of an expression, the count of (
must be greater than or equal to the count of ), and at the end of the expression the two counts
must be the same.
Here is a simple trick to make the counting easier without using pencil and paper. It is diffi-
cult for the brain to keep two counts simultaneously. Keep only one count when scanning the
expression. Start with 1 at the first opening parenthesis, add 1 whenever you see an opening
parenthesis, and subtract one whenever you see a closing parenthe sis. Say the numbers aloud
as you scan the expression. If the count ever drops below zero, or is not zero at the end, the
parentheses are unbalanced. For example, when scanning the previous expression, you would
mutter
-(b * b - (4 * a * c ) ) ) / (2 * a
1 2 1 0 –1
and you would find the error.
Common error 2.4
Forgetting header Files
Every program that carries out input or output needs the
ical functions such as sqrt, you need to include
header file, the compiler will not know sym bols such as cout or sqrt. If the compiler complains
about an undefined function or symbol, check your header files.
Sometimes you may not know which header file to include. Suppose you want to compute
the absolute value of an integer using the abs function. As it happens, this version of abs is not
defined in the
need to locate the documentation of the abs function, preferably using the online help of your
development environment or a reference site on the Internet such as http://www.cplusplus.com
(see Figure 3). The documentation includes a short description of the function and the name of
the header file that you must include.
roundoff errors
Roundoff errors are a fact of life when calculating with floating-point numbers. You probably
have encountered that phenomenon yourself with manual calculations. If you calculate to
two decimal places, you get 0.33. Multiply ing again by 3, you obtain 0.99, not 1.00.
In the processor hardware, numbers are represented in the binary number system, not in
decimal. You still get roundoff errors when binary digits are lost. They just may crop up at dif-
ferent places than you might expect. Here is an example.
#include
using namespace std;
Common error 2.5
Figure 3 online documentation
Common error 2.6
cfe2_ch02_p29_74.indd 44 10/27/10 2:29 PM
2.2 arithmetic 45
Forgetting header Files
Every program that carries out input or output needs the
ical functions such as sqrt, you need to include
header file, the compiler will not know sym bols such as cout or sqrt. If the compiler complains
about an undefined function or symbol, check your header files.
Sometimes you may not know which header file to include. Suppose you want to compute
the absolute value of an integer using the abs function. As it happens, this version of abs is not
defined in the
need to locate the documentation of the abs function, preferably using the online help of your
development environment or a reference site on the Internet such as http://www.cplusplus.com
(see Figure 3). The documentation includes a short description of the function and the name of
the header file that you must include.
roundoff errors
Roundoff errors are a fact of life when calculating with floating-point numbers. You probably
have encountered that phenomenon yourself with manual calculations. If you calculate 1 3 to
two decimal places, you get 0.33. Multiply ing again by 3, you obtain 0.99, not 1.00.
In the processor hardware, numbers are represented in the binary number system, not in
decimal. You still get roundoff errors when binary digits are lost. They just may crop up at dif-
ferent places than you might expect. Here is an example.
#include
using namespace std;
Common error 2.5
Figure 3 online documentation
Common error 2.6
cfe2_ch02_p29_74.indd 45 10/27/10 2:29 PM
http://www.cplusplus.com
46 Chapter 2 Fundamental data types
int main()
{
double price = 4.35;
int cents = 100 * price; // Should be 100 * 4.35 = 435
cout << cents << endl; // Prints 434!
return 0;
}
Of course, one hundred times 4.35 is 435, but the program prints 434.
Most computers represent numbers in the binary system. In the binary system, there is
no exact representation for 4.35, just as there is no exact representation for 1/3 in the decimal
system. The representation used by the com puter is just a little less than 4.35, so 100 times that
value is just a little less than 435. When a floating-point value is converted to an integer, the
entire fractional part, which is almost 1, is thrown away, and the integer 434 is stored in cents.
The remedy is to add 0.5 in order to round to the nearest integer:
int cents = 100 * price + 0.5;
spaces in expressions
It is easier to read
x1 = (-b + sqrt(b * b - 4 * a * c)) / (2 * a);
than
x1=(-b+sqrt(b*b-4*a*c))/(2*a);
Simply put spaces around all operators + - * / % =. However, don’t put a space after a unary
minus: a – used to negate a single quantity, such as -b. That way, it can be easily distinguished
from a binary minus, as in a - b.
It is customary not to put a space after a function name. That is, write sqrt(x) and not
sqrt (x).
casts
Occasionally, you need to store a value into a variable of a different type. Whenever there is
the risk of information loss, the compiler issues a warning. For example, if you store a double
value into an int variable, you can lose informa tion in two ways:
• The fractional part is lost.
• The magnitude may be too large.
For example,
int n = 1.0E100; // NO
is not likely to work, because 10100 is larger than the largest representable integer.
Nevertheless, sometimes you do want to convert a floating-point value into an integer
value. If you are prepared to lose the fractional part and you know that this particular floating-
point number is not larger than the largest pos sible integer, then you can turn off the warning
by using a cast. A cast is a conversion from one type (such as double) to another type (such as
int) that is not safe in general, but that you know to be safe in a particular circumstance. You
express a cast in C++ as follows:
int cents = static_cast
programming tip 2.3
special topic 2.3
combining assignment and arithmetic
In C++, you can combine arithmetic and assignment. For example, the instruction
total += cans * CAN_VOLUME;
is a shortcut for
total = total + cans * CAN_VOLUME;
Similarly,
total *= 2;
is another way of writing
total = total * 2;
Many programmers find this a convenient shortcut. If you like it, go ahead and use it in your
own code. For simplic ity, we won’t use it in this book, though.
special topic 2.4
In 1994, Intel Corporation released what
was then its most powerful processor, the
pentium. unlike previous generations of its processors, it
had a very fast floating-point unit. Intel’s goal was to com-
pete aggressively with the makers of higher-end proces sors
for engineering workstations. the pentium was a huge suc-
cess immediately.
In the summer of 1994, dr. thomas nicely of lynchburg
College, Virginia, ran an extensive set of computations to
analyze the sums of reciprocals of certain sequences of
prime numbers. the results were not always what his the-
ory predicted, even after he took into account the inevita ble
roundoff errors. then dr. nicely noted that the same pro-
gram did produce the correct results when running on the
slower 486 processor that preceded the pentium in Intel’s
lineup. this should not have happened. the optimal round-
off behavior of floating-point calculations were standard-
ized by the Institute for electrical and electronic engineers
(Ieee) and Intel claimed to adhere to the Ieee standard in
both the 486 and the pentium processors. upon further
checking, dr. nicely discovered that there was a very small
set of numbers for which the prod uct of two numbers was
computed differently on the two processors. For example,
4 195 835 4 195 835 3 145 727 3 145 727, , , , , , , ,− ( ) ×( )
is mathematically equal to 0, and it did compute as 0 on a
486 processor. on his pentium processor the result was 256.
as it turned out, Intel had independently discovered
the bug in its testing and had started to produce chips that
fixed it. the bug was caused by an error in a table that was
used to speed up the processor’s floating-point multipli-
cation algo rithm. Intel determined that the problem was
exceedingly rare. they claimed that under normal use, a
typical consumer would only notice the problem once every
27,000 years. unfortunately for Intel, dr. nicely had not
been a normal user.
now Intel had a real problem on its hands. It figured
that the cost of replacing all pentium processors that it had
sold so far would cost a great deal of money. Intel already
had more orders for the chip than it could produce, and it
would be particularly galling to have to give out the scarce
chips as free replacements instead of selling them. Intel’s
management decided to punt and initially offered to replace
the processors only for those customers who could prove
that their work required absolute precision in mathematical
calculations. naturally, that did not go over well with the
hundreds of thousands of customers who had paid retail
prices of $700 and more for a pentium chip and did not
want to live with the nagging feeling that perhaps, one day,
their income tax program would produce a faulty return.
ultimately, Intel caved in to public demand and replaced all
defective chips, at a cost of about 475 million dollars.
1
.4
0
1
.2
0
1
.0
0
0
.8
0
0
.6
0
0
.4
0
0
.2
0
0
.0
0
-0
.2
0
-0
.4
0
-0
.6
0
-0
.8
0
-1
.0
0
-1
.2
0
-1
.4
0
-1
.6
0
-1
.8
0
-2
.0
0
1.40
1.10
0.80
0.50
0.20
-0.10
-0.40
-0.70
-1.00
-1.30
1.333680000
1.333700000
1.333720000
1.333740000
1.333760000
1.333780000
1.333800000
1.333820000
1.333840000
x
/y
4195835+
3145727+
Pentium FDIV error
This graph shows a set of numbers for which the original
Pentium processor obtained the wrong quotient.
Random Fact 2.1 the pentium Floating-point Bug
cfe2_ch02_p29_74.indd 46 10/27/10 2:29 PM
2.2 arithmetic 47
combining assignment and arithmetic
In C++, you can combine arithmetic and assignment. For example, the instruction
total += cans * CAN_VOLUME;
is a shortcut for
total = total + cans * CAN_VOLUME;
Similarly,
total *= 2;
is another way of writing
total = total * 2;
Many programmers find this a convenient shortcut. If you like it, go ahead and use it in your
own code. For simplic ity, we won’t use it in this book, though.
special topic 2.4
In 1994, Intel Corporation released what
was then its most powerful processor, the
pentium. unlike previous generations of its processors, it
had a very fast floating-point unit. Intel’s goal was to com-
pete aggressively with the makers of higher-end proces sors
for engineering workstations. the pentium was a huge suc-
cess immediately.
In the summer of 1994, dr. thomas nicely of lynchburg
College, Virginia, ran an extensive set of computations to
analyze the sums of reciprocals of certain sequences of
prime numbers. the results were not always what his the-
ory predicted, even after he took into account the inevita ble
roundoff errors. then dr. nicely noted that the same pro-
gram did produce the correct results when running on the
slower 486 processor that preceded the pentium in Intel’s
lineup. this should not have happened. the optimal round-
off behavior of floating-point calculations were standard-
ized by the Institute for electrical and electronic engineers
(Ieee) and Intel claimed to adhere to the Ieee standard in
both the 486 and the pentium processors. upon further
checking, dr. nicely discovered that there was a very small
set of numbers for which the prod uct of two numbers was
computed differently on the two processors. For example,
4 195 835 4 195 835 3 145 727 3 145 727, , , , , , , ,− ( ) ×( )
is mathematically equal to 0, and it did compute as 0 on a
486 processor. on his pentium processor the result was 256.
as it turned out, Intel had independently discovered
the bug in its testing and had started to produce chips that
fixed it. the bug was caused by an error in a table that was
used to speed up the processor’s floating-point multipli-
cation algo rithm. Intel determined that the problem was
exceedingly rare. they claimed that under normal use, a
typical consumer would only notice the problem once every
27,000 years. unfortunately for Intel, dr. nicely had not
been a normal user.
now Intel had a real problem on its hands. It figured
that the cost of replacing all pentium processors that it had
sold so far would cost a great deal of money. Intel already
had more orders for the chip than it could produce, and it
would be particularly galling to have to give out the scarce
chips as free replacements instead of selling them. Intel’s
management decided to punt and initially offered to replace
the processors only for those customers who could prove
that their work required absolute precision in mathematical
calculations. naturally, that did not go over well with the
hundreds of thousands of customers who had paid retail
prices of $700 and more for a pentium chip and did not
want to live with the nagging feeling that perhaps, one day,
their income tax program would produce a faulty return.
ultimately, Intel caved in to public demand and replaced all
defective chips, at a cost of about 475 million dollars.
1
.4
0
1
.2
0
1
.0
0
0
.8
0
0
.6
0
0
.4
0
0
.2
0
0
.0
0
-0
.2
0
-0
.4
0
-0
.6
0
-0
.8
0
-1
.0
0
-1
.2
0
-1
.4
0
-1
.6
0
-1
.8
0
-2
.0
0
1.40
1.10
0.80
0.50
0.20
-0.10
-0.40
-0.70
-1.00
-1.30
1.333680000
1.333700000
1.333720000
1.333740000
1.333760000
1.333780000
1.333800000
1.333820000
1.333840000
x
/y
4195835+
3145727+
Pentium FDIV error
This graph shows a set of numbers for which the original
Pentium processor obtained the wrong quotient.
Random Fact 2.1 the pentium Floating-point Bug
cfe2_ch02_p29_74.indd 47 10/27/10 2:30 PM
48 Chapter 2 Fundamental data types
2.3 Input and output
2.3.1 Input
In this section, you will see how to place user input into a variable. Consider for
example the volume1.cpp program on page 36. Rather than assuming that the price for
the two-liter bottle and the six-pack of cans are identical, we can ask the program user
for the prices.
When a program asks for user input, it should first print a message that tells the
user which input is expected. Such a message is called a prompt.
cout << "Please enter the number of bottles: "; // Display prompt
Do not add an endl after the prompt. You want the input to appear after the colon, not
on the following line.
Next, the program issues a command to read the input. The cin object reads input
from the console window. You use the >> operator (sometimes called the extraction
operator) to place an input value into a variable, like this:
int bottles;
cin >> bottles;
When the program executes the input statement, it waits for the user to provide input.
The user also needs to hit the Enter key so that the program accepts the input. After
the user supplies the input, the number is placed into the bottles variable, and the pro-
gram continues.
Note that in this code segment, there was no need to initialize the bottles variable
because it is being filled by the very next statement. As a rule of thumb, you should
initialize a variable when you declare it unless it is filled in an input statement that fol-
lows immediately.
You can read more than one value in a single input statement:
cout << "Please enter the number of bottles and cans: ";
cin >> bottles >> cans;
The user can supply both inputs on the same line:
Please enter the number of bottles and cans: 2 6
Alternatively, the user can press the Enter key after each input:
Please enter the number of bottles and cans: 2
6
use the >> operator
to read a value and
place it in a variable.
syntax 2.3 Input statement
cout << "Enter the number of bottles: ";
int bottles;
cin >> bottles;
Display a prompt in the console window.
The program waits for user input,
then places the input into the variable.
Define a variable to hold the input value.
Don’t use endl here.
2.3.2 Formatted output
When you print the result of a computation, you often want some control over its
appearance. For exam ple, when you print an amount in dollars and cents, you usually
want it to be rounded to two significant digits. That is, you want the output to look
like
Price per ounce: 0.04
instead of
Price per ounce: 0.0409722
The following command instructs cout to use two digits after the decimal point for all
floating-point numbers:
cout << fixed << setprecision(2);
This command does not produce any output; it just manipulates cout so that it will
change the output for mat. The values fixed and setprecision are called manipulators.
We will discuss manipulators in detail in Chapter 8. For now, just remember to
include the statement given above whenever you want currency values displayed
neatly.
To use manipulators, you must include the
#include
You can combine the manipulators and the values to be displayed into a single
statement.
cout << fixed << setprecision(2)
<< "Price per ounce: "
<< price_per_ounce << endl;
There is another manipulator that is sometimes handy. When you display several
rows of data, you usu ally want the columns to line up.
You use the setw manipulator to set the width of the next output field. The width is
the total number of characters used for showing the value, including digits, the deci-
mal point, and spaces. Controlling the width is important when you want columns of
numbers to line up.
For example, if you want a number to be printed in a column that is eight charac-
ters wide, you use
cout << setw(8) << price_per_ounce;
you use manipulators
to specify how values
should be formatted.
You use manipulators to line up
your output in neat columns.
cfe2_ch02_p29_74.indd 48 10/27/10 2:30 PM
2.3 Input and output 49
2.3.2 Formatted output
When you print the result of a computation, you often want some control over its
appearance. For exam ple, when you print an amount in dollars and cents, you usually
want it to be rounded to two significant digits. That is, you want the output to look
like
Price per ounce: 0.04
instead of
Price per ounce: 0.0409722
The following command instructs cout to use two digits after the decimal point for all
floating-point numbers:
cout << fixed << setprecision(2);
This command does not produce any output; it just manipulates cout so that it will
change the output for mat. The values fixed and setprecision are called manipulators.
We will discuss manipulators in detail in Chapter 8. For now, just remember to
include the statement given above whenever you want currency values displayed
neatly.
To use manipulators, you must include the
#include
You can combine the manipulators and the values to be displayed into a single
statement.
cout << fixed << setprecision(2)
<< "Price per ounce: "
<< price_per_ounce << endl;
There is another manipulator that is sometimes handy. When you display several
rows of data, you usu ally want the columns to line up.
You use the setw manipulator to set the width of the next output field. The width is
the total number of characters used for showing the value, including digits, the deci-
mal point, and spaces. Controlling the width is important when you want columns of
numbers to line up.
For example, if you want a number to be printed in a column that is eight charac-
ters wide, you use
cout << setw(8) << price_per_ounce;
you use manipulators
to specify how values
should be formatted.
You use manipulators to line up
your output in neat columns.
cfe2_ch02_p29_74.indd 49 10/27/10 2:30 PM
50 Chapter 2 Fundamental data types
This command prints the value price_per_ounce in a field of width 8, for example
0 . 0 4
(where each represents a space).
There is a notable difference between the setprecision and setw manipulators. Once
you set the preci sion, that value is used for all floating-point numbers. But the width
affects only the next value. Subse quent values are formatted without added spaces.
Our next example program will prompt for the price of a six-pack and the volume
of each can, then print out the price per ounce. The program puts to work what you
just learned about reading input and formatting output.
ch02/volume2.cpp
1 #include
2 #include
3
4 using namespace std;
5
6 int main()
7 {
8 // Read price per pack
9
10 cout << "Please enter the price for a six-pack: ";
11 double pack_price;
12 cin >> pack_price;
13
14 // Read can volume
15
16 cout << "Please enter the volume for each can (in ounces): ";
17 double can_volume;
18 cin >> can_volume;
19
20 // Compute pack volume
21
22 const double CANS_PER_PACK = 6;
23 double pack_volume = can_volume * CANS_PER_PACK;
24
25 // Compute and print price per ounce
26
27 double price_per_ounce = pack_price / pack_volume;
28
29 cout << fixed << setprecision(2);
30 cout << "Price per ounce: " << price_per_ounce << endl;
31
32 return 0;
33 }
program run
Please enter the price for a six-pack: 2.95
Please enter the volume for each can (in ounces): 12
Price per ounce: 0.04
cfe2_ch02_p29_74.indd 50 10/27/10 2:30 PM
2.3 Input and output 51
15.
table 7 Formatting output
output
statement output Comment
cout << 12.345678; 12.3457 By default, a number is printed with 6 significant digits.
cout << fixed
<< setprecision(2)
<< 12.3;
12.30 Use the fixed and setprecision manipulators to control
the number of digits after the decimal point.
cout << ":" << setw(6)
<< 12;
: 12 Four spaces are printed before the number, for a total
width of 6 characters.
cout << ":" << setw(2)
<< 123;
:123 If the width not sufficient, it is ignored.
cout << setw(6)
<< ":" << 12;
:12.3 The width only refers to the next item. Here, the : is
preceded by five spaces.
What is wrong with the following statement sequence?
cout << "Please enter the unit price: ";
double unit_price;
cin >> unit_price;
int quantity;
cin >> quantity;
16. What is problematic about the following statement sequence?
cout << "Please enter the unit price: ";
int unit_price;
cin >> unit_price;
17. What is the output of the following statement sequence?
double bottles = 10;
cout << "The total volume is" << 2 * bottles;
18. How do you print the floating-point variable total_price in dollars and cents,
like this: $1.22?
19. Using the setw manipulator, improve the output statement
cout << "Bottles: " << bottles << endl
<< "Cans: " << cans << endl;
so that the output looks like this:
Bottles: 8
Cans: 24
The numbers to the right should line up. (You may assume that the numbers
have at most 8 digits.)
practice it Now you can try these exercises at the end of the chapter: R2.7, R2.8, P2.6, P2.7.
s e l F c h e c k
cfe2_ch02_p29_74.indd 51 10/27/10 2:30 PM
52 Chapter 2 Fundamental data types
2.4 problem solving: First do It By hand
A very important step for developing an algorithm is to first carry out the computa-
tions by hand. If you can’t compute a solution yourself, it’s unlikely that you’ll be
able to write a program that automates the computation.
To illustrate the use of hand calculations, consider the following problem.
A row of black and white tiles needs to be placed along a wall. For aesthetic rea-
sons, the architect has specified that the first and last tile shall be black.
Your task is to compute the number of tiles needed and the gap at each end, given
the space available and the width of each tile.
Total width
Gap
To make the problem more concrete, let’s assume the following dimensions:
• Total width: 100 inches
• Tile width: 5 inches
The obvious solution would be to fill the space with 20 tiles, but that would not
work—the last tile would be white.
Instead, look at the problem this way: The first tile must always be black, and then
we add some num ber of white/black pairs:
The first tile takes up 5 inches, leaving 95 inches to be covered by pairs. Each pair is 10
inches wide. Therefore the number of pairs is 95 / 10 = 9.5. However, we need to dis-
card the fractional part since we can’t have fractions of tile pairs.
Therefore, we will use 9 tile pairs or 18 tiles, together with the initial black tile.
Altogether, we require 19 tiles.
The tiles span 19 × 5 = 95 inches, leaving a total gap of 100 – 19 × 5 = 5 inches. The
gap should be evenly distributed at both ends. At each end, the gap is (100 – 19 × 5) / 2
= 2.5 inches.
This computation gives us enough information to devise an algorithm with arbi-
trary values for the total width and tile width.
number of pairs = integer part of (total width - tile width) / (2 x tile width)
number of tiles = 1 + 2 x number of pairs
gap at each end = (total width - number of tiles x tile width) / 2
As you can see, doing a hand calculation gives enough insight into the problem that it
becomes easy to develop an algorithm.
pick concrete values
for a typical situation
to use in a hand
calculation.
20. Translate the pseudocode for computing the number of tiles and the gap width
into C++.
21. Suppose the architect specifies a pattern with black, gray, and white tiles, like
this:
Again, the first and last tile should be black. How do you need to modify the
algorithm?
22. A robot needs to tile a floor with alternating black and white tiles. Develop
an algorithm that yields the color (0 for black, 1 for white), given the row and
column number. Start with specific values for the row and column, and then
generalize.
23. For a particular car, repair and maintenance costs in year 1 are estimated at $100;
in year 10, at $1,500. Assuming that the repair cost increases by the same amount
every year, develop pseudocode to com pute the repair cost in year 3 and then
generalize to year n.
24. The shape of a bottle is approximated by two cylinders of radius r1 and r2 and
heights h1 and h2, joined by a cone section of height h3.
Using the formulas for the volume of a cylinder, , and a cone section,
,
develop pseudocode to compute the volume of the bottle. Using an actual bottle
with known volume as a sample, make a hand calculation of your pseudocode.
practice it Now you can try these exercises at the end of the chapter: R2.13, R2.15, R2.16.
s e l F c h e c k
r2
h2
h1
h3
r1
W o r k e d e x a m p l e 2 . 1 computing travel time
In this Worked Example, we develop a hand calculation to compute
the time that a robot requires to retrieve an item from rocky terrain.
cfe2_ch02_p29_74.indd 52 10/27/10 2:30 PM
2.4 problem solving: First do It By hand 53
20. Translate the pseudocode for computing the number of tiles and the gap width
into C++.
21. Suppose the architect specifies a pattern with black, gray, and white tiles, like
this:
Again, the first and last tile should be black. How do you need to modify the
algorithm?
22. A robot needs to tile a floor with alternating black and white tiles. Develop
an algorithm that yields the color (0 for black, 1 for white), given the row and
column number. Start with specific values for the row and column, and then
generalize.
1 2 3 4
1
2
3
4
23. For a particular car, repair and maintenance costs in year 1 are estimated at $100;
in year 10, at $1,500. Assuming that the repair cost increases by the same amount
every year, develop pseudocode to com pute the repair cost in year 3 and then
generalize to year n.
24. The shape of a bottle is approximated by two cylinders of radius r1 and r2 and
heights h1 and h2, joined by a cone section of height h3.
Using the formulas for the volume of a cylinder, V r h= π 2 , and a cone section,
V
r r r r h
=
+ +( )
π
1
2
1 2 2
2
3
,
develop pseudocode to compute the volume of the bottle. Using an actual bottle
with known volume as a sample, make a hand calculation of your pseudocode.
practice it Now you can try these exercises at the end of the chapter: R2.13, R2.15, R2.16.
s e l F c h e c k
r2
h2
h1
h3
r1
W o r k e d e x a m p l e 2 . 1 computing travel time
In this Worked Example, we develop a hand calculation to compute
the time that a robot requires to retrieve an item from rocky terrain.
Available online at www.wiley.com/college/horstmann.
cfe2_ch02_p29_74.indd 53 10/27/10 2:30 PM
www.wiley.com/college/horstmann
54 Chapter 2 Fundamental data types
step 1 Understand the problem: What are the inputs? What are the desired outputs?
In this problem, there are two inputs:
• The denomination of the bill that the customer inserts
• The price of the purchased item
There are two desired outputs:
• The number of dollar coins that the machine returns
• The number of quarters that the machine returns
step 2 Work out examples by hand.
Let’s assume that a customer purchased an item that cost $2.25 and inserted a $5 bill. The cus-
tomer is due $2.75, or two dollar coins and three quarters.
That is easy for you to see, but how can a C++ program come to the same conclusion? The
computation is sim pler if you work in pennies, not dollars. The amount due the customer is
275 pennies. Dividing by 100 yields 2, the number of dollars. Dividing the remainder (75) by
25 yields 3, the number of quarters.
step 3 Write pseudocode for computing the answers.
In the previous step, you worked out a specific instance of the problem. You now need to
come up with a method that works in general.
Given an arbitrary item price and payment, how can you compute the coins due? First,
compute the amount due in pennies:
amount due = 100 x bill value - item price in pennies
To get the dollars, divide by 100 and discard the remainder:
dollar coins = amount due / 100 (without remainder)
h o W t o 2 . 1 carrying out computations
Many programming problems require that you carry out arithmetic computations. This
How To shows you how to turn a problem statement into pseudocode and, ultimately, a C++
program.
For example, suppose you are asked to write a program that simulates a vending machine.
A customer selects an item for purchase and inserts a bill into the vending machine. The vend-
ing machine dispenses the purchased item and gives change. We will assume that all item prices
are multiples of 25 cents, and the machine gives all change in dollar coins and quarters. Your
task is to compute how many coins of each type to return.
A vending machine takes bills
and gives change in coins.
cfe2_ch02_p29_74.indd 54 10/27/10 2:30 PM
2.4 problem solving: First do It By hand 55
step 1 Understand the problem: What are the inputs? What are the desired outputs?
In this problem, there are two inputs:
• The denomination of the bill that the customer inserts
• The price of the purchased item
There are two desired outputs:
• The number of dollar coins that the machine returns
• The number of quarters that the machine returns
step 2 Work out examples by hand.
Let’s assume that a customer purchased an item that cost $2.25 and inserted a $5 bill. The cus-
tomer is due $2.75, or two dollar coins and three quarters.
That is easy for you to see, but how can a C++ program come to the same conclusion? The
computation is sim pler if you work in pennies, not dollars. The amount due the customer is
275 pennies. Dividing by 100 yields 2, the number of dollars. Dividing the remainder (75) by
25 yields 3, the number of quarters.
step 3 Write pseudocode for computing the answers.
In the previous step, you worked out a specific instance of the problem. You now need to
come up with a method that works in general.
Given an arbitrary item price and payment, how can you compute the coins due? First,
compute the amount due in pennies:
amount due = 100 x bill value - item price in pennies
To get the dollars, divide by 100 and discard the remainder:
dollar coins = amount due / 100 (without remainder)
h o W t o 2 . 1 carrying out computations
Many programming problems require that you carry out arithmetic computations. This
How To shows you how to turn a problem statement into pseudocode and, ultimately, a C++
program.
For example, suppose you are asked to write a program that simulates a vending machine.
A customer selects an item for purchase and inserts a bill into the vending machine. The vend-
ing machine dispenses the purchased item and gives change. We will assume that all item prices
are multiples of 25 cents, and the machine gives all change in dollar coins and quarters. Your
task is to compute how many coins of each type to return.
A vending machine takes bills
and gives change in coins.
The remaining amount due can be computed in two ways. If you are familiar with the modu-
lus operator, you can simply compute
amount due = amount due % 100
Alternatively, subtract the penny value of the dollar coins from the amount due:
amount due = amount due - 100 x dollar coins
To get the quarters due, divide by 25:
quarters = amount due / 25
step 4 Define the variables and constants that you need, and specify their types.
Here, we have five variables:
• bill_value
• item_price
• amount_due
• dollar_coins
• quarters
Should we introduce constants to explain 100 and 25 as PENNIES_PER_DOLLAR and PENNIES_PER_
QUARTER? Doing so will make it easier to convert the program to international markets, so we
will take this step.
It is very important that amount_due and PENNIES_PER_DOLLAR are of type int because the com-
putation of dollar_coins uses integer division. Similarly, the other variables are integers.
step 5 Turn the pseudocode into C++ statements.
If you did a thorough job with the pseudocode, this step should be easy. Of course, you have
to know how to express mathematical operations (such as powers or integer division) in C++.
amount_due = PENNIES_PER_DOLLAR * bill_value - item_price;
dollar_coins = amount_due / PENNIES_PER_DOLLAR;
amount_due = amount_due % PENNIES_PER_DOLLAR;
quarters = amount_due / PENNIES_PER_QUARTER;
step 6 Provide input and output.
Before starting the computation, we prompt the user for the bill value and item price:
cout << "Enter bill value (1 = $1 bill, 5 = $5 bill, etc.): ";
cin >> bill_value;
cout << "Enter item price in pennies: ";
cin >> item_price;
When the computation is finished, we display the result. For extra credit, we use the setw
manipulator to make sure that the output lines up neatly.
cout << "Dollar coins: " << setw(6) << dollar_coins << endl
<< "Quarters: " << setw(6) << quarters << endl;
step 7 Include the required headers and provide a main function.
We need the
tor, we also require
Therefore, we do not include the
In the main function, you need to define constants and variables (Step 4), carry out com-
putations (Step 5), and pro vide input and output (Step 6). Clearly, you will want to first get
the input, then do the computations, and finally show the output. Define the constants at the
beginning of the function, and define each variable just before it is needed.
cfe2_ch02_p29_74.indd 55 10/27/10 2:30 PM
56 Chapter 2 Fundamental data types
Here is the complete program, ch02/vending.cpp:
#include
#include
using namespace std;
int main()
{
const int PENNIES_PER_DOLLAR = 100;
const int PENNIES_PER_QUARTER = 25;
cout << "Enter bill value (1 = $1 bill, 5 = $5 bill, etc.): ";
int bill_value;
cin >> bill_value;
cout << "Enter item price in pennies: ";
int item_price;
cin >> item_price;
int amount_due = PENNIES_PER_DOLLAR * bill_value – item_price;
int dollar_coins = amount_due / PENNIES_PER_DOLLAR;
amount_due = amount_due % PENNIES_PER_DOLLAR;
int quarters = amount_due / PENNIES_PER_QUARTER;
cout << "Dollar coins: " << setw(6) << dollar_coins << endl
<< "Quarters: " << setw(6) << quarters << endl;
}
program run
Enter bill value (1 = $1 bill, 5 = $5 bill, etc.): 5
Enter item price in pennies: 225
Dollar coins: 2
Quarters: 3
2.5 strings
Many programs process text, not numbers. Text
consists of characters: letters, numbers, punc-
tuation, spaces, and so on. A string is a sequence
of characters. For example, the string "Harry" is a
sequence of five char acters.
W o r k e d e x a m p l e 2 . 2 computing the cost of stamps
This Worked Example uses arithmetic functions to simulate a stamp vending machine.
strings are sequences
of characters.
2.5.1 the string type
You can define variables that hold strings.
string name = "Harry";
The string type is a part of the C++ standard. To use it, simply include the header file,
#include
We distinguish between string variables (such as the variable name defined above) and
string literals (char acter sequences enclosed in quotes, such as “Harry”). The string
stored in a string variable can change. A string literal denotes a particular string, just
as a number literal (such as 2) denotes a particular number.
Unlike number variables, string variables are guaranteed to be initialized even if
you do not supply an initial value. By default, a string variable is set to an empty
string: a string containing no characters. An empty string literal is written as “”. The
definition
string response;
has the same effect as
string response = “”;
2.5.2 Concatenation
Given two strings, such as “Harry” and “Morgan”, you can concatenate them to one
long string. The result consists of all characters in the first string, followed by all
characters in the second string. In C++, you use the + operator to concatenate two
strings. For example,
string fname = “Harry”;
string lname = “Morgan”;
string name = fname + lname;
results in the string
“HarryMorgan”
What if you’d like the first and last name separated by a space? No problem:
string name = fname + ” ” + lname;
This statement concatenates three strings: fname, the string literal ” “, and lname. The
result is
“Harry Morgan”
2.5.3 string Input
You can read a string from the console:
cout << "Please enter your name: ";
string name;
cin >> name;
use the + operator to
concatenate strings;
that is, to put them
together to yield a
longer string.
Available online at www.wiley.com/college/horstmann.
cfe2_ch02_p29_74.indd 56 10/27/10 2:30 PM
www.wiley.com/college/horstmann
2.5 strings 57
Here is the complete program, ch02/vending.cpp:
#include
#include
using namespace std;
int main()
{
const int PENNIES_PER_DOLLAR = 100;
const int PENNIES_PER_QUARTER = 25;
cout << "Enter bill value (1 = $1 bill, 5 = $5 bill, etc.): ";
int bill_value;
cin >> bill_value;
cout << "Enter item price in pennies: ";
int item_price;
cin >> item_price;
int amount_due = PENNIES_PER_DOLLAR * bill_value – item_price;
int dollar_coins = amount_due / PENNIES_PER_DOLLAR;
amount_due = amount_due % PENNIES_PER_DOLLAR;
int quarters = amount_due / PENNIES_PER_QUARTER;
cout << "Dollar coins: " << setw(6) << dollar_coins << endl
<< "Quarters: " << setw(6) << quarters << endl;
}
program run
Enter bill value (1 = $1 bill, 5 = $5 bill, etc.): 5
Enter item price in pennies: 225
Dollar coins: 2
Quarters: 3
2.5 strings
Many programs process text, not numbers. Text
consists of characters: letters, numbers, punc-
tuation, spaces, and so on. A string is a sequence
of characters. For example, the string "Harry" is a
sequence of five char acters.
W o r k e d e x a m p l e 2 . 2 computing the cost of stamps
This Worked Example uses arithmetic functions to simulate a stamp vending machine.
strings are sequences
of characters.
2.5.1 the string type
You can define variables that hold strings.
string name = "Harry";
The string type is a part of the C++ standard. To use it, simply include the header file,
#include
We distinguish between string variables (such as the variable name defined above) and
string literals (char acter sequences enclosed in quotes, such as “Harry”). The string
stored in a string variable can change. A string literal denotes a particular string, just
as a number literal (such as 2) denotes a particular number.
Unlike number variables, string variables are guaranteed to be initialized even if
you do not supply an initial value. By default, a string variable is set to an empty
string: a string containing no characters. An empty string literal is written as “”. The
definition
string response;
has the same effect as
string response = “”;
2.5.2 Concatenation
Given two strings, such as “Harry” and “Morgan”, you can concatenate them to one
long string. The result consists of all characters in the first string, followed by all
characters in the second string. In C++, you use the + operator to concatenate two
strings. For example,
string fname = “Harry”;
string lname = “Morgan”;
string name = fname + lname;
results in the string
“HarryMorgan”
What if you’d like the first and last name separated by a space? No problem:
string name = fname + ” ” + lname;
This statement concatenates three strings: fname, the string literal ” “, and lname. The
result is
“Harry Morgan”
2.5.3 string Input
You can read a string from the console:
cout << "Please enter your name: ";
string name;
cin >> name;
use the + operator to
concatenate strings;
that is, to put them
together to yield a
longer string.
cfe2_ch02_p29_74.indd 57 10/27/10 2:30 PM
58 Chapter 2 Fundamental data types
When a string is read with the >> operator, only one word is placed into the string
variable. For example, suppose the user types
Harry Morgan
as the response to the prompt. This input consists of two words. After the call cin >>
name, the string “Harry” is placed into the variable name. Use another input statement to
read the second word.
2.5.4 string Functions
The number of characters in a string is called the length of the string. For example, the
length of “Harry” is 5. You can compute the length of a string with the length function.
Unlike the sqrt or pow function, the length function is invoked with the dot notation.
That is, you write the string whose length you want, then a period, then the name of
the function, followed by parentheses:
int n = name.length();
Many C++ functions require you to use this dot notation, and you must memorize
(or look up) which do and which don’t. These functions are called member func-
tions. We say that the member function length is invoked on the variable name.
Once you have a string, you can extract substrings by using the substr member
function. The member function call
s.substr(start, length)
returns a string that is made from the characters in the string s, starting at character
start, and containing length characters. Here is an example:
string greeting = “Hello, World!”;
string sub = greeting.substr(0, 5);
// sub is “Hello”
The substr operation makes a string that consists of five characters taken from the
string greeting. Indeed, “Hello” is a string of length 5 that occurs inside greeting. A
curious aspect of the substr operation is the starting position. Starting position 0
means “start at the beginning of the string”. The first position in a string is labeled 0,
the second one 1, and so on. For example, here are the position numbers in the greet-
ing string:
0 1 2 3 4 5 6 7 8 9 10 11 12
H e l l o , W o r l d !
The position number of the last character (12) is always one less than the length of the
string.
Let’s figure out how to extract the substring “World”. Count characters starting at 0,
not 1. You find that W, the 8th character, has position number 7. The string you want is
5 characters long. Therefore, the appro priate substring command is
string w = greeting.substr(7, 5);
0 1 2 3 4 5 6 7 8 9 10 11 12
H e l l o , W o r l d !
5
the length member
function yields the
number of characters
in a string.
a member function is
invoked using the
dot notation.
use the substr
member function to
extract a substring of
a string.
cfe2_ch02_p29_74.indd 58 10/27/10 2:30 PM
2.5 strings 59
If you omit the length, you get all characters from the given position to the end of the
string. For exam ple,
greeting.substr(7)
is the string “World!” (including the exclamation mark).
Here is a simple program that puts these concepts to work. The program asks for
your name and that of your significant other. It then prints out your initials.
The operation first.substr(0, 1) makes a string
consisting of one char acter, taken from the start of
first. The program does the same for the second.
Then it concatenates the resulting one-character
strings with the string literal “&” to get a string of
length 3, the initials string. (See Figure 4.)
ch02/initials.cpp
1 #include
2 #include
3
4 using namespace std;
5
6 int main()
7 {
8 cout << "Enter your first name: ";
9 string first;
10 cin >> first;
11 cout << "Enter your significant other's first name: ";
12 string second;
13 cin >> second;
14 string initials = first.substr(0, 1)
15 + “&” + second.substr(0, 1);
16 cout << initials << endl;
17
18 return 0;
19 }
program run
Enter your first name: Rodolfo
Enter your significant other's first name: Sally
R&S
Initials are formed from the first
letter of each name.Figure 4 Building the initials string
0 1 2
R & Sinitials =
0 1 2 3 4
S a l l ysecond =
0 1 2 3 4 5
R o d o l f
6
ofirst =
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60 Chapter 2 Fundamental data types
25.
table 8 string operations
statement result Comment
string str = "C";
str = str + "++";
str is set to "C++" When applied to strings,
+ denotes concatenation.
string str = "C" + "++"; error error: You cannot concatenate
two string literals.
cout << "Enter name: ";
cin >> name;
(User input: Harry Morgan)
name contains “Harry” The >> operator places the next
word into the string variable.
cout << "Enter name: ";
cin >> name >> last_name;
(User input: Harry Morgan)
name contains
“Harry”, last_name
contains “Morgan”
Use multiple >> operators to
read more than one word.
string greeting = “H & S”;
int n = greeting.length();
n is set to 5 Each space counts as one
character.
string str = “Sally”;
string str2 = str.substr(1, 3);
str2 is set to “all” Extracts the substring of length
3 starting at position 1. (The
initial position is 0.)
string str = “Sally”;
string str2 = str.substr(1);
str2 is set to “ally” If you omit the length, all
characters from the position
until the end are included.
string a = str.substr(0, 1); a is set to the initial
letter in str
Extracts the substring of length
1 starting at position 0.
string b = str.substr(str.length() – 1); b is set to the last
letter in str
The last letter has position
str.length() – 1. We need not
specify the length.
What is the length of the string “C++ Program”?
26. Consider this string variable.
string str = “C++ Program”;
Give a call to the substr member function that returns the substring “gram”.
27. Use string concatenation to turn the string variable str from Self Check 26 to
“C++ Programming”.
28. What does the following statement sequence print?
string str = “Harry”;
cout << str.substr(0, 1) + str.substr(str.length() - 1);
29. Give an input statement to read a name of the form “John Q. Public”.
practice it Now you can try these exercises at the end of the chapter: R2.6, R2.9, P2.12, P2.19.
s e l F c h e c k
cfe2_ch02_p29_74.indd 60 10/27/10 2:30 PM
2.5 strings 61
Hebrew, Arabic, and English
the situation is much more dra matic in languages that
use the Chi nese script: the Chinese dialects, Japanese,
and korean. the Chinese script is not alphabetic but ideo-
graphic. a character represents an idea or thing. most words
are made up of one, two, or three of these ideo graphic
characters. (over 50,000 ideo graphs are known, of which
about 20,000 are in active use.) therefore, two bytes are
needed to encode them. China, taiwan, Japan, and korea
have incompatible encoding standards for them. (Japanese
and korean writing uses a mixture of native syllabic and
Chinese ideographic characters.)
the inconsistencies among charac ter encodings have
been a major nui sance for international electronic commu-
nication and for software man ufacturers vying for a global
market. starting in 1988, a consortium of hardware and
software manufacturers developed a uniform 21-bit encod-
ing scheme called unicode that is capable of encoding
text in essentially all writ ten languages of the world. about
100,000 characters have been given codes, including more
than 70,000 Chinese, Japanese, and korean ideo graphs.
there are even plans to add codes for extinct languages,
such as egyptian hieroglyphs.
The Chinese Script
Random Fact 2.2 International alphabets and unicode
the english alphabet is pretty simple:
upper- and lowercase a to z. other euro-
pean languages have accent marks and special characters.
For example, German has three so-called umlaut charac-
ters, ä, ö, ü, and a double-s char acter ß. these are not
optional frills; you couldn’t write a page of German text
without using these characters a few times. German key-
boards have keys for these characters.
The German Keyboard Layout
this poses a problem for computer users and designers.
the american standard character encoding (called asCII,
for american standard Code for Information Interchange)
specifies 128 codes: 52 upper- and lowercase char acters,
10 digits, 32 typographical symbols, and 34 control charac-
ters (such as space, newline, and 32 others for controlling
printers and other devices). the umlaut and double-s are
not among them. some German data processing systems
replace seldom-used asCII characters with German letters:
[ \ ] { | } ~ are replaced with Ä Ö Ü ä ö ü ß. While most people
can live without these characters, C++ pro grammers defi-
nitely cannot. other encoding schemes take advantage of
the fact that one byte can encode 256 different characters,
of which only 128 are standardized by asCII. unfor tunately,
there are multiple incompati ble standards for such encod-
ings, resulting in a certain amount of aggra vation among
european computer users.
many countries don’t use the roman script at all. rus-
sian, Greek, hebrew, arabic, and thai letters, to name just a
few, have completely diff erent shapes. to complicate mat-
ters, hebrew and arabic are typed from right to left. each
of these alphabets has between 30 and 100 letters, and the
countries using them have estab lished encoding standards
for them.
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62 Chapter 2 Fundamental data types
Write variable definitions in c++.
• A variable is a storage location with a name.
• When defining a variable, you usually specify
an initial value.
• When defining a variable, you also specify the
type of its values.
• Use the int type for numbers that cannot have
a fractional part.
• Use the double type for floating-point numbers.
• An assignment statement stores a new value in a variable, replac ing the previously
stored value.
• The assignment operator = does not denote mathematical equality.
• You cannot change the value of a variable that is defined as const.
• Use comments to add explanations for humans who read your code. The com-
piler ignores comments.
use the arithmetic operations in c++.
• Use * for multiplication and / for division.
• The ++ operator adds 1 to a variable; the -- operator subtracts 1.
• If both arguments of / are integers, the remain der is discarded.
• The % operator computes the remainder of an integer division.
• Assigning a floating-point variable to an inte ger drops the
fractional part.
• The C++ library defines many mathematical functions such as sqrt (square root)
and pow (raising to a power).
Write programs that read user input and write formatted output.
• Use the >> operator to read a value and place it in
a variable.
• You use manipulators to specify how val ues should be
formatted.
carry out hand calculations when developing an algorithm.
• Pick concrete values for a typical situation to use in a hand calculation.
Write programs that process strings.
• Strings are sequences of characters.
• Use the + operator to concatenate strings; that is, put them together to yield a
longer string.
• The length member function yields the number of characters in a string.
C h a p t e r s u m m a r y • A member function is invoked using the dot notation.
• Use the substr member function to extract a substring of a string
r2.1 What is the value of mystery after this sequence of statements?
int mystery = 1;
mystery = 1 – 2 * mystery;
mystery = mystery + 1;
r2.2 What is wrong with the following sequence of statements?
int mystery = 1;
mystery = mystery + 1;
int mystery = 1 – 2 * mystery;
r2.3 Write the following mathematical expressions in C++.
r2.4 Write the following C++ expressions in mathematical notation.
a. dm = m * (sqrt(1 + v / c) / sqrt(1 – v / c) – 1);
b. volume = PI * r * r * h;
c. volume = 4 * PI * pow(r, 3) / 3;
d. z = sqrt(x * x + y * y);
r2.5 What are the values of the following expressions? In each line, assume that
double x = 2.5;
double y = -1.5;
int m = 18;
int n = 4;
a. x + n * y – (x + n) * y
b. m / n + m % n
c. 5 * x – n / 5
d. 1 – (1 – (1 – (1 – (1 – n))))
e. sqrt(sqrt(n))
r2.6 What are the values of the following expressions? In each line, assume that
string s = “Hello”;
string t = “World”;
r e V I e W e x e r C I s e s
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review exercises 63
• A member function is invoked using the dot notation.
• Use the substr member function to extract a substring of a string
r2.1 What is the value of mystery after this sequence of statements?
int mystery = 1;
mystery = 1 – 2 * mystery;
mystery = mystery + 1;
r2.2 What is wrong with the following sequence of statements?
int mystery = 1;
mystery = mystery + 1;
int mystery = 1 – 2 * mystery;
r2.3 Write the following mathematical expressions in C++.
s s v t gt
G
a
p m m
= + +
=
+
= ⋅ +
0 0
2
2
3
2
1 2
1
2
4
1
π
( )
FV PV
INT
1000
YRS
= + −c a b ab2 2 2 cosγ
r2.4 Write the following C++ expressions in mathematical notation.
a. dm = m * (sqrt(1 + v / c) / sqrt(1 – v / c) – 1);
b. volume = PI * r * r * h;
c. volume = 4 * PI * pow(r, 3) / 3;
d. z = sqrt(x * x + y * y);
r2.5 What are the values of the following expressions? In each line, assume that
double x = 2.5;
double y = -1.5;
int m = 18;
int n = 4;
a. x + n * y – (x + n) * y
b. m / n + m % n
c. 5 * x – n / 5
d. 1 – (1 – (1 – (1 – (1 – n))))
e. sqrt(sqrt(n))
r2.6 What are the values of the following expressions? In each line, assume that
string s = “Hello”;
string t = “World”;
r e V I e W e x e r C I s e s
cfe2_ch02_p29_74.indd 63 10/27/10 2:30 PM
64 Chapter 2 Fundamental data types
a. s.length() + t.length()
b. s.substr(1, 2)
c. s.substr(s.length() / 2, 1)
d. s + t
e. t + s
r2.7 Find at least five compile-time errors in the following program.
#include iostream
int main();
{
cout << "Please enter two numbers:"
cin << x, y;
cout << "The sum of << x << "and" << y
<< "is: " x + y << endl;
return;
}
r2.8 Find at least four run-time errors in the following program.
#include
using namespace std;
int main()
{
int total;
int x1;
cout << "Please enter a number:";
cin >> x1;
total = total + x1;
cout << "Please enter another number:";
int x2;
cin >> x2;
total = total + x1;
double average = total / 2;
cout << "The average of the two numbers is "
<< average << "endl";
return 0;
}
r2.9 Explain the differences between 2, 2.0, "2", and "2.0".
r2.10 Explain what each of the following program segments computes.
a. int x = 2;
int y = x + x;
b. string s = "2";
string t = s + s;
r2.11 Write pseudocode for a program that reads a word and then prints the first charac ter,
the last character, and the characters in the middle. For example, if the input is Harry,
the program prints H y arr.
r2.12 Write pseudocode for a program that reads a name (such as Harold James Morgan) and
then prints a monogram consisting of the initial letters of the first, middle, and last
names (such as HJM).
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review exercises 65
r2.13 Write pseudocode for a program that computes the first and last digit of a number.
For example, if the input is 23456, the program should print out 2 and 6. Hint: %, log10.
r2.14 Modify the pseudocode for the program in How To 2.1 so that the pro gram gives
change in quarters, dimes, and nickels. You can assume that the price is a multiple of
5 cents. To develop your pseudocode, first work with a couple of spe cific values.
r2.15 A cocktail shaker is composed of three cone sections.
Using realistic values for the radii and heights, compute the total
volume, using the formula given in Self Check 24 for a cone section.
Then develop an algorithm that works for arbitrary dimensions.
r2.16 You are cutting off a piece of pie like this, where c is the length of the
straight part (called the chord length) and h is the height of the piece.
hc
d
There is an approximate formula for the area: A ch
h
c
≈ +2
3
3
2
However, h is not so easy to measure, whereas the diameter d of a pie is usually
well-known. Calculate the area where the diameter of the pie is 12 inches and the
chord length of the segment is 10 inches. Gen eralize to an algorithm that yields the
area for any diameter and chord length.
r2.17 The following pseudocode describes how to obtain the name of a day, given the day
number (0 = Sunday, 1 = Monday, and so on.)
Define a string called names containing "SunMonTueWedThuFriSat".
Compute the starting position as 3 x the day number.
Extract the substring of names at the starting position with length 3.
Check this pseudocode, using the day number 4. Draw a diagram of the string that is
being computed, similar to Figure 4 on page 59.
r2.18 The following pseudocode describes how to swap two letters in a word.
We are given a string str and two positions i and j. (i comes before j)
Set first to the substring from the start of the string to the last position before i.
Set middle to the substring from positions i + 1 to j - 1.
Set last to the substring from position j + 1 to the end of the string.
Concatenate the following five strings: first, the string containing just the character at position j, middle,
the string containing just the character at position i, and last.
Check this pseudocode, using the string "Gateway" and positions 2 and 4. Draw a
diagram of the string that is being computed, similar to Figure 4 on page 59.
cfe2_ch02_p29_74.indd 65 10/27/10 2:30 PM
66 Chapter 2 Fundamental data types
r2.19 Run the following program, and explain the output you get.
#include
using namespace std;
int main()
{
int total;
cout << "Please enter a number: ";
double x1;
cin >> x1;
total = total + x1;
cout << "total: " << total << endl;
cout << "Please enter a number: ";
double x2;
cin >> x2;
total = total + x2;
cout << "total: " << total << endl;
total = total / 2;
cout << "total: " << total << endl;
cout << "The average is " << total << endl;
return 0;
}
Note the trace messages (in blue) that are inserted to show the current contents of
the total variable. How do you fix the program? (The program has two separate
errors.)
r2.20 Write a program that prints the values
3 * 1000 * 1000 * 1000
3.0 * 1000 * 1000 * 1000
Explain the results.
r2.21 This chapter contains a number of recommendations regarding variables and con-
stants that make programs easier to read and maintain. Briefly summarize these
rec ommendations.
p2.1 Write a program that displays the dimensions of a letter-size (8.5 × 11 inches) sheet
of paper in millimeters. There are 25.4 millimeters per inch. Use con stants and com-
ments in your program.
p2.2 Write a program that computes and displays the circumference of a letter-size (8.5 ×
11 inches) sheet of paper and the length of its diagonal.
p2.3 Write a program that reads a number and displays the square, cube, and fourth
power. Use the pow function only for the fourth power.
p2.4 Write a program that prompts the user for two integers and then prints
• The sum
• The difference
p r o G r a m m I n G e x e r C I s e s
cfe2_ch02_p29_74.indd 66 10/27/10 2:30 PM
programming exercises 67
• The product
• The average
p2.5 Write a program that prompts the user for two integers and then prints
• The distance (absolute value of the difference)
• The maximum (the larger of the two)
• The minimum (the smaller of the two)
Hint: The max and min functions are defined in the
p2.6 Write a program that prompts the user for a measurement in meters and then
con verts it to miles, feet, and inches.
p2.7 Write a program that prompts the user for a radius and then prints
• The area and circumference of a circle with that radius
• The volume and surface area of a sphere with that radius
p2.8 Write a program that asks the user for the lengths of the sides of a
rectangle and then prints
• The area and perimeter of the rectangle
• The length of the diagonal (use the Pythagorean theorem)
p2.9 Improve the program discussed in the How To 2.1 to allow input of quarters in
addition to bills.
p2.10 Write a program that helps a person decide whether to buy a hybrid car. Your
pro gram’s inputs should be:
• The cost of a new car
• The estimated miles driven per year
• The estimated gas price
• The estimated resale value after 5 years
Compute the total cost of owning the car
for 5 years. (For simplic ity, we will not take
the cost of financing into account.) Obtain
realistic prices for a new and used hybrid and
a comparable car from the Web. Run your
program twice, using today’s gas price and 15,000 miles per year. Include pseudo-
code and the program runs with your assignment.
p2.11 The following pseudocode describes how a bookstore computes the price of an
order from the total price and the number of the books that were ordered.
Read the total book price and the number of books.
Compute the tax (7.5% of the total book price).
Compute the shipping charge ($2 per book).
The price of the order is the sum of the total book price, the tax, and the shipping charge.
Print the price of the order.
Translate this pseudocode into a C++ program.
cfe2_ch02_p29_74.indd 67 10/27/10 2:30 PM
68 Chapter 2 Fundamental data types
p2.12 The following pseudocode describes how to turn a string containing a ten-digit
phone number (such as “4155551212”) into a more readable string with parentheses
and dashes, like this: “(415) 555-1212”.
Take the substring consisting of the first three characters and surround it with “(” and “)”. This is the
area code.
Concatenate the area code, the substring consisting of the next three characters, a hyphen, and the
substring consisting of the last four characters. This is the formatted number.
Translate this pseudocode into a C++ program that reads a telephone number into a
string variable, com putes the formatted number, and prints it.
p2.13 The following pseudocode describes how to extract the dollars and cents from a
price given as a floating-point value. For example, a price 2.95 yields values 2 and 95
for the dollars and cents.
Assign the price to an integer variable dollars.
Multiply the difference price – dollars by 100 and add 0.5.
Assign the result to an integer variable cents.
Translate this pseudocode into a C++ program. Read a price and print the dollars
and cents. Test your program with inputs 2.95 and 4.35.
p2.14 Giving change. Implement a program that directs a cashier
how to give change. The program has two inputs: the
amount due and the amount received from the customer.
Display the dollars, quarters, dimes, nickels, and pennies
that the customer should receive in return.
p2.15 Write a program that asks the user to input
• The number of gallons of gas in the tank
• The fuel efficiency in miles per gallon
• The price of gas per gallon
Then print the cost per 100 miles and how far the car can go with the gas in the tank.
p2.16 File names and extensions. Write a program that prompts the user for the drive letter
(C), the path (\Windows\System), the file name (Readme), and the extension (txt). Then
print the complete file name C:\Windows\System\Readme.txt. (If you use UNIX or a
Macintosh, skip the drive name and use / instead of \ to separate directories.)
p2.17 Write a program that reads a number between 1,000 and 999,999 from the user and
prints it with a comma separating the thousands. Here is a sample dialog; the user
input is in color:
Please enter an integer between 1000 and 999999: 23456
23,456
p2.18 Write a program that reads a number between 1,000 and 999,999 from the user,
where the user enters a comma in the input. Then print the number without a
comma. Here is a sample dialog; the user input is in color:
Please enter an integer between 1,000 and 999,999: 23,456
23456
Hint: Read the input as a string. Measure the length of the string. Suppose it contains
n characters. Then extract substrings consisting of the first n – 4 characters and the
last three characters.
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programming exercises 69
p2.19 Printing a grid. Write a program that prints the following grid to play tic-tac-toe.
+–+–+–+
| | | |
+–+–+–+
| | | |
+–+–+–+
| | | |
+–+–+–+
Of course, you could simply write seven statements of the form
cout << "+--+--+--+";
You should do it the smart way, though. Define string variables to hold two kinds of
patterns: a comb-shaped pattern
+--+--+--+
| | | |
and the bottom line. Print the comb three times and the bottom line once.
p2.20 Write a program that reads an integer and breaks it into a sequence of individual
digits. For example, the input 16384 is displayed as
1 6 3 8 4
You may assume that the input has no more than five digits and is not negative.
p2.21 Write a program that reads two times in military format (0900, 1730) and prints the
number of hours and minutes between the two times. Here is a sample run. User
input is in color.
Please enter the first time: 0900
Please enter the second time: 1730
8 hours 30 minutes
Extra credit if you can deal with the case where the first time is later than the second:
Please enter the first time: 1730
Please enter the second time: 0900
15 hours 30 minutes
p2.22 Writing large letters. A large letter H can be produced like this:
* *
* *
*****
* *
* *
It can be defined as a string constant like this:
const string LETTER_H =
"* *\n* *\n*****\n* *\n* *\n";
(The \n character is explained in Special Topic 1.1.) Do the same for the letters E, L,
and O. Then write the message
H
E
L
L
O
in large letters.
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70 Chapter 2 Fundamental data types
p2.23 Write a program that transforms numbers 1, 2, 3, …, 12
into the corresponding month names January, February,
March, …, December. Hint: Make a very long string "January
February March ...", in which you add spaces such that
each month name has the same length. Then use substr to
extract the month you want.
engineering p2.24 Consider the following circuit.
R1
R2 R3
Write a program that reads the resistances of the three resistors and computes the
total resistance, using Ohm’s law.
engineering p2.25 The dew point temperature Td can be calculated (approximately) from the relative
humidity RH and the actual temperature T by
T
b f T RH
a f T RH
f T RH
a T
b T
RH
d =
⋅ ( )
− ( )
( ) = ⋅
+
+ ( )
,
,
, ln
where a = 17.27 and b = 237.7° C.
Write a program that reads the relative humidity (between 0 and 1) and the tem-
perature (in degrees C) and prints the dew point value. Use the C++ function log to
compute the natural logarithm.
engineering p2.26 The pipe clip temperature sensors shown here are robust sensors that can be clipped
directly onto copper pipes to measure the temperature of the liquids in the pipes.
Each sensor contains a device called a thermistor. Thermistors are semiconductor
devices that exhibit a temperature-dependent resistance described by:
R R e
T T
=
−
0
1 1
0
β
cfe2_ch02_p29_74.indd 70 10/27/10 2:30 PM
programming exercises 71
where R is the resistance (in Ω) at the temperature T (in °K), and R0 is the resistance
(in Ω) at the temperature T0 (in °K). β is a constant that depends on the material used
to make the thermistor. Thermistors are specified by providing values for R0, T0,
and β.
The thermistors used to make the pipe clip temperature sensors have R0 = 1075 Ω at
T0 = 85 °C, and β = 3969 °K. (Notice that β has units of °K. Recall that the tempera-
ture in °K is obtained by adding 273 to the temperature in °C.) The liquid tempera-
ture, in °C, is determined from the resistance R, in Ω, using
T
T
T
R
R
=
+
−
β
β
0
0
0
273
ln
Write a C++ program that prompts the user for the thermistor resistance R and
prints a message giving the liquid temperature in °C.
engineering p2.27 The circuit shown below illustrates some
important aspects of the connection between a
power company and one of its customers. The
customer is represented by three parameters,
Vt, P, and pf. Vt is the voltage accessed by
plugging into a wall outlet. Customers depend
on having a dependable value of Vt in order for
their appliances to work properly. Accordingly,
the power company regulates the value of Vt
carefully. P describes the amount of power
used by the customer and is the primary factor
in determining the customer’s electric bill. The power factor, pf, is less familiar. (The
power factor is calculated as the cosine of an angle so that its value will always be
between zero and one.) In this problem you will be asked to write a C++ program to
investigate the significance of the power factor.
Vs
Customer
+
–
R = 10 Ω
Power
Lines
Power
Company
R = 10 Ω
P = 260 W
p f = 0.6
Vt = 120 Vrms
+
–
In the figure, the power lines are represented, somewhat simplistically, as resistances
in Ohms. The power company is represented as an AC voltage source. The source
voltage, Vs, required to provide the customer with power P at voltage Vt can be
determined using the formula
V V
RP
V
RP
pf V
pfs t
t t
= +
+
−( )2 2 1
2 2
2
(Vs has units of Vrms.) This formula indicates that the value of Vs depends on the
value of pf. Write a C++ program that prompts the user for a power factor value and
cfe2_ch02_p29_74.indd 71 10/27/10 2:30 PM
72 Chapter 2 Fundamental data types
then prints a message giving the corresponding value of Vs, using the values for P, R,
and Vt shown in the figure above.
engineering p2.28 Consider the following tuning circuit connected to an antenna, where C is a variable
capacitor whose capacitance ranges from Cmin to Cmax.
L C
Antenna
The tuning circuit selects the frequency f
LC
=
2π
. To design this circuit for a given
frequency, take C C C= min max and calculate the required inductance L from f and
C. Now the circuit can be tuned to any frequency in the range f
LC
min
max
=
2π
to
f
LC
max
min
=
2π
.
Write a C++ program to design a tuning circuit for a given frequency, using a vari-
able capacitor with given values for Cmin and Cmax. (A typical input is f = 16.7 MHz,
Cmin = 14 pF, and Cmax = 365 pF.) The program should read in f (in Hz), Cmin and
Cmax (in F), and print the required inductance value and the range of frequencies to
which the circuit can be tuned by varying the capacitance.
engineering p2.29 According to the Coulomb force law, the electric force between two charged par-
ticles of charge Q1 and Q2 Coulombs, that are a distance r meters apart, is
F
Q Q
r
= 1 2
24 π ε
Newtons, where ε = × −8 854 10 12. Farads/meter. Write a program
that calculates the force on a pair of charged particles, based on the user input of Q1
Coulombs, Q2 Coulombs, and r meters, and then computes and displays the electric
force.
1. One possible answer is
int bottles_per_case = 8;
You may choose a different variable name or a different initialization value, but your
variable should have type int.
2. There are three errors:
• You cannot have spaces in variable names.
• The variable type should be double because it holds a fractional value.
• There is a semicolon missing at the end of the statement.
3. double unit_price = 1.95;
int quantity = 2;
4. cout << "Total price: " << unit_price * quantity << endl;
5. Change the definition of cans_per_pack to
int cans_per_pack = 4;
6. You need to use a */ delimiter to close a comment that begins with a /*:
double can_volume = 0.355; /* Liters in a 12-ounce can */
7. The program would compile, and it would display the same result. However, a
per son reading the program might find it confusing that fractional cans are being
con sidered.
8. Its value is modified by the assignment statement.
9. Assignment would occur when one car is replaced by another in the parking space.
10. double interest = balance * p / 100;
11. double side_length = sqrt(area);
12. 4 * PI * pow(radius, 3) / 3
or (4.0 / 3) * PI * pow(radius, 3),
but not (4 / 3) * PI * pow(radius, 3)
13. 172 and 9
14. int leftover = 12 - amount % 12;
15. There is no prompt that alerts the program user to enter the quantity.
16. The unit_price variable is defined as an int. If the user were to enter a price such as
1.95, only the 1 would be placed into the variable.
17. The output is
The total volume is20
Note that there is no space between the is and 20.
18. cout << "$" << fixed << setprecision(2) << total_price;
19. cout << "Bottles: " << setw(8) << bottles << endl
<< "Cans: " << setw(8) << cans << endl;
Note that the setw manipulator appears twice. Also note the added spaces in the
string "Cans: ".
a n s W e r s t o s e l F - C h e C k Q u e s t I o n s
cfe2_ch02_p29_74.indd 72 10/27/10 2:30 PM
answers to self-Check Questions 73
1. One possible answer is
int bottles_per_case = 8;
You may choose a different variable name or a different initialization value, but your
variable should have type int.
2. There are three errors:
• You cannot have spaces in variable names.
• The variable type should be double because it holds a fractional value.
• There is a semicolon missing at the end of the statement.
3. double unit_price = 1.95;
int quantity = 2;
4. cout << "Total price: " << unit_price * quantity << endl;
5. Change the definition of cans_per_pack to
int cans_per_pack = 4;
6. You need to use a */ delimiter to close a comment that begins with a /*:
double can_volume = 0.355; /* Liters in a 12-ounce can */
7. The program would compile, and it would display the same result. However, a
per son reading the program might find it confusing that fractional cans are being
con sidered.
8. Its value is modified by the assignment statement.
9. Assignment would occur when one car is replaced by another in the parking space.
10. double interest = balance * p / 100;
11. double side_length = sqrt(area);
12. 4 * PI * pow(radius, 3) / 3
or (4.0 / 3) * PI * pow(radius, 3),
but not (4 / 3) * PI * pow(radius, 3)
13. 172 and 9
14. int leftover = 12 - amount % 12;
15. There is no prompt that alerts the program user to enter the quantity.
16. The unit_price variable is defined as an int. If the user were to enter a price such as
1.95, only the 1 would be placed into the variable.
17. The output is
The total volume is20
Note that there is no space between the is and 20.
18. cout << "$" << fixed << setprecision(2) << total_price;
19. cout << "Bottles: " << setw(8) << bottles << endl
<< "Cans: " << setw(8) << cans << endl;
Note that the setw manipulator appears twice. Also note the added spaces in the
string "Cans: ".
a n s W e r s t o s e l F - C h e C k Q u e s t I o n s
cfe2_ch02_p29_74.indd 73 10/27/10 2:30 PM
74 Chapter 2 Fundamental data types
20. int pairs = (total_width - tile_width) / (2 * tile_width);
int tiles = 1 + 2 * pairs;
double gap = (total_width - tiles * tile_width) / 2;
Be sure that pairs is declared as an int.
21. Now there are groups of four tiles (gray/white/gray/black) following the initial
black tile. Therefore, the algorithm is now
number of groups = integer part of (total width - tile width) / (4 x tile width)
number of tiles = 1 + 4 x number of groups
The formula for the gap is not changed.
22. Clearly, the answer depends only on whether the row and column numbers are even
or odd, so let’s first take the remainder after dividing by 2. Then we can enumerate
all expected answers:
Row % 2 Column % 2 Color
0 0 0
0 1 1
1 0 1
1 1 0
In the first three entries of the table, the color is simply the sum of the remainders.
In the fourth entry, the sum would be 2, but we want a zero. We can achieve that by
taking another remainder operation:
color = ((row % 2) + (column % 2)) % 2
23. In nine years, the repair costs increased by $1,400. Therefore, the increase per year
is $1,400 / 9 ≈ $156. The repair cost in year 3 would be $100 + 2 × $156 = $412. The
repair cost in year n is $100 + n × $156. To avoid accumulation of roundoff errors, it is
actually a good idea to use the original expression that yielded $156, that is,
Repair cost in year n = 100 + n x 1400 / 9
24. The pseudocode follows easily from the equations.
bottom volume = p x r1
2 x h1
top volume = p x r2
2 x h2
middle volume = p x (r1
2 + r1 x r2 + r2
2) x h3 / 3
total volume = bottom volume + top volume + middle volume
Measuring a typical wine bottle yields
r1 = 3.6, r2 = 1.2, h1 = 15, h2 = 7, h3 = 6 (all in centimeters). Therefore,
bottom volume = 610.73
top volume = 31.67
middle volume = 135.72
total volume = 778.12
The actual volume is 750 ml, which is close enough to our computation to give confi-
dence that it is cor rect.
25. The length is 11. The space counts as a character.
26. str.substr(7, 4)
27. str = str + "ming";
28. Hy
29. cin >> first_name >> middle_initial >> last_name;
A robot needs to retrieve an item that is located
in rocky terrain adjacent to a road. The robot can
travel at a faster speed on the road than on the
rocky terrain, so it will want to do so for a certain
distance before moving on a straight line to the
item.
Your task is to compute the total time taken
by the robot to reach its goal, given the following
inputs:
• The distance between the robot and the item in
the x- and y-direction (dx and dy)
• The speed of the robot on the road and the
rocky terrain (s1 and s2)
• The length l1 of the first segment (on the road)
To make the problem more concrete, let’s assume the following dimen sions:
The total time is the time for traversing both segments. The time to traverse the first seg-
ment is simply the length of the segment divided by the speed: 6 km divided by 5 km/h, or 1.2
hours.
W o r k e d e x a m p l e 2 . 1 computing travel time
In this example, we develop a hand calculation to compute the time
that a robot requires to retrieve an item from rocky terrain.
dy
dx Item
Robot
l1
Speed = s1
Speed = s2
l2
10 km
3 km Item
Robot
6 km
Speed = 5 km/h
Speed = 2 km/h
cfe2_ch02_p29_74.indd 74 10/27/10 2:30 PM
3C h a p t e r
75
D e C i s i o n s
to be able to implement decisions
using if statements
to learn how to compare integers,
floating-point numbers, and strings
to understand the Boolean data type
to develop strategies for validating user input
C h a p t e r G o a l s
C h a p t e r C o n t e n t s
3.1 The if STaTemenT 76
Syntax 3.1: if statement 78
Programming Tip 3.1: Brace layout 79
Programming Tip 3.2: always Use Braces 80
Common Error 3.1: a semicolon after the
if Condition 80
Programming Tip 3.3: tabs 81
Special Topic 3.1: the selection operator 81
Programming Tip 3.4: avoid Duplication in
Branches 82
3.2 Comparing numberS and
STringS 82
Syntax 3.2: Comparisons 83
Common Error 3.2: Confusing = and == 85
Programming Tip 3.5: Compile with Zero
Warnings 85
Common Error 3.3: exact Comparison of
Floating-point numbers 86
Special Topic 3.2: lexicographic ordering
of strings 86
How To 3.1: implementing an if statement 87
Worked Example 3.1: extracting the Middle
Random Fact 3.1: the Denver airport luggage
handling system 89
3.3 mulTiple alTernaTiveS 90
Special Topic 3.3: the switch statement 93
3.4 neSTed branCheS 94
Programming Tip 3.6: hand-tracing 97
Common Error 3.4: the Dangling else problem 98
3.5 problem Solving: flowCharTS 99
3.6 problem Solving: TeST CaSeS 102
Programming Tip 3.7: Make a schedule and Make
time for Unexpected problems 103
3.7 boolean variableS and
operaTorS 103
Common Error 3.5: Combining Multiple
relational operators 107
Common Error 3.6: Confusing && and ||
Conditions 107
Special Topic 3.4: short-Circuit evaluation of
Boolean operators 108
Special Topic 3.5: De Morgan’s law 108
3.8 appliCaTion: inpuT
validaTion 109
Random Fact 3.2: artificial intelligence 112
cfe2_ch03_p75_130.indd 75 10/28/10 7:51 PM
76
one of the essential features of computer programs is
their ability to make decisions. like a train that changes
tracks depending on how the switches are set, a program
can take different actions, depending on inputs and other
circumstances.
in this chapter, you will learn how to program simple and
complex decisions. You will apply what you learn to the
task of checking user input.
3.1 the if statement
The if statement is used to implement a decision. When a condition is fulfilled, one
set of statements is executed. Otherwise, another set of statements is executed (see
Syntax 3.1).
Here is an example using the if statement. In
many countries, the number 13 is considered
unlucky. Rather than offending supersti tious ten-
ants, building owners sometimes skip the thirteenth
floor; floor 12 is immediately followed by floor 14.
Of course, floor 13 is not usually left empty or, as
some conspiracy theorists believe, filled with secret
offices and research labs. It is simply called floor
14. The computer that controls the building eleva-
tors needs to compensate for this foible and adjust
all floor numbers above 13.
Let’s simulate this process in C++. We will ask
the user to type in the desired floor number and
then compute the actual floor. When the input is
above 13, then we need to decrement the input to
obtain the actual floor. For example, if the user
provides an input of 20, the program determines
the actual floor as 19. Otherwise, we simply use the
supplied floor number.
int actual_floor;
if (floor > 13)
{
actual_floor = floor – 1;
}
else
{
actual_floor = floor;
}
The flowchart in Figure 1 shows the branching behavior.
In our example, each branch of the if statement contains a single statement. You
can include as many statements in each branch as you like. Sometimes, it happens that
the if statement
allows a program to
carry out different
actions depending on
the nature of the data
to be processed.
This elevator panel “skips” the
thirteenth floor. The floor is not
actually missing—the computer
that controls the elevator adjusts
the floor numbers above 13.
cfe2_ch03_p75_130.indd 76 10/28/10 7:51 PM
3.1 the if statement 77
figure 1
Flowchart for if statement
floor > 13?
True False
actual_floor =
floor – 1
actual_floor =
floor
Condition
figure 2
Flowchart for if statement with no else Branch
floor > 13?
True False
actual_floor–
No else branch
there is nothing to do in the else branch of the statement. In that case, you can omit it
entirely, such as in this example:
int actual_floor = floor;
if (floor > 13)
{
actual_floor–;
} // No else needed
See Figure 2 for the flowchart.
The following program puts the if statement to work. This program asks for the
desired floor and then prints out the actual floor.
An if statement is like a fork in
the road. Depending upon a
decision, different parts of the
program are executed.
cfe2_ch03_p75_130.indd 77 10/28/10 7:51 PM
78 Chapter 3 Decisions
1. In some Asian countries, the number 14 is considered unlucky. Some building
owners play it safe and skip both the thirteenth and the fourteenth floor. How
would you modify the sample program to handle such a building?
2. Consider the following if statement to compute a discounted price:
if (original_price > 100)
{
discounted_price = original_price – 20;
}
else
{
discounted_price = original_price – 10;
}
What is the discounted price if the original price is 95? 100? 105?
3. Compare this if statement with the one in Self Check 2:
if (original_price < 100)
{
discounted_price = original_price - 10;
}
else
{
discounted_price = original_price - 20;
}
Do the two statements always compute the same value? If not, when do the
values differ?
4. Consider the following statements to compute a discounted price:
discounted_price = original_price;
if (original_price > 100)
{
discounted_price = original_price – 10;
}
What is the discounted price if the original price is 95? 100? 105?
5. The variables fuel_amount and fuel_capacity hold the actual amount of fuel and the
size of the fuel tank of a vehicle. If less than 10 percent is remaining in the tank, a
status light should show a red color; otherwise it shows a green color. Simulate
this process by printing out either “red” or “green”.
practice it Now you can try these exercises at the end of the chapter: R3.3, R3.4, P3.16.
brace layout
Programmers vary in how they align braces
in their code. In this book, we follow the
simple rule of making { and } line up.
if (floor > 13)
{
floor–;
}
This style makes it easy to spot matching
braces.
S e l f C h e C k
programming tip 3.1
Properly lining up your code makes your
programs easier to read.
ch03/elevator1.cpp
1 #include
2
3 using namespace std;
4
5 int main()
6 {
7 int floor;
8 cout << "Floor: ";
9 cin >> floor;
10 int actual_floor;
11 if (floor > 13)
12 {
13 actual_floor = floor – 1;
14 }
15 else
16 {
17 actual_floor = floor;
18 }
19
20 cout << "The elevator will travel to the actual floor "
21 << actual_floor << endl;
22
23 return 0;
24 }
program run
Floor: 20
The elevator will travel to the actual floor 19
syntax 3.1 if statement
if (floor > 13)
{
actual_floor = floor – 1;
}
else
{
actual_floor = floor;
}
A condition that is true or false.
Often uses relational operators:
== != < <= > >= (See page 83.)
If the condition is true, the statement(s)
in this branch are executed in sequence;
if the condition is false, they are skipped.
Braces are not required
if the branch contains a
single statement, but it’s
good to always use them.
See page 80.
If the condition is false, the statement(s)
in this branch are executed in sequence;
if the condition is true, they are skipped.
Lining up braces
is a good idea.
See page 79.
Omit the else branch
if there is nothing to do.
Don’t put a semicolon here!
See page 80.
cfe2_ch03_p75_130.indd 78 10/28/10 7:51 PM
3.1 the if statement 79
1. In some Asian countries, the number 14 is considered unlucky. Some building
owners play it safe and skip both the thirteenth and the fourteenth floor. How
would you modify the sample program to handle such a building?
2. Consider the following if statement to compute a discounted price:
if (original_price > 100)
{
discounted_price = original_price – 20;
}
else
{
discounted_price = original_price – 10;
}
What is the discounted price if the original price is 95? 100? 105?
3. Compare this if statement with the one in Self Check 2:
if (original_price < 100)
{
discounted_price = original_price - 10;
}
else
{
discounted_price = original_price - 20;
}
Do the two statements always compute the same value? If not, when do the
values differ?
4. Consider the following statements to compute a discounted price:
discounted_price = original_price;
if (original_price > 100)
{
discounted_price = original_price – 10;
}
What is the discounted price if the original price is 95? 100? 105?
5. The variables fuel_amount and fuel_capacity hold the actual amount of fuel and the
size of the fuel tank of a vehicle. If less than 10 percent is remaining in the tank, a
status light should show a red color; otherwise it shows a green color. Simulate
this process by printing out either “red” or “green”.
practice it Now you can try these exercises at the end of the chapter: R3.3, R3.4, P3.16.
brace layout
Programmers vary in how they align braces
in their code. In this book, we follow the
simple rule of making { and } line up.
if (floor > 13)
{
floor–;
}
This style makes it easy to spot matching
braces.
S e l f C h e C k
programming tip 3.1
Properly lining up your code makes your
programs easier to read.
cfe2_ch03_p75_130.indd 79 10/28/10 7:51 PM
80 Chapter 3 Decisions
Tabs
Block-structured code has the property that nested statements are indented by one or more
levels:
int main()
{
| int floor;
| …
| if (floor > 13)
| {
| | floor–;
| } |
| …
| return 0;
} | |
0 1 2 Indentation level
How do you move the cursor from the leftmost
column to the appropriate indentation level? A
perfectly reasonable strategy is to hit the space bar
a suf ficient number of times. However, many pro-
grammers use the Tab key instead. A tab moves the
cursor to the next indentation level.
While the Tab key is nice, some editors use tab characters for alignment, which is not so
nice. Tab characters can lead to problems when you send your file to another person or a
printer. There is no universal agreement on the width of a tab character, and some soft ware will
ignore tabs altogether. It is therefore best to save your files with spaces instead of tabs. Most
editors have a setting to auto matically convert all tabs to spaces. Look at the documentation of
your development environment to find out how to activate this useful setting.
The Selection operator
C++ has a selection operator of the form
condition ? value1 : value2
The value of that expression is either value1 if the test passes or value2 if it fails. For example,
we can compute the actual floor number as
actual_floor = floor > 13 ? floor – 1 : floor;
which is equivalent to
if (floor > 13)
{
actual_floor = floor – 1;
}
else
{
actual_floor = floor;
}
You can use the selection operator anywhere that a value is expected, for example:
cout << "Actual floor: " << (floor > 13 ? floor – 1 : floor);
We don’t use the selection operator in this book, but it is a convenient construct that you will
find in many C++ pro grams.
programming tip 3.3
You use
the Tab key
to move the
cursor to the next
indentation level.
special topic 3.1
Some programmers put the opening brace on the same line as the if:
if (floor > 13) {
floor–;
}
This style makes it harder to match the braces, but it saves a line of code, allowing you to view
more code on the screen without scroll ing. There are passionate advocates of both styles.
It is important that you pick a layout style and stick with it con sistently within a given
programming project. Which style you choose may depend on your personal preference or a
coding style guide that you need to follow.
always use braces
When the body of an if statement consists of a single statement, you need not use braces. For
example, the following is legal:
if (floor > 13)
floor–;
However, it is a good idea to always include the braces:
if (floor > 13)
{
floor–;
}
The braces makes your code easier to read, and you are less likely to make errors such as the
one described in Common Error 3.1.
a Semicolon after the if Condition
The following code fragment has an unfortunate error:
if (floor > 13) ; // ERROR
{
floor–;
}
There should be no semicolon after the if condition. The compiler interprets this statement as
follows: If floor is greater than 13, execute the statement that is denoted by a single semicolon,
that is, the do-nothing statement. The statement enclosed in braces is no longer a part of the if
statement. It is always executed. Even if the value of floor is not above 13, it is decremented.
Placing a semicolon after the else reserved word is also wrong:
if (floor > 13)
{
actual_floor = floor – 1;
}
else ;
{
actual_floor = floor;
}
In this case, the do-nothing statement is executed if floor > 13 is not fulfilled. This is the end
of the if statement. The next statement, enclosed in braces, is executed in both cases; that is,
actual_floor is always set to floor.
programming tip 3.2
Common error 3.1
cfe2_ch03_p75_130.indd 80 10/28/10 7:51 PM
3.1 the if statement 81
Tabs
Block-structured code has the property that nested statements are indented by one or more
levels:
int main()
{
| int floor;
| …
| if (floor > 13)
| {
| | floor–;
| } |
| …
| return 0;
} | |
0 1 2 Indentation level
How do you move the cursor from the leftmost
column to the appropriate indentation level? A
perfectly reasonable strategy is to hit the space bar
a suf ficient number of times. However, many pro-
grammers use the Tab key instead. A tab moves the
cursor to the next indentation level.
While the Tab key is nice, some editors use tab characters for alignment, which is not so
nice. Tab characters can lead to problems when you send your file to another person or a
printer. There is no universal agreement on the width of a tab character, and some soft ware will
ignore tabs altogether. It is therefore best to save your files with spaces instead of tabs. Most
editors have a setting to auto matically convert all tabs to spaces. Look at the documentation of
your development environment to find out how to activate this useful setting.
The Selection operator
C++ has a selection operator of the form
condition ? value1 : value2
The value of that expression is either value1 if the test passes or value2 if it fails. For example,
we can compute the actual floor number as
actual_floor = floor > 13 ? floor – 1 : floor;
which is equivalent to
if (floor > 13)
{
actual_floor = floor – 1;
}
else
{
actual_floor = floor;
}
You can use the selection operator anywhere that a value is expected, for example:
cout << "Actual floor: " << (floor > 13 ? floor – 1 : floor);
We don’t use the selection operator in this book, but it is a convenient construct that you will
find in many C++ pro grams.
programming tip 3.3
You use
the Tab key
to move the
cursor to the next
indentation level.
special topic 3.1
cfe2_ch03_p75_130.indd 81 10/28/10 7:51 PM
82 Chapter 3 Decisions
avoid duplication in branches
Look to see whether you duplicate code in each branch. If so, move it out of the if statement.
Here is an example of such duplication:
if (floor > 13)
{
actual_floor = floor – 1;
cout << "Actual floor: " << actual_floor << endl;
}
else
{
actual_floor = floor;
cout << "Actual floor: " << actual_floor << endl;
}
The output statement is exactly the same in both branches. This is not an error—the program
will run correctly. However, you can simplify the program by moving the duplicated state-
ment, like this:
if (floor > 13)
{
actual_floor = floor – 1;
}
else
{
actual_floor = floor;
}
cout << "Actual floor: " << actual_floor << endl;
Removing duplication is particularly important when programs are maintained for a long
time. When there are two sets of statements with the same effect, it can easily happen that a
programmer modifies one set but not the other.
3.2 Comparing numbers and strings
Every if statement contains a condi-
tion. In many cases, the condition
involves comparing two values. For
example, in the previous examples we
tested floor > 13. The comparison > is
called a relational operator. C++ has
six relational operators (see Table 1).
As you can see, only two C++ rela-
tional operators (> and <) look as you
would expect from the mathe matical
notation. Computer keyboards do not
have keys for ≥, ≤, or ≠, but the >=, <=,
and != operators are easy to remem-
ber because they look similar. The ==
operator is initially confusing to most
newcomers to C++. In C++, = already
has a meaning, namely assignment.
programming tip 3.4
In C++, you use a relational operator to check
whether one value is greater than another.
relational operators
(< <= > >= == !=)
are used to compare
numbers and strings.
cfe2_ch03_p75_130.indd 82 10/28/10 7:51 PM
3.2 Comparing numbers and strings 83
table 1 relational operators
C++ Math notation Description
> > Greater than
>= ≥ Greater than or equal
< < Less than
<= ≤ Less than or equal
== = Equal
!= ≠ Not equal
The == operator denotes equality testing:
floor = 13; // Assign 13 to floor
if (floor == 13) // Test whether floor equals 13
You must remember to use == inside tests and to use = outside tests. (See Common
Error 3.2 on page 85 for more information.)
You can compare strings as well:
if (input == "Quit") ...
Use != to check whether two strings are different. In C++, letter case matters. For
example, "Quit" and "quit" are not the same string.
syntax 3.2 Comparisons
floor > 13
floor == 13
string input;
if (input == “Y”)
double x; double y; const double EPSILON = 1E-14;
if (fabs(x – y) < EPSILON)
These quantities are compared.
Checks for equality.
Check that you have
the right direction:
> (greater) or < (less)
Use ==, not =.
See page 85.
One of: == != < <= > >= (See page 83.)
Ok to compare strings. (See page 86.)
Checks that these floating-point numbers are very close.
See page 86.
Check the boundary condition:
Do you want to include (>=) or exclude (>)?
cfe2_ch03_p75_130.indd 83 10/28/10 7:51 PM
84 Chapter 3 Decisions
10. How do you test that a string str is not the empty string?
practice it Now you can try these exercises at the end of the chapter: R3.2, R3.5, P3.14.
Confusing = and ==
The rule for the correct usage of = and == is very simple: In tests, always use == and never
use =. If it is so simple, why can’t the compiler be helpful and flag any errors?
Actually, the C++ language allows the use of = inside tests. To understand this, we have to
go back in time. The creators of C, the predecessor to C++, were very frugal. They did not
want to have special values true and false. Instead, they allowed any numeric value inside a
condition, with the convention that 0 denotes false and any non-0 value denotes true. Fur-
thermore, in C and C++ assignments have values. For example, the value of the assignment
expression floor = 13 is 13.
These two features—namely that numbers can be used as truth values and that assignments
are expressions with values—conspire to make a horrible pitfall. The test
if (floor = 13) // ERROR
is legal C++, but it does not test whether floor and 13 are equal. Instead, the code sets floor to
13, and since that value is not zero, the condition of the if statement is always fulfilled.
Fortunately, most compilers issue a warning when they encounter such a statement. You
should take such warn ings seriously. (See Programming Tip 3.5 for more advice about com-
piler warnings.)
Some shell-shocked programmers are so nervous about using = that they use == even when
they want to make an assignment:
floor == floor – 1; // ERROR
This statement tests whether floor equals floor – 1. It doesn’t do anything with the outcome
of the test, but that is not an error. Some compilers will warn that “the code has no effect”, but
others will quietly accept the code.
Compile with Zero warnings
There are two kinds of messages that the compiler gives you: errors and warnings. Error mes-
sages are fatal; the com piler will not translate a program with one or more errors. Warning
messages are advisory; the compiler will translate the program, but there is a good chance that
the program will not do what you expect it to do.
It is a good idea to learn how to activate warnings with your compiler, and to write code
that emits no warnings at all. For example, consider the test
if (floor = 13)
One C++ compiler emits a curious warning message: “Suggest parentheses around assignment
used as truth value”. Sadly, the message is misleading because it was not written for students.
Nevertheless, such a warning gives you another chance to look at the offending statement and
fix it, in this case, by replacing the = with an ==.
In order to make warnings more visible, many compilers require you to take some special
action. This might involve clicking a checkbox in an integrated environment or supplying a
special option on the command line. Ask your instructor or lab assistant how to turn on warn-
ings for your compiler.
Common error 3.2
programming tip 3.5
table 2 relational operator examples
expression Value Comment
3 <= 4 true 3 is less than 4; <= tests for “less than or equal”.
3 =< 4 error The “less than or equal” operator is <=, not =<.
The “less than” symbol comes first.
3 > 4 false > is the opposite of <=.
4 < 4 false The left-hand side must be strictly smaller than the
right-hand side.
4 <= 4 true Both sides are equal; <= tests for “less than or equal”.
3 == 5 - 2 true == tests for equality.
3 != 5 - 1 true != tests for inequality. It is true that 3 is not 5 – 1.
3 = 6 / 2 error Use == to test for equality.
1.0 / 3.0 == 0.333333333 false Although the values are very close to one another,
they are not exactly equal. See Common Error 3.3
on page 86.
"10" > 5 error You cannot compare a string to a number.
Table 2 summarizes how to use relational operators in C++.
6. Which of the following conditions are true, provided a is 3 and b is 4?
a. a + 1 <= b
b. a + 1 >= b
c. a + 1 != b
7. Give the opposite of the condition
floor > 13
8. What is the error in this statement?
if (score_a = score_b)
{
cout << "Tie" << endl;
}
9. Supply a condition in this if statement to test whether the user entered a Y:
string input;
cout << "Enter Y to quit." << endl;
cin >> input;
if (…)
{
cout << "Goodbye." << endl;
return 0;
}
S e l f C h e C k
cfe2_ch03_p75_130.indd 84 10/28/10 7:51 PM
3.2 Comparing numbers and strings 85
10. How do you test that a string str is not the empty string?
practice it Now you can try these exercises at the end of the chapter: R3.2, R3.5, P3.14.
Confusing = and ==
The rule for the correct usage of = and == is very simple: In tests, always use == and never
use =. If it is so simple, why can’t the compiler be helpful and flag any errors?
Actually, the C++ language allows the use of = inside tests. To understand this, we have to
go back in time. The creators of C, the predecessor to C++, were very frugal. They did not
want to have special values true and false. Instead, they allowed any numeric value inside a
condition, with the convention that 0 denotes false and any non-0 value denotes true. Fur-
thermore, in C and C++ assignments have values. For example, the value of the assignment
expression floor = 13 is 13.
These two features—namely that numbers can be used as truth values and that assignments
are expressions with values—conspire to make a horrible pitfall. The test
if (floor = 13) // ERROR
is legal C++, but it does not test whether floor and 13 are equal. Instead, the code sets floor to
13, and since that value is not zero, the condition of the if statement is always fulfilled.
Fortunately, most compilers issue a warning when they encounter such a statement. You
should take such warn ings seriously. (See Programming Tip 3.5 for more advice about com-
piler warnings.)
Some shell-shocked programmers are so nervous about using = that they use == even when
they want to make an assignment:
floor == floor - 1; // ERROR
This statement tests whether floor equals floor - 1. It doesn’t do anything with the outcome
of the test, but that is not an error. Some compilers will warn that “the code has no effect”, but
others will quietly accept the code.
Compile with Zero warnings
There are two kinds of messages that the compiler gives you: errors and warnings. Error mes-
sages are fatal; the com piler will not translate a program with one or more errors. Warning
messages are advisory; the compiler will translate the program, but there is a good chance that
the program will not do what you expect it to do.
It is a good idea to learn how to activate warnings with your compiler, and to write code
that emits no warnings at all. For example, consider the test
if (floor = 13)
One C++ compiler emits a curious warning message: “Suggest parentheses around assignment
used as truth value”. Sadly, the message is misleading because it was not written for students.
Nevertheless, such a warning gives you another chance to look at the offending statement and
fix it, in this case, by replacing the = with an ==.
In order to make warnings more visible, many compilers require you to take some special
action. This might involve clicking a checkbox in an integrated environment or supplying a
special option on the command line. Ask your instructor or lab assistant how to turn on warn-
ings for your compiler.
Common error 3.2
programming tip 3.5
cfe2_ch03_p75_130.indd 85 10/28/10 7:51 PM
86 Chapter 3 Decisions
• The space character comes before all printable characters.
• Numbers come before letters.
• For the ordering of punctuation marks, see Appendix D.
When comparing two strings, you compare the first letters of
each word, then the sec ond letters, and so on, until one of the
strings ends or you find the first letter pair that doesn’t match.
If one of the strings ends, the longer string is considered
the “larger” one. For example, compare "car" with "cart".
The first three letters match, and we reach the end of the first
string. Therefore "car" comes before "cart" in lexicographic
ordering.
When you reach a mismatch, the string containing the
“larger” character is considered “larger”. For example, let’s
compare "cat" with "cart". The first two letters match. Since t
comes after r, the string "cat" comes after "cart" in the lexico-
graphic ordering.
Step 1 Decide upon the branching condition.
In our sample problem, the obvious choice for the
condition is:
original price < 128?
That is just fine, and we will use that condition in our
solution.
But you could equally well come up with a cor-
rect solution if you choose the opposite condition: Is
the original price at least (≥) $128? You might choose
this condition if you put yourself into the position
of a shopper who wants to know when the bigger
discount applies.
Step 2 Give pseudocode for the work that needs to be done
when the condition is true.
In this step, you list the action or actions that are taken in the “positive” branch. The details
depend on your problem. You may want to print a message, compute values, or even exit the
program.
In our example, we need to apply an 8 percent discount:
discounted price = 0.92 x original price
lexicographic order is
used to compare strings.
c a r t
c a r
c a t
Letters
match
r comes
before t
Lexicographic
Ordering
h o W t o 3 . 1 implementing an if Statement
This How To walks you through the process of implementing an if statement. We will illus-
trate the steps with the following example problem:
The university bookstore has a Kilobyte Day sale every October 24, giving an 8 percent
discount on all computer accessory purchases if the price is less than $128, and a 16 percent
discount if the price is at least $128. Write a program that asks the cashier for the original price
and then prints the discounted price.
Sales discounts are often higher for
expensive products. Use the if state
ment to implement such a decision.
exact Comparison of floating-point numbers
Floating-point numbers have only a limited precision, and
calculations can introduce roundoff errors. You must take
these inevitable roundoffs into account when comparing
floating-point numbers. For example, the following code
multiplies the square root of 2 by itself. Ideally, we expect to
get the answer 2:
double r = sqrt(2.0);
if (r * r == 2)
{
cout << "sqrt(2) squared is 2" << endl;
}
else
{
cout << "sqrt(2) squared is not 2 but "
<< setprecision(18) << r * r << endl;
}
This program displays
sqrt(2) squared is not 2 but 2.00000000000000044
It does not make sense in most circumstances to compare floating-point numbers exactly.
Instead, we should test whether they are close enough. That is, the magnitude of their differ-
ence should be less than some threshold. Mathe matically, we would write that x and y are close
enough if
x y− < ε
for a very small number, e. e is the Greek letter epsilon, a letter used to denote a very small
quantity. It is common to set e to 10–14 when comparing double numbers:
const double EPSILON = 1E-14;
double r = sqrt(2.0);
if (fabs(r * r - 2) < EPSILON)
{
cout << "sqrt(2) squared is approximately 2";
}
Include the
lexicographic ordering of Strings
If you compare strings using < <= > >=, they are compared
in “lexicographic” order. This ordering is very similar to
the way in which words are sorted in a dictionary.
For example, consider this code fragment.
string name = “Tom”;
if (name < "Dick") ...
The condition is not fulfilled, because in the diction-
ary Dick comes before Tom. There are a few differences
between the ordering in a dictionary and in C++. In C++:
• All uppercase letters come before the lowercase
letters. For example, "Z" comes before "a".
Common error 3.3
Take limited precision into
account when comparing
floatingpoint numbers.
special topic 3.2
To see which of two terms comes
first in the dictionary, consider the
first letter in which they differ.
cfe2_ch03_p75_130.indd 86 10/28/10 7:51 PM
3.2 Comparing numbers and strings 87
• The space character comes before all printable characters.
• Numbers come before letters.
• For the ordering of punctuation marks, see Appendix D.
When comparing two strings, you compare the first letters of
each word, then the sec ond letters, and so on, until one of the
strings ends or you find the first letter pair that doesn’t match.
If one of the strings ends, the longer string is considered
the “larger” one. For example, compare "car" with "cart".
The first three letters match, and we reach the end of the first
string. Therefore "car" comes before "cart" in lexicographic
ordering.
When you reach a mismatch, the string containing the
“larger” character is considered “larger”. For example, let’s
compare "cat" with "cart". The first two letters match. Since t
comes after r, the string "cat" comes after "cart" in the lexico-
graphic ordering.
Step 1 Decide upon the branching condition.
In our sample problem, the obvious choice for the
condition is:
original price < 128?
That is just fine, and we will use that condition in our
solution.
But you could equally well come up with a cor-
rect solution if you choose the opposite condition: Is
the original price at least (≥) $128? You might choose
this condition if you put yourself into the position
of a shopper who wants to know when the bigger
discount applies.
Step 2 Give pseudocode for the work that needs to be done
when the condition is true.
In this step, you list the action or actions that are taken in the “positive” branch. The details
depend on your problem. You may want to print a message, compute values, or even exit the
program.
In our example, we need to apply an 8 percent discount:
discounted price = 0.92 x original price
lexicographic order is
used to compare strings.
c a r t
c a r
c a t
Letters
match
r comes
before t
Lexicographic
Ordering
h o W t o 3 . 1 implementing an if Statement
This How To walks you through the process of implementing an if statement. We will illus-
trate the steps with the following example problem:
The university bookstore has a Kilobyte Day sale every October 24, giving an 8 percent
discount on all computer accessory purchases if the price is less than $128, and a 16 percent
discount if the price is at least $128. Write a program that asks the cashier for the original price
and then prints the discounted price.
Sales discounts are often higher for
expensive products. Use the if state
ment to implement such a decision.
cfe2_ch03_p75_130.indd 87 10/28/10 7:51 PM
88 Chapter 3 Decisions
and fill it in, as shown in Syntax 3.1 on page 78. Omit the else branch if it is not needed.
In our example, the completed statement is
if (original_price < 128)
{
discount_rate = 0.92;
}
else
{
discount_rate = 0.84;
}
discounted_price = discount_rate * original_price;
W o r k e D e x a M p l e 3 . 1 extracting the middle
This Worked Example shows how to extract the middle character from a
string, or the two middle characters if the length of the string is even. 0 1 2 3 4
c r a t e
Making decisions is
an essential part of
any computer program. nowhere is
this more obvious than in a computer
system that helps sort luggage at an
airport. after scanning the luggage
identification codes, the system sorts
the items and routes them to different
conveyor belts. human operators then
place the items onto trucks. When the
city of Denver built a huge airport to
replace an outdated and congested
facility, the luggage system contractor
went a step further. the new system
was designed to replace the human
operators with robotic carts. Unfortu-
nately, the system plainly did not work.
it was plagued by mechanical prob-
lems, such as luggage falling onto the
tracks and jamming carts. equally frus-
trating were the software glitches.
Carts would uselessly accu mulate at
some locations when they were needed
elsewhere.
the airport had been scheduled
to open in 1993, but without a func-
tioning luggage system, the opening
was delayed for over a year while the
contractor tried to fix the problems.
the contractor never succeeded,
and ultimately a manual system was
installed. the delay cost the city and
airlines close to a billion dollars, and
the contractor, once the leading lug-
gage systems vendor in the United
states, went bankrupt.
Clearly, it is very risky to build a
large system based on a technology
that has never been tried on a smaller
scale. as robots and the software that
controls them get better over time,
they will take on a larger share of lug-
gage handling in the future. But it is
likely that this will happen in an incre-
mental fashion.
The Denver airport originally had a
fully automatic system for moving lug
gage, replacing human operators with
robotic carts. Unfortunately, the sys
tem never worked and was dismantled
before the airport was opened.
Random Fact 3.1 the Denver airport luggage handling system
Step 3 Give pseudocode for the work (if any) that needs to be done when the condition is not true.
What do you want to do in the case that the condition of Step 1 is not fulfilled? Sometimes,
you want to do nothing at all. In that case, use an if statement without an else branch.
In our example, the condition tested whether the price was less than $128. If that condition
is not true, the price is at least $128, so the higher discount of 16 percent applies to the sale:
discounted price = 0.84 x original price
Step 4 Double-check relational operators.
First, be sure that the test goes in the right direction. It is a common error to confuse > and <.
Next, consider whether you should use the < operator or its close cousin, the <= operator.
What should happen if the original price is exactly $128? Reading the problem carefully, we
find that the lower discount applies if the original price is less than $128, and the higher dis-
count applies when it is at least $128. A price of $128 should therefore not fulfill our condition,
and we must use <, not <=.
Step 5 Remove duplication.
Check which actions are common to both branches, and move them outside. (See Program-
ming Tip 3.4 on page 82.)
In our example, we have two statements of the form
discounted price = ___ x original price
They only differ in the discount rate. It is best to just set the rate in the branches, and to do the
computation after wards:
If original price < 128
discount rate = 0.92
Else
discount rate = 0.84
discounted price = discount rate x original price
Step 6 Test both branches.
Formulate two test cases, one that fulfills the condition of the if statement, and one that does
not. Ask yourself what should happen in each case. Then follow the pseudocode and act each
of them out.
In our example, let us consider two scenarios for the original price: $100 and $200. We
expect that the first price is discounted by $8, the second by $32.
When the original price is 100, then the condition 100 < 128 is true, and we get
discount rate = 0.92
discounted price = 0.92 x 100 = 92
When the original price is 200, then the condition 200 < 128 is false, and
discount rate = 0.84
discounted price = 0.84 x 200 = 168
In both cases, we get the expected answer.
Step 7 Assemble the if statement in C++.
Type the skeleton
if ()
{
}
else
{
}
cfe2_ch03_p75_130.indd 88 10/28/10 7:51 PM
3.2 Comparing numbers and strings 89
and fill it in, as shown in Syntax 3.1 on page 78. Omit the else branch if it is not needed.
In our example, the completed statement is
if (original_price < 128)
{
discount_rate = 0.92;
}
else
{
discount_rate = 0.84;
}
discounted_price = discount_rate * original_price;
W o r k e D e x a M p l e 3 . 1 extracting the middle
This Worked Example shows how to extract the middle character from a
string, or the two middle characters if the length of the string is even. 0 1 2 3 4
c r a t e
Making decisions is
an essential part of
any computer program. nowhere is
this more obvious than in a computer
system that helps sort luggage at an
airport. after scanning the luggage
identification codes, the system sorts
the items and routes them to different
conveyor belts. human operators then
place the items onto trucks. When the
city of Denver built a huge airport to
replace an outdated and congested
facility, the luggage system contractor
went a step further. the new system
was designed to replace the human
operators with robotic carts. Unfortu-
nately, the system plainly did not work.
it was plagued by mechanical prob-
lems, such as luggage falling onto the
tracks and jamming carts. equally frus-
trating were the software glitches.
Carts would uselessly accu mulate at
some locations when they were needed
elsewhere.
the airport had been scheduled
to open in 1993, but without a func-
tioning luggage system, the opening
was delayed for over a year while the
contractor tried to fix the problems.
the contractor never succeeded,
and ultimately a manual system was
installed. the delay cost the city and
airlines close to a billion dollars, and
the contractor, once the leading lug-
gage systems vendor in the United
states, went bankrupt.
Clearly, it is very risky to build a
large system based on a technology
that has never been tried on a smaller
scale. as robots and the software that
controls them get better over time,
they will take on a larger share of lug-
gage handling in the future. But it is
likely that this will happen in an incre-
mental fashion.
The Denver airport originally had a
fully automatic system for moving lug
gage, replacing human operators with
robotic carts. Unfortunately, the sys
tem never worked and was dismantled
before the airport was opened.
Random Fact 3.1 the Denver airport luggage handling system
Available online at www.wiley.com/college/horstmann.
cfe2_ch03_p75_130.indd 89 10/28/10 7:51 PM
www.wiley.com/college/horstmann
90 Chapter 3 Decisions
3.3 Multiple alternatives
In Section 3.1, you saw how to program a two-way branch with an if statement. In
many situations, there are more than two cases. In this section, you will see how to
implement a decision with multiple alterna tives. For example, consider a program that
displays the effect of an earthquake, as measured by the Rich ter scale (see Table 3).
The 1989 Loma
Prieta earth quake
that damaged the Bay
Bridge in San Fran
cisco and destroyed
many buildings
measured 7.1 on the
Richter scale.
table 3 richter scale
Value effect
8 Most structures fall
7 Many buildings destroyed
6 Many buildings considerably
damaged, some collapse
4.5 Damage to poorly constructed
buildings
The Richter scale is a measurement of the strength of an earthquake. Every step in the
scale, for example from 6.0 to 7.0, signifies a tenfold increase in the strength of the
quake.
In this case, there are five branches: one each for the four descriptions of damage,
and one for no destruction. Figure 3 shows the flowchart for this multiple-branch
statement.
You use multiple if statements to implement multiple alternatives, like this:
if (richter >= 8.0)
{
cout << "Most structures fall";
}
else if (richter >= 7.0)
{
cout << "Many buildings destroyed";
}
else if (richter >= 6.0)
{
cout << "Many buildings considerably damaged, some collapse";
}
else if (richter >= 4.5)
{
cout << "Damage to poorly constructed buildings";
}
else
{
cout << "No destruction of buildings";
}
As soon as one of the four tests succeeds, the effect is displayed, and no further tests
are attempted. If none of the four cases applies, the final else clause applies, and a
default message is printed. (See the ch03/rich ter.cpp file for the full program.)
Multiple alternatives
are required for
decisions that have
more than two cases.
cfe2_ch03_p75_130.indd 90 10/28/10 7:51 PM
3.3 Multiple alternatives 91
figure 3
Multiple alternatives
richter ≥ 8.0?
richter ≥ 7.0?
richter ≥ 6.0?
richter ≥ 4.5?
No destruction
of buildings
False
False
False
False
True
True
True
True
Most
structures
fall
Many
buildings
destroyed
Many buildings
considerably
damaged,
some collapse
Damage to
poorly constructed
buildings
Here you must sort the conditions and test against the largest cutoff first. Suppose
we reverse the order of tests:
if (richter >= 4.5) // Tests in wrong order
{
cout << "Damage to poorly constructed buildings";
}
else if (richter >= 6.0)
{
cout << "Many buildings considerably damaged, some collapse";
}
else if (richter >= 7.0)
{
cout << "Many buildings destroyed";
}
else if (richter >= 8.0)
{
cfe2_ch03_p75_130.indd 91 10/28/10 7:51 PM
92 Chapter 3 Decisions
The switch Statement
A sequence of if statements that compares a single integer value against several constant alter-
natives can be imple mented as a switch statement. For example,
int digit;
…
switch (digit)
{
case 1: digit_name = “one”; break;
case 2: digit_name = “two”; break;
case 3: digit_name = “three”; break;
case 4: digit_name = “four”; break;
case 5: digit_name = “five”; break;
case 6: digit_name = “six”; break;
case 7: digit_name = “seven”; break;
case 8: digit_name = “eight”; break;
case 9: digit_name = “nine”; break;
default: digit_name = “”; break;
}
This is a shortcut for
int digit;
if (digit == 1) { digit_name = “one”; }
else if (digit == 2) { digit_name = “two”; }
else if (digit == 3) { digit_name = “three”; }
else if (digit == 4) { digit_name = “four”; }
else if (digit == 5) { digit_name = “five”; }
else if (digit == 6) { digit_name = “six”; }
else if (digit == 7) { digit_name = “seven”; }
else if (digit == 8) { digit_name = “eight”; }
else if (digit == 9) { digit_name = “nine”; }
else { digit_name = “”; }
Well, it isn’t much of a shortcut, but it has one advan-
tage—it is obvious that all branches test the same value,
namely digit.
It is possible to have multiple case clauses for a
branch, such as
case 1: case 3: case 5: case 7: case 9:
odd = true; break;
The default branch is chosen if none of the case clauses
match.
Every branch of the switch must be terminated by a
break instruction. If the break is missing, execution falls
through to the next branch, and so on, until finally a
break or the end of the switch is reached. In practice, this
fall-through behavior is rarely useful, but it is a com-
mon cause of errors. If you accidentally forget the break statement, your program compiles but
exe cutes unwanted code. Many programmers consider the switch statement somewhat dan-
gerous and prefer the if statement.
We leave it to you to use the switch statement for your own code or not. At any rate, you
need to have a reading knowledge of switch in case you find it in other programmers’ code.
special topic 3.3
The switch statement lets you
choose from a fixed set of
alternatives.
cout << "Most structures fall";
}
This does not work. Suppose the value of richter is 7.1. That value is at least 4.5,
matching the first case. The other tests will never be attempted.
In this example, it is also important that we use a sequence of else if clauses, not
just multiple indepen dent if statements. Consider this sequence of independent tests:
if (richter >= 8.0) // Didn’t use else
{
cout << "Most structures fall";
}
if (richter >= 7.0)
{
cout << "Many buildings destroyed";
}
if (richter >= 6.0)
{
cout << "Many buildings considerably damaged, some collapse";
}
if (richter >= 4.5)
{
cout << "Damage to poorly constructed buildings";
}
Now the alternatives are no longer exclusive. If richter is 7.1, then the last three tests
all match, and three messages are printed.
11. In a game program, the scores of players A and B are stored in variables score_a
and score_b. Assum ing that the player with the larger score wins, write a
sequence of conditional statements that prints out "A won", "B won", or "Game tied".
12. Write a conditional statement with three branches that sets s to 1 if x is positive,
to –1 if x is negative, and to 0 if x is zero.
13. How could you achieve the task of Self Check 12 with only two branches?
14. Beginners sometimes write statements such as the following:
if (price > 100)
{
discounted_price = price – 20;
}
else if (price <= 100)
{
discounted_price = price - 10;
}
Explain how this code can be improved.
15. Suppose the user enters -1 into the richter.cpp program. What is printed?
16. Suppose we want to have the richter.cpp program check whether the user
entered a negative number. What branch would you add to the if statement, and
where?
practice it Now you can try these exercises at the end of the chapter: R3.20, P3.1, P3.9, P3.10.
When using multiple
if statements, pay
attention to the order
of the conditions.
S e l f C h e C k
cfe2_ch03_p75_130.indd 92 10/28/10 7:51 PM
3.3 Multiple alternatives 93
The switch Statement
A sequence of if statements that compares a single integer value against several constant alter-
natives can be imple mented as a switch statement. For example,
int digit;
...
switch (digit)
{
case 1: digit_name = "one"; break;
case 2: digit_name = "two"; break;
case 3: digit_name = "three"; break;
case 4: digit_name = "four"; break;
case 5: digit_name = "five"; break;
case 6: digit_name = "six"; break;
case 7: digit_name = "seven"; break;
case 8: digit_name = "eight"; break;
case 9: digit_name = "nine"; break;
default: digit_name = ""; break;
}
This is a shortcut for
int digit;
if (digit == 1) { digit_name = "one"; }
else if (digit == 2) { digit_name = "two"; }
else if (digit == 3) { digit_name = "three"; }
else if (digit == 4) { digit_name = "four"; }
else if (digit == 5) { digit_name = "five"; }
else if (digit == 6) { digit_name = "six"; }
else if (digit == 7) { digit_name = "seven"; }
else if (digit == 8) { digit_name = "eight"; }
else if (digit == 9) { digit_name = "nine"; }
else { digit_name = ""; }
Well, it isn’t much of a shortcut, but it has one advan-
tage—it is obvious that all branches test the same value,
namely digit.
It is possible to have multiple case clauses for a
branch, such as
case 1: case 3: case 5: case 7: case 9:
odd = true; break;
The default branch is chosen if none of the case clauses
match.
Every branch of the switch must be terminated by a
break instruction. If the break is missing, execution falls
through to the next branch, and so on, until finally a
break or the end of the switch is reached. In practice, this
fall-through behavior is rarely useful, but it is a com-
mon cause of errors. If you accidentally forget the break statement, your program compiles but
exe cutes unwanted code. Many programmers consider the switch statement somewhat dan-
gerous and prefer the if statement.
We leave it to you to use the switch statement for your own code or not. At any rate, you
need to have a reading knowledge of switch in case you find it in other programmers’ code.
special topic 3.3
The switch statement lets you
choose from a fixed set of
alternatives.
cfe2_ch03_p75_130.indd 93 10/28/10 7:51 PM
94 Chapter 3 Decisions
3.4 nested Branches
It is often necessary to include an if statement inside another. Such an arrangement is
called a nested set of statements. Here is a typical example.
In the United States, different tax rates are used depending on the taxpayer’s mari-
tal status. There are different tax sched ules for single and for married taxpayers. Mar-
ried taxpayers add their income together and pay taxes on the total. Table 4 gives the
tax rate computations, using a simplification of the schedules in effect for the 2008 tax
year. A different tax rate applies to each “bracket”. In this schedule, the income at the
first bracket is taxed at 10 percent, and the income at the sec ond bracket is taxed at 25
percent. The income limits for each bracket depend on the marital status.
Now compute the taxes due, given a filing status and an income figure. The key
point is that there are two levels of decision making. First, you must branch on the
marital status. Then, for each filing status, you must have another branch on income
level.
table 4 Federal tax rate schedule
if your status is single and
if the taxable income is the tax is of the amount over
at most $32,000 10% $0
over $32,000 $3,200 + 25% $32,000
if your status is Married and
if the taxable income is the tax is of the amount over
at most $64,000 10% $0
over $64,000 $6,400 + 25% $64,000
The two-level decision process is reflected in two levels of if statements in the pro-
gram at the end of this section. (See Figure 4 for a flowchart.)
When a decision
statement is
contained inside the
branch of another
decision statement,
the statements are
nested.
nested decisions
are required for
problems that
have two levels of
decision making.
Computing income taxes
requires multiple levels of decisions.
cfe2_ch03_p75_130.indd 94 10/28/10 7:51 PM
3.4 nested Branches 95
figure 4 income tax Computation
10%
bracket
25%
bracket
Single
income
≤ 32,000
10%
bracket
25%
bracket
income
≤ 64,000
False
True
True
False
True
False
In theory, nesting can go deeper than two levels. A three-level decision process
(first by state, then by filing status, then by income level) requires three nesting
levels.
ch03/tax.cpp
1 #include
2 #include
3
4 using namespace std;
5
6 int main()
7 {
8 const double RATE1 = 0.10;
9 const double RATE2 = 0.25;
10 const double RATE1_SINGLE_LIMIT = 32000;
11 const double RATE1_MARRIED_LIMIT = 64000;
12
13 double tax1 = 0;
14 double tax2 = 0;
15
16 double income;
17 cout << "Please enter your income: ";
18 cin >> income;
19
20 cout << "Please enter s for single, m for married: ";
21 string marital_status;
22 cin >> marital_status;
23
cfe2_ch03_p75_130.indd 95 10/28/10 7:51 PM
96 Chapter 3 Decisions
hand-Tracing
A very useful technique for understanding whether a program
works correctly is called hand-tracing. You simulate the pro-
gram’s activity on a sheet of paper. You can use this method with
pseudocode or C++ code.
Get an index card, a cocktail napkin, or whatever sheet of
paper is within reach. Make a column for each variable. Have the
program code ready. Use a marker, such as a paper clip, to mark
the current statement. In your mind, execute statements one at a
time. Every time the value of a variable changes, cross out the old
value and write the new value below the old one.
For example, let’s trace the tax program with the data from the
program run on page 95. In lines 13 and 14, tax1 and tax2 are initial-
ized to 0.
6 int main()
7 {
8 const double RATE1 = 0.10;
9 const double RATE2 = 0.25;
10 const double RATE1_SINGLE_LIMIT = 32000;
11 const double RATE1_MARRIED_LIMIT = 64000;
12
13 double tax1 = 0;
14 double tax2 = 0;
15
In lines 18 and 22, income and marital_status are
initialized by input statements.
16 double income;
17 cout << "Please enter your income: ";
18 cin >> income;
19
20 cout << "Please enter s for single, m for married: ";
21 string marital_status;
22 cin >> marital_status;
23
Because marital_status is not “s”, we move to the
else branch of the outer if statement (line 36).
24 if (marital_status == “s”)
25 {
26 if (income <= RATE1_SINGLE_LIMIT)
27 {
28 tax1 = RATE1 * income;
29 }
30 else
31 {
32 tax1 = RATE1 * RATE1_SINGLE_LIMIT;
33 tax2 = RATE2 * (income - RATE1_SINGLE_LIMIT);
34 }
35 }
36 else
37 {
Since income is not <= 64000, we move to the else branch of the inner if statement (line 42).
38 if (income <= RATE1_MARRIED_LIMIT)
39 {
40 tax1 = RATE1 * income;
41 }
42 else
43 {
44 tax1 = RATE1 * RATE1_MARRIED_LIMIT;
45 tax2 = RATE2 * (income - RATE1_MARRIED_LIMIT);
46 }
programming tip 3.6
Handtracing helps you
understand whether a
program works correctly.
marital
tax1 tax2 income status
0 0
marital
tax1 tax2 income status
0 0 80000 m
24 if (marital_status == "s")
25 {
26 if (income <= RATE1_SINGLE_LIMIT)
27 {
28 tax1 = RATE1 * income;
29 }
30 else
31 {
32 tax1 = RATE1 * RATE1_SINGLE_LIMIT;
33 tax2 = RATE2 * (income - RATE1_SINGLE_LIMIT);
34 }
35 }
36 else
37 {
38 if (income <= RATE1_MARRIED_LIMIT)
39 {
40 tax1 = RATE1 * income;
41 }
42 else
43 {
44 tax1 = RATE1 * RATE1_MARRIED_LIMIT;
45 tax2 = RATE2 * (income - RATE1_MARRIED_LIMIT);
46 }
47 }
48
49 double total_tax = tax1 + tax2;
50
51 cout << "The tax is $" << total_tax << endl;
52 return 0;
53 }
program run
Please enter your income: 80000
Please enter s for single, m for married: m
The tax is $10400
17. What is the amount of tax that a single taxpayer pays on an income of $32,000?
18. Would that amount change if the first nested if statement changed from
if (income <= RATE1_SINGLE_LIMIT)
to
if (income < RATE1_SINGLE_LIMIT)
19. Suppose Harry and Sally each make $40,000 per year. Would they save taxes if
they married?
20. How would you modify the tax.cpp program in order to check that the user
entered a correct value for the marital status (i.e., s or m)?
21. Some people object to higher tax rates for higher incomes, claiming that you
might end up with less money after taxes when you get a raise for working hard.
What is the flaw in this argument?
practice it Now you can try these exercises at the end of the chapter: R3.7, R3.19, P3.13, P3.19.
S e l f C h e C k
cfe2_ch03_p75_130.indd 96 10/28/10 7:51 PM
3.4 nested Branches 97
hand-Tracing
A very useful technique for understanding whether a program
works correctly is called hand-tracing. You simulate the pro-
gram’s activity on a sheet of paper. You can use this method with
pseudocode or C++ code.
Get an index card, a cocktail napkin, or whatever sheet of
paper is within reach. Make a column for each variable. Have the
program code ready. Use a marker, such as a paper clip, to mark
the current statement. In your mind, execute statements one at a
time. Every time the value of a variable changes, cross out the old
value and write the new value below the old one.
For example, let’s trace the tax program with the data from the
program run on page 95. In lines 13 and 14, tax1 and tax2 are initial-
ized to 0.
6 int main()
7 {
8 const double RATE1 = 0.10;
9 const double RATE2 = 0.25;
10 const double RATE1_SINGLE_LIMIT = 32000;
11 const double RATE1_MARRIED_LIMIT = 64000;
12
13 double tax1 = 0;
14 double tax2 = 0;
15
In lines 18 and 22, income and marital_status are
initialized by input statements.
16 double income;
17 cout << "Please enter your income: ";
18 cin >> income;
19
20 cout << "Please enter s for single, m for married: ";
21 string marital_status;
22 cin >> marital_status;
23
Because marital_status is not “s”, we move to the
else branch of the outer if statement (line 36).
24 if (marital_status == “s”)
25 {
26 if (income <= RATE1_SINGLE_LIMIT)
27 {
28 tax1 = RATE1 * income;
29 }
30 else
31 {
32 tax1 = RATE1 * RATE1_SINGLE_LIMIT;
33 tax2 = RATE2 * (income - RATE1_SINGLE_LIMIT);
34 }
35 }
36 else
37 {
Since income is not <= 64000, we move to the else branch of the inner if statement (line 42).
38 if (income <= RATE1_MARRIED_LIMIT)
39 {
40 tax1 = RATE1 * income;
41 }
42 else
43 {
44 tax1 = RATE1 * RATE1_MARRIED_LIMIT;
45 tax2 = RATE2 * (income - RATE1_MARRIED_LIMIT);
46 }
programming tip 3.6
Handtracing helps you
understand whether a
program works correctly.
marital
tax1 tax2 income status
0 0
marital
tax1 tax2 income status
0 0 80000 m
cfe2_ch03_p75_130.indd 97 10/28/10 7:51 PM
98 Chapter 3 Decisions
3.5 problem solving: Flowcharts
You have seen examples of flowcharts earlier in this chapter. A flowchart shows the
structure of decisions and tasks that are required to solve a problem. When you have
to solve a complex problem, it is a good idea to draw a flowchart to visualize the flow
of control.
The basic flowchart elements are shown in Figure 5.
The basic idea is simple enough. Link tasks and input/output boxes in the sequence in
which they should be executed. Whenever you need to make a decision, draw a dia-
mond with two outcomes (see Figure 6).
Each branch can contain a sequence of tasks and even additional decisions. If there
are multiple choices for a value, lay them out as in Figure 7.
Flow charts are made
up of elements for
tasks, input/outputs,
and decisions.
figure 5
Flowchart elements
True
False
ConditionSimple task Input/output
each branch of a
decision can contain
tasks and further
decisions.
figure 6 Flowchart with two outcomes
True
False
False branch True branch
Condition
figure 7 Flowchart with Multiple Choices
True
False
Choice 1
“Choice 1”
branch
True
False
Choice 2
“Choice 2”
branch
True
False
Choice 3
“Choice 3”
branch
“Other”
branch
The values of tax1 and tax2 are updated.
43 {
44 tax1 = RATE1 * RATE1_MARRIED_LIMIT;
45 tax2 = RATE2 * (income - RATE1_MARRIED_LIMIT);
46 }
47 }
Their sum total_tax is computed and printed.
Then the program ends.
48
49 double total_tax = tax1 + tax2;
50
51 cout << "The tax is $" << total_tax << endl;
52 return 0;
53 }
Because the program trace shows the expected
output ($10,400), it suc cessfully demonstrated
that this test case works correctly.
The dangling else problem
When an if statement is nested inside another if statement, the following error may occur.
double shipping_charge = 5.00; // $5 inside continental U.S.
if (country == "USA")
if (state == "HI")
shipping_charge = 10.00; // Hawaii is more expensive
else // Pitfall!
shipping_charge = 20.00; // As are foreign shipments
The indentation level seems to suggest that the else is grouped with the test country == "USA".
Unfortunately, that is not the case. The compiler ignores all indentation and matches the else
with the preceding if. That is, the code is actually
double shipping_charge = 5.00; // $5 inside continental U.S.
if (country == "USA")
if (state == "HI")
shipping_charge = 10.00; // Hawaii is more expensive
else // Pitfall!
shipping_charge = 20.00; // As are foreign shipments
That isn’t what you want. You want to group the else with the first if.
The ambiguous else is called a dangling else. You can avoid this pitfall if you always use
braces, as recommended in Programming Tip 3.2 on page 80:
double shipping_charge = 5.00; // $5 inside continental U.S.
if (country == "USA")
{
if (state == "HI")
{
shipping_charge = 10.00; // Hawaii is more expensive
}
}
else
{
shipping_charge = 20.00; // As are foreign shipments
}
marital
tax1 tax2 income status
0 0 80000 m
6400 4000
marital total
tax1 tax2 income status tax
0 0 80000 m
6400 4000 10400
Common error 3.4
cfe2_ch03_p75_130.indd 98 10/28/10 7:51 PM
3.5 problem solving: Flowcharts 99
3.5 problem solving: Flowcharts
You have seen examples of flowcharts earlier in this chapter. A flowchart shows the
structure of decisions and tasks that are required to solve a problem. When you have
to solve a complex problem, it is a good idea to draw a flowchart to visualize the flow
of control.
The basic flowchart elements are shown in Figure 5.
The basic idea is simple enough. Link tasks and input/output boxes in the sequence in
which they should be executed. Whenever you need to make a decision, draw a dia-
mond with two outcomes (see Figure 6).
Each branch can contain a sequence of tasks and even additional decisions. If there
are multiple choices for a value, lay them out as in Figure 7.
Flow charts are made
up of elements for
tasks, input/outputs,
and decisions.
figure 5
Flowchart elements
True
False
ConditionSimple task Input/output
each branch of a
decision can contain
tasks and further
decisions.
figure 6 Flowchart with two outcomes
True
False
False branch True branch
Condition
figure 7 Flowchart with Multiple Choices
True
False
Choice 1
“Choice 1”
branch
True
False
Choice 2
“Choice 2”
branch
True
False
Choice 3
“Choice 3”
branch
“Other”
branch
cfe2_ch03_p75_130.indd 99 10/28/10 7:51 PM
100 Chapter 3 Decisions
There is one issue that you need to be aware of
when drawing flow charts. Unconstrained branching
and merging can lead to “spaghetti code”, a messy net-
work of possible pathways through a program.
There is a simple rule for avoiding spaghetti code:
Never point an arrow inside another branch.
To understand the rule, consider this example:
Shipping costs are $5 inside the United States, except
that to Hawaii and Alaska they are $10. Inter national
shipping costs are also $10. You might start out with a
flowchart like the following:
False
True
Shipping
cost = $10
Inside US?
True
False
Continental US?
Shipping
cost = $5
Now you may be tempted to reuse the “shipping cost = $10” task:
False
True
Shipping
cost = $10
Inside US?
True
False
Continental US?
Shipping
cost = $5
Spaghetti code has so many
pathways that it becomes
impossible to understand.
never point an
arrow inside
another branch.
cfe2_ch03_p75_130.indd 100 10/28/10 7:51 PM
3.5 problem solving: Flowcharts 101
Don’t do that! The red arrow points inside a different branch. Instead, add another
task that sets the ship ping cost to $10, like this:
False
True
Shipping
cost = $10
Inside US?
True
False
Continental US?
Shipping
cost = $10
Shipping
cost = $5
Not only do you avoid spaghetti code, but it is also a better design. In the future it
may well happen that the cost for international shipments is different from that to
Alaska and Hawaii.
Flowcharts can be very useful for getting an intuitive understanding of the flow of
an algorithm. However, they get large rather quickly when you add more details. At
that point, it makes sense to switch from flowcharts to pseudocode.
22. Draw a flowchart for a program that reads a value temp and prints “Frozen” if it
is less than zero.
23. What is wrong with the flowchart at right?
24. How do you fix the flowchart of
Self Check 23?
25. Draw a flowchart for a program that reads a
value x. If it is less than zero, print “Error”.
Otherwise, print its square root.
26. Draw a flowchart for a program that reads
a value temp. If it is less than zero, print
“Ice”. If it is greater than 100, print “Steam”.
Otherwise, print “Liquid”.
practice it Now you can try these exercises at the end of the
chapter: R3.10, R3.11, R3.12.
S e l f C h e C k
True
False
Input < 0?
True
False
Input > 100?
Status = “OK” Status = “Error”
cfe2_ch03_p75_130.indd 101 10/28/10 7:51 PM
102 Chapter 3 Decisions
make a Schedule and make Time for unexpected problems
Commercial software is notorious for being delivered later than promised. For example,
Microsoft originally prom ised that its Windows Vista operating system would be available late
in 2003, then in 2005, then in March 2006; it finally was released in January 2007. Some of the
early promises might not have been realistic. It was in Microsoft’s interest to let prospective
customers expect the imminent availability of the product. Had customers known the actual
delivery date, they might have switched to a different product in the meantime. Undeniably,
though, Microsoft had not anticipated the full complexity of the tasks it had set itself to solve.
Microsoft can delay the delivery of its product, but it is likely that you cannot. As a student
or a programmer, you are expected to manage your time wisely and to finish your assignments
on time. You can probably do simple pro gramming exercises the night before the due date, but
an assignment that looks twice as hard may well take four times as long, because more things can
go wrong. You should therefore make a schedule whenever you start a pro gramming project.
First, estimate realistically how much time it
will take you to:
• Design the program logic.
• Develop test cases.
• Type in the program and fix syntax errors.
• Test and debug the program.
For example, for the income tax program I might
estimate an hour for the design; 30 minutes for
developing test cases; an hour for data entry and
fixing syntax errors; and an hour for testing and
debugging. That is a total of 3.5 hours. If I work
two hours a day on this project, it will take me
almost two days.
Then think of things that can go wrong. Your
computer might break down. You might be
stumped by a problem with the computer sys-
tem. (That is a particularly important concern for
beginners. It is very common to lose a day over a trivial problem just because it takes time to
track down a person who knows the magic command to over come it.) As a rule of thumb,
double the time of your estimate. That is, you should start four days, not two days, before the
due date. If nothing went wrong, great; you have the program done two days early. When the
inevitable problem occurs, you have a cushion of time that protects you from embar rassment
and failure.
3.7 Boolean Variables and operators
Sometimes, you need to evaluate a logical condition in one part of a program and use
it elsewhere. To store a condition that can be true or false, you use a Boolean variable.
Boolean variables are named after the mathematician George Boole (1815–1864), a
pioneer in the study of logic.
In C++, the bool data type represents the Boolean type. Variables of type bool can
hold exactly two val ues, denoted false and true. These values are not strings or inte-
gers; they are special values, just for Bool ean variables.
programming tip 3.7
Make a schedule for your programming
work and build in time for problems.
the Boolean type
bool has two values,
false and true.
3.6 problem solving: test Cases
Consider how to test the tax computation program from Section 3.4. Of course, you
cannot try out all possible inputs of filing status and income level. Even if you could,
there would be no point in trying them all. If the program correctly computes one or
two tax amounts in a given bracket, then we have a good reason to believe that all
amounts will be correct.
You want to aim for complete coverage of all decision points. Here is a plan for
obtaining a compre hensive set of test cases:
• There are two possibilities for the filing status and two tax brackets for each
status, yielding four test cases.
• Test a handful of boundary conditions, such as an income that is at the boundary
between two brack ets, and a zero income.
• If you are responsible for error checking (which is discussed in Section 3.8), also
test an invalid input, such as a negative income.
Make a list of the test cases and the expected outputs:
Test Case Expected Output Comment
30,000 s 3,000 10% bracket
72,000 s 13,200 3,200 + 25% of 40,000
50,000 m 5,000 10% bracket
104,000 m 16,400 6,400 + 25% of 40,000
32,000 s 3,200 boundary case
0 0 boundary case
When you develop a set of test cases, it is helpful to have a flowchart of your program
(see Section 3.5). Check off each branch that has a test case. Include test cases for the
boundary cases of each decision. For example, if a decision checks whether an input is
less than 100, test with an input of 100.
It is always a good idea to design test cases before starting to code. Working
through the test cases gives you a better understanding of the algorithm that you are
about to implement.
27. Using Figure 1 on page 77 as a guide, follow the process described in Section 3.6 to
design a set of test cases for the elevator.cpp program in Section 3.1.
28. What is a boundary test case for the algorithm in How To 3.1 on page 87? What is
the expected out put?
29. Using Figure 3 on page 91 as a guide, follow the process described in Section 3.6 to
design a set of test cases for the richter.cpp program in Section 3.3.
30. Suppose you are designing a part of a program for a medical
robot that has a sensor returning an x and y location (measured in
cm). You need to check whether the sensor location is inside the
circle, outside the circle, on the boundary (specifically, having a
distance of less than 1 mm from the boundary). Assume the circle
has center (0, 0) and radius 2 cm. Give a set of test cases.
practice it Now you can try these exercises at the end of the chapter: R3.13, R3.14.
each branch of your
program should
be covered by a
test case.
it is a good idea to
design test cases
before implementing
a program.
S e l f C h e C k
cfe2_ch03_p75_130.indd 102 10/28/10 7:51 PM
3.7 Boolean Variables and operators 103
make a Schedule and make Time for unexpected problems
Commercial software is notorious for being delivered later than promised. For example,
Microsoft originally prom ised that its Windows Vista operating system would be available late
in 2003, then in 2005, then in March 2006; it finally was released in January 2007. Some of the
early promises might not have been realistic. It was in Microsoft’s interest to let prospective
customers expect the imminent availability of the product. Had customers known the actual
delivery date, they might have switched to a different product in the meantime. Undeniably,
though, Microsoft had not anticipated the full complexity of the tasks it had set itself to solve.
Microsoft can delay the delivery of its product, but it is likely that you cannot. As a student
or a programmer, you are expected to manage your time wisely and to finish your assignments
on time. You can probably do simple pro gramming exercises the night before the due date, but
an assignment that looks twice as hard may well take four times as long, because more things can
go wrong. You should therefore make a schedule whenever you start a pro gramming project.
First, estimate realistically how much time it
will take you to:
• Design the program logic.
• Develop test cases.
• Type in the program and fix syntax errors.
• Test and debug the program.
For example, for the income tax program I might
estimate an hour for the design; 30 minutes for
developing test cases; an hour for data entry and
fixing syntax errors; and an hour for testing and
debugging. That is a total of 3.5 hours. If I work
two hours a day on this project, it will take me
almost two days.
Then think of things that can go wrong. Your
computer might break down. You might be
stumped by a problem with the computer sys-
tem. (That is a particularly important concern for
beginners. It is very common to lose a day over a trivial problem just because it takes time to
track down a person who knows the magic command to over come it.) As a rule of thumb,
double the time of your estimate. That is, you should start four days, not two days, before the
due date. If nothing went wrong, great; you have the program done two days early. When the
inevitable problem occurs, you have a cushion of time that protects you from embar rassment
and failure.
3.7 Boolean Variables and operators
Sometimes, you need to evaluate a logical condition in one part of a program and use
it elsewhere. To store a condition that can be true or false, you use a Boolean variable.
Boolean variables are named after the mathematician George Boole (1815–1864), a
pioneer in the study of logic.
In C++, the bool data type represents the Boolean type. Variables of type bool can
hold exactly two val ues, denoted false and true. These values are not strings or inte-
gers; they are special values, just for Bool ean variables.
programming tip 3.7
Make a schedule for your programming
work and build in time for problems.
the Boolean type
bool has two values,
false and true.
cfe2_ch03_p75_130.indd 103 10/28/10 7:51 PM
104 Chapter 3 Decisions
Conversely, let’s test whether water is not liquid at a given temperature. That is
the case when the tem perature is at most 0 or at least 100. Use the || (or) operator to
combine the expressions:
if (temp <= 0 || temp >= 100) { cout << "Not liquid"; }
Figure 9 shows flowcharts for these examples.
Sometimes you need to invert a condition with the not logical operator. The !
operator takes a single condition and evaluates to true if that condition is false and to
false if the condition is true. In this exam ple, output occurs if the value of the Boolean
variable frozen is false:
if (!frozen)
{
cout << "Not frozen";
}
to invert a condition,
use the ! (not)
operator.
figure 9 Flowcharts for and and or Combinations
True True True
True
False
False
False False
Temperature
> 0?
Temperature
< 100?
Water is
liquid
Water is
not liquid
Temperature
≤ 0?
Temperature
≥ 100?
Both conditions
must be true
At least
one condition
must be true
and or
Here is a definition of a Boolean variable:
bool failed = true;
You can use the value later in your program to make a decision:
if (failed) // Only executed if failed has been set to true
{
...
}
When you make complex decisions, you often need to combine Boolean values. An
operator that com bines Boolean conditions is called a Boolean operator. In C++, the
&& operator (called and) yields true only when both conditions are true. The || opera-
tor (called or) yields the result true if at least one of the conditions is true.
Suppose you write a program that processes temperature values, and you want
to test whether a given temperature corresponds to liquid water. (At sea level, water
freezes at 0 degrees Celsius and boils at 100 degrees.) Water is liquid if the tempera-
ture is greater than zero and less than 100:
if (temp > 0 && temp < 100) { cout << "Liquid"; }
The condition of the test has two parts, joined by the && operator. (As shown in Table
5 and Appendix C, the > and < operators have higher precedence than the && operator.)
Each part is a Boolean value that can be true or false. The combined expression is
true if both individ ual expressions are true. If either one of the expressions is false,
then the result is also false (see Figure 8).
A Boolean variable
is also called a flag
because it can be
either up (true) or
down (false).
C++ has two Boolean
operators that
combine conditions:
&& (and) and || (or).
a B a && B
true true true
true false false
false true false
false false false
a B a || B
true true true
true false true
false true true
false false false
a !a
true false
false true
figure 8 Boolean truth tables
At this geyser in Iceland,
you can see ice, liquid
water, and steam.
cfe2_ch03_p75_130.indd 104 10/28/10 7:51 PM
3.7 Boolean Variables and operators 105
table 5 selected operators and their precedence
operator Description
++ -- + (unary)
– (unary) !
Increment, decrement, positive, negative,
Boolean not
* / % Multiplication, division, remainder
+ - Addition, subtraction
< <= > >= Comparisons
== != Equal, not equal
&& Boolean and
|| Boolean or
Conversely, let’s test whether water is not liquid at a given temperature. That is
the case when the tem perature is at most 0 or at least 100. Use the || (or) operator to
combine the expressions:
if (temp <= 0 || temp >= 100) { cout << "Not liquid"; }
Figure 9 shows flowcharts for these examples.
Sometimes you need to invert a condition with the not logical operator. The !
operator takes a single condition and evaluates to true if that condition is false and to
false if the condition is true. In this exam ple, output occurs if the value of the Boolean
variable frozen is false:
if (!frozen)
{
cout << "Not frozen";
}
to invert a condition,
use the ! (not)
operator.
figure 9 Flowcharts for and and or Combinations
True True True
True
False
False
False False
Temperature
> 0?
Temperature
< 100?
Water is
liquid
Water is
not liquid
Temperature
≤ 0?
Temperature
≥ 100?
Both conditions
must be true
At least
one condition
must be true
and or
cfe2_ch03_p75_130.indd 105 10/28/10 7:51 PM
106 Chapter 3 Decisions
Combining multiple relational operators
Consider the expression
if (0 <= temp <= 100) // Error
This looks just like the mathematical test 0 ≤ temp ≤ 100. Unfortunately, it is not.
Let us dissect the expression 0 <= temp <= 100. The first half, 0 <= temp, is a test with out-
come true or false, depending on the value of temp. The outcome of that test (true or false) is
then compared against 100. Can one com pare truth values and floating-point numbers? Is true
larger than 100 or not? Unfortunately, to stay compatible with the C language, C++ converts
false to 0 and true to 1. Therefore, the expression will always evaluate to true.
You must be careful not to mix logical and arithmetic expressions in your programs.
Instead, use and to combine two separate tests:
if (0 <= temp && temp <= 100) ...
Another common error, along the same lines, is to write
if (x && y > 0) … // Error
instead of
if (x > 0 && y > 0) …
Unfortunately, the compiler will not issue an error message. Instead, it converts x to true or
false. Zero is converted to false, and any nonzero value is converted to true. If x is not zero,
then it tests whether y is greater than 0, and finally it computes the and of these two truth val-
ues. Naturally, that computation makes no sense.
Confusing && and || Conditions
It is a surprisingly common error to confuse and and or conditions. A value lies between 0 and
100 if it is at least 0 and at most 100. It lies outside that range if it is less than 0 or greater than
100. There is no golden rule; you just have to think carefully.
Often the and or or is clearly stated, and then it isn’t too hard to implement it. But some-
times the wording isn’t as explicit. It is quite common that the individual conditions are nicely
set apart in a bulleted list, but with little indica tion of how they should be combined. Consider
these instructions for filing a tax return. You can claim single filing status if any one of the fol-
lowing is true:
• You were never married.
• You were legally separated or divorced on the last day of the tax year.
• You were widowed, and did not remarry.
Since the test passes if any one of the conditions is true, you must combine the conditions with
or. Elsewhere, the same instructions state that you may use the more advantageous status of
married filing jointly if all five of the fol lowing conditions are true:
• Your spouse died less than two years ago and you did not remarry.
• You have a child whom you can claim as dependent.
• That child lived in your home for all of the tax year.
• You paid over half the cost of keeping up your home for this child.
• You filed a joint return with your spouse the year he or she died.
Because all of the conditions must be true for the test to pass, you must combine them with
an and.
Common error 3.5
Common error 3.6
table 6 Boolean operators
expression Value Comment
0 < 200 && 200 < 100 false Only the first condition is true. Note that
the < operator has a higher precedence than
the && operator.
0 < 200 || 200 < 100 true The first condition is true.
0 < 200 || 100 < 200 true The || is not a test for “either-or”.
If both conditions are true, the result is true.
0 < 200 < 100 true error: The expression 0 < 200 is true, which is
converted to 1. The expression 1 < 100 is true.
You never want to write such an expression;
see Common Error 3.5 on page 107.
-10 && 10 > 0 true error: –10 is not zero. It is converted to true.
You never want to write such an expression;
see Common Error 3.5 on page 107.
0 < x && x < 100 || x == -1 (0 < x && x < 100)
|| x == -1
The && operator has a higher precedence than
the || operator.
!(0 < 200) false 0 < 200 is true, therefore its negation is false.
frozen == true frozen There is no need to compare a Boolean
variable with true.
frozen == false !frozen It is clearer to use ! than to compare with
false.
Table 6 illustrates additional examples of evaluating Boolean operators.
31. Suppose x and y are two integers. How do you test whether both of them are
zero?
32. How do you test whether at least one of them is zero?
33. How do you test whether exactly one of them is zero?
34. What is the value of !!frozen?
35. What is the advantage of using the type bool rather than strings "false"/"true" or
integers 0/1?
practice it Now you can try these exercises at the end of the chapter: R3.27, P3.22, P3.24.
S e l f C h e C k
cfe2_ch03_p75_130.indd 106 10/28/10 7:51 PM
3.7 Boolean Variables and operators 107
Combining multiple relational operators
Consider the expression
if (0 <= temp <= 100) // Error
This looks just like the mathematical test 0 ≤ temp ≤ 100. Unfortunately, it is not.
Let us dissect the expression 0 <= temp <= 100. The first half, 0 <= temp, is a test with out-
come true or false, depending on the value of temp. The outcome of that test (true or false) is
then compared against 100. Can one com pare truth values and floating-point numbers? Is true
larger than 100 or not? Unfortunately, to stay compatible with the C language, C++ converts
false to 0 and true to 1. Therefore, the expression will always evaluate to true.
You must be careful not to mix logical and arithmetic expressions in your programs.
Instead, use and to combine two separate tests:
if (0 <= temp && temp <= 100) ...
Another common error, along the same lines, is to write
if (x && y > 0) … // Error
instead of
if (x > 0 && y > 0) …
Unfortunately, the compiler will not issue an error message. Instead, it converts x to true or
false. Zero is converted to false, and any nonzero value is converted to true. If x is not zero,
then it tests whether y is greater than 0, and finally it computes the and of these two truth val-
ues. Naturally, that computation makes no sense.
Confusing && and || Conditions
It is a surprisingly common error to confuse and and or conditions. A value lies between 0 and
100 if it is at least 0 and at most 100. It lies outside that range if it is less than 0 or greater than
100. There is no golden rule; you just have to think carefully.
Often the and or or is clearly stated, and then it isn’t too hard to implement it. But some-
times the wording isn’t as explicit. It is quite common that the individual conditions are nicely
set apart in a bulleted list, but with little indica tion of how they should be combined. Consider
these instructions for filing a tax return. You can claim single filing status if any one of the fol-
lowing is true:
• You were never married.
• You were legally separated or divorced on the last day of the tax year.
• You were widowed, and did not remarry.
Since the test passes if any one of the conditions is true, you must combine the conditions with
or. Elsewhere, the same instructions state that you may use the more advantageous status of
married filing jointly if all five of the fol lowing conditions are true:
• Your spouse died less than two years ago and you did not remarry.
• You have a child whom you can claim as dependent.
• That child lived in your home for all of the tax year.
• You paid over half the cost of keeping up your home for this child.
• You filed a joint return with your spouse the year he or she died.
Because all of the conditions must be true for the test to pass, you must combine them with
an and.
Common error 3.5
Common error 3.6
cfe2_ch03_p75_130.indd 107 10/28/10 7:51 PM
108 Chapter 3 Decisions
Short-Circuit evaluation of boolean operators
When the && and || operators are computed, evaluation stops as soon
as the truth value is determined. When an && is evaluated and the first
condition is false, the second condition is not evaluated, because it
does not matter what the outcome of the second test is.
For example, consider the expression
quantity > 0 && price / quantity < 10
Suppose the value of quantity is zero. Then the test quantity > 0 fails,
and the second test is not attempted. That is just as well, because it is
illegal to divide by zero.
Similarly, when the first condition of an || expression is true, then the remainder is not
evaluated since the result must be true.
This process is called short-circuit evaluation.
In a short circuit, electricity travels along the path of least resistance.
Similarly, shortcircuit evaluation takes the fastest path for
computing the result of a Boolean expression.
de morgan’s law
Humans generally have a hard time comprehending logical conditions with not operators
applied to and/or expres sions. De Morgan’s Law, named after the logician Augustus De Mor-
gan (1806–1871), can be used to simplify these Boolean expressions.
Suppose we want to charge a higher shipping rate if we don’t ship within the continental
United States.
if (!(country == “USA”
&& state != “AK”
&& state != “HI”))
shipping_charge = 20.00;
This test is a little bit complicated, and you have to think carefully through the logic. When it
is not true that the country is USA and the state is not Alaska and the state is not Hawaii, then
charge $20.00. Huh? It is not true that some people won’t be confused by this code.
The computer doesn’t care, but it takes human programmers to write and maintain the
code. Therefore, it is use ful to know how to simplify such a condition.
De Morgan’s Law has two forms: one for the negation of an and expression and one for the
negation of an or expression:
!(A && B) is the same as !A || !B
!(A || B) is the same as !A && !B
Pay particular attention to the fact that the and and or operators are reversed by moving the
not inward. For exam ple, the negation of “the state is Alaska or it is Hawaii”,
!(state == “AK” || state == “HI”)
is “the state is not Alaska and it is not Hawaii”:
!(state == “AK”) && !(state == “HI”)
special topic 3.4
the && and ||
operators are
computed using
shortcircuit
evaluation: as soon
as the truth value is
determined, no
further conditions
are evaluated.
special topic 3.5
De Morgan’s law tells
you how to negate
&& and || conditions.
cfe2_ch03_p75_130.indd 108 10/28/10 7:52 PM
3.8 application: input Validation 109
That is, of course, the same as
state != “AK” && state != “HI”
Now apply the law to our shipping charge computation:
!(country == “USA”
&& state != “AK”
&& state != “HI”)
is equivalent to
!(country == “USA”)
|| !(state != “AK”)
|| !(state != “HI”)
That yields the simpler test
country != “USA”
|| state == “AK”
|| state == “HI”
To simplify conditions with negations of and or or expressions, it is usually a good idea to
apply De Morgan’s Law to move the negations to the innermost level.
3.8 application: input Validation
An important application for the if statement is input validation.
Whenever your program accepts user input, you need to make sure
that the user-supplied values are valid before you use them in your
computations.
Consider our elevator program. Assume that the elevator panel
has buttons labeled 1 through 20 (but not 13). The following are
illegal inputs:
• The number 13
• Zero or a negative number
• A number larger than 20
• An input that is not a sequence of digits, such as five
In each of these cases, we will want to give an error message and exit
the program.
It is simple to guard against an input of 13:
if (floor == 13)
{
cout << "Error: There is no thirteenth floor." << endl;
return 1;
}
The statement
return 1;
immediately exits the main function and therefore terminates the program. It is a con-
vention to return with the value 0 if the program completed normally, and with a
non-zero value when an error was encountered.
Like a quality control worker, you want
to make sure that user input is correct
before processing it.
cfe2_ch03_p75_130.indd 109 10/28/10 7:52 PM
110 Chapter 3 Decisions
Here is how you ensure that the user doesn’t enter a number outside the valid
range:
if (floor <= 0 || floor > 20)
{
cout << "Error: The floor must be between 1 and 20." << endl;
return 1;
}
However, dealing with an input that is not a valid integer is a more serious problem.
When the statement
cin >> floor;
is executed, and the user types in an input that is not an integer (such as five), then the
integer variable floor is not set. Instead, the input stream cin is set to a failed state. You
call the fail member function to test for that failed state.
if (cin.fail())
{
cout << "Error: Not an integer." << endl;
return 1;
}
The order of the if statements is important. You must first test for cin.fail(). After
all, if the input failed, no value has been assigned to floor, and it makes no sense to
compare it against other values.
Input failure is quite serious in C++. Once input has failed, all subsequent attempts
at input will fail as well. You will learn in Chapter 8 how to write programs that are
more tolerant of bad input. For now, our goal is simply to detect bad input and to exit
the program when it occurs.
Here is the complete elevator program with input validation.
ch03/elevator2.cpp
1 #include
2
3 using namespace std;
4
5 int main()
6 {
7 int floor;
8 cout << "Floor: ";
9 cin >> floor;
10
11 // The following statements check various input errors
12 if (cin.fail())
13 {
14 cout << "Error: Not an integer." << endl;
15 return 1;
16 }
17 if (floor == 13)
18 {
19 cout << "Error: There is no thirteenth floor." << endl;
20 return 1;
21 }
22 if (floor <= 0 || floor > 20)
23 {
24 cout << "Error: The floor must be between 1 and 20." << endl;
25 return 1;
When reading a
value, check that
it is within the
required range.
Use the fail function
to test whether
the input stream
has failed.
cfe2_ch03_p75_130.indd 110 10/28/10 7:52 PM
3.8 application: input Validation 111
26 }
27
28 // Now we know that the input is valid
29 int actual_floor;
30 if (floor > 13)
31 {
32 actual_floor = floor – 1;
33 }
34 else
35 {
36 actual_floor = floor;
37 }
38
39 cout << "The elevator will travel to the actual floor "
40 << actual_floor << endl;
41
42 return 0;
43 }
program run
Floor: 13
Error: There is no thirteenth floor.
36. Consider the elevator2.cpp program. What output do you get when the input is
a. 100?
b. –1?
c. 20?
d. thirteen?
37. Your task is to rewrite the elevator2.cpp program so that there is a single if state-
ment with a complex condition:
if (...)
{
cout << "Error: Bad input" << endl;
return 1;
}
What is the condition?
38. In the Sherlock Holmes story “The Adventure of the Sussex Vampire”, the
inimitable detective uttered these words: “Matilda Briggs was not the name of
a young woman, Watson, … It was a ship which is associated with the giant rat
of Sumatra, a story for which the world is not yet prepared.” Over a hundred
years later, researchers found giant rats in Western New Guinea, another part of
Indonesia.
Suppose you are charged with writing a program that processes rat weights. It
contains the state ments
double weight;
cout << "Enter weight in kg: ";
cin >> weight;
What input checks should you supply?
When processing inputs, you want to reject values that are too large. But how large is too large?
These giant rats, found in Western New Guinea, are about five times the size of a city rat.
S e l f C h e C k
cfe2_ch03_p75_130.indd 111 10/28/10 7:52 PM
112 Chapter 3 Decisions
use the if statement to implement a decision.
• The if statement allows a program to carry out
dif ferent actions depending on the nature of the
data to be processed.
implement comparisons of numbers and objects.
• Relational operators (< <= > >= == !=) are used to compare numbers and strings.
• Lexicographic order is used to compare strings.
implement complex decisions that require multiple if statements.
• Multiple alternatives are required for decisions that have more than two cases.
• When using multiple if statements, pay attention to the order of the conditions.
C h a p t e r s U M M a r Y
are necessary to actually understand
the sentences. For exam ple, consider
the sentence “last fall she enrolled in
Michigan state”. the reader automati-
cally realizes that “fall” is not related
to falling down in this context, but
refers to the season. While there is a
state of Michigan, here Michigan state
denotes the university. a priori, a com-
puter program has none of this knowl-
edge. the goal of the CYC project is to
extract and store the requisite facts—
that is, (1) people enroll in universities;
(2) Michigan is a state; (3) many states
have universi ties named x state Uni-
versity, often abbreviated as x state;
(4) most peo ple enroll in a university in
the fall. By 1995, the project had codi-
fied about 100,000 common-sense
concepts and about a million facts of
knowledge relating them. even this
massive amount of data has not proven
suffi cient for useful applications.
in recent years, artificial intelli-
gence technology has seen substantial
advances. one of the most astound-
ing examples is the out-
come of a series of “grand
challenges” for autono-
mous vehicles posed by
the Defense advanced
research projects agency
(Darpa). Competitors were
invited to submit a com-
puter-controlled vehicle that
had to complete an obstacle
course without a human
driver or remote control.
the first event, in 2004,
was a disap pointment, with
none of the entrants finish-
ing the route. in 2005, five
vehicles completed a gru-
eling 212 km course in the
Mojave desert. stan ford’s stanley came
in first, with an average speed of 30
km/h. in 2007, Darpa moved the com-
petition to an “urban” environment, an
abandoned air force base. Vehicles had
to be able to interact with each other,
following California traffic laws. as
stanford’s sebastian thrun explained:
“in the last Grand Challenge, it didn’t
really mat ter whether an obstacle was
a rock or a bush, because either way
you’d just drive around it. the current
challenge is to move from just sensing
the envi ronment to understanding the
envi ronment.”
Winner of the 2007 DARPA Urban Challenge
39. Consider the following test program:
int main()
{
int m = 1;
cout << "Enter an integer: ";
cin >> m;
int n = 2;
cout << "Enter another integer: ";
cin >> n;
cout << m << " " << n << endl;
return 0;
}
Run this program and enter three at the first prompt. What happens? Why?
practice it Now you can try these exercises at the end of the chapter: R3.1, R3.30, P3.26.
When one uses a
sophisticated com-
puter program such as a tax prepara-
tion package, one is bound to attribute
some intelligence to the computer.
the computer asks sensible questions
and makes computations that we find
a mental challenge. after all, if doing
one’s taxes were easy, we wouldn’t
need a computer to do it for us.
as programmers, however, we
know that all this apparent intelli gence
is an illusion. human program mers
have carefully “coached” the software
in all possible scenarios, and it simply
replays the actions and deci sions that
were programmed into it.
Would it be possible to write com-
puter programs that are genuinely
intelligent in some sense? From the
earliest days of computing, there was
a sense that the human brain might
be nothing but an immense computer,
and that it might well be feasible to
program computers to imitate some
processes of human thought. seri-
ous research into artificial intelligence
began in the mid-1950s, and the first
twenty years brought some impres-
sive successes. programs that play
chess—surely an activity that appears
to require remarkable intellectual pow-
ers—have become so good that they
now routinely beat all but the best
human players. as far back as 1975,
an expertsystem program called Mycin
gained fame for being better in diag-
nosing meningitis in patients than the
average physician.
however, there were serious set-
backs as well. From 1982 to 1992,
the Japanese government embarked
on a massive research project, funded
at over 40 billion Japanese yen. it was
known as the FifthGeneration Project.
its goal was to develop new hardware
and software to greatly improve the
performance of expert system soft-
ware. at its outset, the project created
fear in other countries that the Japa-
nese computer industry was about to
become the undisputed leader in the
field. however, the end results were
disappointing and did little to bring
artificial intelligence applications to
market.
From the very outset, one of the
stated goals of the ai community was
to produce software that could trans-
late text from one language to another,
for example from english to russian.
that undertaking proved to be enor-
mously complicated. human language
appears to be much more subtle and
interwoven with the human experience
than had originally been thought. even
the grammar-checking tools that come
with word-processing programs today
are more of a gim mick than a useful
tool, and analyzing grammar is just the
first step in trans lating sentences.
the CYC (from encyclopedia) proj-
ect, started by Douglas lenat in 1984,
tries to codify the implicit assump-
tions that underlie human speech and
writing. the team members started
out analyzing news articles and asked
themselves what unmen tioned facts
Random Fact 3.2 artificial intelligence
cfe2_ch03_p75_130.indd 112 10/28/10 7:52 PM
Chapter summary 113
use the if statement to implement a decision.
• The if statement allows a program to carry out
dif ferent actions depending on the nature of the
data to be processed.
implement comparisons of numbers and objects.
• Relational operators (< <= > >= == !=) are used to compare numbers and strings.
• Lexicographic order is used to compare strings.
implement complex decisions that require multiple if statements.
• Multiple alternatives are required for decisions that have more than two cases.
• When using multiple if statements, pay attention to the order of the conditions.
C h a p t e r s U M M a r Y
are necessary to actually understand
the sentences. For exam ple, consider
the sentence “last fall she enrolled in
Michigan state”. the reader automati-
cally realizes that “fall” is not related
to falling down in this context, but
refers to the season. While there is a
state of Michigan, here Michigan state
denotes the university. a priori, a com-
puter program has none of this knowl-
edge. the goal of the CYC project is to
extract and store the requisite facts—
that is, (1) people enroll in universities;
(2) Michigan is a state; (3) many states
have universi ties named x state Uni-
versity, often abbreviated as x state;
(4) most peo ple enroll in a university in
the fall. By 1995, the project had codi-
fied about 100,000 common-sense
concepts and about a million facts of
knowledge relating them. even this
massive amount of data has not proven
suffi cient for useful applications.
in recent years, artificial intelli-
gence technology has seen substantial
advances. one of the most astound-
ing examples is the out-
come of a series of “grand
challenges” for autono-
mous vehicles posed by
the Defense advanced
research projects agency
(Darpa). Competitors were
invited to submit a com-
puter-controlled vehicle that
had to complete an obstacle
course without a human
driver or remote control.
the first event, in 2004,
was a disap pointment, with
none of the entrants finish-
ing the route. in 2005, five
vehicles completed a gru-
eling 212 km course in the
Mojave desert. stan ford’s stanley came
in first, with an average speed of 30
km/h. in 2007, Darpa moved the com-
petition to an “urban” environment, an
abandoned air force base. Vehicles had
to be able to interact with each other,
following California traffic laws. as
stanford’s sebastian thrun explained:
“in the last Grand Challenge, it didn’t
really mat ter whether an obstacle was
a rock or a bush, because either way
you’d just drive around it. the current
challenge is to move from just sensing
the envi ronment to understanding the
envi ronment.”
Winner of the 2007 DARPA Urban Challenge
39. Consider the following test program:
int main()
{
int m = 1;
cout << "Enter an integer: ";
cin >> m;
int n = 2;
cout << "Enter another integer: ";
cin >> n;
cout << m << " " << n << endl;
return 0;
}
Run this program and enter three at the first prompt. What happens? Why?
practice it Now you can try these exercises at the end of the chapter: R3.1, R3.30, P3.26.
When one uses a
sophisticated com-
puter program such as a tax prepara-
tion package, one is bound to attribute
some intelligence to the computer.
the computer asks sensible questions
and makes computations that we find
a mental challenge. after all, if doing
one’s taxes were easy, we wouldn’t
need a computer to do it for us.
as programmers, however, we
know that all this apparent intelli gence
is an illusion. human program mers
have carefully “coached” the software
in all possible scenarios, and it simply
replays the actions and deci sions that
were programmed into it.
Would it be possible to write com-
puter programs that are genuinely
intelligent in some sense? From the
earliest days of computing, there was
a sense that the human brain might
be nothing but an immense computer,
and that it might well be feasible to
program computers to imitate some
processes of human thought. seri-
ous research into artificial intelligence
began in the mid-1950s, and the first
twenty years brought some impres-
sive successes. programs that play
chess—surely an activity that appears
to require remarkable intellectual pow-
ers—have become so good that they
now routinely beat all but the best
human players. as far back as 1975,
an expertsystem program called Mycin
gained fame for being better in diag-
nosing meningitis in patients than the
average physician.
however, there were serious set-
backs as well. From 1982 to 1992,
the Japanese government embarked
on a massive research project, funded
at over 40 billion Japanese yen. it was
known as the FifthGeneration Project.
its goal was to develop new hardware
and software to greatly improve the
performance of expert system soft-
ware. at its outset, the project created
fear in other countries that the Japa-
nese computer industry was about to
become the undisputed leader in the
field. however, the end results were
disappointing and did little to bring
artificial intelligence applications to
market.
From the very outset, one of the
stated goals of the ai community was
to produce software that could trans-
late text from one language to another,
for example from english to russian.
that undertaking proved to be enor-
mously complicated. human language
appears to be much more subtle and
interwoven with the human experience
than had originally been thought. even
the grammar-checking tools that come
with word-processing programs today
are more of a gim mick than a useful
tool, and analyzing grammar is just the
first step in trans lating sentences.
the CYC (from encyclopedia) proj-
ect, started by Douglas lenat in 1984,
tries to codify the implicit assump-
tions that underlie human speech and
writing. the team members started
out analyzing news articles and asked
themselves what unmen tioned facts
Random Fact 3.2 artificial intelligence
cfe2_ch03_p75_130.indd 113 10/28/10 7:52 PM
114 Chapter 3 Decisions
implement decisions whose branches require further decisions.
• When a decision statement is contained inside the branch of another decision
statement, the statements are nested.
• Nested decisions are required for problems that have two levels of decision
mak ing.
draw flowcharts for visualizing the control flow of a program.
• Flow charts are made up of elements for tasks, input/
outputs, and decisions.
• Each branch of a decision can contain tasks and further
decisions.
• Never point an arrow inside another branch.
design test cases for your programs.
• Each branch of your program should be covered by a test case.
• It is a good idea to design test cases before implementing a program.
use the boolean data type to store and combine conditions that can be true or false.
• The Boolean type bool has two values, false and true.
• C++ has two Boolean operators that combine conditions: && (and) and || (or).
• To invert a condition, use the ! (not) operator.
• The && and || operators are computed using short-circuit evaluation: As soon as
the truth value is determined, no further conditions are evaluated.
• De Morgan’s law tells you how to negate && and || conditions.
apply if statements to detect whether user input is valid.
• When reading a value, check that it is within the required range.
• Use the fail function to test whether the input stream has failed.
r3.1 Find the errors in the following if statements.
a. if x > 0 then cout << x;
b. if (x > 0) ; { y = 1; } else ; { y = -1; }
c. if (1 + x > pow(x, sqrt(2)) { y = y + x; }
d. if (x = 1) { y++; }
e. cin >> x; if (cin.fail()) { y = y + x; }
r3.2 What do these code fragments print?
a. int n = 1; int m = -1;
if (n < -m) { cout << n; } else { cout << m; }
b. int n = 1; int m = -1;
if (-n >= m) { cout << n; } else { cout << m; }
True
False
Condition
r e V i e W e x e r C i s e s
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review exercises 115
c. double x = 0; double y = 1;
if (fabs(x - y) < 1) { cout << x; } else { cout << y; }
d. double x = sqrt(2); double y = 2;
if (x * x == y) { cout << x; } else { cout << y; }
r3.3 Suppose x and y are variables of type double. Write a code fragment that sets y to x if x
is positive and to 0 otherwise.
r3.4 Suppose x and y are variables of type double. Write a code fragment that sets y to the
absolute value of x without calling the fabs function. Use an if statement.
r3.5 Explain why it is more difficult to compare floating-point numbers than integers.
Write C++ code to test whether an integer n equals 10 and whether a floating-point
number x equals 10.
r3.6 Common Error 3.2 on page 85 explains that a C++ compiler will not report an error
when you use an assignment operator instead of a test for equality, but it may issue a
warning. Write a test program containing a statement
if (floor = 13)
What does your compiler do when you compile the program?
r3.7 Each square on a chess board can be described by a letter and number, such as g5 in
this example:
1
2
4
6
8
3
5
7
1
2
4
6
8
3
5
7
a
a
b
b
d
d
f
f
h
h
c
c
e
e
g5
g
g
The following pseudocode describes an algorithm that determines whether a square
with a given letter and number is dark (black) or light (white).
If the letter is an a, c, e, or g
If the number is odd
color = "black"
Else
color = "white"
Else
If the number is even
color = "black"
Else
color = "white"
Using the procedure in Programming Tip 3.6 on page 97, trace this pseudocode with
input g5.
r3.8 Give a set of four test cases for the algorithm of Exercise R3.7 that covers all
branches.
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116 Chapter 3 Decisions
r3.9 In a scheduling program, we want to check whether two appointments overlap. For
simplicity, appointments start at a full hour, and we use military time (with hours
0–23). The following pseudocode describes an algorithm that determines whether
the appointment with start time start1 and end time end1 overlaps with the appoint-
ment with start time start2 and end time end2.
If start1 > start2
s = start1
Else
s = start2
If end1 < end2
e = end1
Else
e = end2
If s < e
The appointments overlap.
Else
The appointments don’t overlap.
Trace this algorithm with an appointment from 10–12 and one from 11–13, then with
an appointment from 10–11 and one from 12–13.
r3.10 Draw a flow chart for the algorithm in Exercise R3.9.
r3.11 Draw a flow chart for the algorithm in Exercise P3.12.
r3.12 Draw a flow chart for the algorithm in Exercise P3.13.
r3.13 Develop a set of test cases for the algorithm in Exercise R3.9.
r3.14 Develop a set of test cases for the algorithm in Exercise P3.13.
r3.15 Write pseudocode for a program that prompts the user for a month and day and
prints out whether it is one of the following four holidays:
• New Year’s Day (January 1)
• Independence Day (July 4)
• Veterans Day (November 11)
• Christmas Day (December 25)
r3.16 Write pseudocode for a program that assigns letter grades for a quiz, according to the
following table:
Score Grade
90-100 A
80-89 B
70-79 C
60-69 D
< 60 F
r3.17 Explain how the lexicographic ordering of strings in C++ differs from the ordering
of words in a dictionary or telephone book. Hint: Consider strings such as IBM,
wiley.com, Century 21, and While-U-Wait.
r3.18 Of the following pairs of strings, which comes first in lexicographic order?
a. "Tom", "Dick"
b. "Tom", "Tomato"
c. "church", "Churchill"
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review exercises 117
d. "car manufacturer", "carburetor"
e. "Harry", "hairy"
f. "C++", " Car"
g. "Tom", "Tom"
h. "Car", "Carl"
i. "car", "bar"
r3.19 Explain the difference between a sequence of else if clauses and nested if state-
ments. Give an example for each.
r3.20 Give an example of a sequence of else if clauses where the order of the tests does not
matter. Give an example where the order of the tests matters.
r3.21 Rewrite the condition in Section 3.3 to use < operators instead of >= operators. What
is the impact on the order of the comparisons?
r3.22 Give a set of test cases for the tax program in Exercise P3.18. Manually compute the
expected results.
r3.23 Make up another C++ code example that shows the dangling else problem, using the
following statement. A student with a GPA of at least 1.5, but less than 2, is on
probation. With less than 1.5, the student is failing.
r3.24 Complete the following truth table by finding the truth values of the Boolean
expressions for all combinations of the Boolean inputs p, q, and r.
p q r (p && q) || !r !(p && (q || !r))
false false false
false false true
false true false
. . .
5 more combinations
. . .
r3.25 True or false? A && B is the same as B && A for any Boolean conditions A and B.
r3.26 The “advanced search” feature of many search engines allows you to use Boolean
operators for complex queries, such as “(cats OR dogs) AND NOT pets”. Contrast
these search operators with the Boolean operators in C++.
r3.27 Suppose the value of b is false and the value of x is 0. What is the value of each of the
fol lowing expressions?
a. b && x == 0
b. b || x == 0
c. !b && x == 0
d. !b || x == 0
e. b && x != 0
f. b || x != 0
g. !b && x != 0
h. !b || x != 0
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118 Chapter 3 Decisions
r3.28 Simplify the following expressions. Here, b is a variable of type bool.
a. b == true
b. b == false
c. b != true
d. b != false
r3.29 Simplify the following statements. Here, b is a variable of type bool and n is a vari able
of type int.
a. if (n == 0) { b = true; } else { b = false; }
(Hint: What is the value of n == 0?)
b. if (n == 0) { b = false; } else { b = true; }
c. b = false; if (n > 1) { if (n < 2) { b = true; } }
d. if (n < 1) { b = true; } else { b = n > 2; }
r3.30 What is wrong with the following program?
cout << "Enter the number of quarters: ";
cin >> quarters;
total = total + quarters * 0.25;
cout << "Total: " << total << endl;
if (cin.fail()) { cout << "Input error."; }
r3.31 Reading numbers is surprisingly difficult because a C++ input stream looks at the
input one character at a time. First, white space is skipped. Then the stream con-
sumes those input characters that can be a part of a number. Once the stream has
recognized a number, it stops reading if it finds a character that cannot be a part of a
number. However, if the first non-white space character is not a digit or a sign, or if
the first character is a sign and the second one is not a digit, then the stream fails.
Consider a program reading an integer:
cout << "Enter the number of quarters: ";
int quarters;
cin >> quarters;
For each of the following user inputs, circle how many characters have been read and
whether the stream is in the failed state or not.
a. 15.9
b. 15 9
c. +159
d. -15A9
e. Fifteen
f. -Fifteen
g. + 15
h. 1.5E3
i. +1+5
p3.1 Write a program that reads a temperature value and the letter C for Celsius or F for
Fahrenheit. Print whether water is liquid, solid, or gaseous at the given temperature
at sea level.
p r o G r a M M i n G e x e r C i s e s
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programming exercises 119
p3.2 The boiling point of water drops by about one degree centigrade for every 300
meters (or 1,000 feet) of altitude. Improve the program of Exercise P3.1 to allow the
user to supply the altitude in meters or feet.
p3.3 Write a program that reads in three floating-point numbers and prints the largest of
the three inputs. For example:
Please enter three numbers: 4 9 2.5
The largest number is 9.
p3.4 Write a program that reads in three strings and sorts them lexicographically.
Enter three strings: Charlie Able Baker
Able
Baker
Charlie
p3.5 Write a program that reads an integer and prints how many digits the number has, by
checking whether the number is ≥ 10, ≥ 100, and so on. (Assume that all integers are
less than ten billion.) If the number is negative, first multiply it with –1.
p3.6 Write a program that reads three numbers and prints “all the same” if they are all the
same, “all different” if they are all different, and “neither” otherwise.
p3.7 Write a program that reads three numbers and prints “increasing” if they are in
increasing order, “decreasing” if they are in decreasing order, and “neither” other-
wise. Here, “increasing” means “strictly increasing”, with each value larger than its
pre decessor. The sequence 3 4 4 would not be considered increasing.
p3.8 Repeat Exercise P3.7, but before reading the numbers, ask the user whether increas-
ing/decreasing should be “strict” or “lenient”. In lenient mode, the sequence 3 4 4 is
increasing and the sequence 4 4 4 is both increasing and decreasing.
p3.9 Write a program that translates a letter grade into a number grade. Letter grades are
A, B, C, D, and F, possibly followed by + or –. Their numeric values are 4, 3, 2, 1,
and 0. There is no F+ or F–. A + increases the numeric value by 0.3, a – decreases it
by 0.3. However, an A+ has value 4.0.
Enter a letter grade: B-
The numeric value is 2.7.
p3.10 Write a program that translates a number between 0 and 4 into the closest letter
grade. For example, the number 2.8 (which might have been the average of several
grades) would be converted to B–. Break ties in favor of the better grade; for exam ple
2.85 should be a B.
p3.11 Write a program that takes user input describing a playing card in the following
shorthand notation:
A Ace
2 … 10 Card values
J Jack
Q Queen
K King
D Diamonds
H Hearts
S Spades
C Clubs
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120 Chapter 3 Decisions
Your program should print the full description of the card. For example,
Enter the card notation: QS
Queen of Spades
p3.12 When two points in time are compared, each given as hours (in military time,
rang ing from 0 and 23) and minutes, the following pseudocode determines which
comes first.
If hour1 < hour2
time1 comes first.
Else if hour1 and hour2 are the same
If minute1 < minute2
time1 comes first.
Else if minute1 and minute2 are the same
time1 and time2 are the same.
Else
time2 comes first.
Else
time2 comes first.
Write a program that prompts the user for two points in time and prints the time that
comes first, then the other time.
p3.13 The following algorithm yields the season (Spring, Summer, Fall, or Winter) for a
given month and day.
If month is 1, 2, or 3, season = "Winter"
Else if month is 4, 5, or 6, season = "Spring"
Else if month is 7, 8, or 9, season = "Summer"
Else if month is 10, 11, or 12, season = "Fall"
If month is divisible by 3 and day >= 21
If season is “Winter”, season = “Spring”
Else if season is “Spring”, season = “Summer”
Else if season is “Summer”, season = “Fall”
Else season = “Winter”
Write a program that prompts the user for a month
and day and then prints the season, as determined
by this algorithm.
p3.14 Write a program that reads in two floating-point numbers and tests whether they are
the same up to two decimal places. Here are two sample runs.
Enter two floating-point numbers: 2.0 1.99998
They are the same up to two decimal places.
Enter two floating-point numbers: 2.0 1.98999
They are different.
p3.15 Write a program to simulate a bank transaction. There are two bank accounts: check-
ing and savings. First, ask for the initial balances of the bank accounts; reject negative
balances. Then ask for the transactions; options are deposit, withdrawal, and trans-
fer. Then ask for the account; options are checking and savings. Then ask for the
amount; reject transactions that overdraw an account. At the end, print the balances
of both accounts.
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programming exercises 121
p3.16 Write a program that reads in the name and salary of an employee. Here the salary
will denote an hourly wage, such as $9.25. Then ask how many hours the employee
worked in the past week. Be sure to accept fractional hours. Any overtime work
(over 40 hours per week) is paid at 150 percent of the regular wage. Compute the
pay. Print a paycheck for the employee.
p3.17 Write a program that prompts for the day and month of the user’s birthday and then
prints a horoscope. Make up fortunes for programmers, like this:
Please enter your birthday (month and day): 6 16
Gemini are experts at figuring out the behavior of complicated programs.
You feel where bugs are coming from and then stay one step ahead. Tonight,
your style wins approval from a tough critic.
Each fortune should contain the name of the astrological sign. (You will find the
names and date ranges of the signs at a distressingly large number of sites on the
Internet.)
p3.18 Write a program that computes taxes for the following schedule:
if your status is single and
if the taxable income is over but not over the tax is of the amount over
$0 $8,000 10% $0
$8,000 $32,000 $800 + 15% $8,000
$32,000 $4,400 + 25% $32,000
if your status is Married and
if the taxable income is over but not over the tax is of the amount over
$0 $16,000 10% $0
$16,000 $64,000 $1,600 + 15% $16,000
$64,000 $8,800 + 25% $64,000
p3.19 The original U.S. income tax of 1913 was quite simple. The tax was
• 1 percent on the first $50,000.
• 2 percent on the amount over $50,000 up to $75,000.
• 3 percent on the amount over $75,000 up to $100,000.
• 4 percent on the amount over $100,000 up to $250,000.
• 5 percent on the amount over $250,000 up to $500,000.
• 6 percent on the amount over $500,000.
There was no separate schedule for single or married taxpayers. Write a program that
computes the income tax according to this schedule.
p3.20 The tax.cpp program uses a simplified version of the 2008 U.S. income tax schedule.
Look up the tax brackets and rates for the current year, for both single and married
filers, and implement a program that computes the actual income tax.
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122 Chapter 3 Decisions
p3.21 Unit conversion. Write a unit conversion program that asks the users from which
unit they want to convert (fl. oz, gal, oz, lb, in, ft, mi) and to which unit they want to
convert (ml, l, g, kg, mm, cm, m, km). Reject incompatible conversions (such as gal
→ km). Ask for the value to be converted, then display the result:
Convert from? gal
Convert to? ml
Value? 2.5
2.5 gal = 9462.5 ml
p3.22 Write a program that prompts the user to provide a single character from the alpha-
bet. Print Vowel or Consonant, depending on the user input. If the user input is not a
letter (between a and z or A and Z), or is a string of length > 1, print an error message.
p3.23 Roman numbers. Write a program that converts a positive integer into the Roman
number system. The Roman number system has digits
I 1
V 5
X 10
L 50
C 100
D 500
M 1,000
Numbers are formed according to the following rules. (1) Only numbers up to 3,999
are represented. (2) As in the decimal system, the thousands, hundreds, tens, and
ones are expressed separately. (3) The num bers 1 to 9 are expressed as
I 1
II 2
III 3
IV 4
V 5
VI 6
VII 7
VIII 8
IX 9
As you can see, an I preceding a V or X is
subtracted from the value, and you can
never have more than three I’s in a row. (4)
Tens and hundreds are done the same way,
except that the let ters X, L, C and C, D, M
are used instead of I, V, X, respec tively.
Your program should take an input, such as
1978, and convert it to Roman numerals,
MCMLXXVIII.
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programming exercises 123
p3.24 Write a program that asks the user to enter a month (1 for January, 2 for February,
and so on) and then prints the number of days in the month. For February, print “28
or 29 days”.
Enter a month: 5
30 days
Do not use a separate if/else branch for each month. Use Boolean operators.
p3.25 A year with 366 days is called a leap year. A year is a leap year if it is divisible by four
(for example, 1980), except that it is not a leap year if it is divisible by 100 (for
example, 1900); however, it is a leap year if it is divisible by 400 (for example, 2000).
There were no exceptions before the introduction of the Gregorian calendar on
October 15, 1582 (1500 was a leap year). Write a program that asks the user for a year
and computes whether that year is a leap year.
p3.26 Add error handling to Exercise P3.2. If the user does not enter a number when
expected, or provides an invalid unit for the altitude, print an error message and end
the program.
engineering p3.27 Write a program that prompts the user for a wavelength value and prints a descrip-
tion of the corresponding part of the electromagnetic spectrum, as given in Table 7.
table 7 electromagnetic spectrum
type Wavelength (m) Frequency (hz)
Radio Waves > 10–1 < 3 × 109
Microwaves 10–3 to 10–1 3 × 109 to 3 × 1011
Infrared 7 × 10–7 to 10–3 3 × 1011 to 4 × 1014
Visible light 4 × 10–7 to 7 × 10–7 4 × 1014 to 7.5 × 1014
Ultraviolet 10–8 to 4 × 10–7 7.5 × 1014 to 3 × 1016
X-rays 10–11 to 10–8 3 × 1016 to 3 × 1019
Gamma rays < 10–11 > 3 × 1019
engineering p3.28 Repeat Exercise P3.27, modifying the program so that it prompts for the frequency
instead.
engineering p3.29 Repeat Exercise P3.27, modifying the program so that it first asks the user whether
the input will be a wavelength or a frequency.
engineering p3.30 A minivan has two sliding doors. Each door can be
opened by either a dashboard switch, its inside handle, or
its outside handle. However, the inside handles do not
work if a child lock switch is activated. In order for the
sliding doors to open, the gear shift must be in park, and
the master unlock switch must be activated. (This book’s
author is the long-suffering owner of just such a vehicle.)
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124 Chapter 3 Decisions
Your task is to simulate a portion of the control software for the vehicle. The input is
a sequence of values for the switches and the gear shift, in the following order:
• Dashboard switches for left and right sliding door, child lock, and master
unlock (0 for off or 1 for activated)
• Inside and outside handles on the left and right sliding doors (0 or 1)
• The gear shift setting (one of P N D 1 2 3 R).
A typical input would be 0 0 0 1 0 1 0 0 P.
Print “left door opens” and/or “right door opens” as appropriate. If neither door
opens, print “both doors stay closed”.
engineering p3.31 Sound level L in units of decibel (dB) is determined by
L = 20 log10(p/p0)
where p is the sound pressure of the sound (in Pascals, abbreviated Pa), and p0 is a
reference sound pres sure equal to 20 × 10–6 Pa (where L is 0 dB). The following table
gives descriptions for certain sound lev els.
Threshold of pain 130 dB
Possible hearing damage 120 dB
Jack hammer at 1 m 100 dB
Traffic on a busy roadway at 10 m 90 dB
Normal conversation 60 dB
Calm library 30 dB
Light leaf rustling 0 dB
Write a program that reads a value and a unit, either dB or Pa, and then prints the
closest description from the list above.
engineering p3.32 The electric circuit shown below is designed to measure the temperature of the gas in
a chamber.
+
–Vs = 20 V
Rs = 75 Ω
R Vm
+
–
Voltmeter
11.43 V
The resistor R represents a temperature sensor enclosed in the chamber. The resis-
tance R, in Ω, is related to the temperature T, in °C, by the equation
R R kT= +0
In this device, assume R0 = 100 Ω and k = 0.5. The voltmeter displays the value of the
voltage, Vm , across the sensor. This voltage Vm indicates the temperature, T, of the
gas according to the equation
T
R
k
R
k
R
k
V
V V
R
k
s m
s m
= − =
−
−0 0
cfe2_ch03_p75_130.indd 124 10/28/10 7:52 PM
programming exercises 125
Suppose the voltmeter voltage is constrained to the range Vmin = 12 volts ≤ Vm ≤
Vmax = 18 volts. Write a program that accepts a value of Vm and checks that it’s
between 12 and 18. The program should return the gas temperature in degrees
Celsius when Vm is between 12 and 18 and an error message when it isn’t.
engineering p3.33 Crop damage due to frost is one of the many risks confronting farmers. The figure
below shows a simple alarm circuit designed to warn of frost. The alarm circuit uses
a device called a thermistor to sound a buzzer when the temperature drops below
freezing. Thermistors are semiconductor devices that exhibit a temperature depen-
dent resistance described by the equation
R R e T T=
−
0
1 1
0
β
where R is the resistance, in Ω, at the temperature T, in °K, and R0 is the resistance,
in Ω, at the temperature T0, in°K. β is a constant that depends on the material used to
make the thermistor.
–
+
9 V
R 3
R 4R 2
RThermistor
9 V
Comparator
Buzzer
The circuit is designed so that the alarm will sound when
R
R R
R
R R
2
2
4
3 4+
<
+
The thermistor used in the alarm circuit has R0 = 33,192 Ω at T0 = 40 °C, and β =
3,310 °K. (Notice that β has units of °K. Recall that the temperature in °K is obtained
by adding 273° to the temperature in °C.) The resistors R2, R3, and R4 have a resis-
tance of 156.3 kΩ = 156,300 Ω.
Write a C++ program that prompts the user for a temperature in °F and prints a
message indicating whether or not the alarm will sound at that temperature.
engineering p3.34 A mass m = 2 kilograms is attached to the end of a rope of length r = 3 meters. The
mass is whirled around at high speed. The rope can withstand a maximum tension of
T = 60 Newtons. Write a program that accepts a rotation speed v and determines if
such a speed will cause the rope to break. Hint: T m v r= 2 .
engineering p3.35 A mass m is attached to the end of a rope of length r = 3 meters. The rope can only be
whirled around at speeds of 1, 10, 20, or 40 meters per second. The rope can with-
stand a maximum tension of T = 60 Newtons. Write a program where the user enters
the value of the mass m, and the program determines the greatest speed at which it
can be whirled without breaking the rope. Hint: T m v r= 2 .
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126 Chapter 3 Decisions
engineering p3.36 The average person can jump off the ground with
a velocity of 7 mph without fear of leaving the
planet. However, if an astronaut jumps with this
velocity while standing on Halley’s Comet, will
the astronaut ever come back down? Create a
program that allows the user to input a launch
velocity (in mph) from the surface of Halley’s
Comet and determine whether a jumper will
return to the surface. If not, the program should
calculate how much more massive the comet must
be in order to return the jumper to the surface.
Hint: Escape velocity is v
G M
Rescape
= 2 , where G N m kg= × −6 67 10 11 2 2. is
the gravitational constant, M kg= ×1 3 1022. is the mass of Halley’s comet, and
R m= ×1 153 106. is its radius.
1. Change the if statement to
if (floor > 14)
{
actual_floor = floor – 2;
}
2. 85. 90. 85.
3. The only difference is if original_price is 100. The statement in Self Check 2 sets
discounted_price to 90; this one sets it to 80.
4. 95. 100. 95.
5. if (fuel_amount < 0.10 * fuel_capacity)
{
cout << "red" << endl;
}
else
{
cout << "green" << endl;
}
6. (a) and (b) are both true, (c) is false.
7. floor <= 13
8. The values should be compared with ==, not =.
9. input == "Y"
10. str != "" or str.length() > 0
11. if (score_a > score_b)
{
cout << "A won";
}
else if (score_a < score_b)
{
cout << "B won";
a n s W e r s t o s e l F - C h e C k Q U e s t i o n s
cfe2_ch03_p75_130.indd 126 10/28/10 7:52 PM
answers to self-Check Questions 127
}
else
{
cout << "Game tied";
}
12. if (x > 0)
{
s = 1;
}
else if (x < 0)
{
s = -1;
}
else
{
s = 0;
}
13. You could first set s to one of the three values:
s = 0;
if (x > 0) { s = 1; }
else if (x < 0) { s = -1; }
14. The if (price <= 100) can be omitted, making it clear that the else branch is the sole
alternative.
15. No destruction of buildings
16. Add a branch before the final else:
else if (richter < 0) { cout << "Error: Negative input" << endl; }
17. $3,200
18. No. Then the computation is 0.10 × 32000 + 0.25 (32000 – 32000).
19. No. Their individual tax is $5,200 each, and if they married, they would pay $10,400.
Actually, taxpayers in higher tax brackets (which our program does not model) may
pay higher taxes when they marry, a phenomenon known as the mar riage penalty.
20. Change else in line 36 to else if (marital_status == "m"), and add another branch after
line 47:
else { cout << "Error: marital status should be s or m." << endl; }
21. The higher tax rate is only applied on the income in the higher bracket. Suppose you
are single and make $31,900. Should you try to get a $200 raise? Absolutely: you get
to keep 90 percent of the first $100 and 75 percent of the next $100.
22.
True
False
temp < 0? Print “Frozen”
Read temp
cfe2_ch03_p75_130.indd 127 10/28/10 7:52 PM
128 Chapter 3 Decisions
23. The “True” arrow from the first decision points into the “True” branch of the sec-
ond decision, creating spaghetti code.
24. Here is one solution. In Section 3.7, you will see how you can combine the condi-
tions for a more elegant solution.
True
False
Input < 0? Status = “Error”
True
False
Input > 100?
Status = “OK”
Status = “Error”
25.
True
False
Print Print “Error”
x < 0?
Read x
cfe2_ch03_p75_130.indd 128 10/28/10 7:52 PM
answers to self-Check Questions 129
26.
27. Test Case Expected Output Comment
12 12 Below 13th floor
14 13 Above 13th floor
13 ? The specification is not clear— See Section 3.8 for a
version of this program with error handling
28. A boundary test case is a price of $128. A 16 percent discount should apply because
the problem statement states that the larger discount applies if the price is at least
$128. Thus, the expected output is $107.52.
29. Test Case Expected Output Comment
9 Most structures fall
7.5 Many buildings destroyed
6.5 Many buildings considerably...
5 Damage to poorly...
3 No destruction...
8.0 Most structures fall Boundary case. In this program, boundary cases are not as
significant since the behavior of an earthquake changes
gradually.
-1 The specification is not clear—see Self Check 16 for a
version of this program with error handling.
True
False
temp < 0? Print “Ice”
True
False
temp > 100? Print “Steam”
Print “Liquid”
Read temp
cfe2_ch03_p75_130.indd 129 10/28/10 7:52 PM
130 Chapter 3 Decisions
Step 1 Decide on the branching condition.
We need to take different actions for strings of odd and even length. Therefore, the condition is
Is the length of the string odd?
In C++, you use the remainder of division by 2 to find out whether a value is even or odd.
Then the test becomes
str.length() % 2 == 1?
Step 2 Give pseudocode for the work that needs to be done when the condition is true.
We need to find the position of the middle character. If the length is 5, the position is 2.
In general,
position = str.length() / 2 (with the remainder discarded)
result = str.substr(position, 1)
Step 3 Give pseudocode for the work (if any) that needs to be done when the condition is not true.
Again, we need to find the position of the middle character. If the length is 6, the starting posi-
tion is 2, and the ending position is 3.
In general,
position = str.length() / 2 – 1
result = str.substr(position, 2)
Step 4 Double-check relational operators.
Do we really want str.length() % 2 == 1? For example, when the length is 5, 5 % 2 is the remainder
of the division 5 / 2, which is 1. In general, dividing an odd number by 2 leaves a remainder
of 1. (Actually, dividing a negative odd num ber by 2 leaves a remainder of –1, but the string
length is never negative.) Therefore, our condition is correct.
Step 5 Remove duplication.
Here is the statement that we have developed:
If str.length() % 2 == 1
position = str.length() / 2 (with remainder discarded)
result = str.substr(position, 1)
Else
position = str.length() / 2 – 1
result = str.substr(position, 2)
W o r k e D e x a M p l e 3 . 1 extracting the middle
Your task is to extract a string containing the middle character from a given string str. For
example, if the string is “crate”, the result is the string “a”. However, if the string has an even
number of letters, extract the middle two characters. If the string is “crates”, the result is “at”.
30. Test Case Expected Output Comment
(0.5, 0.5) inside
(4, 2) outside
(0, 2) on the boundary Exactly on the boundary
(1.414, 1.414) on the boundary Close to the boundary
(0, 1.9) inside Not less than 1 mm from the boundary
(0, 2.1) outside Not less than 1 mm from the boundary
31. x == 0 && y == 0
32. x == 0 || y == 0
33. (x == 0 && y != 0) || (y == 0 && x != 0)
34. The same as the value of frozen.
35. You are guaranteed that there are no other values. With strings or integers, you
would need to check that no values such as “maybe” or –1 enter your calculations.
36. (a) Error: The floor must be between 1 and 20. (b) Error: The floor must be between 1 and
20. (c) The elevator will travel to the actual floor 19 (d) Error: Not an integer.
37. cin.fail() || floor == 13 || floor <= 0 || floor > 20
38. Check for cin.fail(), to make sure a researcher didn’t supply an input such as oh my.
Check for weight <= 0, since any rat must surely have a positive weight. We don’t
know how giant a rat could be, but the New Guinea rats weighed no more than 2 kg.
A regular house rat (rattus rattus) weighs up to 0.2 kg, so we’ll say that any weight
> 10 kg was surely an input error, perhaps confusing grams and kilograms. Thus, the
checks are
if (cin.fail())
{
cout << "Error: Not a number" << endl;
return 1;
}
if (weight < 0)
{
cout << "Error: Weight cannot be negative." << endl;
return 1;
}
if (weight > 10)
{
cout << "Error: Weight > 10 kg.” << endl;
return 1;
}
39. The first input fails. The value of m is unchanged. Because a previous input failed, the
next input doesn’t even try to get additional keystrokes. It also fails, and n is un-
changed. The program prints 1 2.
cfe2_ch03_p75_130.indd 130 10/28/10 7:52 PM
4C h a p t e r
131
L o o p s
to implement while, for, and do loops
to avoid infinite loops and off-by-one
errors
to understand nested loops
to implement programs that read and process data sets
to use a computer for simulations
C h a p t e r G o a L s
C h a p t e r C o n t e n t s
4.1 The while loop 132
Syntax 4.1: while statement 133
Common Error 4.1: Infinite Loops 136
Common Error 4.2: Don’t think “are We
there Yet?” 137
Common Error 4.3: off-by-one errors 137
Random Fact 4.1: the First Bug 138
4.2 problem Solving:
hand-Tracing 139
4.3 The for loop 142
Syntax 4.2: for statement 144
Programming Tip 4.1: Use for Loops for their
Intended purpose only 147
Programming Tip 4.2: Choose Loop Bounds that
Match Your task 147
Programming Tip 4.3: Count Iterations 147
4.4 The do loop 148
Programming Tip 4.4: Flowcharts for Loops 149
4.5 proceSSing inpuT 150
Special Topic 4.1: Clearing the Failure state 153
Special Topic 4.2: the Loop-and-a-half problem
and the break statement 153
Special Topic 4.3: redirection of Input and
output 154
4.6 problem Solving:
SToryboardS 154
4.7 common loop algoriThmS 157
How To 4.1: Writing a Loop 162
Worked Example 4.1: Credit Card processing
4.8 neSTed loopS 165
4.9 random numberS and
SimulaTionS 168
Random Fact 4.2: software piracy 172
cfe2_ch04_p131_192.indd 131 10/28/10 8:12 PM
132
In a loop, a part of a program is repeated over and over,
until a specific goal is reached. Loops are important for
calculations that require repeated steps and for processing
input consisting of many data items. In this chapter you will
learn about loop statements in C++, as well as techniques
for writing programs that process input and simulate
activities in the real world.
4.1 the while Loop
In this section, you will learn how to repeatedly
execute statements until a goal has been reached.
Recall the investment problem from Chapter 1.
You put $10,000 into a bank account that earns 5
percent interest per year. How many years does it
take for the account balance to be double the origi-
nal investment?
In Chapter 1 we developed the following algo-
rithm for this problem:
Start with a year value of 0, a column for the interest, and a
balance of $10,000.
year interest balance
0 $10,000
Repeat the following steps while the balance is less than $20,000
Add 1 to the year value.
Compute the interest as balance x 0.05 (i.e., 5 percent interest).
Add the interest to the balance.
Report the final year value as the answer.
You now know how to define and update the variables in C++. What you don’t yet
know is how to carry out “Repeat steps while the balance is less than $20,000”.
Because the interest earned also
earns interest, a bank balance
grows exponentially.
In a particle accelerator, subatomic particles
traverse a loop-shaped tunnel multiple times,
gaining the speed required for physical experiments.
Similarly, in computer science, statements in a
loop are executed while a condition is true.
In C++, the while statement implements such a
repetition (see Syntax 4.1). The code
while (condition)
{
statements
}
keeps executing the statements while the condition
is true. In our case, we want to increment the year
counter and add interest while the balance is less
than the target balance of $20,000:
while (balance < TARGET)
{
year++;
double interest = balance * RATE / 100;
balance = balance + interest;
}
A while statement is an example of a loop. If you draw a flowchart, the flow of execu-
tion loops again to the point where the condition is tested (see Figure 1).
Loops execute a
block of code
repeatedly while
a condition
remains true.
syntax 4.1 while statement
double balance = 0;
.
.
.
while (balance < TARGET)
{
year++;
double interest = balance * RATE / 100;
balance = balance + interest;
}
If the condition
never becomes false,
an infinite loop occurs.
See page 136.
These statements
are executed while
the condition is true.
Braces are not required if the body contains a
single statement, but it’s good to always use them.
See page 80.
Don’t put a semicolon here!
See page 80.
Beware of “off-by-one”
errors in the loop condition.
See page 137.
Lining up braces
is a good idea.
See page 79.
This variable is defined outside the loop
and updated in the loop.
This variable is created
in each loop iteration.
cfe2_ch04_p131_192.indd 132 10/28/10 8:12 PM
4.1 the while Loop 133
figure 1 Flowchart of a while Loop
False
True
Add interest
to balance
Increment
year
balance <
TARGET?
In C++, the while statement implements such a
repetition (see Syntax 4.1). The code
while (condition)
{
statements
}
keeps executing the statements while the condition
is true. In our case, we want to increment the year
counter and add interest while the balance is less
than the target balance of $20,000:
while (balance < TARGET)
{
year++;
double interest = balance * RATE / 100;
balance = balance + interest;
}
A while statement is an example of a loop. If you draw a flowchart, the flow of execu-
tion loops again to the point where the condition is tested (see Figure 1).
Loops execute a
block of code
repeatedly while
a condition
remains true.
syntax 4.1 while statement
double balance = 0;
.
.
.
while (balance < TARGET)
{
year++;
double interest = balance * RATE / 100;
balance = balance + interest;
}
If the condition
never becomes false,
an infinite loop occurs.
See page 136.
These statements
are executed while
the condition is true.
Braces are not required if the body contains a
single statement, but it’s good to always use them.
See page 80.
Don’t put a semicolon here!
See page 80.
Beware of “off-by-one”
errors in the loop condition.
See page 137.
Lining up braces
is a good idea.
See page 79.
This variable is defined outside the loop
and updated in the loop.
This variable is created
in each loop iteration.
cfe2_ch04_p131_192.indd 133 10/28/10 8:12 PM
134 Chapter 4 Loops
When you define a variable inside the loop body, the variable is created for each
iteration of the loop and removed after the end of each iteration. For example, con-
sider the interest variable in this loop:
while (balance < TARGET)
{
year++;
double interest = balance * RATE / 100;
balance = balance + interest;
} // interest no longer defined here
In contrast, the balance and years variables were defined outside the loop body. That
way, the same vari able is used for all iterations of the loop.
A new interest variable
is created in each iteration.
figure 2
execution of the
doublinv Loop
while (balance < TARGET)
{
year++;
double interest = balance * RATE / 100;
balance = balance + interest;
}
cout << year << endl;
while (balance < TARGET)
{
year++;
double interest = balance * RATE / 100;
balance = balance + interest;
}
Check the loop condition1 The condition is true
while (balance < TARGET)
{
year++;
double interest = balance * RATE / 100;
balance = balance + interest;
}
Execute the statements in the loop2
while (balance < TARGET)
{
year++;
double interest = balance * RATE / 100;
balance = balance + interest;
}
Check the loop condition again3 The condition is still true
while (balance < TARGET)
{
year++;
double interest = balance * RATE / 100;
balance = balance + interest;
}
After 15 iterations4 The condition is
no longer true
Execute the statement following the loop5
.
.
.
year = 0
balance = 10000
year = 1
interest = 500
balance = 10500
year = 1
balance = 10500
year = 15
balance = 20789.28
year = 15
balance = 20789.28
cfe2_ch04_p131_192.indd 134 10/28/10 8:12 PM
4.1 the while Loop 135
Here is the program that solves the investment problem. Figure 2 illustrates the
program’s execution.
ch04/doublinv.cpp
1 #include
2
3 using namespace std;
4
5 int main()
6 {
7 const double RATE = 5;
8 const double INITIAL_BALANCE = 10000;
9 const double TARGET = 2 * INITIAL_BALANCE;
10
11 double balance = INITIAL_BALANCE;
12 int year = 0;
13
14 while (balance < TARGET)
15 {
16 year++;
17 double interest = balance * RATE / 100;
18 balance = balance + interest;
19 }
20
21 cout << "The investment doubled after "
22 << year << " years." << endl;
23
24 return 0;
25 }
program run
The investment doubled after 15 years.
1. How many years does it take for the investment to triple? Modify the program
and run it.
2. If the interest rate is 10 percent per year, how many years does it take for the
investment to double? Modify the program and run it.
3. Modify the program so that the balance after each year is printed. How did you
do that?
4. Suppose we change the program so that the condition of the while loop is
while (balance <= TARGET)
What is the effect on the program? Why?
5. What does the following loop print?
int n = 1;
while (n < 100)
{
n = 2 * n;
cout << n << " ";
}
practice it Now you can try these exercises at the end of the chapter: R4.3, P4.11, P4.16.
S e l f c h e c k
cfe2_ch04_p131_192.indd 135 10/28/10 8:12 PM
136 Chapter 4 Loops
table 1 while Loop examples
Loop output explanation
i = 5;
while (i > 0)
{
cout << i << " ";
i--;
}
5 4 3 2 1 When i is 0, the loop condition is false,
and the loop ends.
i = 5;
while (i > 0)
{
cout << i << " ";
i++;
}
5 6 7 8 9 10 11 ... The i++ statement is an error causing an
“infinite loop” (see Common Error 4.1
on page 136).
i = 5;
while (i > 5)
{
cout << i << " ";
i--;
}
(No output) The statement i > 5 is false,
and the loop is never executed.
i = 5;
while (i < 0)
{
cout << i << " ";
i--;
}
(No output) The programmer probably thought,
“Stop when i is less than 0”. However,
the loop condition controls when the
loop is executed, not when it ends (see
Common Error 4.2 on page 137).
i = 5;
while (i > 0) ;
{
cout << i << " ";
i--;
}
(No output, program
does not terminate)
Note the semicolon before the {. This
loop has an empty body. It runs
forever, checking whether i > 0 and
doing nothing in the body.
infinite loops
A very annoying loop error is an infinite loop: a loop that runs forever and can be stopped
only by killing the pro gram or restarting the computer. If there are output statements in the
program, then line after line of output flashes by on the screen. Otherwise, the program just
sits there and hangs, seeming to do nothing. On some systems, you can terminate a hanging
program by hitting Ctrl + C. On others, you can close the window in which the program runs.
A common reason for infinite loops is forgetting to update the variable that controls the
loop:
year = 1;
while (year <= 20)
{
Common error 4.1
balance = balance * (1 + RATE / 100);
}
Here the programmer forgot to add a year++ command in the loop. As a result, the year always
stays at 1, and the loop never comes to an end.
Another common reason for an infinite loop is accidentally incrementing a counter that
should be decremented (or vice versa). Consider this example:
year = 20;
while (year > 0)
{
balance = balance * (1 + RATE / 100);
year++;
}
The year variable really should have been decremented, not
incremented. This is a common error because increment ing
counters is so much more common than decrementing that
your fingers may type the ++ on autopilot. As a con sequence,
year is always larger than 0, and the loop never ends. (Actually,
year may eventually exceed the largest representable positive
integer and wrap around to a negative number. Then the loop
ends—of course, with a com pletely wrong result.)
don’t Think “are we There yet?”
When doing something repetitive, most of us want to know when
we are done. For exam ple, you may think, “I want to get at least
$20,000,” and set the loop condition to
balance >= TARGET
But the while loop thinks the opposite: How long am I allowed to
keep going? The correct loop condition is
while (balance < TARGET)
In other words: “Keep at it while the balance is less than the target.”
When writing a loop condition, don’t ask, “Are we there yet?”
The condition determines how long the loop will keep going.
off-by-one errors
Consider our computation of the number of years that are required to double an investment:
int year = 0;
while (balance < TARGET)
{
year++;
double interest = balance * RATE / 100;
balance = balance + interest;
}
cout << "The investment doubled after " << year << " years." << endl;
Like this hamster who can’t
stop running in the tread mill,
an infinite loop never ends.
Common error 4.2
Common error 4.3
cfe2_ch04_p131_192.indd 136 10/28/10 8:12 PM
4.1 the while Loop 137
balance = balance * (1 + RATE / 100);
}
Here the programmer forgot to add a year++ command in the loop. As a result, the year always
stays at 1, and the loop never comes to an end.
Another common reason for an infinite loop is accidentally incrementing a counter that
should be decremented (or vice versa). Consider this example:
year = 20;
while (year > 0)
{
balance = balance * (1 + RATE / 100);
year++;
}
The year variable really should have been decremented, not
incremented. This is a common error because increment ing
counters is so much more common than decrementing that
your fingers may type the ++ on autopilot. As a con sequence,
year is always larger than 0, and the loop never ends. (Actually,
year may eventually exceed the largest representable positive
integer and wrap around to a negative number. Then the loop
ends—of course, with a com pletely wrong result.)
don’t Think “are we There yet?”
When doing something repetitive, most of us want to know when
we are done. For exam ple, you may think, “I want to get at least
$20,000,” and set the loop condition to
balance >= TARGET
But the while loop thinks the opposite: How long am I allowed to
keep going? The correct loop condition is
while (balance < TARGET)
In other words: “Keep at it while the balance is less than the target.”
When writing a loop condition, don’t ask, “Are we there yet?”
The condition determines how long the loop will keep going.
off-by-one errors
Consider our computation of the number of years that are required to double an investment:
int year = 0;
while (balance < TARGET)
{
year++;
double interest = balance * RATE / 100;
balance = balance + interest;
}
cout << "The investment doubled after " << year << " years." << endl;
Like this hamster who can’t
stop running in the tread mill,
an infinite loop never ends.
Common error 4.2
Common error 4.3
cfe2_ch04_p131_192.indd 137 10/28/10 8:12 PM
138 Chapter 4 Loops
Should year start at 0 or at 1? Should you test for balance < TARGET or for balance <= TARGET? It is
easy to be off by one in these expressions.
Some people try to solve off-by-one errors by randomly inserting +1 or -1 until the pro-
gram seems to work—a terrible strategy. It can take a long time to compile and test all the vari-
ous possibilities. Expending a small amount of mental effort is a real time saver.
Fortunately, off-by-one errors are easy to avoid, simply by working through a couple of
test cases and using the information from the test cases to come up
with a rationale for your decisions.
Should year start at 0 or at 1? Look at a scenario with simple val-
ues: an initial balance of $100 and an interest rate of 50 percent. After
year 1, the balance is $150, and after year 2 it is $225, or over $200. So
the investment doubled after 2 years. The loop executed two times,
incrementing year each time. Hence year must start at 0, not at 1.
year balance
0 $100
1 $150
2 $225
In other words, the balance variable denotes the balance after the end of the year. At the outset,
the balance variable contains the balance after year 0 and not after year 1.
Next, should you use a < or <= comparison in the test? If you want to settle this question
with an example, you need to find a scenario in which the final balance is exactly twice the
initial balance. This happens when the interest is 100 percent. The loop executes once. Now
year is 1, and balance is exactly equal to 2 * INITIAL_BALANCE. Has the investment doubled after
one year? It has. Therefore, the loop should not execute again. If the test condition is balance
< TARGET, the loop stops, as it should. If the test condition had been balance <= TARGET, the loop
would have executed once more.
In other words, you keep adding interest while the balance has not yet doubled.
an off-by-one error is
a common error
when programming
loops. think through
simple test cases
to avoid this type
of error.
according to legend,
the first bug was
found in the Mark II, a huge electrome-
chanical computer at harvard Univer-
sity. It really was caused by a bug—a
moth was trapped in a relay switch.
actually, from the note that the
operator left in the log book next to
the moth (see the photo), it appears as
if the term “bug” had already been in
active use at the time.
The First Bug
the pioneering computer scientist
Maurice Wilkes wrote, “somehow, at
the Moore school and afterwards, one
had always assumed there would be
no particular difficulty in getting pro-
grams right. I can remember the exact
instant in time at which it dawned on
me that a great part of my future life
would be spent finding mistakes in my
own programs.”
Random Fact 4.1 the First Bug
4.2 problem solving: hand-tracing
In Programming Tip 3.6, you learned about the method of hand-tracing. When you
hand-trace code or pseudocode, you write the names of the variables on a sheet of
paper, mentally execute each step of the code and update the variables.
It is best to have the code written or printed on a sheet of paper. Use a marker,
such as a paper clip, to mark the current line. Whenever a variable changes, cross out
the old value and write the new value below. When a program produces output, also
write down the output in another column.
Consider this example. What value is displayed?
int n = 1729;
int sum = 0;
while (n > 0)
{
int digit = n % 10;
sum = sum + digit;
n = n / 10;
}
cout << sum << endl;
There are three variables: n, sum, and digit.
The first two variables are initialized with 1729 and 0 before the loop is entered.
int n = 1729;
int sum = 0;
while (n > 0)
{
int digit = n % 10;
sum = sum + digit;
n = n / 10;
}
cout << sum << endl;
Because n is greater than zero, enter the loop. The variable digit is set to 9 (the remain-
der of dividing 1729 by 10). The variable sum is set to 0 + 9 = 9.
int n = 1729;
int sum = 0;
while (n > 0)
{
int digit = n % 10;
sum = sum + digit;
n = n / 10;
}
cout << sum << endl;
hand-tracing is a
simulation of code
execution in which
you step through
instructions and
track the values of
the variables.
n sum digit
1729 0
n sum digit
1729 0
9 9
cfe2_ch04_p131_192.indd 138 10/28/10 8:13 PM
4.2 problem solving: hand-tracing 139
4.2 problem solving: hand-tracing
In Programming Tip 3.6, you learned about the method of hand-tracing. When you
hand-trace code or pseudocode, you write the names of the variables on a sheet of
paper, mentally execute each step of the code and update the variables.
It is best to have the code written or printed on a sheet of paper. Use a marker,
such as a paper clip, to mark the current line. Whenever a variable changes, cross out
the old value and write the new value below. When a program produces output, also
write down the output in another column.
Consider this example. What value is displayed?
int n = 1729;
int sum = 0;
while (n > 0)
{
int digit = n % 10;
sum = sum + digit;
n = n / 10;
}
cout << sum << endl;
There are three variables: n, sum, and digit.
n sum digit
The first two variables are initialized with 1729 and 0 before the loop is entered.
int n = 1729;
int sum = 0;
while (n > 0)
{
int digit = n % 10;
sum = sum + digit;
n = n / 10;
}
cout << sum << endl;
Because n is greater than zero, enter the loop. The variable digit is set to 9 (the remain-
der of dividing 1729 by 10). The variable sum is set to 0 + 9 = 9.
int n = 1729;
int sum = 0;
while (n > 0)
{
int digit = n % 10;
sum = sum + digit;
n = n / 10;
}
cout << sum << endl;
hand-tracing is a
simulation of code
execution in which
you step through
instructions and
track the values of
the variables.
n sum digit
1729 0
n sum digit
1729 0
9 9
cfe2_ch04_p131_192.indd 139 10/28/10 8:13 PM
140 Chapter 4 Loops
Finally, n becomes 172. (Recall that the remainder in the division 1729 / 10 is dis-
carded because both argu ments are integers.)
Cross out the old values and write the new ones under the old ones.
int n = 1729;
int sum = 0;
while (n > 0)
{
int digit = n % 10;
sum = sum + digit;
n = n / 10;
}
cout << sum << endl;
Now check the loop condition again.
int n = 1729;
int sum = 0;
while (n > 0)
{
int digit = n % 10;
sum = sum + digit;
n = n / 10;
}
cout << sum << endl;
Because n is still greater than zero, repeat
the loop. Now digit becomes 2, sum is set
to 9 + 2 = 11, and n is set to 17.
Repeat the loop once again, setting digit
to 7, sum to 11 + 7 = 18, and n to 1.
Enter the loop for one last time. Now
digit is set to 1, sum to 19, and n becomes
zero.
n sum digit
1729 0
172 9 9
n sum digit
1729 0
172 9 9
17 11 2
n sum digit
1729 0
172 9 9
17 11 2
1 18 7
n sum digit
1729 0
172 9 9
17 11 2
1 18 7
0 19 1
int n = 1729;
int sum = 0;
while (n > 0)
{
int digit = n % 10;
sum = sum + digit;
n = n / 10;
}
cout << sum << endl;
The condition n > 0 is now false. Continue with the statement after the loop.
int n = 1729;
int sum = 0;
while (n > 0)
{
int digit = n % 10;
sum = sum + digit;
n = n / 10;
}
cout << sum << endl;
This statement is an output statement. The value that is output is the value of sum,
which is 19.
Of course, you can get the same answer by just running the code. However, hand-
tracing can give you an insight that you would not get if you simply ran the code.
Consider again what happens in each itera tion:
• We extract the last digit of n.
• We add that digit to sum.
• We strip the digit off n.
In other words, the loop forms the sum of the digits in n. You now know what the
loop does for any value of n, not just the one in the example. (Why would anyone
want to form the sum of the digits? Operations of this kind are useful for checking the
validity of credit card numbers and other forms of ID numbers—see Exercise P4.5.)
Hand-tracing does not just help you understand code that works correctly. It is
a powerful technique for finding errors in your code. When a program behaves in a
way that you don’t expect, get out a sheet of paper and track the values of the vari-
ables as you mentally step through the code.
You don’t need a working program to do hand-tracing. You can hand-trace
pseudocode. In fact, it is an excellent idea to hand-trace your pseudocode before you
go to the trouble of translating it into actual code, to confirm that it works correctly.
6. Hand-trace the following code, showing the value of n and the output.
int n = 5;
while (n >= 0)
{
n–;
cout << n << endl;
}
n sum digit output
1729 0
172 9 9
17 11 2
1 18 7
0 19 1 19
hand-tracing can
help you understand
how an unfamiliar
algorithm works.
hand-tracing can
show errors in code
or pseudocode.
S e l f c h e c k
cfe2_ch04_p131_192.indd 140 10/28/10 8:13 PM
4.2 problem solving: hand-tracing 141
int n = 1729;
int sum = 0;
while (n > 0)
Because n equals zero,
this condition is not true.
{
int digit = n % 10;
sum = sum + digit;
n = n / 10;
}
cout << sum << endl;
The condition n > 0 is now false. Continue with the statement after the loop.
int n = 1729;
int sum = 0;
while (n > 0)
{
int digit = n % 10;
sum = sum + digit;
n = n / 10;
}
cout << sum << endl;
This statement is an output statement. The value that is output is the value of sum,
which is 19.
Of course, you can get the same answer by just running the code. However, hand-
tracing can give you an insight that you would not get if you simply ran the code.
Consider again what happens in each itera tion:
• We extract the last digit of n.
• We add that digit to sum.
• We strip the digit off n.
In other words, the loop forms the sum of the digits in n. You now know what the
loop does for any value of n, not just the one in the example. (Why would anyone
want to form the sum of the digits? Operations of this kind are useful for checking the
validity of credit card numbers and other forms of ID numbers—see Exercise P4.5.)
Hand-tracing does not just help you understand code that works correctly. It is
a powerful technique for finding errors in your code. When a program behaves in a
way that you don’t expect, get out a sheet of paper and track the values of the vari-
ables as you mentally step through the code.
You don’t need a working program to do hand-tracing. You can hand-trace
pseudocode. In fact, it is an excellent idea to hand-trace your pseudocode before you
go to the trouble of translating it into actual code, to confirm that it works correctly.
6. Hand-trace the following code, showing the value of n and the output.
int n = 5;
while (n >= 0)
{
n–;
cout << n << endl;
}
n sum digit output
1729 0
172 9 9
17 11 2
1 18 7
0 19 1 19
hand-tracing can
help you understand
how an unfamiliar
algorithm works.
hand-tracing can
show errors in code
or pseudocode.
S e l f c h e c k
cfe2_ch04_p131_192.indd 141 10/28/10 8:13 PM
142 Chapter 4 Loops
7. Hand-trace the following code, showing the value of n and the output. What
potential error do you notice?
int n = 1;
while (n <= 3)
{
cout << n << ", ";
n++;
}
8. Hand-trace the following code, assuming that a is 2 and n is 4. Then explain what
the code does for arbitrary values of a and n.
int r = 1;
int i = 1;
while (i <= n)
{
r = r * a;
i++;
}
9. Trace the following code. What error do you observe?
int n = 1;
while (n != 50)
{
cout << n << endl;
n = n + 10;
}
10. The following pseudocode is intended to count the number of digits in the
number n:
count = 1
temp = n
while (temp > 10)
Increment count.
Divide temp by 10.
Trace the pseudocode for n = 123 and n = 100. What error do you find?
practice it Now you can try these exercises at the end of the chapter: R4.1, R4.5.
4.3 the for Loop
It often happens that you want to execute a sequence of statements a given number of
times. You can use a while loop that is controlled by a counter, as in the following
example:
counter = 1; // Initialize the counter
while (counter <= 10) // Check the counter
{
cout << counter << endl;
counter++; // Update the counter
}
Because this loop type is so common, there is a special form for it, called the for loop
(see Syntax 4.2).
the for loop is
used when a
value runs from a
starting point to an
ending point with a
constant increment
or decrement.
cfe2_ch04_p131_192.indd 142 10/28/10 8:13 PM
4.3 the for Loop 143
for (counter = 1; counter <= 10; counter++)
{
cout << counter << endl;
}
Some people call this loop count-controlled. In contrast,
the while loop of the preceding section can be called an
event-controlled loop because it executes until an event
occurs (for example, when the balance reaches the target).
Another commonly-used term for a count-controlled
loop is definite. You know from the outset that the loop
body will be executed a definite number of times––ten
times in our example. In contrast, you do not know how
many iterations it takes to accumulate a target balance.
Such a loop is called indefinite.
The for loop neatly groups the initialization, condi-
tion, and update expressions together. However, it is
important to realize that these expressions are not exe-
cuted together (see Figure 3).
• The initialization is executed once, before the loop is entered. 1
• The condition is checked before each iteration. 2 5
• The update is executed after each iteration. 4
You can visualize the
for loop as an orderly
sequence of steps.
figure 3
execution of a
for Loop
for (counter = 1; counter <= 10; counter++)
{
cout << counter << endl;
}
Initialize counter1
for (counter = 1; counter <= 10; counter++)
{
cout << counter << endl;
}
Check condition2
for (counter = 1; counter <= 10; counter++)
{
cout << counter << endl;
}
Execute loop body3
for (counter = 1; counter <= 10; counter++)
{
cout << counter << endl;
}
Update counter4
for (counter = 1; counter <= 10; counter++)
{
cout << counter << endl;
}
Check condition again5
counter = 1
counter = 1
counter = 1
counter = 2
counter = 2
cfe2_ch04_p131_192.indd 143 10/28/10 8:13 PM
144 Chapter 4 Loops
syntax 4.2 for statement
for (int i = 5; i <= 10; i++)
{
sum = sum + i;
}
This loop executes 6 times.
See page 147.
These three
expressions should be related.
See page 147.
This initialization
happens once
before the loop starts.
The loop is
executed while
this condition is true.
This update is
executed after
each iteration.
The variable i is
defined only in this for loop.
See page 144.
A for loop can count down instead of up:
for (counter = 10; counter >= 0; counter–) …
The increment or decrement need not be in steps of 1:
for (counter = 0; counter <= 10; counter = counter + 2) ...
See Table 2 on page 146 for additional variations.
So far, we assumed that the counter variable had already been defined before the for
loop. Alternatively, you can define a variable in the loop initialization. Such a variable
is defined only in the loop:
for (int counter = 1; counter <= 10; counter++)
{
...
} // counter no longer defined here
Here is a typical use of the for loop. We want to print the balance of our savings
account over a period of years, as shown in this table:
The for loop pattern applies because the variable year starts at 1 and then moves in
constant increments until it reaches the target:
Year Balance
1 10500.00
2 11025.00
3 11576.25
4 12155.06
5 12762.82
cfe2_ch04_p131_192.indd 144 10/28/10 8:13 PM
4.3 the for Loop 145
for (int year = 1; year <= nyears; year++)
{
Update balance.
Print year and balance.
}
Here is the complete program. Figure 4 shows the
corresponding flowchart.
figure 4 Flowchart of a for Loop
ch04/invtable.cpp
1 #include
2 #include
3
4 using namespace std;
5
6 int main()
7 {
8 const double RATE = 5;
9 const double INITIAL_BALANCE = 10000;
10 double balance = INITIAL_BALANCE;
11 int nyears;
12 cout << "Enter number of years: ";
13 cin >> nyears;
14
15 cout << fixed << setprecision(2);
16 for (int year = 1; year <= nyears; year++)
17 {
18 balance = balance * (1 + RATE / 100);
19 cout << setw(4) << year << setw(10) << balance << endl;
20 }
21
22 return 0;
23 }
True
False
year++
year ≤ nyears ?
year = 1
Update balance;
Print year and
balance
cfe2_ch04_p131_192.indd 145 10/28/10 8:13 PM
146 Chapter 4 Loops
program run
Enter number of years: 10
1 10500.00
2 11025.00
3 11576.25
4 12155.06
5 12762.82
6 13400.96
7 14071.00
8 14774.55
9 15513.28
10 16288.95
11. Write the for loop of the invtable.cpp program as a while loop.
table 2 for Loop examples
Loop Values of i Comment
for (i = 0; i <= 5; i++) 0 1 2 3 4 5 Note that the loop is executed 6 times.
(See Programming Tip 4.3 on page 147.)
for (i = 5; i >= 0; i–) 5 4 3 2 1 0 Use i– for decreasing values.
for (i = 0; i < 9; i = i + 2) 0 2 4 6 8 Use i = i + 2 for a step size of 2.
for (i = 0; i != 9; i = i + 2) 0 2 4 6 8 10 12 14 ...
(infinite loop)
You can use < or <= instead of != to avoid
this problem.
for (i = 1; i <= 20; i = i * 2) 1 2 4 8 16 You can specify any rule for modifying i,
such as doubling it in every step.
for (i = 0; i < str.length(); i++) 0 1 2 … until the last valid
index of the string str
In the loop body, use the expression
str.substr(i, 1) to get a string containing
the ith character.
12. How many numbers does this loop print?
for (int n = 10; n >= 0; n–)
{
cout << n << endl;
}
13. Write a for loop that prints all even numbers between 10 and 20 (inclusive).
14. Write a for loop that computes the sum of the integers from 1 to n.
15. How would you modify the for loop of the invtable.cpp program to print all bal-
ances until the invest ment has doubled?
practice it Now you can try these exercises at the end of the chapter: R4.2, R4.7, P4.12.
S e l f c h e c k
use for loops for Their intended purpose only
A for loop is an idiom for a loop of a particular form. A value runs from the start to the end,
with a constant incre ment or decrement.
The compiler won’t check whether the initialization, condition, and update expressions are
related. For example, the following loop is legal:
// Confusing—unrelated expressions
for (cout << "Inputs: "; cin >> x; sum = sum + x)
{
count++;
}
However, programmers reading such a for loop will be confused because it does not match
their expectations. Use a while loop for iterations that do not follow the for idiom.
choose loop bounds That match your Task
Suppose you want to print line numbers that go from 1 to 10. Of course, you will want to use
a loop
for (int i = 1; i <= 10; i++)
The values for i are bounded by the relation 1 ≤ i ≤ 10. Because there are ≤ on both bounds, the
bounds are called symmetric.
When traversing the characters in a string, it is more natural to use the bounds
for (int i = 0; i < str.length(); i++)
In this loop, i traverses all valid positions in the string. You can access the ith character as str.
substr(i, 1). The val ues for i are bounded by 0 ≤ i < str.length(), with a ≤ to the left and a <
to the right. That is appropriate, because str.length() is not a valid position. Such bounds are
called asymmetric.
In this case, it is not a good idea to use symmetric bounds:
for (int i = 0; i <= str.length() - 1; i++) // Use < instead
The asymmetric form is easier to understand.
count iterations
Finding the correct lower and upper bounds for an iteration can be confusing. Should you
start at 0 or at 1? Should you use <= or < in the termination condition?
Counting the number of iterations is a very useful device for better understanding a loop.
Counting is easier for loops with asymmetric bounds. The loop
for (int i = a; i < b; i++)
is executed b - a times. For example, the loop
for (int i = 0; i < 10; i++)
runs ten times, with values 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
The loop with symmetric bounds,
for (int i = a; i <= b; i++)
is executed b - a + 1 times. That “+1” is the source of many programming errors.
programming tip 4.1
programming tip 4.2
programming tip 4.3
cfe2_ch04_p131_192.indd 146 10/28/10 8:13 PM
4.3 the for Loop 147
use for loops for Their intended purpose only
A for loop is an idiom for a loop of a particular form. A value runs from the start to the end,
with a constant incre ment or decrement.
The compiler won’t check whether the initialization, condition, and update expressions are
related. For example, the following loop is legal:
// Confusing—unrelated expressions
for (cout << "Inputs: "; cin >> x; sum = sum + x)
{
count++;
}
However, programmers reading such a for loop will be confused because it does not match
their expectations. Use a while loop for iterations that do not follow the for idiom.
choose loop bounds That match your Task
Suppose you want to print line numbers that go from 1 to 10. Of course, you will want to use
a loop
for (int i = 1; i <= 10; i++)
The values for i are bounded by the relation 1 ≤ i ≤ 10. Because there are ≤ on both bounds, the
bounds are called symmetric.
When traversing the characters in a string, it is more natural to use the bounds
for (int i = 0; i < str.length(); i++)
In this loop, i traverses all valid positions in the string. You can access the ith character as str.
substr(i, 1). The val ues for i are bounded by 0 ≤ i < str.length(), with a ≤ to the left and a <
to the right. That is appropriate, because str.length() is not a valid position. Such bounds are
called asymmetric.
In this case, it is not a good idea to use symmetric bounds:
for (int i = 0; i <= str.length() - 1; i++) // Use < instead
The asymmetric form is easier to understand.
count iterations
Finding the correct lower and upper bounds for an iteration can be confusing. Should you
start at 0 or at 1? Should you use <= or < in the termination condition?
Counting the number of iterations is a very useful device for better understanding a loop.
Counting is easier for loops with asymmetric bounds. The loop
for (int i = a; i < b; i++)
is executed b - a times. For example, the loop
for (int i = 0; i < 10; i++)
runs ten times, with values 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
The loop with symmetric bounds,
for (int i = a; i <= b; i++)
is executed b - a + 1 times. That “+1” is the source of many programming errors.
programming tip 4.1
programming tip 4.2
programming tip 4.3
cfe2_ch04_p131_192.indd 147 10/28/10 8:13 PM
148 Chapter 4 Loops
For example,
for (int i = 0; i <= 10; i++)
runs 11 times. Maybe that is what you want; if not, start at 1 or use < 10.
One way to visualize this “+1” error is by looking at a fence. Each section has one fence
post to the left, and there is a final post on the right of the last section. Forgetting to count the
last value is often called a “fence post error”.
4.4 the do Loop
Sometimes you want to execute the body of a loop at least once and perform the loop
test after the body is executed. The do loop serves that purpose:
do
{
statements
}
while (condition);
The body of the do loop is executed first, then the condition is tested.
Some people call such a loop a post-test loop because the condition is tested after
completing the loop body. In contrast, while and for loops are pre-test loops. In those
loop types, the condition is tested before entering the loop body.
A typical example for such a loop is input validation. Suppose you ask a user to
enter a value < 100. If the user didn’t pay attention and entered a larger value, you ask
again, until the value is correct. Of course, you cannot test the value until the user has
entered it. This is a perfect fit for the do loop (see Figure 5):
int value;
do
{
cout << "Enter a value < 100: ";
cin >> value;
}
while (value >= 100);
How many posts do you need
for a fence with four sections?
It is easy to be “off by one” with
problems such as this one.
the do loop is
appropriate when
the loop body
must be executed
at least once.
figure 5 Flowchart of a do Loop
16. Suppose that we want to check for inputs that
are at least 0 and at most 100. Modify the do loop
for this check.
17. Rewrite the input check, using a while loop.
What is the disadvantage of your solution?
18. Suppose C++ didn’t have a do loop. Could you
rewrite any do loop as a while loop?
19. Write a do loop that reads integers and computes
their sum. Stop when reading the value 0.
20. Write a do loop that reads positive integers and
computes their sum. Stop when reading the
same value twice in a row. For example, if the
input is 1 2 3 4 4, then the sum is 14 and the loop
stops.
practice it Now you can try these exercises at the end of the chapter: R4.8, R4.12, R4.13.
flowcharts for loops
In Section 3.5, you learned how to use flowcharts to visualize the flow of control in a program.
There are two types of loops that you can include in a flowchart; they correspond to a while
loop and a do loop in C++. They differ in the placement of the condition—either before or
after the loop body.
As described in Section 3.5, you want to avoid “spaghetti code” in your flowcharts. For loops,
that means that you never want to have an arrow that points inside a loop body.
True
False
value ≥ 100?
Prompt user
to enter
a value < 100
Copy the input
to value
S e l f c h e c k
programming tip 4.4
cfe2_ch04_p131_192.indd 148 10/28/10 8:13 PM
4.4 the do Loop 149
figure 5 Flowchart of a do Loop
16. Suppose that we want to check for inputs that
are at least 0 and at most 100. Modify the do loop
for this check.
17. Rewrite the input check, using a while loop.
What is the disadvantage of your solution?
18. Suppose C++ didn’t have a do loop. Could you
rewrite any do loop as a while loop?
19. Write a do loop that reads integers and computes
their sum. Stop when reading the value 0.
20. Write a do loop that reads positive integers and
computes their sum. Stop when reading the
same value twice in a row. For example, if the
input is 1 2 3 4 4, then the sum is 14 and the loop
stops.
practice it Now you can try these exercises at the end of the chapter: R4.8, R4.12, R4.13.
flowcharts for loops
In Section 3.5, you learned how to use flowcharts to visualize the flow of control in a program.
There are two types of loops that you can include in a flowchart; they correspond to a while
loop and a do loop in C++. They differ in the placement of the condition—either before or
after the loop body.
False
True
Loop body
Condition?
True
False
Loop body
Condition?
As described in Section 3.5, you want to avoid “spaghetti code” in your flowcharts. For loops,
that means that you never want to have an arrow that points inside a loop body.
True
False
value ≥ 100?
Prompt user
to enter
a value < 100
Copy the input
to value
S e l f c h e c k
programming tip 4.4
cfe2_ch04_p131_192.indd 149 10/28/10 8:13 PM
150 Chapter 4 Loops
4.5 processing Input
In this section, you will learn how to read and pro-
cess a sequence of input values.
Whenever you read a sequence of inputs, you
need to have some method of indicating the end
of the sequence. Sometimes you are lucky and no
input value can be zero. Then you can prompt the
user to keep entering numbers, or 0 to finish the
sequence. If zero is allowed but negative numbers
are not, you can use –1 to indicate termination. A
value that serves as a signal for termination is called
a sentinel.
Let’s put this technique to work in a program
that computes the average of a set of salary values.
In our sample program, we will use –1 as a sentinel.
An employee would surely not work for a negative
salary, but there may be volunteers who work for
free.
Inside the loop, we read an input. If the input is
not –1, we process it. In order to compute the aver-
age, we need the total sum of all salaries, and the
number of inputs.
while (...)
{
cin >> salary;
if (salary != -1)
{
sum = sum + salary;
count++;
}
}
We stay in the loop while the sentinel value is not detected.
while (salary != -1)
{
…
}
There is just one problem: When the loop is entered for the first time, no data value
has been read. Be sure to initialize salary with some value other than the sentinel:
double salary = 0; // Any value other than –1 will do
Alternatively, use a do loop
do
{
…
}
while (salary != -1)
In the military, a sentinel guards
a border or passage. In computer
science, a sentinel value denotes
the end of an input sequence or the
border between input sequences.
a sentinel value
denotes the end of a
data set, but it is not
part of the data.
cfe2_ch04_p131_192.indd 150 10/28/10 8:13 PM
4.5 processing Input 151
The following program reads inputs until the user enters the sentinel, and then com-
putes and prints the average.
ch04/sentinel.cpp
1 #include
2
3 using namespace std;
4
5 int main()
6 {
7 double sum = 0;
8 int count = 0;
9 double salary = 0;
10 cout << "Enter salaries, -1 to finish: ";
11 while (salary != -1)
12 {
13 cin >> salary;
14 if (salary != -1)
15 {
16 sum = sum + salary;
17 count++;
18 }
19 }
20 if (count > 0)
21 {
22 double average = sum / count;
23 cout << "Average salary: " << average << endl;
24 }
25 else
26 {
27 cout << "No data" << endl;
28 }
29 return 0;
30 }
program run
Enter salaries, -1 to finish: 10 10 40 -1
Average salary: 20
Numeric sentinels only work if there is some restriction on the input. In many cases,
though, there isn’t. Suppose you want to compute the average of a data set that may
contain 0 or negative values. Then you cannot use 0 or –1 to indicate the end of the
input.
In such a situation, you can read input data until input fails. As you have seen in
Section 3.8, the condi tion
cin.fail()
is true if the preceding input has failed. For example, suppose that the input was read
with these state ments:
double value;
cin >> value;
If the user enters a value that is not a number (such as Q), then the input fails.
cfe2_ch04_p131_192.indd 151 10/28/10 8:13 PM
152 Chapter 4 Loops
We now encounter an additional complexity. You only know that input failed after
you have entered the loop and attempted to read it. To remember the failure, use a
Boolean variable:
cout << "Enter values, Q to quit: ";
bool more = true;
while (more)
{
cin >> value;
if (cin.fail())
{
more = false;
}
else
{
Process value.
}
}
Some programmers dislike the introduction of a Boolean variable to control a loop.
Special Topic 4.2 on page 153 shows an alternative mechanism for leaving a loop. How-
ever, when reading input, there is an easier way. The expression
cin >> value
can be used in a condition. It evaluates to true if cin has not failed after reading value.
Therefore, you can read and process a set of inputs with the following loop:
cout << "Enter values, Q to quit: ";
while (cin >> value)
{
Process value.
}
This loop is suitable for processing a single sequence of inputs. You will learn more
about reading inputs in Chapter 8.
21. What does the sentinel.cpp program print when the user immediately types –1
when prompted for a value?
22. Why does the sentinel.cpp program have two checks of the form
salary != -1
23. What would happen if the definition of the salary variable in sentinel.cpp was
changed to
double salary = -1;
24. We prompt the user “Enter values, Q to quit.” What happens when the user
enters a different letter?
25. What is wrong with the following loop for reading a sequence of values?
cout << "Enter values, Q to quit: ";
while (!cin.fail())
{
double value;
cin >> value;
sum = sum + value;
count++;
}
You can use a
Boolean variable to
control a loop. set
the variable to true
before entering the
loop, then set it to
false to leave
the loop.
S e l f c h e c k
practice it Now you can try these exercises at the end of the chapter: R4.10, P4.6, P4.7.
clearing the failure State
When an input operation has failed, all further input operations also fail. If you want to read
two number sequences and use a letter as a sentinel, you need to clear the failure state after
reading the first sentinel. Call the clear function:
cout << "Enter values, Q to quit.\n";
while (cin >> values)
{
Process input.
}
cin.clear();
Suppose the user has entered 30 10 5 Q. The input of Q has caused the failure. Because only suc-
cessfully processed characters are removed from the input, the Q character is still present. Read
it into a dummy variable:
string sentinel;
cin >> sentinel;
Now you can go on and read more inputs.
The loop-and-a-half problem and the break Statement
Some programmers dislike loops that are controlled by a Boolean variable, such as:
bool more = true;
while (more)
{
cin >> value;
if (cin.fail())
{
more = false;
}
else
{
Process value.
}
}
The actual test for loop termination is in the middle of the loop, not at the top. This is called a
loop and a half because one must go halfway into the loop before knowing whether one needs
to terminate.
As an alternative, you can use the break reserved word.
while (true)
{
cin >> value;
if (cin.fail()) { break; }
Process value.
}
The break statement breaks out of the enclosing loop, independent of the loop condition.
In the loop-and-a-half case, break statements can be beneficial. But it is difficult to lay down
clear rules as to when they are safe and when they should be avoided. We do not use the break
statement in this book.
special topic 4.1
special topic 4.2
cfe2_ch04_p131_192.indd 152 10/28/10 8:13 PM
4.5 processing Input 153
practice it Now you can try these exercises at the end of the chapter: R4.10, P4.6, P4.7.
clearing the failure State
When an input operation has failed, all further input operations also fail. If you want to read
two number sequences and use a letter as a sentinel, you need to clear the failure state after
reading the first sentinel. Call the clear function:
cout << "Enter values, Q to quit.\n";
while (cin >> values)
{
Process input.
}
cin.clear();
Suppose the user has entered 30 10 5 Q. The input of Q has caused the failure. Because only suc-
cessfully processed characters are removed from the input, the Q character is still present. Read
it into a dummy variable:
string sentinel;
cin >> sentinel;
Now you can go on and read more inputs.
The loop-and-a-half problem and the break Statement
Some programmers dislike loops that are controlled by a Boolean variable, such as:
bool more = true;
while (more)
{
cin >> value;
if (cin.fail())
{
more = false;
}
else
{
Process value.
}
}
The actual test for loop termination is in the middle of the loop, not at the top. This is called a
loop and a half because one must go halfway into the loop before knowing whether one needs
to terminate.
As an alternative, you can use the break reserved word.
while (true)
{
cin >> value;
if (cin.fail()) { break; }
Process value.
}
The break statement breaks out of the enclosing loop, independent of the loop condition.
In the loop-and-a-half case, break statements can be beneficial. But it is difficult to lay down
clear rules as to when they are safe and when they should be avoided. We do not use the break
statement in this book.
special topic 4.1
special topic 4.2
cfe2_ch04_p131_192.indd 153 10/28/10 8:13 PM
154 Chapter 4 Loops
redirection of input and output
Consider the sentinel program that computes the average value of an input sequence. If you
use such a program, then it is quite likely that you already have the values in a file, and it seems
a shame that you have to type them all in again. The command line interface of your operating
system provides a way to link a file to the input of a pro gram, as if all the characters in the file
had actually been typed by a user. If you type
sentinel < numbers.txt
the program is executed. Its input instructions no longer expect
input from the keyboard. All input commands get their input
from the file numbers.txt. This process is called input redirection.
Input redirection is an excellent tool for testing programs.
When you develop a program and fix its bugs, it is bor ing to keep
entering the same input every time you run the program. Spend a
few minutes putting the inputs into a file, and use redirection.
You can also redirect output. In this program, that is not terribly useful. If you run
sentinel < numbers.txt > output.txt
the file output.txt contains the input prompts and the output, such as
Enter a value, -1 to finish: Enter a value, -1 to finish:
Enter a value, -1 to finish: Enter a value, -1 to finish:
Average: 15
However, redirecting output is obviously useful for programs that produce lots of output.
You can print the file con taining the output or edit it before you turn it in for grading.
4.6 problem solving: storyboards
When you design a program that interacts with a user, you need to make a plan for
that interaction. What information does the user provide, and in which order? What
information will your program display, and in which format? What should happen
when there is an error? When does the program quit?
This planning is similar to the development of a movie or a computer game, where
storyboards are used to plan action sequences. A storyboard is made up of panels that
show a sketch of each step. Annota tions explain what is happening and note any spe-
cial situations. Storyboards are also used to develop software—see Figure 6.
Making a storyboard is very helpful when you begin designing a program. You
need to ask yourself which information you need in order to compute the answers
that the program user wants. You need to decide how to present those answers. These
are important considerations that you want to settle before you design an algorithm
for computing the answers.
Let’s look at a simple example. We want to write a program that helps users with
questions such as “How many tablespoons are in a pint?” or “How many inches are
30 centimeters?”
What information does the user provide?
• The quantity and unit to convert from
• The unit to convert to
special topic 4.3
Use input redirection to
read input from a file.
Use output redirection to
capture program output
in a file.
a storyboard consists
of annotated
sketches for each
step in an action
sequence.
Developing a
storyboard helps you
understand the
inputs and outputs
that are required for
a program.
cfe2_ch04_p131_192.indd 154 10/28/10 8:13 PM
4.6 problem solving: storyboards 155
figure 6
storyboard for the
Design of a Web
application
What if there is more than one quantity? A user may have a whole table of centimeter
values that should be converted into inches.
What if the user enters units that our program doesn’t know how to handle, such
as angstrom?
What if the user asks for impossible conversions, such as inches to gallons?
Let’s get started with a storyboard panel. It is a good idea to write the user inputs in
a different color. (Underline them if you don’t have a color pen handy.)
What unit do you want to convert from? cm
What unit do you want to convert to? in
Enter values, terminated by zero
30
30 cm = 11.81 in
100
100 cm = 39.37 in
0
What unit do you want to convert from?
Format makes clear what got converted
Allows conversion of multiple values
Converting a Sequence of Values
The storyboard shows how we deal with a potential confusion. A user who wants to
know how many inches are 30 centimeters may not read the first prompt carefully
and specify inches. But then the output is “30 in = 76.2 cm”, alerting the user to the
problem.
The storyboard also raises an issue. How is the user supposed to know that “cm”
and “in” are valid units? Would “centimeter” and “inches” also work? What happens
cfe2_ch04_p131_192.indd 155 10/28/10 8:13 PM
156 Chapter 4 Loops
when the user enters a wrong unit? Let’s make another storyboard to demonstrate
error handling.
What unit do you want to convert from? cm
What unit do you want to convert to? inches
Sorry, unknown unit.
What unit do you want to convert to? inch
Sorry, unknown unit.
What unit do you want to convert to? grrr
Handling Unknown Units (needs improvement)
To eliminate frustration, it is better to list the units that the user can supply.
From unit (in, ft, mi, mm, cm, m, km, oz, lb, g, kg, tsp, tbsp, pint, gal): cm
To unit: in
No need to list the units again
We switched to a shorter prompt to make room for all the unit names. Exercise R4.17
explores a different alternative.
There is another issue that we haven’t addressed yet. How does the user quit the
program? The first storyboard gives the impression that the program will go on
forever.
We can ask the user after seeing the sentinel that terminates an input sequence.
From unit (in, ft, mi, mm, cm, m, km, oz, lb, g, kg, tsp, tbsp, pint, gal): cm
To unit: in
Enter values, terminated by zero
30
30 cm = 11.81 in
0
More conversions (y, n)? n
(Program exits)
Sentinel triggers the prompt to exit
Exiting the Program
As you can see from this case study, a storyboard is essential for developing a work-
ing program. You need to know the flow of the user interaction in order to structure
your program.
26. Provide a storyboard panel for a program that reads a number of test scores and
prints the average score. The program only needs to process one set of scores.
Don’t worry about error handling.
S e l f c h e c k
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4.7 Common Loop algorithms 157
27. Google has a simple interface for converting units. You just type the question,
and you get the answer.
Make storyboards for an equivalent interface in a C++ program. Show the
“happy day” scenario in which all goes well, and show the handling of two kinds
of errors.
28. Consider a modification of the program in Self Check 26. Drop the lowest score
before computing the average. Provide a storyboard for the situation in which a
user only provides one score.
29. What is the problem with implementing the following storyboard in C++?
Enter scores: 90 80 90 100 80
The average is 88
Enter scores: 100 70 70 100 80
The average is 88
Enter scores: -1
(Program exits)
-1 is used as a sentinel to exit the program
Computing Multiple Averages
30. Produce a storyboard for a program that compares the growth of a $10,000
investment for a given number of years under two interest rates.
practice it Now you can try these exercises at the end of the chapter: R4.17, R4.18, R4.19.
4.7 Common Loop algorithms
In the following sections, we discuss some of the most common algorithms that are
implemented as loops. You can use them as starting points for your loop designs.
4.7.1 sum and average Value
Computing the sum of a number of inputs is a very common task. Keep a running
total: a variable to which you add each input value. Of course, the total should be ini-
tialized with 0.
double total = 0;
double input;
while (cin >> input)
{
total = total + input;
}
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158 Chapter 4 Loops
To compute an average, count how many values you have, and divide by the count.
Be sure to check that the count is not zero.
double total = 0;
int count = 0;
double input;
while (cin >> input)
{
total = total + input;
count++;
}
double average = 0;
if (count > 0) { average = total / count; }
4.7.2 Counting Matches
You often want to know how many values fulfill a particular condition. For example,
you may want to count how many spaces are in a string. Keep a counter, a variable
that is initialized with 0 and incremented whenever there is a match.
int spaces = 0;
for (int i = 0; i < str.length(); i++)
{
string ch = str.substr(i, 1);
if (ch == " ")
{
spaces++;
}
}
For example, if str is the string "My Fair Lady", spaces is incremented twice (when i is 2
and 7).
Note that the spaces variable is declared outside the loop. We want the loop to
update a single variable. The ch variable is declared inside the loop. A separate vari-
able is created for each iteration and removed at the end of each loop iteration.
This loop can also be used for scanning inputs. The following loop reads text, a
word at a time, and counts the number of words with at most three letters:
int short_words = 0;
string input;
while (cin >> input)
{
if (input.length() <= 3)
{
short_words++;
}
}
to compute an
average, keep a total
and a count of all
values.
to count values that
fulfill a condition,
check all values and
increment a counter
for each match.
In a loop that counts matches,
a counter is incremented
whenever a match is found.
4.7.3 Finding the First Match
When you count the values that fulfill a con-
dition, you need to look at all values. How-
ever, if your task is to find a match, then you
can stop as soon as the condition is fulfilled.
Here is a loop that finds the first space in a
string. Because we do not visit all elements in
the string, a while loop is a better choice than
a for loop:
bool found = false;
int position = 0;
while (!found && position < str.length())
{
string ch = str.substr(position, 1);
if (ch == " ") { found = true; }
else { position++; }
}
If a match was found, then found is true and position is the index of the first match. If
the loop did not find a match, then found remains false after the end of the loop.
In the preceding example, we searched a string for a character that matches a con-
dition. You can apply the same process for user input. Suppose you are asking a user
to enter a positive value < 100. Keep asking until the user provides a correct input:
bool valid = false;
double input;
while (!valid)
{
cout << "Please enter a positive value < 100: ";
cin >> input;
if (0 < input && input < 100) { valid = true; }
else { cout << "Invalid input." << endl; }
}
Note that the variable input is declared outside the while loop because you will want to
use the input after the loop has finished. If it had been declared inside the loop body,
you would not be able to use it outside the loop.
4.7.4 Maximum and Minimum
To compute the largest value in a sequence, keep a variable that stores the largest ele-
ment that you have encountered, and update it when you find a larger one:
double largest;
cin >> largest;
double input;
while (cin >> input)
{
if (input > largest)
{
largest = input;
}
}
This algorithm requires that there is at least one input.
When searching, you look at items until a
match is found.
If your goal is to find
a match, exit the loop
when the match
is found.
to find the largest
value, update the
largest value seen so
far whenever you see
a larger one.
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4.7 Common Loop algorithms 159
4.7.3 Finding the First Match
When you count the values that fulfill a con-
dition, you need to look at all values. How-
ever, if your task is to find a match, then you
can stop as soon as the condition is fulfilled.
Here is a loop that finds the first space in a
string. Because we do not visit all elements in
the string, a while loop is a better choice than
a for loop:
bool found = false;
int position = 0;
while (!found && position < str.length())
{
string ch = str.substr(position, 1);
if (ch == " ") { found = true; }
else { position++; }
}
If a match was found, then found is true and position is the index of the first match. If
the loop did not find a match, then found remains false after the end of the loop.
In the preceding example, we searched a string for a character that matches a con-
dition. You can apply the same process for user input. Suppose you are asking a user
to enter a positive value < 100. Keep asking until the user provides a correct input:
bool valid = false;
double input;
while (!valid)
{
cout << "Please enter a positive value < 100: ";
cin >> input;
if (0 < input && input < 100) { valid = true; }
else { cout << "Invalid input." << endl; }
}
Note that the variable input is declared outside the while loop because you will want to
use the input after the loop has finished. If it had been declared inside the loop body,
you would not be able to use it outside the loop.
4.7.4 Maximum and Minimum
To compute the largest value in a sequence, keep a variable that stores the largest ele-
ment that you have encountered, and update it when you find a larger one:
double largest;
cin >> largest;
double input;
while (cin >> input)
{
if (input > largest)
{
largest = input;
}
}
This algorithm requires that there is at least one input.
When searching, you look at items until a
match is found.
If your goal is to find
a match, exit the loop
when the match
is found.
to find the largest
value, update the
largest value seen so
far whenever you see
a larger one.
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160 Chapter 4 Loops
To find the height of the tallest bus rider,
remember the largest value so far, and
update it whenever you see a taller one.
To compute the smallest value, simply reverse the comparison:
double smallest;
cin >> smallest;
double input;
while (cin >> input)
{
if (input < smallest)
{
smallest = input;
}
}
4.7.5 Comparing adjacent Values
When processing a sequence of values in a loop, you sometimes need to compare a
value with the value that just preceded it. For example, suppose you want to check
whether a sequence of inputs contains adjacent duplicates such as 1 7 2 9 9 4 9.
Now you face a challenge. Consider the typical loop for reading a value:
double input;
while (cin >> input)
{
// Now input contains the current input
…
}
How can you compare the current input with the preceding one? At any time, input
contains the current input, overwriting the previous one.
The answer is to store the previous input, like this:
double input;
double previous;
while (cin >> input)
{
if (input == previous) { cout << "Duplicate input" << endl; }
previous = input;
}
to compare adjacent
inputs, store the
preceding input in
a variable.
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4.7 Common Loop algorithms 161
When comparing adjacent values,
store the previous value in a variable.
One problem remains. When the loop is entered for the first time, previous has not yet
been set. You can solve this problem with an initial input operation outside the loop:
double input;
double previous;
cin >> previous;
while (cin >> input)
{
if (input == previous) { cout << "Duplicate input" << endl; }
previous = input;
}
31. What total is computed when no user input is provided in the algorithm in
Section 4.7.1?
32. How do you compute the total of all positive inputs?
33. What is the value of position when no match is found in the algorithm in
Section 4.7.3?
34. What is wrong with the following loop for finding the position of the first space
in a string?
bool found = false;
for (int position = 0; !found && position < str.length(); position++)
{
string ch = str.substr(position, 1);
if (ch == " ") { found = true; }
}
35. How do you find the last space in a string?
36. What is wrong with the following loop for finding the smallest input value?
double smallest = 0;
double input;
while (cin >> input)
{
if (input < smallest)
{
smallest = input;
}
}
37. What happens with the algorithm in Section 4.7.5 when no input is provided
at all?
practice it Now you can try these exercises at the end of the chapter: P4.8, P4.13, P4.14.
S e l f c h e c k
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162 Chapter 4 Loops
Step 1 Decide what work must be done inside the loop.
Every loop needs to do some kind of repetitive work, such as
• Reading another item.
• Updating a value (such as a bank balance or total).
• Incrementing a counter.
If you can’t figure out what needs to go inside the loop, start by writing down the steps that
you would take if you solved the problem by hand. For example, with the temperature reading
problem, you might write
Read first value.
Read second value.
If second value is higher than the first, set highest temperature to that value, highest month to 2.
Read next value.
If value is higher than the first and second, set highest temperature to that value, highest month to 3.
Read next value.
If value is higher than the highest temperature seen so far, set highest temperature to that value,
highest month to 4.
. . .
Now look at these steps and reduce them to a set of uniform actions that can be placed into the
loop body. The first action is easy:
Read next value.
The next action is trickier. In our description, we used tests “higher than the first”, “higher
than the first and sec ond”, “higher than the highest temperature seen so far”. We need to settle
on one test that works for all iterations. The last formulation is the most general.
Similarly, we must find a general way of setting the highest month. We need a variable that
stores the current month, running from 1 to 12. Then we can formulate the second loop action:
If value is higher than the highest temperature, set highest temperature to that value,
highest month to current month.
h o W t o 4 . 1 writing a loop
This How To walks you through the process
of implementing a loop statement. We will
illustrate the steps with the following example
problem:
Read twelve temperature values (one for
each month), and dis play the number of the
month with the highest temperature. For
example, according to http://worldclimate.com,
the average maxi mum temperatures for Death
Valley are (in order by month):
18.2 22.6 26.4 31.1 36.6 42.2
45.7 44.5 40.2 33.1 24.2 17.6
In this case, the month with the highest tem-
perature (45.7 degrees Celsius) is July, and the
program should display 7.
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http://worldclimate.com
4.7 Common Loop algorithms 163
Step 1 Decide what work must be done inside the loop.
Every loop needs to do some kind of repetitive work, such as
• Reading another item.
• Updating a value (such as a bank balance or total).
• Incrementing a counter.
If you can’t figure out what needs to go inside the loop, start by writing down the steps that
you would take if you solved the problem by hand. For example, with the temperature reading
problem, you might write
Read first value.
Read second value.
If second value is higher than the first, set highest temperature to that value, highest month to 2.
Read next value.
If value is higher than the first and second, set highest temperature to that value, highest month to 3.
Read next value.
If value is higher than the highest temperature seen so far, set highest temperature to that value,
highest month to 4.
. . .
Now look at these steps and reduce them to a set of uniform actions that can be placed into the
loop body. The first action is easy:
Read next value.
The next action is trickier. In our description, we used tests “higher than the first”, “higher
than the first and sec ond”, “higher than the highest temperature seen so far”. We need to settle
on one test that works for all iterations. The last formulation is the most general.
Similarly, we must find a general way of setting the highest month. We need a variable that
stores the current month, running from 1 to 12. Then we can formulate the second loop action:
If value is higher than the highest temperature, set highest temperature to that value,
highest month to current month.
h o W t o 4 . 1 writing a loop
This How To walks you through the process
of implementing a loop statement. We will
illustrate the steps with the following example
problem:
Read twelve temperature values (one for
each month), and dis play the number of the
month with the highest temperature. For
example, according to http://worldclimate.com,
the average maxi mum temperatures for Death
Valley are (in order by month):
18.2 22.6 26.4 31.1 36.6 42.2
45.7 44.5 40.2 33.1 24.2 17.6
In this case, the month with the highest tem-
perature (45.7 degrees Celsius) is July, and the
program should display 7.
Altogether our loop is
Loop
Read next value.
If value is higher than the highest temperature, set highest temperature to that value,
highest month to current month.
Increment current month.
Step 2 Specify the loop condition.
What goal do you want to reach in your loop? Typical examples are:
• Has the counter reached the final value?
• Have you read the last input value?
• Has a value reached a given threshold?
In our example, we simply want the current month to reach 12.
Step 3 Determine the loop type.
We distinguish between two major loop types. A definite or count-controlled loop is executed
a definite number of times. In an indefinite or event-controlled loop, the number of iterations
is not known in advance—the loop is exe cuted until some event happens. A typical example of
the latter is a loop that reads data until a sentinel is encoun tered.
If you know in advance how many times a loop is repeated, use a for statement. For other
loops, consider the loop condition. Do you need to complete one iteration of the loop body
before you can tell when to terminate the loop? In that case, you should choose a do loop. Oth-
erwise, use a while loop.
In our example, we read 12 temperature values. Therefore, we choose a for loop.
Step 4 Set up variables for entering the loop for the first time.
List all variables that are used and updated in the loop, and determine how to initialize them.
Commonly, counters are initialized with 0 or 1, totals with 0.
In our example, the variables are
current month
highest value
highest month
We need to be careful how we set up the highest temperature value. We can’t simply set it to
0. After all, our program needs to work with temperature values from Antarctica, all of which
may be negative.
A good option is to set the highest temperature value to the first input value. Of course,
then we need to remem ber to only read in another 11 values, with the current month starting
at 2.
We also need to initialize the highest month with 1. After all, in an Australian city, we may
never find a month that is warmer than January.
Step 5 Process the result after the loop has finished.
In many cases, the desired result is simply a variable that was updated in the loop body. For
example, in our temper ature program, the result is the highest month. Sometimes, the loop
computes values that contribute to the final result. For example, suppose you are asked to
average the temperatures. Then the loop should compute the sum, not the average. After the
loop has completed, you are ready to compute the average: divide the sum by the number of
inputs.
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164 Chapter 4 Loops
Here is our complete loop:
Read first value; store as highest value.
highest month = 1
for (current month = 2; current month <= 12; current month++)
Read next value.
If value is higher than the highest value, set highest value to that value,
highest month to current month.
Step 6 Trace the loop with typical examples.
Hand-trace your loop code, as described in Section 4.2. Choose example values that are not
too complex—executing the loop 3–5 times is enough to check for the most common errors.
Pay special attention when entering the loop for the first and last time.
Sometimes, you want to make a slight modification to make tracing feasible. For example,
when hand-tracing the investment doubling problem, use an interest rate of 20 percent rather
than 5 percent. When hand-tracing the tem perature loop, use 4 data values, not 12.
Let’s say the data are 22.6 36.6 44.5 24.2. Here is the walkthrough:
current month current value highest month highest value
1 22.6
2 36.6 2 36.6
3 44.5 3 44.5
4 24.2
The trace demonstrates that highest month and highest value are properly set.
Step 7 Implement the loop in C++.
Here’s the loop for our example. Exercise P4.4 asks you to complete the program.
double highest_value;
cin >> highest_value;
int highest_month = 1;
for (int current_month = 2; current_month <= 12; current_month++)
{
double next_value;
cin >> next_value;
if (next_value > highest_value)
{
highest_value = next_value;
highest_month = current_month;
}
}
cout << highest_month << endl;
W o r k e D e x a M p L e 4 . 1 credit card processing
This Worked Example uses a loop to remove spaces from a
credit card number.
4.8 nested Loops
In Section 3.4, you saw how to nest two if statements. Similarly, complex iterations
sometimes require a nested loop: a loop inside another loop statement. When pro-
cessing tables, nested loops occur naturally. An outer loop iterates over all rows of the
table. An inner loop deals with the columns in the current row.
In this section you will see how to print a table. For simplicity, we will simply print
powers xn, as in the table at right.
Here is the pseudocode for printing the table:
Print table header.
For x from 1 to 10
Print table row.
Print endl.
How do you print a table row? You need to print a
value for each expo nent. This requires a second loop:
For n from 1 to 4
Print xn.
This loop must be placed inside the preceding loop. We say that the inner loop is
nested inside the outer loop (see Figure 7).
When the body of a
loop contains
another loop, the
loops are nested. a
typical use of nested
loops is printing a
table with rows
and columns.
x1 x2 x3 x4
1 1 1 1
2 4 8 16
3 9 27 81
… … … …
10 100 1000 10000
figure 7
Flowchart of a nested Loop
True
False
x ≤ 10 ?
x = 1
True
False n ≤ 4 ?
n = 1
n++
Print xn
x++
Print new line
This loop is nested
in the outer loop.
Available online at www.wiley.com/college/horstmann.
cfe2_ch04_p131_192.indd 164 10/28/10 8:13 PM
www.wiley.com/college/horstmann
4.8 nested Loops 165
Here is our complete loop:
Read first value; store as highest value.
highest month = 1
for (current month = 2; current month <= 12; current month++)
Read next value.
If value is higher than the highest value, set highest value to that value,
highest month to current month.
Step 6 Trace the loop with typical examples.
Hand-trace your loop code, as described in Section 4.2. Choose example values that are not
too complex—executing the loop 3–5 times is enough to check for the most common errors.
Pay special attention when entering the loop for the first and last time.
Sometimes, you want to make a slight modification to make tracing feasible. For example,
when hand-tracing the investment doubling problem, use an interest rate of 20 percent rather
than 5 percent. When hand-tracing the tem perature loop, use 4 data values, not 12.
Let’s say the data are 22.6 36.6 44.5 24.2. Here is the walkthrough:
The trace demonstrates that highest month and highest value are properly set.
Step 7 Implement the loop in C++.
Here’s the loop for our example. Exercise P4.4 asks you to complete the program.
double highest_value;
cin >> highest_value;
int highest_month = 1;
for (int current_month = 2; current_month <= 12; current_month++)
{
double next_value;
cin >> next_value;
if (next_value > highest_value)
{
highest_value = next_value;
highest_month = current_month;
}
}
cout << highest_month << endl;
W o r k e D e x a M p L e 4 . 1 credit card processing
This Worked Example uses a loop to remove spaces from a
credit card number.
4.8 nested Loops
In Section 3.4, you saw how to nest two if statements. Similarly, complex iterations
sometimes require a nested loop: a loop inside another loop statement. When pro-
cessing tables, nested loops occur naturally. An outer loop iterates over all rows of the
table. An inner loop deals with the columns in the current row.
In this section you will see how to print a table. For simplicity, we will simply print
powers xn, as in the table at right.
Here is the pseudocode for printing the table:
Print table header.
For x from 1 to 10
Print table row.
Print endl.
How do you print a table row? You need to print a
value for each expo nent. This requires a second loop:
For n from 1 to 4
Print xn.
This loop must be placed inside the preceding loop. We say that the inner loop is
nested inside the outer loop (see Figure 7).
When the body of a
loop contains
another loop, the
loops are nested. a
typical use of nested
loops is printing a
table with rows
and columns.
x1 x2 x3 x4
1 1 1 1
2 4 8 16
3 9 27 81
… … … …
10 100 1000 10000
figure 7
Flowchart of a nested Loop
True
False
x ≤ 10 ?
x = 1
True
False n ≤ 4 ?
n = 1
n++
Print xn
x++
Print new line
This loop is nested
in the outer loop.
cfe2_ch04_p131_192.indd 165 10/28/10 8:13 PM
166 Chapter 4 Loops
There are 10 rows in the outer loop. For each x, the program prints four columns
in the inner loop. Thus, a total of 10 × 4 = 40 values are printed.
Following is the complete program. Note that we also use loops to print the table
header. However, those loops are not nested.
ch04/powtable.cpp
1 #include
2 #include
3 #include
4
5 using namespace std;
6
7 int main()
8 {
9 const int NMAX = 4;
10 const double XMAX = 10;
11
12 // Print table header
13
14 for (int n = 1; n <= NMAX; n++)
15 {
16 cout << setw(10) << n;
17 }
18 cout << endl;
19 for (int n = 1; n <= NMAX; n++)
20 {
21 cout << setw(10) << "x ";
22 }
23 cout << endl << endl;
24
25 // Print table body
26
27 for (double x = 1; x <= XMAX; x++)
28 {
29 // Print table row
30
31 for (int n = 1; n <= NMAX; n++)
32 {
33 cout << setw(10) << pow(x, n);
34 }
35 cout << endl;
36 }
37
38 return 0;
39 }
The hour and minute displays in a digital clock are
an exam ple of nested loops. The hours loop 12 times,
and for each hour, the minutes loop 60 times.
cfe2_ch04_p131_192.indd 166 10/28/10 8:13 PM
4.8 nested Loops 167
program run
1 2 3 4
x x x x
1 1 1 1
2 4 8 16
3 9 27 81
4 16 64 256
5 25 125 625
6 36 216 1296
7 49 343 2401
8 64 512 4096
9 81 729 6561
10 100 1000 10000
table 3 nested Loop examples
nested Loops output explanation
for (i = 1; i <= 3; i++)
{
for (j = 1; j <= 4; j++) { cout << "*"; }
cout << endl;
}
****
****
****
Prints 3 rows of 4
asterisks each.
for (i = 1; i <= 4; i++)
{
for (j = 1; j <= 3; j++) { cout << "*"; }
cout << endl;
}
***
***
***
***
Prints 4 rows of 3
asterisks each.
for (i = 1; i <= 4; i++)
{
for (j = 1; j <= i; j++) { cout << "*"; }
cout << endl;
}
*
**
***
****
Prints 4 rows of
lengths 1, 2, 3, and 4.
for (i = 1; i <= 3; i++)
{
for (j = 1; j <= 5; j++)
{
if (j % 2 == 0) { cout << "*"; }
else { cout << "-"; }
}
cout << endl;
}
-*-*-
-*-*-
-*-*-
Prints asterisks in
even columns,
dashes in odd
columns.
for (i = 1; i <= 3; i++)
{
for (j = 1; j <= 5; j++)
{
if ((i + j) % 2 == 0) { cout << "*"; }
else { cout << " "; }
}
cout << endl;
}
* * *
* *
* * *
Prints a
checkerboard
pattern.
cfe2_ch04_p131_192.indd 167 10/28/10 8:13 PM
168 Chapter 4 Loops
38. Why is there a statement cout << endl in the outer loop but not in the inner loop?
39. How would you change the program so that all powers from x0 to x5 are
displayed?
40. If you make the change in Self Check 39, how many values are displayed?
41. What do the following nested loops display?
for (int i = 0; i < 3; i++)
{
for (int j = 0; j < 4; j++)
{
cout << i + j;
}
cout << endl;
}
42. Write nested loops that make the following pattern of brackets:
[][][][]
[][][][]
[][][][]
practice it Now you can try these exercises at the end of the chapter: R4.23, P4.21, P4.22.
4.9 random numbers and simulations
A simulation program uses the computer to simulate an activity in the real world (or
an imaginary one). Simulations are commonly used for predicting climate change,
analyzing traffic, picking stocks, and many other applications in science and busi-
ness. In the following sections, you will learn how to imple ment simulations that
model phenomena with a degree of randomness.
4.9.1 Generating random numbers
Many events in the real world are difficult to predict with absolute precision, yet we
can sometimes know the average behavior quite well. For example, a store may know
from experience that a customer arrives every five minutes. Of course, that is an aver-
age—customers don’t arrive in five minute intervals. To accurately model customer
traffic, you want to take that random fluctuation into account. Now, how can you
run such a simulation in the computer?
The C++ library has a random number generator, which produces numbers that
appear to be com pletely random. Calling rand() yields a random integer between
0 and RAND_MAX (which is an implementa tion-dependent constant, typically, but not
always, the largest valid int value). Call rand() again, and you get a different number.
The rand function is declared in the
The following program calls the rand function ten times.
ch04/random.cpp
1 #include
2 #include
3
4 using namespace std;
S e l f c h e c k
In a simulation, you
use the computer to
simulate an activity.
You can introduce
randomness by
calling the random
number generator.
cfe2_ch04_p131_192.indd 168 10/28/10 8:13 PM
4.9 random numbers and simulations 169
5
6 int main()
7 {
8 for (int i = 1; i <= 10; i++)
9 {
10 int r = rand();
11 cout << r << endl;
12 }
13 return 0;
14 }
program run
1804289383
846930886
1681692777
1714636915
1957747793
424238335
719885386
1649760492
596516649
118964142
Actually, the numbers are not completely random. They are drawn from sequences
of numbers that don’t repeat for a long time. These sequences are actually computed
from fairly simple formulas; they just behave like random numbers. For that reason,
they are often called pseudorandom numbers.
Try running the program again. You will get the exact same output! This confirms
that the random numbers are generated by formulas. However, when running simu-
lations, you don’t always want to get the same results. To overcome this problem,
specify a seed for the random number sequence. Every time you use a new seed, the
random number generator starts generating a new sequence. The seed is set with the
srand function. A simple value to use as a seed is the current time:
srand(time(0));
Simply make this call once in your program, before generating any random numbers.
Then the random numbers will be different in every program run. Also include the
4.9.2 simulating Die tosses
In actual applications, you need to transform the output from
the random num ber generator into different ranges. For exam-
ple, to simulate the throw of a die, you need random numbers
between 1 and 6.
Here is the general recipe for computing random integers
between two bounds a and b. As you know from Program-
ming Tip 4.3 on page 147, there are b – a + 1 values between a
and b, including the bounds themselves. First compute
rand() % (b – a + 1) to obtain a random value between 0 and
b – a, then add a, yielding a random value between a and b:
int r = rand() % (b – a + 1) + a;
cfe2_ch04_p131_192.indd 169 10/28/10 8:13 PM
170 Chapter 4 Loops
Here is a program that simulates the throw of a pair of dice.
ch04/dice.cpp
1 #include
2 #include
3 #include
4 #include
5
6 using namespace std;
7
8 int main()
9 {
10 srand(time(0));
11
12 for (int i = 1; i <= 10; i++)
13 {
14 int d1 = rand() % 6 + 1;
15 int d2 = rand() % 6 + 1;
16 cout << d1 << " " << d2 << endl;
17 }
18 cout << endl;
19 return 0;
20 }
program run
5 1
2 1
1 2
5 1
1 2
6 4
4 4
6 1
6 3
5 2
4.9.3 the Monte Carlo Method
The Monte Carlo method is an inge-
nious method for finding approxi-
mate solutions to problems that can-
not be precisely solved. (The method
is named after the famous casino in
Monte Carlo.) Here is a typical
example: It is difficult to compute
the number p, but you can approxi-
mate it quite well with the following
simulation.
Simulate shooting a dart into a
square surrounding a circle of radius
1. That is easy: generate random x
and y coordinates between –1 and 1.
cfe2_ch04_p131_192.indd 170 10/28/10 8:13 PM
4.9 random numbers and simulations 171
If the generated point lies inside the circle, we count it as a hit. That is the case
when x2 + y2 ≤ 1. Because our shots are entirely random, we expect that the ratio of
hits ̸ tries is approximately equal to the ratio of the areas
of the circle and the square, that is, p ̸ 4. Therefore, our
estimate for p is 4 × hits ̸ tries. This method yields an esti-
mate for p, using nothing but simple arithmetic.
To run the Monte Carlo simulation, you have to work
a little harder with random number generation. When
you throw a die, it has to come up with one of six faces.
When throwing a dart, however, there are many possible
outcomes. You must generate a random floating-point
number.
First, generate the following value:
double r = rand() * 1.0 / RAND_MAX; // Between 0 and 1
The value r is a random floating-point value between 0 and 1. (You have to multiply
by 1.0 to ensure that one of the operands of the / operator is a floating-point number.
The division rand() / RAND_MAX would be an integer division—see Common Error 2.3.)
To generate a random value between –1 and 1, you compute:
double x = -1 + 2 * r; // Between –1 and 1
As r ranges from 0 to 1, x ranges from –1 + 2 × 0 = –1 to –1 + 2 × 1 = 1.
Here is the program that carries out the simulation.
ch04/montecarlo.cpp
1 #include
2 #include
3 #include
4 #include
5
6 using namespace std;
7
8 int main()
9 {
10 const int TRIES = 10000;
11
12 srand(time(0));
13
14 int hits = 0;
15 for (int i = 1; i <= TRIES; i++)
16 {
17 double r = rand() * 1.0 / RAND_MAX; // Between 0 and 1
18 double x = -1 + 2 * r; // Between –1 and 1
19 r = rand() * 1.0 / RAND_MAX;
20 double y = -1 + 2 * r;
21 if (x * x + y * y <= 1) { hits++; }
22 }
23 double pi_estimate = 4.0 * hits / TRIES;
24 cout << "Estimate for pi: " << pi_estimate << endl;
25 return 0;
26 }
program run
Estimate for pi: 3.1504
x
y
1–1
1
–1
cfe2_ch04_p131_192.indd 171 10/28/10 8:13 PM
172 Chapter 4 Loops
43. How do you simulate a coin toss with the rand function?
44. How do you simulate the picking of a random playing card?
45. Why does the dice.cpp file include the
46. In many games, you throw a pair of dice to get a value between 2 and 12. What is
wrong with this simulated throw of a pair of dice?
int sum = rand() % 11 + 2;
47. How do you generate a random floating-point number between 0 and 100?
practice it Now you can try these exercises at the end of the chapter: R4.24, P4.10, P4.25.
S e l f c h e c k
as you read this,
you have written a
few computer programs, and you have
experienced firsthand how much effort
it takes to write even the hum blest of
programs. Writing a real soft ware prod-
uct, such as a financial application or a
computer game, takes a lot of time and
money. Few people, and fewer compa-
nies, are going to spend that kind of
time and money if they don’t have a rea-
sonable chance to make more money
from their effort. (actually, some com-
panies give away their software in the
hope that users will upgrade to more
elaborate paid versions or pay for con-
sulting. other companies give away the
software that enables users to read and
use files but sell the software needed
to create those files. Finally, there are
individuals who donate their time, out
of enthusiasm, and produce programs
that you can copy freely. see random
Fact 9.2 for more information.)
When selling software, a company
must rely on the honesty of its cus-
tomers. It is an easy matter for an
unscrupulous person to make copies
of computer programs without paying
for them. In most countries that is ille-
gal. Most governments provide legal
protection, such as copyright laws and
patents, to encourage the develop-
ment of new products. Countries that
tolerate widespread piracy have found
that they have an ample cheap supply
of foreign software, but no local man-
ufacturers willing to design good soft-
ware for their own citizens, such as
word processors in the local script or
financial programs adapted to the local
tax laws.
When a mass market for software
first appeared, vendors were enraged
by the money they lost through piracy.
they tried to fight back by var ious
schemes to ensure that only the legiti-
mate owner could use the soft ware,
such as dongles—devices that must
be attached to a printer port before
the software will run. Legitimate users
hated these mea sures. they paid for
the software, but they had to suffer
through inconve niences, such as hav-
ing multiple don gles stick out from
their computer. In the United states,
market pressures forced most vendors
to give up on these copy protection
schemes, but they are still common-
place in other parts of the world.
Because it is so easy and inexpen-
sive to pirate software, and the chance
of being found out is minimal, you
have to make a moral choice for your-
self. If a package that you would really
like to have is too expensive for your
budget, do you steal it, or do you stay
honest and get by with a more afford-
able product?
of course, piracy
is not limited to
software. the same
issues arise for other
digital products as
well. You may have
had the opportunity
to obtain copies of
songs or movies
with out payment. or you may have
been frustrated by a copy protec-
tion device on your music player that
made it diffi cult for you to listen to
songs that you paid for. admittedly,
it can be diffi cult to have a lot of sym-
pathy for a musical ensemble whose
publisher charges a lot of money for
what seems to have been very little
effort on their part, at least when
compared to the effort that goes into
designing and implementing a soft-
ware package. nevertheless, it seems
only fair that artists and authors
receive some compensation for their
efforts. how to pay artists, authors,
and programmers fairly, without
bur dening honest customers, is an
unsolved problem at the time of this
writing, and many computer scientists
are engaged in research in this area.
Random Fact 4.2 software piracy
explain the flow of execution in a loop.
• Loops execute a block of code repeatedly while
a condition remains true.
• An off-by-one error is a common error when programming loops. Think through
simple test cases to avoid this type of error.
use the technique of hand-tracing to analyze the behavior of a pro gram.
• Hand-tracing is a simulation of code execution in which you step through
instructions and track the values of the variables.
• Hand-tracing can help you understand how an unfamiliar algorithm works.
• Hand-tracing can show errors in code or pseudocode.
use for loops for implementing counting loops.
• The for loop is used when a value runs from a starting point to an ending point
with a constant increment or decrement.
choose between the while loop and the do loop.
• The do loop is appropriate when the loop body must be executed at least once.
implement loops that read sequences of input data.
• A sentinel value denotes the end of a data set, but it is not part of
the data.
• You can use a Boolean variable to control a loop. Set the
variable to true before entering the loop, then set it to false
to leave the loop.
• Use input redirection to read input from a file. Use output
redirection to capture program output in a file.
use the technique of storyboarding for planning user interactions.
• A storyboard consists of annotated sketches for each step in an action sequence.
• Developing a storyboard helps you understand the inputs and outputs that are
required for a program.
C h a p t e r s U M M a r Y
cfe2_ch04_p131_192.indd 172 10/28/10 8:13 PM
Chapter summary 173
explain the flow of execution in a loop.
• Loops execute a block of code repeatedly while
a condition remains true.
• An off-by-one error is a common error when programming loops. Think through
simple test cases to avoid this type of error.
use the technique of hand-tracing to analyze the behavior of a pro gram.
• Hand-tracing is a simulation of code execution in which you step through
instructions and track the values of the variables.
• Hand-tracing can help you understand how an unfamiliar algorithm works.
• Hand-tracing can show errors in code or pseudocode.
use for loops for implementing counting loops.
• The for loop is used when a value runs from a starting point to an ending point
with a constant increment or decrement.
choose between the while loop and the do loop.
• The do loop is appropriate when the loop body must be executed at least once.
implement loops that read sequences of input data.
• A sentinel value denotes the end of a data set, but it is not part of
the data.
• You can use a Boolean variable to control a loop. Set the
variable to true before entering the loop, then set it to false
to leave the loop.
• Use input redirection to read input from a file. Use output
redirection to capture program output in a file.
use the technique of storyboarding for planning user interactions.
• A storyboard consists of annotated sketches for each step in an action sequence.
• Developing a storyboard helps you understand the inputs and outputs that are
required for a program.
C h a p t e r s U M M a r Y
cfe2_ch04_p131_192.indd 173 10/28/10 8:13 PM
174 Chapter 4 Loops
know the most common loop algorithms.
• To compute an average, keep a total and a count of all values.
• To count values that fulfill a condition, check all values and increment a counter
for each match.
• If your goal is to find a match, exit the loop when the match is found.
• To find the largest value, update the largest value seen so far whenever you see a
larger one.
• To compare adjacent inputs, store the preceding input in a variable.
use nested loops to implement multiple levels of iteration.
• When the body of a loop contains another loop, the loops are nested. A typical
use of nested loops is printing a table with rows and columns.
apply loops to the implementation of simulations.
• In a simulation, you use the computer to simulate an activity. You can
introduce randomness by calling the random number generator.
r4.1 Provide trace tables for these loops.
a. int i = 0; int j = 10; int n = 0;
while (i < j) { i++; j--; n++; }
b. int i = 0; int j = 0; int n = 0;
while (i < 10) { i++; n = n + i + j; j++; }
c. int i = 10; int j = 0; int n = 0;
while (i > 0) { i–; j++; n = n + i – j; }
d. int i = 0; int j = 10; int n = 0;
while (i != j) { i = i + 2; j = j – 2; n++; }
r4.2 What do these loops print?
a. for (int i = 1; i < 10; i++) { cout << i << " "; }
b. for (int i = 1; i < 10; i += 2) { cout << i << " "; }
c. for (int i = 10; i > 1; i–) { cout << i << " "; }
d. for (int i = 0; i < 10; i++) { cout << i << " "; }
e. for (int i = 1; i < 10; i = i * 2) { cout << i << " "; }
f. for (int i = 1; i < 10; i++) { if (i % 2 == 0) { cout << i << " "; } }
r4.3 What is an infinite loop? On your computer, how can you terminate a program that
executes an infinite loop?
r4.4 What is an “off-by-one” error? Give an example from your own programming
experience.
r e V I e W e x e r C I s e s
cfe2_ch04_p131_192.indd 174 10/28/10 8:13 PM
review exercises 175
r4.5 Write a program trace for the pseudocode in Exercise P4.9, assuming the input
val ues are 4 7 –2 –5 0.
r4.6 Is the following code legal?
for (int i = 0; i < 10; i++)
{
for (int i = 0; i < 10; i++)
{
cout << i << " ";
}
cout << endl;
}
What does it print? Is it good coding style? If not, how would you improve it?
r4.7 How often do the following loops execute? Assume that i is not changed in the loop
body.
a. for (int i = 1; i <= 10; i++) ...
b. for (int i = 0; i < 10; i++) ...
c. for (int i = 10; i > 0; i–) …
d. for (int i = -10; i <= 10; i++) ...
e. for (int i = 10; i >= 0; i++) …
f. for (int i = -10; i <= 10; i = i + 2) ...
g. for (int i = -10; i <= 10; i = i + 3) ...
r4.8 Write pseudocode for a program that prints a calendar such as the following:
Su M T W Th F Sa
1 2 3 4
5 6 7 8 9 10 11
12 13 14 15 16 17 18
19 20 21 22 23 24 25
26 27 28 29 30 31
r4.9 Write pseudocode for a program that prints a Celsius/Fahrenheit conversion table
such as the following:
Celsius | Fahrenheit
--------+-----------
0 | 32
10 | 50
20 | 68
... ...
100 | 212
r4.10 Write pseudocode for a program that reads a sequence of student records and prints
the total score for each student. Each record has the student’s first and last name,
followed by a sequence of test scores and a sentinel of –1. The sequence is termi-
nated by the word END. Here is a sample sequence:
Harry Morgan 94 71 86 95 -1
Sally Lin 99 98 100 95 90 -1
END
Provide a trace table for this sample input.
cfe2_ch04_p131_192.indd 175 10/28/10 8:13 PM
176 Chapter 4 Loops
r4.11 Rewrite the following for loop into a while loop.
int s = 0;
for (int i = 1; i <= 10; i++)
{
s = s + i;
}
r4.12 Rewrite the following do/while loop into a while loop.
int n;
cin >> n;
double x = 0;
double s;
do
{
s = 1.0 / (1 + n * n);
n++;
x = x + s;
}
while (s > 0.01);
r4.13 Provide trace tables of the following loops.
a. int s = 1;
int n = 1;
while (s < 10) { s = s + n; }
n++;
b. int s = 1;
for (int n = 1; n < 5; n++) { s = s + n; }
c. int s = 1;
int n = 1;
do
{
s = s + n;
n++;
}
while (s < 10 * n);
r4.14 What do the following loops print? Work out the answer by tracing the code, not by
using the computer.
a. int s = 1;
for (int n = 1; n <= 5; n++)
{
s = s + n;
cout << s << " ";
}
b. int s = 1;
for (int n = 1; s <= 10; cout << s << " ")
{
n = n + 2;
s = s + n;
}
c. int s = 1;
int n;
for (n = 1; n <= 5; n++)
{
s = s + n;
n++;
cfe2_ch04_p131_192.indd 176 10/28/10 8:13 PM
review exercises 177
}
cout << s << " " << n;
r4.15 What do the following program segments print? Find the answers by tracing the
code, not by using the computer.
a. int n = 1;
for (int i = 2; i < 5; i++) { n = n + i; }
cout << n;
b. int i;
double n = 1 / 2;
for (i = 2; i <= 5; i++) { n = n + 1.0 / i; }
cout << i;
c. double x = 1;
double y = 1;
int i = 0;
do
{
y = y / 2;
x = x + y;
i++;
}
while (x < 1.8);
cout << i;
d. double x = 1;
double y = 1;
int i = 0;
while (y >= 1.5)
{
x = x / 2;
y = x + y;
i++;
}
cout << i;
r4.16 Give an example of a for loop where symmetric bounds are more natural. Give an
example of a for loop where asymmetric bounds are more natural.
r4.17 Add a storyboard panel for the conversion program in Section 4.6 on page 154 that
shows a sce nario where a user enters incompatible units.
r4.18 In Section 4.6, we decided to show users a list of all valid units in the prompt. If the
program supports many more units, this approach is unworkable. Give a story board
panel that illustrates an alternate approach: If the user enters an unknown unit, a list
of all known units is shown.
r4.19 Change the storyboards in Section 4.6 to support a menu that asks users whether
they want to convert units, see program help, or quit the program. The menu should
be displayed at the beginning of the program, when a sequence of values has been
converted, and when an error is displayed.
r4.20 Draw a flow chart for a program that carries out unit conversions as described in
Section 4.6.
r4.21 In Section 4.7.4, the code for finding the largest and smallest input initializes the
largest and smallest variables with an input value. Why can’t you initialize them
with zero?
r4.22 What are nested loops? Give an example where a nested loop is typically used.
cfe2_ch04_p131_192.indd 177 10/28/10 8:13 PM
178 Chapter 4 Loops
r4.23 The nested loops
for (int i = 1; i <= height; i++)
{
for (int j = 1; j <= width; j++) { cout << "*"; }
cout << endl;
}
display a rectangle of a given width and height, such as
****
****
****
Write a single for loop that displays the same rectangle.
r4.24 Suppose you design an educational game to teach children how to read a clock. How
do you generate random values for the hours and minutes?
r4.25 In a travel simulation, Harry will visit one of his friends that are located in three
states. He has ten friends in California, three in Nevada, and two in Utah. How do
you produce a random number between 1 and 3, denoting the destination state, with
a probability that is proportional to the number of friends in each state?
p4.1 Write programs with loops that compute
a. The sum of all even numbers between 2 and 100 (inclusive).
b. The sum of all squares between 1 and 100 (inclusive).
c. All powers of 2 from 20 up to 220.
d. The sum of all odd numbers between a and b (inclusive), where a and b are
inputs.
e. The sum of all odd digits of an input. (For example, if the input is 32677, the
sum would be 3 + 7 + 7 = 17.)
p4.2 Write programs that read a sequence of integer inputs and print
a. The smallest and largest of the inputs.
b. The number of even and odd inputs.
c. Cumulative totals. For example, if the input is 1 7 2 9, the program should print
1 8 10 19.
d. All adjacent duplicates. For example, if the input is 1 3 3 4 5 5 6 6 2, the program
should print 3 5 6.
p4.3 Write programs that read a line of input as a string and print
a. Only the uppercase letters in the string.
b. Every second letter of the string.
c. The string, with all vowels replaced by an underscore.
d. The number of vowels in the string.
e. The positions of all vowels in the string.
p r o G r a M M I n G e x e r C I s e s
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programming exercises 179
p4.4 Complete the program in How To 4.1 on page 162. Your program should read twelve
temperature values and print the month with the highest temperature.
p4.5 Credit Card Number Check. The last digit of a credit card number is the check digit,
which protects against transcription errors such as an error in a single digit or
switching two digits. The following method is used to verify actual credit card
numbers but, for simplicity, we will describe it for numbers with 8 digits instead
of 16:
• Starting from the rightmost digit, form the sum of every other digit. For
example, if the credit card number is 43589795, then you form the sum
5 + 7 + 8 + 3 = 23.
• Double each of the digits that were not included in the preceding step. Add all
digits of the resulting numbers. For example, with the number given above,
doubling the digits, starting with the next-to-last one, yields 18 18 10 8. Adding
all digits in these values yields 1 + 8 + 1 + 8 + 1 + 0 + 8 = 27.
• Add the sums of the two preceding steps. If the last digit of the result is 0, the
number is valid. In our case, 23 + 27 = 50, so the number is valid.
Write a program that implements this algorithm. The user should supply an 8-digit
number, and you should print out whether the number is valid or not. If it is not
valid, you should print out the value of the check digit that would make the number
valid.
p4.6 Currency conversion. Write a program that first asks the user to type today’s ex-
change rate between U.S. dollars and Japanese yen, then reads U.S. dollar values and
converts each to yen. Use 0 as a sentinel.
p4.7 Write a program that first asks the user to type in today’s exchange rate between U.S.
dollars and Japanese yen, then reads U.S. dollar values and converts each to Japanese
yen. Use 0 as the sentinel value to denote the end of dollar inputs. Then the program
reads a sequence of yen amounts and converts them to dollars. The second sequence
is terminated by another zero value.
p4.8 Write a program that reads a set of floating-point values. Ask the user to enter the
values, then print
• the average of the values.
• the smallest of the values.
• the largest of the values.
• the range, that is the difference between the smallest and largest.
Of course, you may only prompt for the values once.
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180 Chapter 4 Loops
p4.9 Translate the following pseudocode for finding the minimum value from a set of
inputs into a C++ program.
Set a Boolean variable "first" to true.
While another value has been read successfully
If first is true
Set the minimum to the value.
Set first to false.
Else if the value is less than the minimum
Set the minimum to the value.
Print the minimum.
p4.10 Translate the following pseudocode for randomly permuting the characters in a
string into a C++ program.
Read a word.
Repeat word.length() times
Pick a random position i in the word.
Pick a random position j > i in the word.
Swap the letters at positions j and i.
Print the word.
To swap the letters, construct substrings as follows:
first middle lasti j
Then replace the string with
first + word.substr(j, 1) + middle + word.substr(i, 1) + last
p4.11 Write a program that reads a word and prints each character of the word on a sepa-
rate line. For example, if the user provides the input “Harry”, the program prints
H
a
r
r
y
p4.12 Write a program that reads a word and prints the word in reverse. For example, if the
user provides the input “Harry”, the program prints
yrraH
p4.13 Write a program that reads a word and prints the number of vowels in the word. For
this exercise, assume that a e i o u y are vowels. For example, if the user pro vides the
input “Harry”, the program prints 2 vowels.
p4.14 Write a program that reads a word and prints the number of syllables in the word.
For this exercise, assume that syllables are determined as follows: Each sequence of
vowels a e i o u y, except for the last e in a word, is a vowel. However, if that algo-
rithm yields a count of 0, change it to 1. For example,
Word Syllables
Harry 2
hairy 2
hare 1
the 1
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programming exercises 181
p4.15 Write a program that reads a word and prints all substrings, sorted by length. For
example, if the user provides the input “rum”, the program prints
r
u
m
ru
um
rum
p4.16 Write a program that reads a number and prints all of its binary digits: Print the
remainder number % 2, then replace the number with number / 2. Keep going until the
number is 0. For example, if the user provides the input 13, the output should be
1
0
1
1
p4.17 Mean and standard deviation. Write a program that reads a set of floating-point data
values. Choose an appropriate mechanism for prompting for the end of the data set.
When all values have been read, print out the count of the values, the average, and
the standard deviation. The average of a data set {x1, . . ., xn} is x x ni= ∑ , where
∑ = + +x x xi n1 … is the sum of the input values. The stan dard deviation is
s
x x
n
i=
−( )
−
∑ 2
1
However, this formula is not suitable for the task. By the time the program has
computed x , the individ ual xi are long gone. Until you know how to save these
values, use the numerically less stable formula
s
x x
n
i n i=
− ( )
−
∑∑ 2 1
2
1
You can compute this quantity by keeping track of the count, the sum, and the sum
of squares as you process the input values.
p4.18 The Fibonacci numbers are defined by the sequence
f
f
f f fn n n
1
2
1 2
1
1
=
=
= +− −
Reformulate that as
fold1 = 1;
fold2 = 1;
fnew = fold1 + fold2;
After that, discard fold2, which is no longer needed, and set fold2 to fold1 and fold1 to
fnew. Repeat fnew an appropriate number of times.
Implement a program that computes the Fibonacci numbers in that way.
Fibonacci numbers describe the
growth of a rabbit population.
cfe2_ch04_p131_192.indd 181 10/28/10 8:13 PM
182 Chapter 4 Loops
p4.19 Factoring of integers. Write a program that asks the user for an integer and then
prints out all its factors. For example, when the user enters 150, the program should
print
2
3
5
5
p4.20 Prime numbers. Write a program that prompts the user for an integer and then prints
out all prime numbers up to that integer. For example, when the user enters 20, the
program should print
2
3
5
7
11
13
17
19
Recall that a number is a prime number if it is not divisible by any number except 1
and itself.
p4.21 Write a program that prints a multiplication table, like this:
1 2 3 4 5 6 7 8 9 10
2 4 6 8 10 12 14 16 18 20
3 6 9 12 15 18 21 24 27 30
…
10 20 30 40 50 60 70 80 90 100
p4.22 Write a program that reads an integer and displays, using asterisks, a filled and
hol low square, placed next to each other. For example if the side length is 5, the
pro gram should display
***** *****
***** * *
***** * *
***** * *
***** *****
p4.23 Write a program that reads an integer and displays, using asterisks, a filled diamond
of the given side length. For example, if the side length is 4, the program should
dis play
*
***
*****
*******
*****
***
*
p4.24 The game of Nim. This is a well-known game with a number of variants. The
fol lowing variant has an interesting winning strategy. Two players alternately take
marbles from a pile. In each move, a player chooses how many marbles to take. The
player must take at least one but at most half of the marbles. Then the other player
takes a turn. The player who takes the last marble loses.
cfe2_ch04_p131_192.indd 182 10/28/10 8:13 PM
programming exercises 183
You will write a program in which the computer plays against a human opponent.
Generate a random integer between 10 and 100 to denote the initial size of the pile.
Generate a random integer between 0 and 1 to decide whether the computer or the
human takes the first turn. Generate a random integer between 0 and 1 to decide
whether the computer plays smart or stupid. In stupid mode the computer simply
takes a random legal value (between 1 and n̸2) from the pile whenever it has a turn.
In smart mode the computer takes off enough marbles to make the size of the pile a
power of two minus 1—that is, 3, 7, 15, 31, or 63. That is always a legal move, except
when the size of the pile is currently one less than a power of two. In that case, the
computer makes a random legal move.
You will note that the computer cannot be beaten in smart mode when it has the first
move, unless the pile size happens to be 15, 31, or 63. Of course, a human player who
has the first turn and knows the win ning strategy can win against the computer.
p4.25 The Drunkard’s Walk. A drunkard in a grid of streets randomly picks one of four
directions and stumbles to the next intersection, then again randomly picks one of
four directions, and so on. You might think that on average the drunkard doesn’t
move very far because the choices cancel each other out, but that is actually not
the case.
Represent locations as integer pairs (x, y). Implement the drunkard’s walk over 100
intersections and print the beginning and ending location.
p4.26 The Monty Hall Paradox. Marilyn vos Savant described the following problem
(loosely based on a game show hosted by Monty Hall) in a popular magazine:
“Suppose you’re on a game show, and you’re given the choice of three doors: Behind
one door is a car; behind the others, goats. You pick a door, say No. 1, and the host,
who knows what’s behind the doors, opens another door, say No. 3, which has a
goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advan-
tage to switch your choice?”
Ms. vos Savant proved that it is to your advantage, but many of her readers, includ-
ing some mathematics professors, disagreed, arguing that the probability would not
change because another door was opened.
Your task is to simulate this game show. In each iteration, ran domly pick a door
number between 1 and 3 for placing the car. Randomly have the player pick a door.
Randomly have the game show host pick one of the two doors having a goat. Now
incre ment a counter for strategy 1 if the player wins by switching to the third door,
and increment a counter for strategy 2 if the player wins by sticking with the original
choice. Run 1,000 iterations and print both counters.
p4.27 The Buffon Needle Experiment. The
follow ing experiment was devised by
Comte Georges-Louis Leclerc de
Buffon (1707–1788), a French
naturalist. A needle of length 1 inch is
dropped onto paper that is ruled with
lines 2 inches apart. If the needle drops
onto a line, we count it as a hit. (See
Figure 8.) Buffon conjectured that the
quotient tries/hits approximates p. figure 8 the Buffon needle experiment
cfe2_ch04_p131_192.indd 183 10/28/10 8:13 PM
184 Chapter 4 Loops
figure 9
a hit in the Buffon needle experiment 2
0
yhighyy
ylowyy α
For the Buffon needle experiment, you must generate two random numbers: one to
describe the starting position and one to describe the angle of the needle with the
x-axis. Then you need to test whether the needle touches a grid line.
Generate the lower point of the needle. Its x-coordinate is irrelevant, and you may
assume its y-coordi nate ylow to be any random number between 0 and 2. The angle α
between the needle and the x-axis can be any value between 0 degrees and 180
degrees (p radians). The upper end of the needle has y-coordinate
y yhigh low= + sinα
The needle is a hit if yhigh is at least 2, as shown in Figure 9.
Stop after 10,000 tries and print the quotient tries/hits. (This program is not suitable
for computing the value of p. You need p in the computation of the angle.)
engineering p4.28 In a predator-prey simulation, you compute the populations of predators and prey,
using the following equations:
prey prey A B pred
pred pred C D
n n n
n n
+
+
= × + − ×( )
= × − + ×
1
1
1
1 ppreyn( )
Here, A is the rate at which prey birth exceeds natural
death, B is the rate of predation, C is the rate at which
predator deaths exceed births without food, and D
represents predator increase in the presence of food.
Write a program that prompts users for these rates, the
initial population sizes, and the number of peri ods. Then
print the populations for the given number of periods.
As inputs, try A = 0.1, B = C = 0.01, and D = 0.00002
with initial prey and predator populations of 1,000 and 20.
engineering p4.29 Projectile flight. Suppose a cannonball is propelled straight into the air with a start ing
velocity v0. Any calculus book will state that the position of the ball after t sec onds
is s t gt v t( ) = − +1
2
2
0 , where g = 9 81. m sec
2 is the gravi tational force of the earth.
No calculus book ever mentions why someone would want to carry out such an
obviously dangerous experiment, so we will do it in the safety of the computer.
In fact, we will confirm the theorem from calculus by a simulation. In our simula-
tion, we will consider how the ball moves in very short time intervals Δt. In a short
cfe2_ch04_p131_192.indd 184 10/28/10 8:13 PM
programming exercises 185
time interval the velocity v is nearly con stant, and we can compute the distance the
ball moves as Δs = vΔt. In our program, we will simply set
const double DELTA_T = 0.01;
and update the position by
s = s + v * DELTA_T;
The velocity changes constantly—in fact, it is reduced by the gravitational force of
the earth. In a short time interval, Δv = –gΔt, we must keep the velocity updated as
v = v – g * DELTA_T;
In the next iteration the new velocity is used to update the distance.
Now run the simulation until the cannonball falls back to the earth. Get the initial
velocity as an input (100 m̸sec is a good value). Update the position and velocity 100
times per second, but print out the posi tion only every full second. Also printout the
values from the exact formula s t gt v t( ) = − +1
2
2
0 for com parison.
Note: You may wonder whether there is
a benefit to this simulation when an
exact formula is available. Well, the
formula from the calculus book is not
exact. Actually, the gravitational force
diminishes the farther the cannonball is
away from the surface of the earth. This
complicates the algebra sufficiently that
it is not possible to give an exact formula
for the actual motion, but the computer
simulation can simply be extended to
apply a variable gravitational force. For cannonballs, the calculus-book formula is
actually good enough, but computers are necessary to compute accurate trajectories
for higher-flying objects such as ballistic mis siles.
engineering p4.30 A simple model for the hull of a ship is given by
y
B x
L
z
T
= −
−
2
1
2
1
2 2
where B is the beam, L is the length, and T is the draft.
cfe2_ch04_p131_192.indd 185 10/28/10 8:13 PM
186 Chapter 4 Loops
(Note: There are two values of y for each x and z because the
hull is symmetric from starboard to port.)
The cross-sectional area at a point x is called the “section” in
nautical parlance. To compute it, let z go from 0 to –T in n
increments, each of size T n. For each value of z, compute
the value for y. Then sum the areas of trapezoidal strips. At
right are the strips where n = 4.
Write a program that reads in values for B, L, T, x, and n and
then prints out the cross-sectional area at x.
engineering p4.31 Radioactive decay of radioactive materials can be
modeled by the equation A = A0e
-t (log 2̸h), where A is
the amount of the material at time t, A0 is the amount
at time 0, and h is the half-life.
Technetium-99 is a radioisotope that is used in imaging
of the brain. It has a half-life of 6 hours. Your program
should display the relative amount A ̸ A0 in a patient
body every hour for 24 hours after receiving a dose.
engineering p4.32 The photo at left shows an electric device called a “transformer”. Transformers are
often constructed by wrapping coils of wire around a ferrite core. The figure below
illustrates a situation that occurs in various audio devices such as cell phones and
music players. In this circuit, a transformer is used to connect a speaker to the output
of an audio amplifier.
Vs = 40 V
Speaker
+
–
R0 = 20 Ω
Rs = 8 Ω
TransformerAmplifier
1 : n
The symbol used to represent the transformer is intended to suggest two coils of
wire. The parameter n of the transformer is called the “turns ratio” of the trans-
former. (The number of times that a wire is wrapped around the core to form a coil is
called the number of turns in the coil. The turns ratio is literally the ratio of the
number of turns in the two coils of wire.)
When designing the circuit, we are concerned primarily with the value of the power
delivered to the speakers—that power causes the speakers to produce the sounds we
want to hear. Suppose we were to connect the speakers directly to the amplifier
without using the transformer. Some fraction of the power available from the
amplifier would get to the speakers. The rest of the available power would be lost in
the amplifier itself. The transformer is added to the circuit to increase the fraction of
the amplifier power that is delivered to the speakers.
The power, Ps, delivered to the speakers is calculated using the formula
P R
nV
n R R
s s
s
s
=
+
2
0
2
Write a C++ program that models the circuit shown and varies the turns ratio from
0.01 to 2 in 0.01 increments, then determines the value of the turns ratio that maxi-
mizes the power delivered to the speakers.
1. 23 years.
2. 7 years.
3. Add a statement
cout << balance << endl;
as the last statement in the while loop.
4. The program prints the same output. This is because the balance after 14 years is
slightly below $20,000, and after 15 years, it is slightly above $20,000.
5. 2 4 8 16 32 64 128
Note that the value 128 is printed even though it is larger than 100.
6. n output
5
4 4
3 3
2 2
1 1
0 0
-1 -1
7. n output
1 1,
2 1, 2,
3 1, 2, 3,
4
There is a comma after the last value. Usually, commas are between values only.
8. a n r i
2 4 1 1
2 2
4 3
8 4
16 5
The code computes an.
9. n output
1 1
11 11
21 21
31 31
41 41
51 51
61 61
...
This is an infinite loop. n is never equal to 50.
10. count n
1 123
2 12.3
3 1.23
a n s W e r s t o s e L F - C h e C k Q U e s t I o n s
cfe2_ch04_p131_192.indd 186 10/28/10 8:13 PM
answers to self-Check Questions 187
Write a C++ program that models the circuit shown and varies the turns ratio from
0.01 to 2 in 0.01 increments, then determines the value of the turns ratio that maxi-
mizes the power delivered to the speakers.
1. 23 years.
2. 7 years.
3. Add a statement
cout << balance << endl;
as the last statement in the while loop.
4. The program prints the same output. This is because the balance after 14 years is
slightly below $20,000, and after 15 years, it is slightly above $20,000.
5. 2 4 8 16 32 64 128
Note that the value 128 is printed even though it is larger than 100.
6. n output
5
4 4
3 3
2 2
1 1
0 0
-1 -1
7. n output
1 1,
2 1, 2,
3 1, 2, 3,
4
There is a comma after the last value. Usually, commas are between values only.
8. a n r i
2 4 1 1
2 2
4 3
8 4
16 5
The code computes an.
9. n output
1 1
11 11
21 21
31 31
41 41
51 51
61 61
...
This is an infinite loop. n is never equal to 50.
10. count n
1 123
2 12.3
3 1.23
a n s W e r s t o s e L F - C h e C k Q U e s t I o n s
cfe2_ch04_p131_192.indd 187 10/28/10 8:13 PM
188 Chapter 4 Loops
This yields the correct answer. The number 123 has 3 digits.
count n
1 100
2 10
This yields the wrong answer. The number 100 also has 3 digits. The loop condition
should have been
while (temp >= 10)
11. int year = 1;
while (year <= nyears)
{
balance = balance * (1 + RATE / 100);
cout << setw(4) << year << setw(10) << balance << endl;
year++;
}
12. 11 numbers: 10 9 8 7 6 5 4 3 2 1 0
13. for (int i = 10; i <= 20; i = i + 2)
{
cout << n << endl;
}
14. int sum = 0;
for (int i = 1; i <= n; i++)
{
sum = sum + i;
}
15. for (int year = 1; balance <= 2 * INITIAL_BALANCE; year++)
However, it is best not to use a for loop in this case because the loop condition does
not relate to the year variable. A while loop would be a better choice.
16. do
{
cout << "Enter a value between 0 and 100: ";
cin >> value;
}
while (value < 0 || value > 100);
17. int value = 100;
while (value >= 100)
{
cout << "Enter a value < 100: ";
cin >> value;
}
Here, the variable value had to be initialized with an artificial value to ensure that the
loop is entered at least once.
18. Yes. The do loop
do { body } while (condition);
is equivalent to this while loop:
bool first = true;
while (first || condition) { body; first = false; }
19. int x;
int sum = 0;
do
{
cfe2_ch04_p131_192.indd 188 10/28/10 8:13 PM
answers to self-Check Questions 189
cin >> x;
sum = sum + x;
}
while (x != 0);
20. int x = 0;
int previous;
do
{
previous = x;
cin >> x;
sum = sum + x;
}
while (previous != x);
21. No data
22. The first check ends the loop after the sentinel has been read. The second check
ensures that the sentinel is not processed as an input value.
23. The while loop would never be entered. The user would never be prompted for in-
put. Since count stays 0, the program would then print “No data”.
24. The stream also fails. A more accurate prompt would have been: “Enter values, a key
other than a digit to quit.” But that might be more confusing to the program user
who would need now ponder which key to choose.
25. You don’t know whether the input fails until after you try reading input.
26. Computing the aver age
27. Simple conversion
Unknown unit
Program doesn’t understand question syntax
Enter scores, Q to quit: 90 80 90 100 80 Q
The average is 88
(Program exits)
Your conversion question: How many in are 30 cm
30 cm = 11.81 in
(Program exits) Run program again for another question
Only one value can be converted.
Your conversion question: How many inches are 30 cm?
Unknown unit: inches
Known units are in, ft, mi, mm, cm, m, km, oz, lb, g, kg, tsp, tbsp, pint, gal
(Program exits)
Your conversion question: What is an ångström?
Please formulate your question as “How many (unit) are (value) (unit)?”
(Program exits)
cfe2_ch04_p131_192.indd 189 10/28/10 8:13 PM
190 Chapter 4 Loops
28. One score is not enough
29. It would not be possible to implement this interface using the C++ features we have
covered up to this point. There is no way for the program to know when the first set
of inputs ends. (When you read numbers with cin >> value, it is your choice whether
to put them on a single line or multiple lines.)
30. Comparing two inter est rates
31. The total is zero.
32. double total = 0;
double input;
while (cin >> input)
{
if (input > 0) { total = total + input; }
}
33. position is str.length().
34. The loop will stop when a match is found, but you cannot access the match because
position is not defined outside the loop.
35. Start the loop at the end of string:
bool found = false;
int position = str.length() – 1;
while (!found && position >= 0)
{
string ch = str.substr(position, 1);
if (ch == ” “) { found = true; }
else { position–; }
}
36. Unless the input contains zero or negative numbers, the smallest value is incorrectly
computed as 0.
37. When executing cin >> previous, cin fails and previous is unchanged. The statement
cin >> input also fails, and the while loop is never entered.
38. All values in the inner loop should be displayed on the same line.
Enter scores, Q to quit: 90 Q
Error: At least two scores are required.
(Program exits)
First interest rate in percent: 5
Second interest rate in percent: 10
Years: 5
Year 5% 10%
0 10000.00 10000.00
1 10500.00 11000.00
2 11025.00 12100.00
3 11576.25 13310.00
4 12155.06 14641.00
5 12762.82 16105.10
This row clarifies that 1 means
the end of the first year
cfe2_ch04_p131_192.indd 190 10/28/10 8:13 PM
answers to self-Check Questions 191
39. Change lines 14, 19, and 31 to for (int n = 0; n <= NMAX; n++). Change NMAX
to 5.
40. 60: The outer loop is executed 10 times, and the inner loop 6 times.
41. 0123
1234
2345
42. for (int i = 1; i <= 3; i++)
{
for (int j = 1; j <= 4; j++)
{
cout << "[]";
}
cout << endl;
}
43. Compute rand() % 2, and use 0 for heads, 1 for tails, or the other way around.
44. Compute rand() % 4 and associate the numbers 0 ... 3 with the four suits. Then com-
pute rand() % 13 and associate the numbers 0 ... 12 with Jack, Ace, 2 ... 10, Queen, and
King.
45. It is required for calling the time function.
46. The call will produce a value between 2 and 12, but all values have the same proba-
bility. When throwing a pair of dice, the number 7 is six times as likely as the num ber
2. The correct formula is
int sum = rand() % 6 + rand() % 6 + 2;
47. rand() * 100.0 / RAND_MAX
cfe2_ch04_p131_192.indd 191 10/28/10 8:13 PM
This page intentionally left blank
5C h a p t e r
193
F u n C t i o n s
to be able to implement functions
to become familiar with the concept of
parameter passing
to appreciate the importance of function
comments
to develop strategies for decomposing complex tasks into simpler ones
to be able to determine the scope of a variable
to recognize when to use value and reference parameters
C h a p t e r G o a l s
C h a p t e r C o n t e n t s
5.1 Functions as Black Boxes 194
5.2 implementing Functions 196
Syntax 5.1: Function Definition 197
Programming Tip 5.1: Function Comments 199
5.3 parameter passing 199
Programming Tip 5.2: Do not Modify parameter
Variables 201
5.4 return Values 202
Common Error 5.1: Missing return Value 203
Special Topic 5.1: Function Declarations 203
How To 5.1: implementing a Function 204
Worked Example 5.1: Matching and replacing
parts of a string
Worked Example 5.2: using a Debugger
5.5 Functions Without return
Values 206
5.6 proBlem solVing: reusaBle
Functions 208
5.7 proBlem solVing: stepWise
reFinement 210
Programming Tip 5.3: Keep Functions short 216
Programming Tip 5.4: tracing Functions 216
Programming Tip 5.5: stubs 217
Worked Example 5.3: Calculating a
Course Grade
5.8 VariaBle scope and gloBal
VariaBles 218
Programming Tip 5.6: avoid Global Variables 220
5.9 reFerence parameters 220
Programming Tip 5.7: prefer return Values to
reference parameters 225
Special Topic 5.2: Constant references 225
5.10 recursiVe Functions
(optional) 226
How To 5.2: thinking recursively 229
Random Fact 5.1: the explosive Growth of
personal Computers 230
cfe2_ch05_p193_248.indd 193 10/26/10 6:14 PM
194
Functions are a fundamental building block of C++
programs. a function packages a computation into a form
that can be easily understood and reused. (the person in the
image to the left is executing the function “make two cups
of espresso”.) in this chapter, you will learn how to design
and implement your own functions. using the process of
stepwise refinement, you will be able to break up complex
tasks into sets of cooperating functions.
5.1 Functions as Black Boxes
A function is a sequence of instructions with a name. You have already encountered
several functions. For example, the function named pow, which was introduced in
Chapter 2, contains instructions to com pute a power xy. Moreover, every C++ pro-
gram has a function called main.
You call a function in order to execute its instructions. For example, consider the
following program:
int main()
{
double z = pow(2, 3);
...
}
By using the expression pow(2, 3), main calls the pow function, asking it to compute the
power 23. The main function is temporarily suspended. The instructions of the pow
function execute and compute the result. The pow function returns its result (that is,
the value 8) back to main, and the main function resumes execu tion (see Figure 1).
a function is a
named sequence
of instructions.
Figure 1 execution Flow During a Function Call
Wait
Pass 2 and 3 to
pow
Use result
main
Pass result
to caller
Compute 23
pow
When another function calls the pow function, it provides “inputs”, such as the
expressions 2 and 3 in the call pow(2, 3). These expressions are called arguments. This
terminology avoids confusion with other inputs, such as those pro vided by a human
user. Similarly, the “output” that the pow function computes is called the return value.
Functions can have multiple arguments, but they have only one return value.
Note that the return value of a function is returned to the calling function, not dis-
played on the screen. For example, suppose your program contains a statement
double z = pow(2, 3);
When the pow function returns its result, the return value is stored in the variable z. If
you want the value to be displayed, you need to add a statement such as cout << z.
At this point, you may wonder how the pow function performs its job. For example,
how does pow(2, 3) compute that 23 is 8? By multiplying 2 × 2 × 2? With logarithms?
Fortunately, as a user of the function, you don’t need to know how the function is
implemented. You just need to know the specification of the function: If you provide
arguments x and y, the function returns xy. Engineers use the term black box for a
device with a given specification but unknown implementation. You can think of pow
as a black box, as shown in Figure 2.
When you design your own functions, you will want to make them appear as black
boxes to other pro grammers. Those programmers want to use your functions with-
out knowing what goes on inside. Even if you are the only person working on a pro-
gram, making each function into a black box pays off: there are fewer details that you
need to keep in mind.
arguments are
supplied when a
function is called.
the return value is
the result that the
function computes.
Although a thermostat is usually white, you
can think of it as a black box. The input is the
desired temperature, and the output is a signal
to the heater or air conditioner.
cfe2_ch05_p193_248.indd 194 10/26/10 6:14 PM
5.1 Functions as Black Boxes 195
Figure 2 the pow Function as a Black Box
2, 3
8
pow
Inputs
Output
When another function calls the pow function, it provides “inputs”, such as the
expressions 2 and 3 in the call pow(2, 3). These expressions are called arguments. This
terminology avoids confusion with other inputs, such as those pro vided by a human
user. Similarly, the “output” that the pow function computes is called the return value.
Functions can have multiple arguments, but they have only one return value.
Note that the return value of a function is returned to the calling function, not dis-
played on the screen. For example, suppose your program contains a statement
double z = pow(2, 3);
When the pow function returns its result, the return value is stored in the variable z. If
you want the value to be displayed, you need to add a statement such as cout << z.
At this point, you may wonder how the pow function performs its job. For example,
how does pow(2, 3) compute that 23 is 8? By multiplying 2 × 2 × 2? With logarithms?
Fortunately, as a user of the function, you don’t need to know how the function is
implemented. You just need to know the specification of the function: If you provide
arguments x and y, the function returns xy. Engineers use the term black box for a
device with a given specification but unknown implementation. You can think of pow
as a black box, as shown in Figure 2.
When you design your own functions, you will want to make them appear as black
boxes to other pro grammers. Those programmers want to use your functions with-
out knowing what goes on inside. Even if you are the only person working on a pro-
gram, making each function into a black box pays off: there are fewer details that you
need to keep in mind.
arguments are
supplied when a
function is called.
the return value is
the result that the
function computes.
Although a thermostat is usually white, you
can think of it as a black box. The input is the
desired temperature, and the output is a signal
to the heater or air conditioner.
cfe2_ch05_p193_248.indd 195 10/26/10 6:14 PM
196 Chapter 5 Functions
1. Consider the function call pow(3, 2). What are the arguments and return value?
2. What is the return value of the function call pow(pow(2, 2), 2)?
3. The ceil function in the C++ standard library takes a single argument x and
returns the smallest inte ger ≥ x. What is the return value of ceil(2.3)?
4. It is possible to determine the answer to Self Check 3 without knowing how the
ceil function is implemented. Use an engineering term to describe this aspect of
the ceil function.
practice it Now you can try these exercises at the end of the chapter: R5.1, P5.1.
5.2 implementing Functions
In this section, you will learn how to implement a
function from a given specification. We will use a
very simple example: a function to compute the
volume of a cube with a given side length.
The cube_volume function uses a given side
length to compute the volume of a cube.
When writing this function, you need to
• Pick a name for the function (cube_volume).
• Declare a variable for each argument (double side_length). These variables are
called parameter vari ables.
• Specify the type of the return value (double).
Put all this information together to form the first line of the function’s definition:
double cube_volume(double side_length)
Next, specify the body of the function: the statements that are executed when the
function is called.
The volume of a cube of side length s is s × s × s. However, for greater clarity, our
parameter variable has been called side_length, not s, so we need to compute side_
length * side_length * side_length.
We will store this value in a variable called volume:
double volume = side_length * side_length * side_length;
In order to return the result of the function, use the return statement:
return volume;
The body of a function is enclosed in braces. Here is the complete function:
double cube_volume(double side_length)
{
double volume = side_length * side_length * side_length;
return volume;
}
s e l F c h e c k
When defining a
function, you provide
a name for the
function, a variable
for each argument,
and a type for
the result.
cfe2_ch05_p193_248.indd 196 10/26/10 6:14 PM
5.2 implementing Functions 197
The return statement gives the
function’s result to the caller.
Let’s put this function to use. We’ll supply a main function that calls the cube_volume
function twice.
int main()
{
double result1 = cube_volume(2);
double result2 = cube_volume(10);
cout << "A cube with side length 2 has volume " << result1 << endl;
cout << "A cube with side length 10 has volume " << result2 << endl;
return 0;
}
When the function is called with different arguments, the function returns different
results. Consider the call cube_volume(2). The argument 2 corresponds to the side_
length parameter variable. Therefore, in this call, side_length is 2. The function com-
putes side_length * side_length * side_length, or 2 * 2 * 2. When the function is called
with a different argument, say 10, then the function computes 10 * 10 * 10.
Now we combine both functions into a test program. Because main calls cube_
volume, the cube_volume function must be known before the main function is defined.
This is easily achieved by placing cube_volume first and main last in the source file. (See
Special Topic 5.1 on page 203 for an alternative.)
syntax 5.1 Function Definition
double cube_volume(double side_length)
{
double volume = side_length * side_length * side_length;
return volume;
}
Type of return value
Name of function
Type of parameter variable
Name of parameter variable
return statement exits
function and returns result.
See page 202.
Function body,
executed when
function is called.
cfe2_ch05_p193_248.indd 197 10/26/10 6:14 PM
198 Chapter 5 Functions
Here is the com plete program. Note the comment that describes the behavior of
the function. (Programming Tip 5.1 on page 199 describes the format of the comment.)
ch05/cube.cpp
1 #include
2
3 using namespace std;
4
5 /**
6 Computes the volume of a cube.
7 @param side_length the side length of the cube
8 @return the volume
9 */
10 double cube_volume(double side_length)
11 {
12 double volume = side_length * side_length * side_length;
13 return volume;
14 }
15
16 int main()
17 {
18 double result1 = cube_volume(2);
19 double result2 = cube_volume(10);
20 cout << "A cube with side length 2 has volume " << result1 << endl;
21 cout << "A cube with side length 10 has volume " << result2 << endl;
22
23 return 0;
24 }
program run
A cube with side length 2 has volume 8
A cube with side length 10 has volume 1000
5. What is the value of cube_volume(3)?
6. What is the value of cube_volume(cube_volume(2))?
7. Provide an alternate implementation of the body of the cube_volume function by
calling the pow func tion.
8. Define a function square_area that computes the area of a square of a given side
length.
9. Consider this function:
int mystery(int x, int y)
{
double result = (x + y) / (y - x);
return result;
}
What is the result of the call mystery(2, 3)?
practice it Now you can try these exercises at the end of the chapter: R5.2, P5.2, P5.7.
s e l F c h e c k
Function comments
Whenever you write a function, you should comment its behavior. Comments are for human
readers, not compilers, and there is no universal standard for the layout of a function com-
ment. In this book, we will use the following lay out:
/**
Computes the volume of a cube.
@param side_length the side length of the cube
@return the volume
*/
double cube_volume(double side_length)
{
double volume = side_length * side_length * side_length;
return volume;
}
This particular documentation style is borrowed from the Java programming language. It is
widely supported by C++ tools as well, for example by the Doxygen tool (www.doxygen.org).
The first line of the comment describes the purpose of the function. Each @param clause
describes a parameter vari able and the @return clause describes the return value.
Note that the function comment does not document the implementation (how the function
does what it does) but rather the design (what the function does, its inputs, and its results). The
comment allows other programmers to use the function as a “black box”.
5.3 parameter passing
In this section, we examine the mechanism of passing arguments into functions.
When a function is called, its parameter variables are created (Another commonly
used term for a parameter variable is for mal parameter.) In the function call, an
expression is supplied for each parameter variable, called the argument. (Another
commonly used term for this expression is actual parameter.) Each parameter vari-
able is initialized with the value of the corresponding argument.
Consider the function call illustrated in Figure 3:
double result1 = cube_volume(2);
programming tip 5.1
Function comments
explain the purpose
of the function, the
meaning of the
parameter variables
and return value, as
well as any special
requirements.
parameter variables
hold the argument
values supplied in
the function call.
A recipe for a fruit pie may say to use any kind of fruit.
Here, “fruit” is an example of a parameter variable.
Apples and cherries are examples of arguments.
pie(fruit) pie(fruit)
cfe2_ch05_p193_248.indd 198 10/26/10 6:14 PM
5.3 parameter passing 199
Function comments
Whenever you write a function, you should comment its behavior. Comments are for human
readers, not compilers, and there is no universal standard for the layout of a function com-
ment. In this book, we will use the following lay out:
/**
Computes the volume of a cube.
@param side_length the side length of the cube
@return the volume
*/
double cube_volume(double side_length)
{
double volume = side_length * side_length * side_length;
return volume;
}
This particular documentation style is borrowed from the Java programming language. It is
widely supported by C++ tools as well, for example by the Doxygen tool (www.doxygen.org).
The first line of the comment describes the purpose of the function. Each @param clause
describes a parameter vari able and the @return clause describes the return value.
Note that the function comment does not document the implementation (how the function
does what it does) but rather the design (what the function does, its inputs, and its results). The
comment allows other programmers to use the function as a “black box”.
5.3 parameter passing
In this section, we examine the mechanism of passing arguments into functions.
When a function is called, its parameter variables are created (Another commonly
used term for a parameter variable is for mal parameter.) In the function call, an
expression is supplied for each parameter variable, called the argument. (Another
commonly used term for this expression is actual parameter.) Each parameter vari-
able is initialized with the value of the corresponding argument.
Consider the function call illustrated in Figure 3:
double result1 = cube_volume(2);
programming tip 5.1
Function comments
explain the purpose
of the function, the
meaning of the
parameter variables
and return value, as
well as any special
requirements.
parameter variables
hold the argument
values supplied in
the function call.
A recipe for a fruit pie may say to use any kind of fruit.
Here, “fruit” is an example of a parameter variable.
Apples and cherries are examples of arguments.
pie(fruit) pie(fruit)
cfe2_ch05_p193_248.indd 199 10/26/10 6:14 PM
www.doxygen.org
200 Chapter 5 Functions
•
Figure 3 parameter passing
1 Function call result1 =
side_length =
2 Initializing function parameter variable result1 =
side_length = 2
3 About to return to the caller result1 =
side_length =
volume = 8
2
4 After function call result1 = 8
double result1 = cube_volume(2);
double volume = side_length * side_length * side_length;
return volume;
double result1 = cube_volume(2);
double result1 = cube_volume(2);
The parameter variable side_length of the cube_volume function is created. 1
• The parameter variable is initialized with the value of the argument that was
passed in the call. In our case, side_length is set to 2. 2
• The function computes the expression side_length * side_length * side_length,
which has the value 8. That value is stored in the variable volume. 3
• The function returns. All of its variables are removed. The return value is trans-
ferred to the caller, that is, the function calling the cube_volume function. 4
Now consider what happens in a subsequent call cube_volume(10). A new parameter
variable is created. (Recall that the previous parameter variable was removed when
the first call to cube_volume returned.) It is initialized with the argument 10, and the
process repeats. After the second function call is complete, its variables are again
removed.
Like any other variables, parameter variables can only be set to values of compat-
ible types. For exam ple, the side_length parameter variable of the cube_volume function
has type double. It is valid to call cube_volume(2.0) or cube_volume(2). In the latter call,
the integer 2 is automatically converted to the double value 2.0. However, a call cube_
volume("two") is not legal.
10. What does this program print? Use a diagram like Figure 3 to find the answer.
double mystery(int x, int y)
{
double z = x + y;
z = z / 2.0;
s e l F c h e c k
return z;
}
int main()
{
int a = 4;
int b = 7;
cout << mystery(a, b) << endl;
}
11. What does this program print? Use a diagram like Figure 3 to find the answer.
int mystery(int x)
{
int y = x * x;
return y;
}
int main()
{
int a = 4;
cout << mystery(a + 1) << endl;
}
12. What does the following program print? Use a diagram like Figure 3 to find the
answer.
int mystery(int n)
{
n++;
n++;
return n;
}
int main()
{
int a = 5;
cout << mystery(a) << endl;
}
practice it Now you can try these exercises at the end of the chapter: R5.5, P5.10.
do not modify parameter Variables
In C++, a parameter variable is just like any other variable. You can modify the values of the
parameter variables in the body of a function. For example,
int total_cents(int dollars, int cents)
{
cents = dollars * 100 + cents; // Modifies parameter variable
return cents;
}
However, many programmers find this practice confusing. It mixes the concept of a parameter
(input to the func tion) with that of a variable (storage for a value). To avoid the confusion,
simply introduce a separate variable:
int total_cents(int dollars, int cents)
{
int result = dollars * 100 + cents;
return result;
}
programming tip 5.2
cfe2_ch05_p193_248.indd 200 10/26/10 6:14 PM
5.3 parameter passing 201
return z;
}
int main()
{
int a = 4;
int b = 7;
cout << mystery(a, b) << endl;
}
11. What does this program print? Use a diagram like Figure 3 to find the answer.
int mystery(int x)
{
int y = x * x;
return y;
}
int main()
{
int a = 4;
cout << mystery(a + 1) << endl;
}
12. What does the following program print? Use a diagram like Figure 3 to find the
answer.
int mystery(int n)
{
n++;
n++;
return n;
}
int main()
{
int a = 5;
cout << mystery(a) << endl;
}
practice it Now you can try these exercises at the end of the chapter: R5.5, P5.10.
do not modify parameter Variables
In C++, a parameter variable is just like any other variable. You can modify the values of the
parameter variables in the body of a function. For example,
int total_cents(int dollars, int cents)
{
cents = dollars * 100 + cents; // Modifies parameter variable
return cents;
}
However, many programmers find this practice confusing. It mixes the concept of a parameter
(input to the func tion) with that of a variable (storage for a value). To avoid the confusion,
simply introduce a separate variable:
int total_cents(int dollars, int cents)
{
int result = dollars * 100 + cents;
return result;
}
programming tip 5.2
cfe2_ch05_p193_248.indd 201 10/26/10 6:14 PM
202 Chapter 5 Functions
5.4 return Values
You use the return statement to specify the result of a function. When the return state-
ment is processed, the function exits immediately. This behavior is convenient for
handling exceptional cases at the begin ning:
double cube_volume(double side_length)
{
if (side_length < 0) { return 0; }
double volume = side_length * side_length * side_length;
return volume;
}
If the function is called with a negative value for side_length, then the function returns
0 and the remain der of the function is not executed. (See Figure 4.)
In the preceding example, each return statement returned a constant or a variable.
Actually, the return statement can return the value of any expression. Instead of sav-
ing the return value in a variable and returning the variable, it is often possible to
eliminate the variable and return a more complex expression:
double cube_volume(double side_length)
{
return side_length * side_length * side_length;
}
It is important that every branch of a function return a value. Consider the following
incorrect function:
double cube_volume(double side_length)
{
if (side_length >= 0)
{
return side_length * side_length * side_length;
} // Error
}
Suppose you call cube_volume with a negative value for the side length. Of course, you
aren’t supposed to call that, but it might happen as the result of a coding error. Because
the if condition is not true, the return statement is not executed. However, the func-
tion must return something. Depending on the circumstances, the compiler might
the return statement
terminates a function
call and yields the
function result.
side_length < 0? return 0
return volume
volume =
side_length ×
side_length ×
side_length
True
False
Figure 4 A return statement exits a Function immediately
flag this as an error, or the function might return a random value. Protect against this
problem by returning some safe value:
double cube_volume(double side_length)
{
if (side_length >= 0)
{
return side_length * side_length * side_length;
}
return 0;
}
The last statement of every function ought to be a return statement. This ensures that
some value gets returned when the function reaches the end.
missing return Value
A function always needs to return something. If the code of the function contains several
branches, make sure that each one of them returns a value:
int sign(double x)
{
if (x < 0) { return -1; }
if (x > 0) { return 1; }
// Error: missing return value if x equals 0
}
This function computes the sign of a number: –1 for negative numbers and +1 for positive
numbers. If the argument is zero, however, no value is returned. Most compilers will issue a
warning in this situation, but if you ignore the warning and the function is ever called with an
argument of 0, a random quantity will be returned.
Function declarations
It is a compile-time error to call a function that the compiler does not know, just as it is an
error to use an undefined variable. You can avoid this error if you define all functions before
they are first used. First define lower-level helper functions, then the mid-level workhorse
functions, and finally main in your program.
Some programmers prefer to list the main function first in their programs. If you share that
preference, you need to learn how to declare the other functions at the top of the program. A
declaration of a function lists the return type, function name, and parameter variables, but it
contains no body:
double cube_volume(double side_length);
This is an advertisement that promises that the function is implemented elsewhere. It is easy
to distinguish declara tions from definitions: Declarations end in a semicolon, whereas defini-
tions are followed by a {…} block. Declara tions are also called prototypes.
In a function prototype, the names of the parameters are optional. You could also write
double cube_volume(double);
However, it is a good idea to include parameter names in order to document the purpose of
each parameter.
The declarations of common functions such as pow are contained in header files. If you have
a look inside
Common error 5.1
special topic 5.1
cfe2_ch05_p193_248.indd 202 10/26/10 6:14 PM
5.4 return Values 203
flag this as an error, or the function might return a random value. Protect against this
problem by returning some safe value:
double cube_volume(double side_length)
{
if (side_length >= 0)
{
return side_length * side_length * side_length;
}
return 0;
}
The last statement of every function ought to be a return statement. This ensures that
some value gets returned when the function reaches the end.
missing return Value
A function always needs to return something. If the code of the function contains several
branches, make sure that each one of them returns a value:
int sign(double x)
{
if (x < 0) { return -1; }
if (x > 0) { return 1; }
// Error: missing return value if x equals 0
}
This function computes the sign of a number: –1 for negative numbers and +1 for positive
numbers. If the argument is zero, however, no value is returned. Most compilers will issue a
warning in this situation, but if you ignore the warning and the function is ever called with an
argument of 0, a random quantity will be returned.
Function declarations
It is a compile-time error to call a function that the compiler does not know, just as it is an
error to use an undefined variable. You can avoid this error if you define all functions before
they are first used. First define lower-level helper functions, then the mid-level workhorse
functions, and finally main in your program.
Some programmers prefer to list the main function first in their programs. If you share that
preference, you need to learn how to declare the other functions at the top of the program. A
declaration of a function lists the return type, function name, and parameter variables, but it
contains no body:
double cube_volume(double side_length);
This is an advertisement that promises that the function is implemented elsewhere. It is easy
to distinguish declara tions from definitions: Declarations end in a semicolon, whereas defini-
tions are followed by a {…} block. Declara tions are also called prototypes.
In a function prototype, the names of the parameters are optional. You could also write
double cube_volume(double);
However, it is a good idea to include parameter names in order to document the purpose of
each parameter.
The declarations of common functions such as pow are contained in header files. If you have
a look inside
Common error 5.1
special topic 5.1
cfe2_ch05_p193_248.indd 203 10/26/10 6:14 PM
204 Chapter 5 Functions
Here is an alternate organization of the cube.cpp file:
#include
using namespace std;
// Declaration of cube_volume
double cube_volume(double side_length);
int main()
{
double result1 = cube_volume(2); // Use of cube_volume
double result2 = cube_volume(10);
cout << "A cube with side length 2 has volume " << result1 << endl;
cout << "A cube with side length 10 has volume " << result2 << endl;
return 0;
}
// Definition of cube_volume
double cube_volume(double side_length)
{
return side_length * side_length * side_length;
}
If you prefer this approach, go ahead and use it in your programs. You just need to be aware
of one drawback. Whenever you change the name of a function or one of the parameter types,
you need to fix it in both places: in the declaration and in the definition.
step 1 Describe what the function should do.
Provide a simple English description, such as “Compute the volume of a pyramid whose base
is a square.”
step 2 Determine the function’s “inputs”.
Make a list of all the parameters that can vary. It is
common for beginners to implement functions that
are overly specific. For example, you may know that
the great pyramid of Giza, the largest of the Egyptian
pyramids, has a height of 146 meters and a base length
of 230 meters. You should not use these numbers in
your calculation, even if the original problem only
asked about the great pyramid. It is just as easy—and
far more useful—to write a function that com putes the
volume of any pyramid.
h o W t o 5 . 1 implementing a Function
A function is a computation that can be used multiple times with different parameters, either
in the same program or in different programs. Whenever a computation is needed more than
once, turn it into a function.
To illustrate this process, suppose that you are helping archaeologists who research Egyp-
tian pyramids. You have taken on the task of writing a function that determines the volume of
a pyramid, given its height and base length.
cfe2_ch05_p193_248.indd 204 10/26/10 6:14 PM
5.4 return Values 205
Here is an alternate organization of the cube.cpp file:
#include
using namespace std;
// Declaration of cube_volume
double cube_volume(double side_length);
int main()
{
double result1 = cube_volume(2); // Use of cube_volume
double result2 = cube_volume(10);
cout << "A cube with side length 2 has volume " << result1 << endl;
cout << "A cube with side length 10 has volume " << result2 << endl;
return 0;
}
// Definition of cube_volume
double cube_volume(double side_length)
{
return side_length * side_length * side_length;
}
If you prefer this approach, go ahead and use it in your programs. You just need to be aware
of one drawback. Whenever you change the name of a function or one of the parameter types,
you need to fix it in both places: in the declaration and in the definition.
step 1 Describe what the function should do.
Provide a simple English description, such as “Compute the volume of a pyramid whose base
is a square.”
step 2 Determine the function’s “inputs”.
Make a list of all the parameters that can vary. It is
common for beginners to implement functions that
are overly specific. For example, you may know that
the great pyramid of Giza, the largest of the Egyptian
pyramids, has a height of 146 meters and a base length
of 230 meters. You should not use these numbers in
your calculation, even if the original problem only
asked about the great pyramid. It is just as easy—and
far more useful—to write a function that com putes the
volume of any pyramid.
h o W t o 5 . 1 implementing a Function
A function is a computation that can be used multiple times with different parameters, either
in the same program or in different programs. Whenever a computation is needed more than
once, turn it into a function.
To illustrate this process, suppose that you are helping archaeologists who research Egyp-
tian pyramids. You have taken on the task of writing a function that determines the volume of
a pyramid, given its height and base length.
In our case, the parameters are the pyramid’s height and base length. At this point, we have
enough information to document the function:
/**
Computes the volume of a pyramid whose base is a square.
@param height the height of the pyramid
@param base_length the length of one side of the pyramid’s base
@return the volume of the pyramid
*/
step 3 Determine the types of the parameter variables and the return value.
The height and base length can both be floating-point numbers. Therefore, we will choose the
type double for both parameter variables. The computed volume is also a floating-point num-
ber, yielding a return type of double. There fore, the function will be defined as
double pyramid_volume(double height, double base_length)
step 4 Write pseudocode for obtaining the desired result.
In most cases, a function needs to carry out several steps to find the desired answer. You may
need to use mathemat ical formulas, branches, or loops. Express your function in pseudocode.
An Internet search yields the fact that the volume of a pyramid is computed as
volume = 1/3 x height x base area
Since the base is a square, we have
base area = base length x base length
Using these two equations, we can compute the volume from the parameter variables.
step 5 Implement the function body.
In our example, the function body is quite simple. Note the use of the return statement to
return the result.
{
double base_area = base_length * base_length;
return height * base_area / 3;
}
step 6 Test your function.
After implementing a function, you should test it in isolation. Such a test is called a unit test.
Work out test cases by hand, and make sure that the function produces the correct results. For
example, for a pyramid with height 9 and base length 10, we expect the area to be 1̸ 3 × 9 × 100
= 300. If the height is 0, we expect an area of 0.
int main()
{
cout << "Volume: " << pyramid_volume(9, 10) << endl;
cout << "Expected: 300"
cout << "Volume: " << pyramid_volume(0, 10) << endl;
cout << "Expected: 0"
return 0;
}
The output confirms that the function worked as expected:
Volume: 300
Expected: 300
Volume: 0
Expected: 0
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206 Chapter 5 Functions
5.5 Functions Without return Values
Sometimes, you need to carry out a
sequence of instructions that does
not yield a value. If that instruction
sequence occurs multiple times, you
will want to package it into a function.
In C++, you use the return type void to
indicate the absence of a return value.
Here is a typical example. Your task
is to print a string in a box, like this:
-------
!Hello!
-------
However, different strings can be substituted for Hello. A function for this task can be
defined as follows:
void box_string(string str)
Now you develop the body of the function in the usual way, by formulating a general
method for solving the task.
• Print a line that contains the - character n + 2 times, where n is the length of the string.
• Print a line containing the string, surrounded with a ! to the left and right.
• Print another line containing the - character n + 2 times.
Here is the function implementation:
/**
Prints a string in a box.
@param str the string to print
*/
void box_string(string str)
{
W o r K e D e x a M p l e 5 . 1 matching and replacing parts of a string
This Worked Example creates a function to find the first occurrence of one string in a given
string and replace it with a second string.
W o r K e D e x a M p l e 5 . 2 using a debugger
In this Worked Example, you will learn how to use a debugger to find errors in a program.
A void function returns no value, but it can
produce output.
use a return type of
void to indicate that a
function does not
return a value.
Available online at www.wiley.com/college/horstmann.
cfe2_ch05_p193_248.indd 206 10/26/10 6:14 PM
www.wiley.com/college/horstmann
5.5 Functions Without return Values 207
5.5 Functions Without return Values
Sometimes, you need to carry out a
sequence of instructions that does
not yield a value. If that instruction
sequence occurs multiple times, you
will want to package it into a function.
In C++, you use the return type void to
indicate the absence of a return value.
Here is a typical example. Your task
is to print a string in a box, like this:
-------
!Hello!
-------
However, different strings can be substituted for Hello. A function for this task can be
defined as follows:
void box_string(string str)
Now you develop the body of the function in the usual way, by formulating a general
method for solving the task.
• Print a line that contains the - character n + 2 times, where n is the length of the string.
• Print a line containing the string, surrounded with a ! to the left and right.
• Print another line containing the - character n + 2 times.
Here is the function implementation:
/**
Prints a string in a box.
@param str the string to print
*/
void box_string(string str)
{
W o r K e D e x a M p l e 5 . 1 matching and replacing parts of a string
This Worked Example creates a function to find the first occurrence of one string in a given
string and replace it with a second string.
W o r K e D e x a M p l e 5 . 2 using a debugger
In this Worked Example, you will learn how to use a debugger to find errors in a program.
A void function returns no value, but it can
produce output.
use a return type of
void to indicate that a
function does not
return a value.
int n = str.length();
for (int i = 0; i < n + 2; i++) { cout << "-"; }
cout << endl;
cout << "!" << str << "!" << endl;
for (int i = 0; i < n + 2; i++) { cout << "-"; }
cout << endl;
}
Note that this function doesn’t compute any value. It performs some actions and
then returns to the caller. (See the sample program ch05/box.cpp.)
Because there is no return value, you cannot use box_string in an expression. You
can call
box_string("Hello");
but not
result = box_string("Hello"); // Error: box_string doesn’t return a result.
If you want to return from a void function before reaching the end, you use a return
statement without a value. For example,
void box_string(string str)
{
int n = str.length();
if (n == 0)
{
return; // Return immediately
}
...
}
13. How do you generate the following printout, using the box_string function?
-------
!Hello!
-------
-------
!World!
-------
14. What is wrong with the following statement?
cout << box_string("Hello");
15. Implement a function shout that prints a line consisting of a string followed by
three exclamation marks. For example, shout("Hello") should print Hello!!!. The
function should not return a value.
16. How would you modify the box_string function to leave a space around the
string that is being boxed, like this:
---------
! Hello !
---------
17. The box_string function contains the code for printing a line of - characters twice.
Place that code into a separate function print_line, and use that function to sim-
plify box_string. What is the code of both functions?
practice it Now you can try these exercises at the end of the chapter: R5.4, P5.24.
s e l F c h e c k
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208 Chapter 5 Functions
5.6 problem solving: reusable Functions
You have used many functions from the C++ standard library. These functions have
been provided as a part of standard C++ so that programmers need not recreate them.
Of course, the C++ library doesn’t cover every conceivable need. You will often be
able to save yourself time by designing your own func tions that can be used for mul-
tiple problems.
When you write nearly identical code or pseudocode multiple times, either in the
same program or in separate programs, consider introducing a function. Here is a
typical example of code replication:
int hours;
do
{
cout << "Enter a value between 0 and 23: ";
cin >> hours;
}
while (hours < 0 || hours > 23);
int minutes;
do
{
cout << "Enter a value between 0 and 59: ";
cin >> minutes;
}
while (minutes < 0 || minutes > 59);
This program segment reads two variables, making sure that each of them is within a
certain range. It is easy to extract the common behavior into a function:
/**
Prompts a user to enter a value up to a given maximum until the user
provides a valid input.
@param high the largest allowable input
@return the value provided by the user (between 0 and high, inclusive)
*/
int read_int_up_to(int high)
{
int input;
do
{
cout << "Enter a value between 0 and " << high << ": ";
cin >> input;
}
while (input < 0 || input > high);
return input;
}
Then use this function twice:
int hours = read_int_up_to(23);
int minutes = read_int_up_to(59);
We have now removed the replication of the loop—it only occurs once, inside the
function.
Note that the function can be reused in other programs that need to read integer
values. However, we should consider the possibility that the smallest value need not
always be zero.
eliminate replicated
code or pseudocode
by defining a
function.
Here is a better alterna tive:
/**
Prompts a user to enter a value within a given range until the user
provides a valid input.
@param low the smallest allowable input
@param high the largest allowable input
@return the value provided by the user (between low and high, inclusive)
*/
int read_int_between(int low, int high)
{
int input;
do
{
cout << "Enter a value between " << low << " and " << high << ": ";
cin >> input;
}
while (input < low || input > high);
return input;
}
In our program, we call
int hours = read_int_between(0, 23);
Another program can call
int month = read_int_between(1, 12);
In general, you will want to provide parameter variables for the values that vary when
a function is reused.
18. Consider the following statements:
int total_pennies = static_cast
int total_tax_pennies = static_cast
Introduce a function to reduce code duplication.
19. Consider this code that prints a page number on the left or right side of a page:
if (page % 2 == 0) { cout << page << endl; }
else { cout << setw(80) << page << endl; }
Introduce a function with return type bool to make the condition in the if
statement easier to under stand.
Design your
functions to be
reusable. supply
parameter variables
for the values that
can vary when the
function is reused.
s e l F c h e c k
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5.6 problem solving: reusable Functions 209
When carrying out the same task
multiple times, use a function.
Here is a better alterna tive:
/**
Prompts a user to enter a value within a given range until the user
provides a valid input.
@param low the smallest allowable input
@param high the largest allowable input
@return the value provided by the user (between low and high, inclusive)
*/
int read_int_between(int low, int high)
{
int input;
do
{
cout << "Enter a value between " << low << " and " << high << ": ";
cin >> input;
}
while (input < low || input > high);
return input;
}
In our program, we call
int hours = read_int_between(0, 23);
Another program can call
int month = read_int_between(1, 12);
In general, you will want to provide parameter variables for the values that vary when
a function is reused.
18. Consider the following statements:
int total_pennies = static_cast
int total_tax_pennies = static_cast
Introduce a function to reduce code duplication.
19. Consider this code that prints a page number on the left or right side of a page:
if (page % 2 == 0) { cout << page << endl; }
else { cout << setw(80) << page << endl; }
Introduce a function with return type bool to make the condition in the if
statement easier to under stand.
Design your
functions to be
reusable. supply
parameter variables
for the values that
can vary when the
function is reused.
s e l F c h e c k
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210 Chapter 5 Functions
20. Consider the following function that computes compound interest for an
account with an initial bal ance of $10,000 and an interest rate of 5 percent:
double balance(int years) { return 10000 * pow(1.05, years); }
How can you make this function more reusable?
21. The comment explains what the following loop does. Use a function instead.
// Counts the number of spaces
int spaces = 0;
for (int i = 0; i < input.length(); i++)
{
if (input.substr(i, 1) == " ") { spaces++; }
}
22. In Self Check 21, you were asked to implement a function that counts spaces.
How can you general ize it so that it can count any character? Why would you
want to do this?
practice it Now you can try these exercises at the end of the chapter: R5.7, P5.23.
5.7 problem solving: stepwise refinement
One of the most powerful strategies for
problem solving is the process of stepwise
refinement. To solve a difficult task, break it
down into simpler tasks. Then keep break-
ing down the simpler tasks into even simpler
ones, until you are left with tasks that you
know how to solve.
Here is an application of this process to a
problem of everyday life. You get up in the
morning and simply must get coffee. How do
you get coffee? You see whether you can
get someone else, such as your mother or
mate, to bring you some. If that fails, you
must make coffee. How do you make coffee?
If there is instant coffee available, you can make instant coffee. How do you make instant
coffee? Simply boil water and mix the boiling water with the instant coffee. How do
you boil water? If there is a microwave, then you fill a cup with water, place it in the
microwave and heat it for three minutes. Otherwise, you fill a kettle with water and
heat it on the stove until the water comes to a boil. On the other hand, if you don’t
have instant coffee, you must brew coffee. How do you brew coffee? You add water to
the coffee maker, put in a filter, grind coffee, put the coffee in the filter, and turn the cof-
fee maker on. How do you grind coffee? You add coffee beans to the coffee grinder
and push the button for 60 seconds.
Figure 5 shows a flowchart view of the coffee-making solution. Refinements are
shown as expanding boxes. In C++, you implement a refinement as a function. For
example, a function brew_coffee would call grind_coffee, and it would be called from a
function make_coffee.
Let us apply the process of stepwise refinement to a program ming problem.
A production process is broken down
into sequences of assembly steps.
use the process of
stepwise refinement
to decompose
complex tasks into
simpler ones.
cfe2_ch05_p193_248.indd 210 10/26/10 6:14 PM
5.7 problem solving: stepwise refinement 211
Figure 5 Flowchart of Coffee-Making solution
Yes No
Get
coffee
Ask for
coffee
Can you
ask someone
?
Make
coffee
Yes No
Do you
have instant
coffee?
Brew
coffee
Add coffee
beans to
grinderPut cup
in micro-
wave
Bring to
a boil
Fill cup
with water
Fill kettle
with water
Heat
3 min.
Grind
60 sec.
Add water
to coffee
maker
Add filter
to coffee
maker
Add coffee
beans to
filter
Grind
coffee
beans
Turn coffee
maker on
Make
instant
coffee
Boil
water
Mix water
and instant
coffee
Do you
have a micro-
wave?
Yes No
When printing a check, it is customary to
write the check amount both as a number
(“$274.15”) and as a text string (“two hundred
seventy four dollars and 15 cents”). Doing so
reduces the recipient’s temptation to add a few
digits in front of the amount.
For a human, this isn’t particularly difficult,
but how can a com puter do this? There is no
built-in function that turns 274 into "two hundred seventy four". We need to program
this function. Here is the description of the function we want to write:
/**
Turns a number into its English name.
@param number a positive integer < 1,000
@return the name of number (e.g., “two hundred seventy four”)
*/
string int_name(int number)
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212 Chapter 5 Functions
How can this function do its job? Let’s look at a simple case first. If the number is
between 1 and 9, we need to compute "one" … "nine". In fact, we need the same
computation again for the hundreds (two hun dred). Using the stepwise decomposition
process, we design another function for this simpler task. Again, rather than imple-
menting the function, we first write the comment:
/**
Turns a digit into its English name.
@param digit an integer between 1 and 9
@return the name of digit (“one” ... “nine”)
*/
string digit_name(int digit)
This sounds simple enough to implement, using an if statement with nine branches.
No further functions should be required for completing the digit_name function, so
we will worry about the implementation later.
Numbers between 10 and 19 are special cases. Let’s have a separate function teen_
name that converts them into strings "eleven", "twelve", "thirteen", and so on:
/**
Turns a number between 10 and 19 into its English name.
@param number an integer between 10 and 19
@return the name of the number (“ten” ... “nineteen”)
*/
string teen_name(int number)
Next, suppose that the number is between 20 and 99. Then we show the tens as
"twenty", "thirty", …, "ninety". For simplicity and consistency, put that computation
into a separate function:
/**
Gives the name of the tens part of a number between 20 and 99.
@param number an integer between 20 and 99
@return the name of the tens part of the number (“twenty” ... “ninety”)
*/
string tens_name(int number)
Now suppose the number is at least 20 and at most 99. If the number is evenly divisi-
ble by 10, we use tens_name, and we are done. Otherwise, we print the tens with tens_
name and the ones with digit_name. If the number is between 100 and 999, then we show
a digit, the word "hundred", and the remainder as described previously.
Here is the pseudocode of the algorithm:
part = number (The part that still needs to be converted)
name = "" (The name of the number)
If part >= 100
name = name of hundreds in part + ” hundred”
Remove hundreds from part.
If part >= 20
Append tens_name(part) to name.
Remove tens from part.
Else if part >= 10
Append teen_name(part) to name.
part = 0
When you discover
that you need a
function, write a
description of the
parameter variables
and return values.
a function may
require simpler
functions to carry
out its work.
cfe2_ch05_p193_248.indd 212 10/26/10 6:14 PM
5.7 problem solving: stepwise refinement 213
If (part > 0)
Append digit_name(part) to name.
This pseudocode has a number of important improvements over the verbal descrip-
tion. It shows how to arrange the tests, starting with the comparisons against the
larger numbers, and it shows how the smaller number is subsequently processed in
further if statements.
On the other hand, this pseudocode is vague about the actual conversion of the
pieces, just referring to “name of hundreds” and the like. Furthermore, we were
vague about spaces. As it stands, the code would produce strings with no spaces,
twohundredseventyfour, for example. Compared to the complexity of the main problem,
one would hope that spaces are a minor issue. It is best not to muddy the pseudocode
with minor details.
Now turn the pseudocode into real code. The last three cases are easy, because
helper functions are already developed for them:
if (part >= 20)
{
name = name + ” ” + tens_name(part);
part = part % 10;
}
else if (part >= 10)
{
name = name + ” ” + teen_name(part);
part = 0;
}
if (part > 0)
{
name = name + ” ” + digit_name(part);
}
Finally, let us tackle the case of numbers between 100 and 999. Because part < 1000,
part / 100 is a single digit, and we obtain its name by calling digit_name. Then we add
the “hundred” suffix:
if (part >= 100)
{
name = digit_name(part / 100) + ” hundred”;
part = part % 100;
}
Now you have seen all the important building blocks for the int_name function. Here
is the complete pro gram:
ch05/intname.cpp
1 #include
2 #include
3
4 using namespace std;
5
6 /**
7 Turns a digit into its English name.
8 @param digit an integer between 1 and 9
9 @return the name of digit (“one” … “nine”)
10 */
11 string digit_name(int digit)
12 {
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214 Chapter 5 Functions
13 if (digit == 1) return “one”;
14 if (digit == 2) return “two”;
15 if (digit == 3) return “three”;
16 if (digit == 4) return “four”;
17 if (digit == 5) return “five”;
18 if (digit == 6) return “six”;
19 if (digit == 7) return “seven”;
20 if (digit == 8) return “eight”;
21 if (digit == 9) return “nine”;
22 return “”;
23 }
24
25 /**
26 Turns a number between 10 and 19 into its English name.
27 @param number an integer between 10 and 19
28 @return the name of the given number (“ten” … “nineteen”)
29 */
30 string teen_name(int number)
31 {
32 if (number == 10) return “ten”;
33 if (number == 11) return “eleven”;
34 if (number == 12) return “twelve”;
35 if (number == 13) return “thirteen”;
36 if (number == 14) return “fourteen”;
37 if (number == 15) return “fifteen”;
38 if (number == 16) return “sixteen”;
39 if (number == 17) return “seventeen”;
40 if (number == 18) return “eighteen”;
41 if (number == 19) return “nineteen”;
42 return “”;
43 }
44
45 /**
46 Gives the name of the tens part of a number between 20 and 99.
47 @param number an integer between 20 and 99
48 @return the name of the tens part of the number (“twenty” … “ninety”)
49 */
50 string tens_name(int number)
51 {
52 if (number >= 90) return “ninety”;
53 if (number >= 80) return “eighty”;
54 if (number >= 70) return “seventy”;
55 if (number >= 60) return “sixty”;
56 if (number >= 50) return “fifty”;
57 if (number >= 40) return “forty”;
58 if (number >= 30) return “thirty”;
59 if (number >= 20) return “twenty”;
60 return “”;
61 }
62
63 /**
64 Turns a number into its English name.
65 @param number a positive integer < 1,000
66 @return the name of the number (e.g. “two hundred seventy four”)
67 */
68 string int_name(int number)
69 {
70 int part = number; // The part that still needs to be converted
71 string name; // The return value
72
cfe2_ch05_p193_248.indd 214 10/26/10 6:14 PM
5.7 problem solving: stepwise refinement 215
73 if (part >= 100)
74 {
75 name = digit_name(part / 100) + ” hundred”;
76 part = part % 100;
77 }
78
79 if (part >= 20)
80 {
81 name = name + ” ” + tens_name(part);
82 part = part % 10;
83 }
84 else if (part >= 10)
85 {
86 name = name + ” ” + teen_name(part);
87 part = 0;
88 }
89
90 if (part > 0)
91 {
92 name = name + ” ” + digit_name(part);
93 }
94
95 return name;
96 }
97
98 int main()
99 {
100 cout << "Please enter a positive integer: ";
101 int input;
102 cin >> input;
103 cout << int_name(input) << endl;
104 return 0;
105 }
program run
Please enter a positive integer: 729
seven hundred twenty nine
23. Explain how you can improve the int_name function so that it can handle argu-
ments up to 9,999.
24. Why does line 87 set part = 0?
25. What happens when you call int_name(0)? How can you change the int_name
function to handle this case correctly?
26. Trace the function call int_name(72), as described in Programming Tip 5.4.
27. Use the process of stepwise refinement to break down the task of printing the
following table into simpler tasks.
+-----+-----------+
| i | i * i * i |
+-----+-----------+
| 1 | 1 |
| 2 | 8 |
...
| 20 | 8000 |
+-----+-----------+
s e l F c h e c k
cfe2_ch05_p193_248.indd 215 10/26/10 6:14 PM
216 Chapter 5 Functions
practice it Now you can try these exercises at the end of the chapter: R5.12, P5.16, P5.19.
keep Functions short
There is a certain cost for writing a function. You need to design, code, and test the function.
The function needs to be documented. You need to spend some effort to make the function
reusable rather than tied to a specific context. To avoid this cost, it is always tempting just to
stuff more and more code in one place rather than going through the trouble of breaking up
the code into separate functions. It is quite common to see inexperienced programmers pro-
duce functions that are several hundred lines long.
As a rule of thumb, a function that is so long that its code will not fit on a single screen in
your development envi ronment should probably be broken up.
tracing Functions
When you design a complex set of functions, it is a good idea to carry out a manual walk-
through before entrusting your program to the computer.
Take an index card, or some other piece of paper, and write down the function call that you
want to study. Write the name of the function and the names and values of the parameter vari-
ables, like this:
int_name(number = 416)
Then write the names and initial values of the function variables. Write them in a table, since
you will update them as you walk through the code.
int_name(number = 416)
part name
416 ""
We enter the test part >= 100. part / 100 is 4 and part % 100 is 16. digit_name(4) is easily seen to
be “four”. (Had digit_name been complicated, you would have started another sheet of paper
to figure out that function call. It is quite common to accumulate several sheets in this way.)
Now name has changed to name + ” ” + digit_name(part / 100) + ” hundred”, that is “four hun-
dred”, and part has changed to part % 100, or 16.
int_name(number = 416)
part name
416 “”
16 “four hundred”
programming tip 5.3
programming tip 5.4
Now you enter the branch part >= 10. teen_name(16) is sixteen, so the variables now have the
values
Now it becomes clear why you need to set part to 0 in line 87. Otherwise, you would enter the
next branch and the result would be “four hundred sixteen six”. Tracing the code is an effective
way to understand the subtle aspects of a func tion.
stubs
When writing a larger program, it is not always
feasible to implement and test all functions at
once. You often need to test a function that calls
another, but the other function hasn’t yet been
implemented. Then you can temporarily replace
the missing function with a stub. A stub is a func-
tion that returns a simple value that is sufficient
for testing another function. Here are examples of
stub functions:
/**
Turns a digit into its English name.
@param digit an integer between 1 and 9
@return the name of digit (“one” … “nine”)
*/
string digit_name(int digit)
{
return “mumble”;
}
/**
Gives the name of the tens part of a number between 20 and 99.
@param number an integer between 20 and 99
@return the tens name of the number (“twenty” … “ninety”)
*/
string tens_name(int number)
{
return “mumblety”;
}
If you combine these stubs with the int_name function and test it with an argument of 274, you
will get a result of “mumble hundred mumblety mumble”, which indicates that the basic logic of the
int_name function is working correctly.
programming tip 5.5
Stubs are incomplete functions that can
be used for testing.
cfe2_ch05_p193_248.indd 216 10/26/10 6:14 PM
5.7 problem solving: stepwise refinement 217
Now you enter the branch part >= 10. teen_name(16) is sixteen, so the variables now have the
values
int_name(number = 416)
part name
416 “”
16 “four hundred”
0 “four hundred sixteen”
Now it becomes clear why you need to set part to 0 in line 87. Otherwise, you would enter the
next branch and the result would be “four hundred sixteen six”. Tracing the code is an effective
way to understand the subtle aspects of a func tion.
stubs
When writing a larger program, it is not always
feasible to implement and test all functions at
once. You often need to test a function that calls
another, but the other function hasn’t yet been
implemented. Then you can temporarily replace
the missing function with a stub. A stub is a func-
tion that returns a simple value that is sufficient
for testing another function. Here are examples of
stub functions:
/**
Turns a digit into its English name.
@param digit an integer between 1 and 9
@return the name of digit (“one” … “nine”)
*/
string digit_name(int digit)
{
return “mumble”;
}
/**
Gives the name of the tens part of a number between 20 and 99.
@param number an integer between 20 and 99
@return the tens name of the number (“twenty” … “ninety”)
*/
string tens_name(int number)
{
return “mumblety”;
}
If you combine these stubs with the int_name function and test it with an argument of 274, you
will get a result of “mumble hundred mumblety mumble”, which indicates that the basic logic of the
int_name function is working correctly.
programming tip 5.5
Stubs are incomplete functions that can
be used for testing.
cfe2_ch05_p193_248.indd 217 10/26/10 6:14 PM
218 Chapter 5 Functions
5.8 Variable scope and Global Variables
It is possible to define the same variable name more than once in a program. When the
variable name is used, you need to know to which definition it belongs. In this sec-
tion, we discuss the rules for dealing with multiple definitions of the same name.
A variable that is defined within a function is visible from the point at which it is
defined until the end of the block in which it was defined. This area is called the scope
of the variable.
Consider the volume variables in the following example:
Each volume variable is defined in a separate function, and their scopes do not
overlap.
W o r K e D e x a M p l e 5 . 3 calculating a course grade
This Worked Example uses stepwise refinement to solve
the problem of converting a set of letter grades into an
average grade for a course.
the scope of a
variable is the part of
the program in which
it is visible.
double cube_volume(double side_length)
{
double volume = side_length * side_length * side_length;
return volume;
}
int main()
{
double volume = cube_volume(2);
cout << volume << endl;
return 0;
}
In the same way that there can be a street named “Main Street” in different cities,
a C++ program can have multiple variables with the same name.
It is not legal to define two variables with the same name in the same scope. For
example, the following is not legal:
int main()
{
double volume = cube_volume(2);
double volume = cube_volume(10);
// ERROR: cannot define another volume variable in this scope
...
}
However, you can define another variable with the same name in a nested block.
Here, we define two variables called amount.
The scope of the parameter variable amount is the entire function, except inside the
nested block. Inside the nested block, amount refers to the variable that was defined in
that block. We say that the inner variable shadows the variable that is defined in the
outer block. You should avoid this potentially confusing situation in the functions
that you write, simply by renaming one of the variables.
Variables that are defined inside functions are called local variables. C++ also sup-
ports global vari ables: variables that are defined outside functions. A global variable
is visible to all functions that are defined after it. For example, the
defines global variables cin and cout.
Here is an example of a global variable:
int balance = 10000; // A global variable
void withdraw(double amount)
{
if (balance >= amount)
{
balance = balance – amount;
}
}
int main()
{
withdraw(1000);
cout << balance << endl;
return 0;
}
The scope of the variable balance extends over both the withdraw and the main functions.
Generally, global variables are not a good idea. When multiple functions update
global variables, the result can be difficult to predict. Particularly in larger programs
that are developed by multiple program mers, it is very important that the effect of
each function be clear and easy to understand. You should avoid global variables in
your programs.
double withdraw(double balance, double amount)
{
if (...)
{
double amount = 10; // Another variable named amount
...
}
...
}
a variable in a nested
block shadows a
variable with the
same name in an
outer block.
a local variable is
defined inside a
function. a global
variable is defined
outside a function.
avoid global
variables in your
programs.
Available online at www.wiley.com/college/horstmann.
cfe2_ch05_p193_248.indd 218 10/26/10 6:14 PM
www.wiley.com/college/horstmann
5.8 Variable scope and Global Variables 219
5.8 Variable scope and Global Variables
It is possible to define the same variable name more than once in a program. When the
variable name is used, you need to know to which definition it belongs. In this sec-
tion, we discuss the rules for dealing with multiple definitions of the same name.
A variable that is defined within a function is visible from the point at which it is
defined until the end of the block in which it was defined. This area is called the scope
of the variable.
Consider the volume variables in the following example:
Each volume variable is defined in a separate function, and their scopes do not
overlap.
W o r K e D e x a M p l e 5 . 3 calculating a course grade
This Worked Example uses stepwise refinement to solve
the problem of converting a set of letter grades into an
average grade for a course.
the scope of a
variable is the part of
the program in which
it is visible.
double cube_volume(double side_length)
{
double volume = side_length * side_length * side_length;
return volume;
}
int main()
{
double volume = cube_volume(2);
cout << volume << endl;
return 0;
}
In the same way that there can be a street named “Main Street” in different cities,
a C++ program can have multiple variables with the same name.
It is not legal to define two variables with the same name in the same scope. For
example, the following is not legal:
int main()
{
double volume = cube_volume(2);
double volume = cube_volume(10);
// ERROR: cannot define another volume variable in this scope
...
}
However, you can define another variable with the same name in a nested block.
Here, we define two variables called amount.
The scope of the parameter variable amount is the entire function, except inside the
nested block. Inside the nested block, amount refers to the variable that was defined in
that block. We say that the inner variable shadows the variable that is defined in the
outer block. You should avoid this potentially confusing situation in the functions
that you write, simply by renaming one of the variables.
Variables that are defined inside functions are called local variables. C++ also sup-
ports global vari ables: variables that are defined outside functions. A global variable
is visible to all functions that are defined after it. For example, the
defines global variables cin and cout.
Here is an example of a global variable:
int balance = 10000; // A global variable
void withdraw(double amount)
{
if (balance >= amount)
{
balance = balance – amount;
}
}
int main()
{
withdraw(1000);
cout << balance << endl;
return 0;
}
The scope of the variable balance extends over both the withdraw and the main functions.
Generally, global variables are not a good idea. When multiple functions update
global variables, the result can be difficult to predict. Particularly in larger programs
that are developed by multiple program mers, it is very important that the effect of
each function be clear and easy to understand. You should avoid global variables in
your programs.
double withdraw(double balance, double amount)
{
if (...)
{
double amount = 10; // Another variable named amount
...
}
...
}
a variable in a nested
block shadows a
variable with the
same name in an
outer block.
a local variable is
defined inside a
function. a global
variable is defined
outside a function.
avoid global
variables in your
programs.
cfe2_ch05_p193_248.indd 219 10/26/10 6:14 PM
220 Chapter 5 Functions
Consider this sample program:
1 int x;
2 int mystery(int x)
3 {
4 int s = 0;
5 for (int i = 0; i < x; i++)
6 {
7 int x = i + 1;
8 s = s + x;
9 }
10 return x;
11 }
12 int main()
13 {
14 x = 4;
15 int s = mystery(x);
16 cout << s << endl;
17 }
28. Which line defines a global variable?
29. Which lines define local variables named x?
30. Which lines are in the scope of the definition of x in line 2?
31. Which variable is changed by the assignment in line 14?
32. This program defines two variables with the same name whose scopes don’t
overlap. What are they?
practice it Now you can try these exercises at the end of the chapter: R5.8, R5.9.
avoid global Variables
There are a few cases where global variables are required (such as cin and cout), but they are
quite rare. Programs with global variables are difficult to maintain and extend because you
can no longer view each function as a “black box” that simply receives arguments and returns
a result. When functions modify global variables, it becomes more difficult to understand the
effect of function calls. As programs get larger, this difficulty mounts quickly. Instead of using
global variables, use function parameters to transfer information from one part of a program
to another.
5.9 reference parameters
If you want to write a function that changes the value of an argument, you must use a
reference parame ter in order to allow the change. We first explain why a different
parameter type is necessary, then we show you the syntax for reference parameters.
Consider a function that simulates withdrawing a given amount of money from a
bank account, pro vided that sufficient funds are available. If the amount of money is
insufficient, a $10 penalty is deducted instead. The function would be used as follows:
double harrys_account = 1000;
withdraw(harrys_account, 100); // Now harrys_account is 900
withdraw(harrys_account, 1000); // Insufficient funds. Now harrys_account is 890
s e l F c h e c k
programming tip 5.6
cfe2_ch05_p193_248.indd 220 10/26/10 6:14 PM
5.9 reference parameters 221
Here is a first attempt:
void withdraw(double balance, double amount) // Does not work
{
const double PENALTY = 10;
if (balance >= amount)
{
balance = balance – amount;
}
else
{
balance = balance – PENALTY;
}
}
But this doesn’t work.
Let’s walk through the function call withdraw(harrys_account, 100)—see Figure 6.
As the function starts, the parameter variable balance is created 1 and set to the same
value as harrys_account, and amount is set to 100 2 . Then balance is modified 3 . Of
course, that modification has no effect on harrys_account, because balance is a separate
variable. When the function returns, balance is forgotten, and no money was with-
drawn from harrys_account 4 .
Figure 6 When balance and account are Value parameters
1 Function call harrys_account =
balance =
amount =
1000
2 Initializing function parameter variables harrys_account =
balance =
amount = 100
1000
1000
3 About to return to the caller harrys_account =
balance =
amount = 100
900
1000
4 After function call harrys_account = 1000
withdraw(harrys_account, 100);
withdraw(harrys_account, 100);
balance = balance – amount;
cfe2_ch05_p193_248.indd 221 10/26/10 6:14 PM
222 Chapter 5 Functions
Figure 7 reference and Value parameters
harrys_account =
balance =
amount = 100
900
Reference
parameter
Value
parameter
main function
withdraw function
The parameter variable balance is called a value parameter, because it is initialized
with the value of the supplied argument. All functions that we have written so far use
value parameters. In this situation, though, we don’t just want balance to have the
same value as harrys_account. We want balance to refer to the actual variable harrys_
account (or joes_account or whatever variable is supplied in the call). The contents of
that variable should be updated.
You use a reference parameter when you want to update a variable that was sup-
plied in the function call. When we make balance into a reference parameter, then
balance is not a new variable but a reference to an existing variable. Any change in bal-
ance is actually a change in the variable to which balance refers in that particular call.
Figure 7 shows the difference between value and reference parameters.
To indicate a reference parameter, you place an & after the type name.
void withdraw(double& balance, double amount)
The type double& is read “a reference to a double” or, more briefly, “double ref”.
The withdraw function now has two parameter variables: one of type “double ref”
and the other a value parameter of type double. The body of the function is unchanged.
What has changed is the meaning of the assignments to the balance variable.
The assignment
balance = balance – amount;
now changes the variable that was passed to the function (see Figure 8).
Modifying a value
parameter has no
effect on the caller.
a reference
parameter refers to
a variable that is
supplied in a
function call.
A reference parameter for a bank balance is
like an ATM card—it allows you to change the
balance. In contrast, a value parameter can
only tell you the balance.
For example, the call
withdraw(harrys_account, 100);
modifies the variable harrys_account, and the call
withdraw(sallys_account, 150);
modifies the variable sallys_account.
The argument for a reference parameter must always be a variable. It would be an
error to supply a number:
withdraw(1000, 500); // Error: argument for reference parameter must be a variable
The reason is clear—the function modifies the reference parameter, but it is impossi-
ble to change the value of a number. For the same reason, you cannot supply an
expression:
withdraw(harrys_account + 150, 500);
// Error: argument for reference parameter must be a variable
ch05/account.cpp
1 #include
2
3 using namespace std;
4
Modifying a reference
parameter updates
the variable that was
supplied in the call.
cfe2_ch05_p193_248.indd 222 10/26/10 6:14 PM
5.9 reference parameters 223
Figure 8 When balance is a reference parameter
1 Function call harrys_account =
balance =
amount =
1000
2 Initializing function parameters harrys_account =
balance =
amount = 100
1000
3 About to return to the caller harrys_account =
balance =
amount = 100
900
4 After function call harrys_account = 900
withdraw(harrys_account, 100);
withdraw(harrys_account, 100);
balance = balance – amount;
For example, the call
withdraw(harrys_account, 100);
modifies the variable harrys_account, and the call
withdraw(sallys_account, 150);
modifies the variable sallys_account.
The argument for a reference parameter must always be a variable. It would be an
error to supply a number:
withdraw(1000, 500); // Error: argument for reference parameter must be a variable
The reason is clear—the function modifies the reference parameter, but it is impossi-
ble to change the value of a number. For the same reason, you cannot supply an
expression:
withdraw(harrys_account + 150, 500);
// Error: argument for reference parameter must be a variable
ch05/account.cpp
1 #include
2
3 using namespace std;
4
Modifying a reference
parameter updates
the variable that was
supplied in the call.
cfe2_ch05_p193_248.indd 223 10/26/10 6:14 PM
224 Chapter 5 Functions
5 /**
6 Withdraws the amount from the given balance, or withdraws
7 a penalty if the balance is insufficient.
8 @param balance the balance from which to make the withdrawal
9 @param amount the amount to withdraw
10 */
11 void withdraw(double& balance, double amount)
12 {
13 const double PENALTY = 10;
14 if (balance >= amount)
15 {
16 balance = balance – amount;
17 }
18 else
19 {
20 balance = balance – PENALTY;
21 }
22 }
23
24 int main()
25 {
26 double harrys_account = 1000;
27 double sallys_account = 500;
28 withdraw(harrys_account, 100);
29 // Now harrys_account is 900
30 withdraw(harrys_account, 1000); // Insufficient funds
31 // Now harrys_account is 890
32 withdraw(sallys_account, 150);
33 cout << "Harry's account: " << harrys_account << endl;
34 cout << "Sally's account: " << sallys_account << endl;
35
36 return 0;
37 }
program run
Harry's account: 890
Sally's account: 350
33. Would the withdraw function work correctly if the amount parameter was defined
as double& instead of double?
34. The following function is intended to transfer the given amount of money from
one account to another. Supply the function parameters.
void transfer(...)
{
if (balance1 >= amount)
{
balance1 = balance1 – amount;
balance2 = balance2 + amount;
}
}
35. Change the withdraw function so that it returns a bool value indicating whether
the withdrawal was successful. Do not charge a penalty if the balance was
insufficient.
s e l F c h e c k
36. Write a function minmax so that the call minmax(x, y, a, b) sets a to the smaller of x
and y and b to the larger of x and y.
37. What does this program print?
void mystery(int& a, int& b)
{
a = a – b;
b = b + a;
a = b – a;
}
int main()
{
int x = 4;
int y = 3;
mystery(x, y);
cout << x << " " << y << endl;
}
practice it Now you can try these exercises at the end of the chapter: R5.17, P5.14.
prefer return Values to reference parameters
Some programmers use reference parameters as a mechanism for setting the result of a func-
tion. For example,
void cube_volume(double side_length, double& volume)
{
volume = side_length * side_length * side_length;
}
However, this function is less convenient than our previous cube_volume function. It cannot be
used in expressions such as cout << cube_volume(2).
Use a reference parameter only when a function needs to update a variable.
constant references
It is not very efficient to have a value parameter that is a large object (such as a string value).
Copying the object into a parameter variable is less efficient than using a reference parameter.
With a reference parameter, only the location of the variable, not its value, needs to be trans-
mitted to the function.
You can instruct the compiler to give you the efficiency of a reference parameter and the
meaning of a value parameter, by using a constant reference as shown below. The function
void shout(const string& str)
{
cout << str << "!!!" << endl;
}
works exactly the same as the function
void shout(string str)
{
cout << str << "!!!" << endl;
}
There is just one difference: Calls to the first function execute a bit faster.
programming tip 5.7
special topic 5.2
cfe2_ch05_p193_248.indd 224 10/26/10 6:14 PM
5.9 reference parameters 225
36. Write a function minmax so that the call minmax(x, y, a, b) sets a to the smaller of x
and y and b to the larger of x and y.
37. What does this program print?
void mystery(int& a, int& b)
{
a = a - b;
b = b + a;
a = b - a;
}
int main()
{
int x = 4;
int y = 3;
mystery(x, y);
cout << x << " " << y << endl;
}
practice it Now you can try these exercises at the end of the chapter: R5.17, P5.14.
prefer return Values to reference parameters
Some programmers use reference parameters as a mechanism for setting the result of a func-
tion. For example,
void cube_volume(double side_length, double& volume)
{
volume = side_length * side_length * side_length;
}
However, this function is less convenient than our previous cube_volume function. It cannot be
used in expressions such as cout << cube_volume(2).
Use a reference parameter only when a function needs to update a variable.
constant references
It is not very efficient to have a value parameter that is a large object (such as a string value).
Copying the object into a parameter variable is less efficient than using a reference parameter.
With a reference parameter, only the location of the variable, not its value, needs to be trans-
mitted to the function.
You can instruct the compiler to give you the efficiency of a reference parameter and the
meaning of a value parameter, by using a constant reference as shown below. The function
void shout(const string& str)
{
cout << str << "!!!" << endl;
}
works exactly the same as the function
void shout(string str)
{
cout << str << "!!!" << endl;
}
There is just one difference: Calls to the first function execute a bit faster.
programming tip 5.7
special topic 5.2
cfe2_ch05_p193_248.indd 225 10/26/10 6:14 PM
226 Chapter 5 Functions
5.10 recursive Functions (optional)
A recursive function is a function that calls itself. This is not as
unusual as it sounds at first. Suppose you face the arduous task
of cleaning up an entire house. You may well say to yourself,
“I’ll pick a room and clean it, and then I’ll clean the other
rooms.” In other words, the cleanup task calls itself, but with a
simpler input. Eventually, all the rooms will be cleaned.
In C++, a recursive function uses the same principle. Here is
a typical example. We want to print triangle patterns like this:
[]
[][]
[][][]
[][][][]
Specifically, our task is to provide a function
void print_triangle(int side_length)
The triangle given above is printed by calling print_triangle(4).
To see how recursion helps, consider how a triangle with side length 4 can be
obtained from a triangle with side length 3.
[]
[][]
[][][]
[][][][]
Print the triangle with side length 3.
Print a line with four [].
More generally, for an arbitrary side length:
Print the triangle with side length – 1.
Print a line with side length [].
Here is the pseudocode translated to C++:
void print_triangle(int side_length)
{
print_triangle(side_length - 1);
for (int i = 0; i < side_length; i++)
{
cout << "[]";
}
cout << endl;
}
There is just one problem with this idea. When the side length is 1, we don’t want to
call print_triangle(0), print_triangle(-1), and so on. The solution is simply to treat
this as a special case, and not to print any thing when side_length is less than 1.
void print_triangle(int side_length)
{
if (side_length < 1) { return; }
print_triangle(side_length - 1);
for (int i = 0; i < side_length; i++)
{
cout << "[]";
Cleaning up a house can be solved recursively:
Clean one room, then clean up the rest.
}
cout << endl;
}
Look at the print_triangle function one more time and notice how utterly reasonable
it is. If the side length is 0, nothing needs to be printed. The next part is just as reason-
able. Print the smaller triangle and don’t think about why that works. Then print a
row of []. Clearly, the result is a triangle of the desired size.
There are two key requirements to make sure that the recursion is successful:
• Every recursive call must simplify the task in some way.
• There must be special cases to handle the simplest tasks directly.
The print_triangle function calls itself again with smaller and smaller side lengths.
Eventually the side length must reach 0, and the function stops calling itself.
Here is what happens when we print a triangle with side length 4.
• The call printTriangle(4) calls printTriangle(3).
• The call printTriangle(3) calls printTriangle(2).
• The call printTriangle(2) calls printTriangle(1).
• The call printTriangle(1) calls printTriangle(0).
• The call printTriangle(0) returns, doing nothing.
• The call printTriangle(1) prints [].
• The call printTriangle(2) prints [][].
• The call printTriangle(3) prints [][][].
• The call print_triangle(4) prints [][][][].
The call pattern of a recursive function looks complicated, and the key to the success-
ful design of a recur sive function is not to think about it.
ch05/triangle.cpp
1 #include
2
3 using namespace std;
4
5 /**
6 Prints a triangle with a given side length.
7 @param side_length the side length (number of [] along the base)
8 */
a recursive
computation solves
a problem by using
the solution of the
same problem with
simpler inputs.
For a recursion to
terminate, there must
be special cases for
the simplest inputs.
This set of Russian dolls looks similar to the
call pattern of a recursive function.
cfe2_ch05_p193_248.indd 226 10/26/10 6:14 PM
5.10 recursive Functions (optional) 227
}
cout << endl;
}
Look at the print_triangle function one more time and notice how utterly reasonable
it is. If the side length is 0, nothing needs to be printed. The next part is just as reason-
able. Print the smaller triangle and don’t think about why that works. Then print a
row of []. Clearly, the result is a triangle of the desired size.
There are two key requirements to make sure that the recursion is successful:
• Every recursive call must simplify the task in some way.
• There must be special cases to handle the simplest tasks directly.
The print_triangle function calls itself again with smaller and smaller side lengths.
Eventually the side length must reach 0, and the function stops calling itself.
Here is what happens when we print a triangle with side length 4.
• The call printTriangle(4) calls printTriangle(3).
• The call printTriangle(3) calls printTriangle(2).
• The call printTriangle(2) calls printTriangle(1).
• The call printTriangle(1) calls printTriangle(0).
• The call printTriangle(0) returns, doing nothing.
• The call printTriangle(1) prints [].
• The call printTriangle(2) prints [][].
• The call printTriangle(3) prints [][][].
• The call print_triangle(4) prints [][][][].
The call pattern of a recursive function looks complicated, and the key to the success-
ful design of a recur sive function is not to think about it.
ch05/triangle.cpp
1 #include
2
3 using namespace std;
4
5 /**
6 Prints a triangle with a given side length.
7 @param side_length the side length (number of [] along the base)
8 */
a recursive
computation solves
a problem by using
the solution of the
same problem with
simpler inputs.
For a recursion to
terminate, there must
be special cases for
the simplest inputs.
This set of Russian dolls looks similar to the
call pattern of a recursive function.
cfe2_ch05_p193_248.indd 227 10/26/10 6:14 PM
228 Chapter 5 Functions
9 void print_triangle(int side_length)
10 {
11 if (side_length < 1) { return; }
12 print_triangle(side_length - 1);
13 for (int i = 0; i < side_length; i++)
14 {
15 cout << "[]";
16 }
17 cout << endl;
18 }
19
20 int main()
21 {
22 cout << "Enter the side length: ";
23 int input;
24 cin >> input;
25 print_triangle(input);
26 return 0;
27 }
program run
Enter the side length: 10
[]
[][]
[][][]
[][][][]
[][][][][]
[][][][][][]
[][][][][][][]
[][][][][][][][]
[][][][][][][][][]
[][][][][][][][][][]
Recursion is not really necessary to print triangle shapes. You can use nested loops,
like this:
for (int i = 0; i < side_length; i++)
{
for (int j = 0; j < i; j++)
{
cout << "[]";
}
cout << endl;
}
However, this pair of loops is a bit tricky. Many people find the recursive solution
simpler to understand.
38. Consider this slight modification of the print_triangle function:
void print_triangle(int side_length)
{
if (side_length < 1) { return; }
for (int i = 0; i < side_length; i++)
{
cout << "[]";
s e l F c h e c k
}
cout << endl;
print_triangle(side_length - 1);
}
What is the result of print_triangle(4)?
39. Consider this recursive function:
int mystery(int n)
{
if (n <= 0) { return 0; }
return n + mystery(n - 1);
}
What is mystery(4)?
40. Consider this recursive function:
int mystery(int n)
{
if (n <= 0) { return 0; }
return mystery(n / 2) + 1;
}
What is mystery(20)?
41. Write a recursive function for printing n box shapes [] in a row.
42. The int_name function in Section 5.7 accepted arguments < 1,000. Using a
recursive call, extend its range to 999,999. For example an input of 12,345
should return "twelve thousand three hundred forty five".
practice it Now you can try these exercises at the end of the chapter: R5.20, P5.28, P5.30.
step 1 Break the input into parts that can themselves be inputs to the problem.
In your mind, fix a particular input or set of inputs for the task that
you want to solve, and think how you can simplify the inputs. Look
for simplifications that can be solved by the same task, and whose
solutions are related to the original task.
In the digit sum problem, consider how we can simplify an input
such as n = 1729. Would it help to subtract 1? After all, digit_sum(1729)
= digit_sum(1728) + 1. But consider n = 1000. There seems to be no obvious relationship
between digit_sum(1000) and digit_sum(999).
A much more promising idea is to remove the last digit, that is, compute n / 10 = 172. The
digit sum of 172 is directly related to the digit sum of 1729.
h o W t o 5 . 2 thinking recursively
To solve a problem recursively requires a different mindset than to solve it by programming
loops. In fact, it helps if you are, or pretend to be, a bit lazy and let others do most of the work
for you. If you need to solve a complex problem, pretend that “someone else” will do most
of the heavy lifting and solve the problem for all simpler inputs. Then you only need to figure
out how you can turn the solutions with simpler inputs into a solution for the whole problem.
To illustrate the recursive thinking process, consider the problem of Section 4.2, com puting
the sum of the digits of a number. We want to design a function digit_sum that computes the
sum of the digits of an integer n. For example, digit_sum(1729) = 1 + 7 + 2 + 9 = 19.
the key to finding a
recursive solution is
reducing the input to
a simpler input for
the same problem.
cfe2_ch05_p193_248.indd 228 10/26/10 6:14 PM
5.10 recursive Functions (optional) 229
}
cout << endl;
print_triangle(side_length - 1);
}
What is the result of print_triangle(4)?
39. Consider this recursive function:
int mystery(int n)
{
if (n <= 0) { return 0; }
return n + mystery(n - 1);
}
What is mystery(4)?
40. Consider this recursive function:
int mystery(int n)
{
if (n <= 0) { return 0; }
return mystery(n / 2) + 1;
}
What is mystery(20)?
41. Write a recursive function for printing n box shapes [] in a row.
42. The int_name function in Section 5.7 accepted arguments < 1,000. Using a
recursive call, extend its range to 999,999. For example an input of 12,345
should return "twelve thousand three hundred forty five".
practice it Now you can try these exercises at the end of the chapter: R5.20, P5.28, P5.30.
step 1 Break the input into parts that can themselves be inputs to the problem.
In your mind, fix a particular input or set of inputs for the task that
you want to solve, and think how you can simplify the inputs. Look
for simplifications that can be solved by the same task, and whose
solutions are related to the original task.
In the digit sum problem, consider how we can simplify an input
such as n = 1729. Would it help to subtract 1? After all, digit_sum(1729)
= digit_sum(1728) + 1. But consider n = 1000. There seems to be no obvious relationship
between digit_sum(1000) and digit_sum(999).
A much more promising idea is to remove the last digit, that is, compute n / 10 = 172. The
digit sum of 172 is directly related to the digit sum of 1729.
h o W t o 5 . 2 thinking recursively
To solve a problem recursively requires a different mindset than to solve it by programming
loops. In fact, it helps if you are, or pretend to be, a bit lazy and let others do most of the work
for you. If you need to solve a complex problem, pretend that “someone else” will do most
of the heavy lifting and solve the problem for all simpler inputs. Then you only need to figure
out how you can turn the solutions with simpler inputs into a solution for the whole problem.
To illustrate the recursive thinking process, consider the problem of Section 4.2, com puting
the sum of the digits of a number. We want to design a function digit_sum that computes the
sum of the digits of an integer n. For example, digit_sum(1729) = 1 + 7 + 2 + 9 = 19.
the key to finding a
recursive solution is
reducing the input to
a simpler input for
the same problem.
cfe2_ch05_p193_248.indd 229 10/26/10 6:14 PM
230 Chapter 5 Functions
step 2 Combine solutions with simpler inputs into a solution of the original problem.
In your mind, consider the solutions for the simpler inputs that you have discovered in Step 1.
Don’t worry how those solutions are obtained. Simply have faith that the solutions are readily
available. Just say to yourself: These are simpler inputs, so someone else will solve the problem
for me.
In the case of the digit sum task, ask yourself how you can obtain
digit_sum(1729) if you know digit_sum(172). You simply add the last
digit (9), and you are done. How do you get the last digit? As the
remainder n % 10. The value digit_sum(n) can therefore be obtained as
digit_sum(n / 10) + n % 10
Don’t worry how digit_sum(n / 10) is computed. The input is
smaller, and therefore it just works.
step 3 Find solutions to the simplest inputs.
A recursive computation keeps simplifying its inputs. To make sure that the recursion comes
to a stop, you must deal with the simplest inputs separately. Come up with special solutions
for them. That is usually very easy.
Look at the simplest inputs for the digit_sum test:
• A number with a single digit
• 0
When designing a
recursive solution,
do not worry about
multiple nested calls.
simply focus on
reducing a problem
to a slightly
simpler one.
in 1971, Marcian e.
“ted” hoff, an engi-
neer at intel Corpo-
ration, was working on a chip for a
manufacturer of electronic calculators.
he realized that it would be a better
idea to develop a general-purpose chip
that could be programmed to inter-
face with the keys and display of a cal-
culator, rather than to do yet another
custom design. thus, the microproces-
sor was born. at the time, its primary
application was as a con troller for cal-
culators, washing machines, and the
like. it took years for the computer
industry to notice that a genuine cen-
tral processing unit was now available
as a single chip.
hobbyists were the first to catch
on. in 1974 the first computer kit, the
altair 8800, was available from Mits
electronics for about $350. the kit
consisted of the microprocessor, a cir-
cuit board, a very small amount of
memory, toggle switches, and a row of
display lights. purchasers had to sol-
der and assemble it, then program it in
machine language through the toggle
switches. it was not a big hit.
the first big hit was the apple ii. it
was a real computer with a keyboard,
a monitor, and a floppy disk drive.
When it was first released, users had a
$3,000 machine that could play space
invaders, run a primitive bookkeep-
ing program, or let users program it
in BasiC. the original apple ii did not
even support lowercase letters, mak-
ing it worthless for word processing.
the breakthrough came in 1979 with
a new spreadsheet program, VisiCalc.
in a spreadsheet, you enter financial
data and their relationships into a grid
of rows and columns (see the figure at
right). then you modify some of the
data and watch in real time how the
others change. For example, you can
see how changing the mix of widgets
in a manufacturing plant might affect
estimated costs and profits. Middle
managers in companies, who under-
stood computers and were fed up with
having to wait for hours or days to get
their data runs back from the comput-
ing center, snapped up VisiCalc and the
computer that was needed to run it.
For them, the computer was a spread-
sheet machine.
the next big hit was the iBM per-
sonal Computer, ever after known as
the pC. it was the first widely available
personal computer that used intel’s
16-bit processor, the 8086, whose
successors are still being used in per-
sonal computers today. the success
of the pC was based not on any engi-
neering breakthroughs but on the fact
that it was easy to clone. iBM pub lished
the computer’s specifications in order
to encourage third parties to develop
plug-in cards. perhaps iBM did not
foresee that functionally equiva lent
versions of their computer could be
recreated by others, but a variety of
pC clone vendors emerged, and ulti-
mately iBM stopped selling per sonal
computers.
iBM never produced an operating
system for its pCs—that is, the soft-
ware that organizes the interaction
between the user and the computer,
starts application programs, and man-
ages disk storage and other resources.
instead, iBM offered customers the
option of three separate operating
systems. Most customers couldn’t
care less about the operating system.
Random Fact 5.1 the explosive Growth of personal Computers
they chose the system that was able
to launch most of the few applications
that existed at the time. it happened
to be Dos (Disk operating system) by
Microsoft. Microsoft licensed the same
operating system to other hardware
vendors and encouraged software
companies to write Dos applications.
a huge number of useful application
programs for pC-compatible machines
was the result.
pC applications were certainly use-
ful, but they were not easy to learn.
every vendor developed a different
user interface: the collection of key-
strokes, menu options, and settings
that a user needed to master to use
a software package effectively. Data
exchange between applications was
difficult, because each program used a
different data format. the apple Mac-
intosh changed all that in 1984. the
designers of the Macintosh had the
vision to supply an intuitive user inter-
face with the computer and to force
software developers to adhere to it.
it took Microsoft and pC-compatible
manufacturers years to catch up.
Most personal computers are used
for accessing information from online
sources, entertainment, word process-
ing, and home finance. some analysts
predict that the personal computer will
merge with the television set and cable
network into an entertainment and
information appliance.
The Visicalc Spreadsheet Running on an Apple II
cfe2_ch05_p193_248.indd 230 10/26/10 6:14 PM
5.10 recursive Functions (optional) 231
A number with a single digit is its own digit sum, so you can stop the recursion when n < 10,
and return n in that case. Or, if you prefer, you can be even lazier. If n has a single digit, then
digit_sum(n / 10) + n % 10 equals digit_sum(0) + n. You can simply terminate the recursion
when n is zero.
step 4 Implement the solution by combining the simple cases and the reduction step.
Now you are ready to implement the solution. Make separate cases for the simple inputs that
you considered in Step 3. If the input isn’t one of the simplest cases, then implement the logic
you discovered in Step 2.
Here is the complete digit_sum function:
int digit_sum(int n)
{
// Special case for terminating the recursion
if (n == 0) { return 0; }
// General case
return digit_sum(n / 10) + n % 10;
}
in 1971, Marcian e.
“ted” hoff, an engi-
neer at intel Corpo-
ration, was working on a chip for a
manufacturer of electronic calculators.
he realized that it would be a better
idea to develop a general-purpose chip
that could be programmed to inter-
face with the keys and display of a cal-
culator, rather than to do yet another
custom design. thus, the microproces-
sor was born. at the time, its primary
application was as a con troller for cal-
culators, washing machines, and the
like. it took years for the computer
industry to notice that a genuine cen-
tral processing unit was now available
as a single chip.
hobbyists were the first to catch
on. in 1974 the first computer kit, the
altair 8800, was available from Mits
electronics for about $350. the kit
consisted of the microprocessor, a cir-
cuit board, a very small amount of
memory, toggle switches, and a row of
display lights. purchasers had to sol-
der and assemble it, then program it in
machine language through the toggle
switches. it was not a big hit.
the first big hit was the apple ii. it
was a real computer with a keyboard,
a monitor, and a floppy disk drive.
When it was first released, users had a
$3,000 machine that could play space
invaders, run a primitive bookkeep-
ing program, or let users program it
in BasiC. the original apple ii did not
even support lowercase letters, mak-
ing it worthless for word processing.
the breakthrough came in 1979 with
a new spreadsheet program, VisiCalc.
in a spreadsheet, you enter financial
data and their relationships into a grid
of rows and columns (see the figure at
right). then you modify some of the
data and watch in real time how the
others change. For example, you can
see how changing the mix of widgets
in a manufacturing plant might affect
estimated costs and profits. Middle
managers in companies, who under-
stood computers and were fed up with
having to wait for hours or days to get
their data runs back from the comput-
ing center, snapped up VisiCalc and the
computer that was needed to run it.
For them, the computer was a spread-
sheet machine.
the next big hit was the iBM per-
sonal Computer, ever after known as
the pC. it was the first widely available
personal computer that used intel’s
16-bit processor, the 8086, whose
successors are still being used in per-
sonal computers today. the success
of the pC was based not on any engi-
neering breakthroughs but on the fact
that it was easy to clone. iBM pub lished
the computer’s specifications in order
to encourage third parties to develop
plug-in cards. perhaps iBM did not
foresee that functionally equiva lent
versions of their computer could be
recreated by others, but a variety of
pC clone vendors emerged, and ulti-
mately iBM stopped selling per sonal
computers.
iBM never produced an operating
system for its pCs—that is, the soft-
ware that organizes the interaction
between the user and the computer,
starts application programs, and man-
ages disk storage and other resources.
instead, iBM offered customers the
option of three separate operating
systems. Most customers couldn’t
care less about the operating system.
Random Fact 5.1 the explosive Growth of personal Computers
they chose the system that was able
to launch most of the few applications
that existed at the time. it happened
to be Dos (Disk operating system) by
Microsoft. Microsoft licensed the same
operating system to other hardware
vendors and encouraged software
companies to write Dos applications.
a huge number of useful application
programs for pC-compatible machines
was the result.
pC applications were certainly use-
ful, but they were not easy to learn.
every vendor developed a different
user interface: the collection of key-
strokes, menu options, and settings
that a user needed to master to use
a software package effectively. Data
exchange between applications was
difficult, because each program used a
different data format. the apple Mac-
intosh changed all that in 1984. the
designers of the Macintosh had the
vision to supply an intuitive user inter-
face with the computer and to force
software developers to adhere to it.
it took Microsoft and pC-compatible
manufacturers years to catch up.
Most personal computers are used
for accessing information from online
sources, entertainment, word process-
ing, and home finance. some analysts
predict that the personal computer will
merge with the television set and cable
network into an entertainment and
information appliance.
The Visicalc Spreadsheet Running on an Apple II
cfe2_ch05_p193_248.indd 231 10/26/10 6:14 PM
232 Chapter 5 Functions
understand the concepts of functions, arguments, and return values.
• A function is a named sequence of instructions.
• Arguments are supplied when a function is called. The return value is the result
that the function computes.
Be able to implement functions.
• When defining a function, you provide a name for the func tion, a variable for each
argument, and a type for the result.
• Function comments explain the purpose of the function, the meaning of the
parameter variables and return value, as well as any special requirements.
describe the process of parameter passing.
• Parameter variables hold the argument values
supplied in the function call.
describe the process of returning a value from a function.
• The return statement terminates a function call and yields
the function result.
design and implement functions without return values.
• Use a return type of void to indicate that a function does not return a value.
develop functions that can be reused for multiple problems.
• Eliminate replicated code or pseudocode by defining a function.
• Design your functions to be reusable. Supply parameter variables for the values
that can vary when the function is reused.
apply the design principle of stepwise refinement.
• Use the process of stepwise refinement to decompose complex tasks into
sim pler ones.
• When you discover that you need a function, write a description of the parame ter
variables and return values.
• A function may require simpler functions to carry out its work.
C h a p t e r s u M M a r y
pie(fruit) pie(fruit)
determine the scope of variables in a program.
• The scope of a variable is the part of the program
in which it is visible.
• A variable in a nested block shadows a variable
with the same name in an outer block.
• A local variable is defined inside a function.
A global variable is defined outside a function.
• Avoid global variables in your programs.
describe how reference parameters work.
• Modifying a value parameter has no effect on the caller.
• A reference parameter refers to a variable that is supplied in a function call.
• Modifying a reference parameter updates the variable that was supplied in the call.
understand recursive function calls and implement simple recursive functions.
• A recursive computation solves a problem by using the solution
of the same problem with simpler inputs.
• For a recursion to terminate, there must be special cases for the
simplest inputs.
• The key to finding a recursive solution is reducing the input to a
simpler input for the same problem.
• When designing a recursive solution, do not worry about multiple nested calls.
Simply focus on reducing a problem to a slightly simpler one.
r5.1 What is the difference between an argument and a return value? How many argu-
ments can a function have? How many return values?
r5.2 In which sequence are the lines of the program cube.cpp on page 198 executed, start ing
with the first line of main?
r5.3 Give examples of the following, either from the C++ library or from the functions
discussed in this chapter:
a. A function with two double arguments and a double return value
b. A function with a double argument and a double return value
c. A function with two int arguments and an int return value
d. A function with an int argument and a string return value
e. A function with a string argument and no return value
f. A function with a reference parameter and no return value
g. A function with no arguments and an int return value
r e V i e W e x e r C i s e s
cfe2_ch05_p193_248.indd 232 10/26/10 6:14 PM
review exercises 233
determine the scope of variables in a program.
• The scope of a variable is the part of the program
in which it is visible.
• A variable in a nested block shadows a variable
with the same name in an outer block.
• A local variable is defined inside a function.
A global variable is defined outside a function.
• Avoid global variables in your programs.
describe how reference parameters work.
• Modifying a value parameter has no effect on the caller.
• A reference parameter refers to a variable that is supplied in a function call.
• Modifying a reference parameter updates the variable that was supplied in the call.
understand recursive function calls and implement simple recursive functions.
• A recursive computation solves a problem by using the solution
of the same problem with simpler inputs.
• For a recursion to terminate, there must be special cases for the
simplest inputs.
• The key to finding a recursive solution is reducing the input to a
simpler input for the same problem.
• When designing a recursive solution, do not worry about multiple nested calls.
Simply focus on reducing a problem to a slightly simpler one.
r5.1 What is the difference between an argument and a return value? How many argu-
ments can a function have? How many return values?
r5.2 In which sequence are the lines of the program cube.cpp on page 198 executed, start ing
with the first line of main?
r5.3 Give examples of the following, either from the C++ library or from the functions
discussed in this chapter:
a. A function with two double arguments and a double return value
b. A function with a double argument and a double return value
c. A function with two int arguments and an int return value
d. A function with an int argument and a string return value
e. A function with a string argument and no return value
f. A function with a reference parameter and no return value
g. A function with no arguments and an int return value
r e V i e W e x e r C i s e s
cfe2_ch05_p193_248.indd 233 10/26/10 6:14 PM
234 Chapter 5 Functions
r5.4 True or false?
a. A function has exactly one return statement.
b. A function has at least one return statement.
c. A function has at most one return value.
d. A function with return value void never has a return statement.
e. When executing a return statement, the function exits immediately.
f. A function with return value void must print a result.
g. A function without arguments always returns the same value.
r5.5 Consider these functions:
double f(double x) { return g(x) + sqrt(h(x)); }
double g(double x) { return 4 * h(x); }
double h(double x) { return x * x + k(x) - 1; }
double k(double x) { return 2 * (x + 1); }
Without actually compiling and running a program, determine the results of the
following function calls:
a. double x1 = f(2);
b. double x2 = g(h(2));
c. double x3 = k(g(2) + h(2));
d. double x4 = f(0) + f(1) + f(2);
e. double x5 = f(-1) + g(-1) + h(-1) + k(-1);
r5.6 Write pseudocode for a function that translates a telephone number with letters in it
(such as 1-800-FLOWERS) into the actual phone number. Use the standard letters
on a phone pad.
r5.7 Design a function that prints a floating-point number as a currency value (with a $
sign and two decimal digits).
a. Indicate how the programs ch02/volume2.cpp and ch04/invtable.cpp should
change to use your function.
b. What change is required if the programs should show a different currency, such
as euro?
r5.8 For each of the variables in the following program, indicate the scope. Then deter-
mine what the program prints, without actually running the program.
1 int a = 0;
2 int b = 0;
3 int f(int c)
4 {
cfe2_ch05_p193_248.indd 234 10/26/10 6:14 PM
review exercises 235
5 int n = 0;
6 a = c;
7 if (n < c)
8 {
9 n = a + b;
10 }
11 return n;
12 }
13
14 int g(int c)
15 {
16 int n = 0;
17 int a = c;
18 if (n < f(c))
19 {
20 n = a + b;
21 }
22 return n;
23 }
24
25 int main()
26 {
27 int i = 1;
28 int b = g(i);
29 cout << a + b + i << endl;
30 return 0;
31 }
r5.9 We have seen three kinds of variables in C++: global variables, parameter variables,
and local variables. Classify the variables of Exercise R5.8 according to these
cate gories.
r5.10 Use the process of stepwise refinement to describe the process of making scrambled
eggs. Discuss what you do if you do not find eggs in the refrigerator.
r5.11 How many parameters does the following function have? How many return values
does it have? Hint: The C++ notions of “parameter ” and “return value” are not the
same as the intuitive notions of “input” and “output”.
void average(double& avg)
{
cout << "Please enter two numbers: ";
double x;
double y;
cin >> x >> y;
avg = (x + y) / 2;
}
r5.12 Perform a walkthrough of the int_name function with the following arguments:
a. 5
b. 12
c. 21
d. 301
e. 324
f. 0
g. -2
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236 Chapter 5 Functions
r5.13 Consider the following function:
int f(int n)
{
if (n <= 1) { return 1; }
if (n % 2 == 0) // n is even
{
return f(n / 2);
}
else { return f(3 * n + 1); }
}
Perform traces of the computations f(1), f(2), f(3), f(4), f(5), f(6), f(7), f(8), f(9),
and f(10).
r5.14 Eliminate the global variable in the code at the end of Section 5.8 by
a. passing the balance to the withdraw function and returning the updated balance.
b. passing the balance as a reference parameter to the withdraw function.
r5.15 Given the following functions, trace the function call print_roots(4).
int i;
int isqrt(int n)
{
i = 1;
while (i * i <= n) { i++; }
return i - 1;
}
void print_roots(int n)
{
for (i = 0; i <= n; i++) { cout << isqrt(i) << " "; }
}
How can you fix the code so that the output is as expected (that is, 0 1 1 1 2)?
r5.16 Consider the following function that is intended to swap the values of two integers:
void false_swap1(int& a, int& b)
{
a = b;
b = a;
}
int main()
{
int x = 3;
int y = 4;
false_swap1(x, y);
cout << x << " " << y << endl;
return 0;
}
Why doesn’t the function swap the contents of x and y? How can you rewrite the
function to work cor rectly?
r5.17 Consider the following function that is intended to swap the values of two integers:
void false_swap2(int a, int b)
{
int temp = a;
a = b;
b = temp;
}
int main()
{
int x = 3;
int y = 4;
false_swap2(x, y);
cout << x << " " << y << endl;
return 0;
}
Why doesn’t the function swap the contents of x and y? How can you rewrite the
function to work cor rectly?
r5.18 The following function swaps two integers, without requiring a temporary variable:
void tricky_swap(int& a, int& b)
{
a = a - b;
b = a + b;
a = b - a;
}
However, it fails in one important case, namely when calling tricky_swap(x, x).
Explain what should hap pen and what actually happens.
r5.19 Give pseudocode for a recursive function for printing all substrings of a given string.
For example, the substrings of the string "rum" are "rum" itself, "ru", "um", "r", "u",
"m", and the empty string. You may assume that all letters of the string are dif ferent.
r5.20 Give pseudocode for a recursive function that sorts all letters in a string. For exam-
ple, the string "goodbye" would be sorted into "bdegooy".
p5.1 The max function that is declared in the
arguments. Write a program that reads three floating-point numbers, uses the max
function, and displays
• the larger of the first two inputs.
• the larger of the last two inputs.
• the largest of all three inputs.
p5.2 Write a function that computes the balance of a bank account with a given initial
balance and interest rate, after a given number of years. Assume interest is com-
pounded yearly.
p5.3 Write the following functions and provide a program to test them.
a. double smallest(double x, double y, double z), returning the smallest of the
arguments
b. double average(double x, double y, double z), returning the average of the
arguments
p r o G r a M M i n G e x e r C i s e s
cfe2_ch05_p193_248.indd 236 10/26/10 6:14 PM
programming exercises 237
int temp = a;
a = b;
b = temp;
}
int main()
{
int x = 3;
int y = 4;
false_swap2(x, y);
cout << x << " " << y << endl;
return 0;
}
Why doesn’t the function swap the contents of x and y? How can you rewrite the
function to work cor rectly?
r5.18 The following function swaps two integers, without requiring a temporary variable:
void tricky_swap(int& a, int& b)
{
a = a - b;
b = a + b;
a = b - a;
}
However, it fails in one important case, namely when calling tricky_swap(x, x).
Explain what should hap pen and what actually happens.
r5.19 Give pseudocode for a recursive function for printing all substrings of a given string.
For example, the substrings of the string "rum" are "rum" itself, "ru", "um", "r", "u",
"m", and the empty string. You may assume that all letters of the string are dif ferent.
r5.20 Give pseudocode for a recursive function that sorts all letters in a string. For exam-
ple, the string "goodbye" would be sorted into "bdegooy".
p5.1 The max function that is declared in the
arguments. Write a program that reads three floating-point numbers, uses the max
function, and displays
• the larger of the first two inputs.
• the larger of the last two inputs.
• the largest of all three inputs.
p5.2 Write a function that computes the balance of a bank account with a given initial
balance and interest rate, after a given number of years. Assume interest is com-
pounded yearly.
p5.3 Write the following functions and provide a program to test them.
a. double smallest(double x, double y, double z), returning the smallest of the
arguments
b. double average(double x, double y, double z), returning the average of the
arguments
p r o G r a M M i n G e x e r C i s e s
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238 Chapter 5 Functions
p5.4 Write the following functions:
a. bool all_the_same(double x, double y, double z), returning true if the arguments
are all the same
b. bool all_different(double x, double y, double z), returning true if the arguments
are all different
c. bool sorted(double x, double y, double z), returning true if the arguments are
sorted, with the smallest one coming first
Provide a program that tests your functions.
p5.5 Write the following functions:
a. int first_digit(int n), returning the first digit of the argument
b. int last_digit(int n), returning the last digit of the argument
c. int digits(int n), returning the number of digits of the argument
For example, first_digit(1729) is 1, last_digit(1729) is 9, and digits(1729) is 4. Provide
a program that tests your functions.
p5.6 Write a function
string middle(string str)
that returns a string containing the middle character in str if the length of str is odd,
or the two middle characters if the length is even. For example, middle(“middle”)
returns “dd”.
p5.7 Write a function
string repeat(string str, int n)
that returns the string str repeated n times. For example, repeat(“ho”, 3) returns
“hohoho”.
p5.8 Write a function
int count_vowels(string str)
that returns a count of all vowels in the string str. Vowels are the letters a, e, i, o, and
u, and their upper case variants.
p5.9 Write a function
int count_words(string str)
that returns a count of all words in the string str. Words are separated by spaces. For
example, count_words(“Mary had a little lamb”) should return 5.
p5.10 It is a well-known phenomenon that most people are easily able to read a text whose
words have two characters flipped, provided the first and last letter of each word are
not changed. For example:
I dn’ot gvie a dman for a man taht can olny sepll a wrod one way. (Mrak Taiwn)
Write a function string scramble(string word) that constructs a scrambled version of a
given word, ran domly flipping two characters other than the first and last one. Then
write a program that reads words from cin and prints the scrambled words.
p5.11 Write functions
double sphere_volume(double r)
double sphere_surface(double r)
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programming exercises 239
double cylinder_volume(double r, double h)
double cylinder_surface(double r, double h)
double cone_volume(double r, double h)
double cone_surface(double r, double h)
that compute the volume and surface area of a sphere with radius r, a cylinder with a
circular base with radius r and height h, and a cone with a circular base with radius r
and height h. Then write a program that prompts the user for the values of r and h,
calls the six functions, and prints the results.
p5.12 Write functions
double distance(double x1, double x2, double y1, double y2)
void midpoint(double x1, double x2, double y1, double y2, double& xmid, double& ymid)
void slope(double x1, double x2, double y1, double y2, bool& vertical, double& s)
that compute the distance, midpoint, and slope of the line segment joining the points
(x1, y1) and (x2, y2). The slope function should either set vertical to true and not set s,
or set vertical to false and set s to the slope.
p5.13 Write a function
double read_double(string prompt)
that displays the prompt string, followed by a space, reads a floating-point number
in, and returns it. Here is a typical usage:
salary = read_double(“Please enter your salary:”);
perc_raise = read_double(“What percentage raise would you like?”);
p5.14 Write a function void sort2(int& a, int& b) that swaps the values of a and b if a is
greater than b and otherwise leaves a and b unchanged. For example,
int u = 2;
int v = 3;
int w = 4;
int x = 1;
sort2(u, v); // u is still 2, v is still 3
sort2(w, x); // w is now 1, x is now 4
p5.15 Write a function sort3(int& a, int& b, int& c) that swaps its three arguments to
arrange them in sorted order. For example,
int v = 3;
int w = 4;
int x = 1;
sort3(v, w, x); // v is now 1, w is now 3, x is now 4
Hint: Use sort2 of Exercise P5.14.
p5.16 Enhance the int_name function so that it works correctly for values < 1,000,000,000.
p5.17 Enhance the int_name function so that it works correctly for negative values and zero.
Caution: Make sure the improved function doesn’t print 20 as "twenty zero".
p5.18 For some values (for example, 20), the int_name function returns a string with a
lead ing space (" twenty"). Repair that blemish and ensure that spaces are inserted only
when necessary. Hint: There are two ways of accomplishing this. Either ensure that
leading spaces are never inserted, or remove leading spaces from the result before
returning it.
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240 Chapter 5 Functions
p5.19 Write a program that prints a paycheck. Ask the program user for the name of the
employee, the hourly rate, and the number of hours worked. If the number of hours
exceeds 40, the employee is paid “time and a half”, that is, 150 percent of the hourly
rate on the hours exceeding 40. Your check should look similar to that in the figure
below. Use fictitious names for the payer and the bank. Be sure to use stepwi se
refine ment and break your solution into several functions. Use the int_name function
to print the dollar amount of the check.
AmountDate
CHECK
NUMBER 063331 74-39311 567390
Publishers
,
Bank Minnesota
2000 Prince Blvd
Jonesville, MN 55400
4659484PAY
TWO HUNDRED SEVENTY FOUR AND 15 / 100 ******************************************
TO THE ORDER OF:
John Wiley & Sons, Inc.
111 River Street
Hoboken, NJ 07030-5774
JOHN DOE
1009 Franklin Blvd
Sunnyvale, CA 95014
04/29/12 $*******274.15
p5.20 Write a function that computes the weekday of a given date, using a formula known
as Zeller’s congruence. Let
d = the day of the month
mm = the modified month (3 = March, ..., 12 = December, 13 = January,
14 = February)
w = the weekday (0 = Monday, 1 = Tuesday, ..., 6 = Sunday)
Then
w d
mm year year
= + +
× +( )( )
+
× ( )
+
× ( )
5
26 1
10
5 100
4
21 100
4
%
% 7
Here, all ̸ denote integer division and % denotes the remainder operation.
p5.21 Leap years. Write a function
bool leap_year(int year)
that tests whether a year is a leap year: that is, a year with
366 days. Leap years are necessary to keep the calendar
synchronized with the sun because the earth revolves
around the sun once every 365.25 days. Actually, that
figure is not entirely precise, and for all dates after 1582 the
Gregorian correction applies. Usu ally years that are divisible by 4 are leap years, for
example 1996. However, years that are divisible by 100 (for example, 1900) are not
leap years, but years that are divisible by 400 are leap years (for example, 2000).
p5.22 Write a program that converts a Roman number such as MCMLXXVIII to its
deci mal number representation. Hint: First write a function that yields the numeric
value of each of the letters. Then use the following algorithm:
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programming exercises 241
total = 0
While the roman number string is not empty
If the first character has a larger value than the second, or the string has length 1
Add value(first character) to total.
Remove the character.
Else
Add value(second character) - value(first character) to total.
Remove both characters.
p5.23 In Exercise P3.23 you were asked to write a program to
convert a number to its representation in Roman numerals.
At the time, you did not know how to eliminate duplicate
code, and as a consequence the resulting program was rather
long. Rewrite that program by implementing and using the
following function:
string roman_digit(int n, string one, string five, string ten)
That function translates one digit, using the strings specified for the one, five, and ten
values. You would call the function as follows:
roman_ones = roman_digit(n % 10, "I", "V", "X");
n = n / 10;
roman_tens = roman_digit(n % 10, "X", "L", "C");
...
p5.24 Postal bar codes. For faster sorting of letters, the United States Postal Service encour-
ages companies that send large volumes of mail to use a bar code denoting the zip
code (see Figure 9).
The encoding scheme for a five-digit zip code is shown in Figure 10. There are
full-height frame bars on each side. The five encoded digits are followed by a check
digit, which is computed as follows: Add up all digits, and choose the check digit to
make the sum a multiple of 10. For example, the zip code 95014 has a sum of 19, so
the check digit is 1 to make the sum equal to 20.
Figure 9 a postal Bar Code
*************** ECRLOT ** CO57
CODE C671RTS2
JOHN DOE CO57
1009 FRANKLIN BLVD
SUNNYVALE CA 95014 – 5143
Figure 10 encoding for Five-Digit Bar Codes
Frame bars
Digit 1 Digit 2 Digit 3 Digit 4 Digit 5 Check
Digit
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242 Chapter 5 Functions
Each digit of the zip code, and the check digit, is encoded according to the following
table where 0 denotes a half bar and 1 a full bar.
The digit can be easily computed from the bar code using the column weights 7, 4, 2,
1, 0. For example, 01100 is 0 × 7 + 1 × 4 + 1 × 2 + 0 × 1 × 0 × 0 = 6. The only exception
is 0, which would yield 11 according to the weight formula.
Write a program that asks the user for a zip code and prints the bar code. Use : for
half bars, | for full bars. For example, 95014 becomes
||:|:::|:|:||::::::||:|::|:::|||
p5.25 Write a program that reads in a bar code (with : denoting half bars and | denoting full
bars) and prints out the zip code it represents. Print an error message if the bar code
is not correct.
p5.26 Write a program that prints instructions to get coffee, asking the user for input
whenever a decision needs to be made. Decompose each task into a function, for
example:
void brew_coffee()
{
cout << "Add water to the coffee maker." << endl;
cout << "Put a filter in the coffee maker." << endl;
grind_coffee();
cout << "Put the coffee in the filter." << endl;
...
}
p5.27 Write a recursive function
string reverse(string str)
that computes the reverse of a string. For example, reverse("flow") should return
"wolf". Hint: Reverse the substring starting at the second character, then add the first
Digit
Bar 1
(weight 7)
Bar 2
(weight 4)
Bar 3
(weight 2)
Bar 4
(weight 1)
Bar 5
(weight 0)
1 0 0 0 1 1
2 0 0 1 0 1
3 0 0 1 1 0
4 0 1 0 0 1
5 0 1 0 1 0
6 0 1 1 0 0
7 1 0 0 0 1
8 1 0 0 1 0
9 1 0 1 0 0
0 1 1 0 0 0
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programming exercises 243
character at the end. For example, to reverse "flow", first reverse "low" to "wol", then
add the "f" at the end.
p5.28 Write a recursive function
bool is_palindrome(string str)
that returns true if str is a palindrome, that is, a word that is the same when reversed.
Examples of palin drome are “deed”, “rotor”, or “aibohphobia”. Hint: A word is a
palindrome if the first and last letters match and the remainder is also a palindrome.
p5.29 Use recursion to implement a function bool find(string str, string match) that tests
whether match is contained in str:
bool b = find("Mississippi", "sip"); // Sets b to true
Hint: If str starts with match, then you are done. If not, consider the string that you
obtain by removing the first character.
p5.30 Use recursion to determine the number of digits in a number n. Hint: If n is < 10, it
has one digit. Otherwise, it has one more digit than n / 10.
p5.31 Use recursion to compute an, where n is a positive integer. Hint: If n is 1, then an = a.
Otherwise, an = a × an–1.
engineering p5.32 The effective focal length f of a lens of thickness d
that has surfaces with radii of curva ture R1 and R2
is given by
1
1
1 1 1
1 2 1 2f
n
R R
n d
nR R
= −( ) − + −( )
where n is the refractive index of the lens medium.
Write a function that computes f in terms of the
other parameters.
engineering p5.33 A laboratory container is shaped like the frustum of a cone:
h
R2
R1
Write functions to compute the volume and surface area, using these equations:
V h R R R R= + +( )13 12 22 1 2π
S R R R R h R= +( ) −( ) + +π π1 2 2 1
2 2
1
2
f
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244 Chapter 5 Functions
engineering p5.34 In a movie theater, the angle q at which a viewer sees the picture on the screen
depends on the distance x of the viewer from the screen. For a movie theater with the
dimensions shown in the picture below, write a function that computes the angle for
a given distance.
24 ft.
6 ft.8°
θ
θ
x
Next, provide a more general function that works for theaters with arbitrary
dimensions.
engineering p5.35 Electric wire, like that in the photo, is a cylindrical conductor covered by an insulat-
ing material. The resistance of a piece of wire is given by the formula
R
L
A
L
d
= =
ρ ρ
π
4
2
where ρ is the resistivity of the conductor, and L, A, and d are the length, cross-
sectional area, and diameter of the wire. The resistivity of copper is 1.678 × 10−8 Ω m.
The wire diameter, d, is commonly specified by the American wire gauge (AWG),
which is an integer, n. The diameter of an AWG n wire is given by the formula
d
n
= ×
−
0 127 92
36
39. mm
Write a C++ function
double diameter(int wire_gauge)
that accepts the wire gauge and returns the corresponding wire diameter. Write
another C++ function
double copper_wire_resistance(double length, int wire_gauge)
that accepts the length and gauge of a piece of copper wire and returns the resistance
of that wire. The resistivity of aluminum is 2.82 × 10−8 Ω m. Write a third C++
function
double aluminum_wire_resistance(double length, int wire_gauge)
that accepts the length and gauge of a piece of aluminum wire and returns the
resistance of that wire.
Write a C++ program to test these functions.
engineering p5.36 The drag force on a car is given by
F v ACD D=
1
2
2ρ
where ρ is the density of air (1.23 kg̸m3), v is the velocity in units of m̸s, A is the
projected area of the car (2.5 m2), and CD is the drag coefficient (0.2).
The amount of power in watts required to overcome such drag force is P = FDv, and
the equivalent horsepower required is Hp = P ̸ 746. Write a program that accepts a
car’s velocity and computes the power in watts and in horsepower needed to over-
come the resulting drag force. Note: 1 mph = 0.447 m̸s.
1. The arguments are 3 and 2. The return value is 9.
2. The inner call to pow returns 22 = 4. Therefore, the outer call returns 42 = 16.
3. 3.
4. Users of the function can treat it as a black box.
5. 27.
6. 8 × 8 × 8 = 512.
7. double volume = pow(side_length, 3);
return volume;
8. double square_area(double side_length)
{
double area = side_length * side_length;
return area;
}
9. (2 + 3) / (3 - 2) = 5
10. When the function is called, x is set to 4, y is set to 7, and z becomes 11. Then z is
changed to 5.5, and that value is returned and printed.
11. When the function is called, x is set to 5. Then y is set to 25, and that value is returned
and printed.
12. When the function is called, n is set to 5. Then n is incremented twice, setting it to 7.
That value is returned and printed.
13. box_string("Hello");
box_string("World");
14. The box_string function does not return a value. Therefore, you cannot use it in a
<< expression.
15. void shout(string str)
{
cout << str << "!!!" << endl;
}
16. void box_string(string str)
{
int n = str.length();
for (int i = 0; i < n + 4; i++) { cout << "-"; }
cout << endl;
cout << "! " << str << " !" << endl;
for (int i = 0; i < n + 4; i++) { cout << "-"; }
cout << endl;
}
17. void print_line(int count)
{
for (int i = 0; i < count; i++)
{
a n s W e r s t o s e l F - C h e C K Q u e s t i o n s
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answers to self-Check Questions 245
The amount of power in watts required to overcome such drag force is P = FDv, and
the equivalent horsepower required is Hp = P ̸ 746. Write a program that accepts a
car’s velocity and computes the power in watts and in horsepower needed to over-
come the resulting drag force. Note: 1 mph = 0.447 m̸s.
1. The arguments are 3 and 2. The return value is 9.
2. The inner call to pow returns 22 = 4. Therefore, the outer call returns 42 = 16.
3. 3.
4. Users of the function can treat it as a black box.
5. 27.
6. 8 × 8 × 8 = 512.
7. double volume = pow(side_length, 3);
return volume;
8. double square_area(double side_length)
{
double area = side_length * side_length;
return area;
}
9. (2 + 3) / (3 - 2) = 5
10. When the function is called, x is set to 4, y is set to 7, and z becomes 11. Then z is
changed to 5.5, and that value is returned and printed.
11. When the function is called, x is set to 5. Then y is set to 25, and that value is returned
and printed.
12. When the function is called, n is set to 5. Then n is incremented twice, setting it to 7.
That value is returned and printed.
13. box_string("Hello");
box_string("World");
14. The box_string function does not return a value. Therefore, you cannot use it in a
<< expression.
15. void shout(string str)
{
cout << str << "!!!" << endl;
}
16. void box_string(string str)
{
int n = str.length();
for (int i = 0; i < n + 4; i++) { cout << "-"; }
cout << endl;
cout << "! " << str << " !" << endl;
for (int i = 0; i < n + 4; i++) { cout << "-"; }
cout << endl;
}
17. void print_line(int count)
{
for (int i = 0; i < count; i++)
{
a n s W e r s t o s e l F - C h e C K Q u e s t i o n s
cfe2_ch05_p193_248.indd 245 10/26/10 6:14 PM
246 Chapter 5 Functions
cout << "-";
}
cout << endl;
}
void box_string(string str)
{
int n = str.length();
print_line(n + 2);
cout << "!" << str << "!" << endl;
print_line(n + 2);
}
18. int total_pennies = round_to_pennies(total);
int total_tax_pennies = round_to_pennies(total * tax_rate);
where the function is defined as
/**
@param amount an amount in dollars and cents
@return the amount in pennies, rounded to the nearest penny
*/
int round_to_pennies(double amount)
{
return static_cast
}
19. if (is_even(page)) …
where the function is defined as follows:
bool is_even(int n)
{
return n % 2 == 0;
}
20. Add parameter variables so you can pass the initial balance and interest rate to the
function:
double balance(double initial_balance, double rate, int years)
{
return initial_balance * pow(1 + rate / 100, years);
}
21. int spaces = count_spaces(input);
where the function is defined as follows:
/**
@param str any string
@return the number spaces in str
*/
int count_spaces(string str)
{
int count = 0;
for (int i = 0; i < str.length(); i++)
{
if (str.substr(i, 1) == " ")
{
count++;
}
}
return count;
}
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answers to self-Check Questions 247
22. It is very easy to replace the space with any character.
/**
@param str any string
@param ch a string of length 1
@return the number of times that ch occurs in str
*/
int count(string str, string ch)
{
int count = 0;
for (int i = 0; i < str.length(); i++)
{
if (str.substr(i, 1) == ch) { count++; }
}
return count;
}
This is useful if you want to count other characters. For example, count(input, ",")
counts how many commas are in the input.
23. Change line 75 to
name = name + digit_name(part / 100) + " hundred";
In line 72, add the statement
if (part >= 1000)
{
name = digit_name(part / 1000) + “thousand “;
part = part % 1000;
}
In line 65, change 1000 to 10000 in the comment.
24. In the case of “teens”, we already have the last digit as part of the name.
25. Nothing is printed. One way of dealing with this case is to add the following state-
ment before line 70.
if (number == 0) { return “zero”; }
26. Here is the approximate trace:
int_name(number = 72)
part name
72 ” seventy”
2 ” seventy two”
Note that the string starts with a blank space. Exercise P5.18 asks you to eliminate it.
27. Here is one possible solution. Break up the task print table into print header and
print body. The print header task calls print separator, prints the header cells, and calls print
separator again. The print body task repeatedly calls print row and then calls print separator.
28. 1.
29. 2, 7.
30. Lines 3, 4, 5, 6, 10, 11, but not 7 through 9.
31. The global variable defined in line 1.
cfe2_ch05_p193_248.indd 247 10/26/10 6:14 PM
248 Chapter 5 Functions
32. The variables s defined in lines 4 and 15.
33. Yes, but since the function does not modify the amount parameter variable, there is no
need to do so.
34. void transfer(double& balance1, double& balance2, double amount)
35. bool withdraw(double& balance, double amount)
{
if (balance >= amount)
{
balance = balance – amount;
return true;
}
else
{
return false;
}
}
36. void minmax(double x, double y, double& a, double& b)
{
if (x < y) { a = x; b = y; }
else { a = y; b = x; }
}
37. The program sets x to 1, then y to 4, then x to 3. It prints 3 4.
38. [][][][]
[][][]
[][]
[]
39. 4 + 3 + 2 + 1 + 0 = 10
40. mystery(10) + 1 = mystery(5) + 2 = mystery(2) + 3 = mystery(1) + 4 =
mystery(0) + 5 = 5
41. The idea is to print one [], then print n - 1 of them.
void print_boxes(int n)
{
if (n == 0) { return; }
cout << "[]";
print_boxes(n - 1);
}
42. Simply add the following to the beginning of the function:
if (part >= 1000)
{
return int_name(part / 1000) + ” thousand ” + int_name(part % 1000);
}
step 1 Describe what the function should do.
This has been given to us already.
step 2 Determine the function’s “inputs”.
There are three arguments:
• The string that is being edited (such as “Mississippi”)
• The string that is being replaced (such as “ss”)
• The replacement string (such as “n”)
At this point, we have enough information to document the function:
/**
Replaces the first occurrence of match in str with repl.
@param str the string that is being edited
@param match the string that is being replaced
@param repl the replacement string
*/
step 3 Determine the types of the parameter variables and the return value.
The parameter variables are all strings. However, because the first parameter is being modi-
fied, its type is string&. The func tion modifies the first parameter, and it does not return a
value. Therefore, the return type is void.
The function will be defined as
void replace_first(string& str, string match, string repl)
step 4 Write pseudocode for obtaining the desired result.
We first need to find the position in which match occurs. (If it doesn’t occur anywhere, we do
nothing.) For example, the string “ss” is found in position 2 in “Mississippi”. Let us assume
that we know that position, and call it i. Then the answer is obtained by concatenating
• the substring of str from 0 to i – 1
• the replacement string
• the substring of str from i + match.length() to the end
How do we find that position? In the spirit of stepwise refinement, we will delegate that task
to another function, find_index. That starts a new sequence of steps, which, for greater clarity,
we will place after the steps for this function.
step 5 Implement the function body.
We simply translate the plan into C++:
{
int i = find_index(str, match);
if (i == -1) { return; } // No match
str = str.substr(0, i) + repl + str.substr(i + match.length());
}
W o r K e D e x a M p l e 5 . 1 matching and replacing parts of a string
Searching and replacing text is a common word processor function. Your task is to write a
function that replaces the first occurrence of one string with another in a given string. For
example, when asked to replace “ss” with “n” in “Mississippi”, the result is “Minissippi”.
cfe2_ch05_p193_248.indd 248 10/26/10 6:14 PM
6C h a p t e r
249
a r r ay s a n d
V e C t o r s
to become familiar with using arrays
and vectors to collect values
to learn about common algorithms for processing arrays
and vectors
to write functions that receive and return arrays and vectors
to be able to use two-dimensional arrays
C h a p t e r G o a l s
C h a p t e r C o n t e n t s
6.1 ArrAys 250
Syntax 6.1: defining an array 251
Common Error 6.1: Bounds errors 254
Programming Tip 6.1: Use arrays for sequences
of related Values 255
Random Fact 6.1: an early Internet Worm 255
6.2 Common ArrAy Algorithms 256
Special Topic 6.1: sorting with the
C++ library 263
Special Topic 6.2: a sorting algorithm 263
Special Topic 6.3: Binary search 264
6.3 ArrAys And FunCtions 265
Special Topic 6.4: Constant array parameters 269
6.4 Problem solving: AdAPting
Algorithms 269
How To 6.1: Working with arrays 271
Worked Example 6.1: rolling the dice
6.5 Problem solving: disCovering
Algorithms by mAniPulAting
PhysiCAl objeCts 274
6.6 two-dimensionAl ArrAys 278
Syntax 6.2: two-dimensional array
definition 279
Common Error 6.2: omitting the Column size of
a two-dimensional array parameter 284
Worked Example 6.2: a World population table
6.7 veCtors 284
Syntax 6.3: defining a Vector 285
Programming Tip 6.2: prefer Vectors over
arrays 289
Random Fact 6.2: the First programmer 290
cfe2_ch06_p249_306.indd 249 10/26/10 7:47 PM
250
In many programs, you need to collect large numbers of
values. In standard C++, you use arrays and vectors for this
purpose. arrays are a fundamental structure of the C++
language. the standard C++ library provides the vector
construct as a more convenient alternative when working
with collections whose size is not fixed. In this chapter, you
will learn about arrays, vectors, and common algorithms
for processing them.
6.1 arrays
We start this chapter by introducing the array data type. Arrays are the fundamental
mechanism in C++ for collecting multiple values. In the following sections, you will
learn how to define arrays and how to access array elements.
6.1.1 defining arrays
Suppose you write a program that reads a sequence of values and prints out the
sequence, marking the largest value, like this:
32
54
67.5
29
34.5
80
115 <= largest value
44.5
100
65
You do not know which value to mark as the largest one until you have seen them all.
After all, the last value might be the largest one. Therefore, the program must first
store all values before it can print them.
Could you simply store each value in a separate variable? If you know that there
are ten inputs, then you can store the values in ten variables value1, value2, value3, …,
value10. However, such a sequence of variables is not very practical to use. You would
have to write quite a bit of code ten times, once for each of the variables. To solve this
problem, use an array: a structure for storing a sequence of values.
Figure 1
an array of size 10
values =
10
Here we define an array that can hold ten values:
double values[10];
This is the definition of a variable values whose type is “array of double”. That is, val-
ues stores a sequence of floating-point numbers. The [10] indicates the size of the
array. (See Figure 1.) The array size must be a constant that is known at compile time.
When you define an array, you can specify the initial values. For example,
double values[] = { 32, 54, 67.5, 29, 34.5, 80, 115, 44.5, 100, 65 };
When you supply initial values, you don’t need to specify the array size. The com-
piler determines the size by counting the values.
Use an array to
collect a sequence
of values of the
same type.
table 1 defining arrays
int numbers[10]; An array of ten integers.
const int SIZE = 10;
int numbers[SIZE];
It is a good idea to use a named constant for
the size.
int size = 10;
int numbers[size];
Caution: In standard C++, the size must be a
constant. This array definition will not work
with all compilers.
int squares[5] = { 0, 1, 4, 9, 16 }; An array of five integers, with initial values.
int squares[] = { 0, 1, 4, 9, 16 }; You can omit the array size if you supply
initial values. The size is set to the number of
initial values.
int squares[5] = { 0, 1, 4 }; If you supply fewer initial values than the size,
the remaining values are set to 0. This array
contains 0, 1, 4, 0, 0.
string names[3]; An array of three strings.
!
cfe2_ch06_p249_306.indd 250 10/26/10 7:47 PM
6.1 arrays 251
syntax 6.1 defining an array
double values[5] = { 32, 54, 67.5, 29, 34.5 };
Element type
Optional list
of initial values
SizeName
Size must
be a constant.
Ok to omit size if
initial values are given.
Use brackets to access an element.
values[i] = 0;
The index must be ≥ 0 and
< the size of the array.
See page 254.
Here we define an array that can hold ten values:
double values[10];
This is the definition of a variable values whose type is “array of double”. That is, val-
ues stores a sequence of floating-point numbers. The [10] indicates the size of the
array. (See Figure 1.) The array size must be a constant that is known at compile time.
When you define an array, you can specify the initial values. For example,
double values[] = { 32, 54, 67.5, 29, 34.5, 80, 115, 44.5, 100, 65 };
When you supply initial values, you don’t need to specify the array size. The com-
piler determines the size by counting the values.
Use an array to
collect a sequence
of values of the
same type.
table 1 defining arrays
int numbers[10]; An array of ten integers.
const int SIZE = 10;
int numbers[SIZE];
It is a good idea to use a named constant for
the size.
int size = 10;
int numbers[size];
Caution: In standard C++, the size must be a
constant. This array definition will not work
with all compilers.
int squares[5] = { 0, 1, 4, 9, 16 }; An array of five integers, with initial values.
int squares[] = { 0, 1, 4, 9, 16 }; You can omit the array size if you supply
initial values. The size is set to the number of
initial values.
int squares[5] = { 0, 1, 4 }; If you supply fewer initial values than the size,
the remaining values are set to 0. This array
contains 0, 1, 4, 0, 0.
string names[3]; An array of three strings.
!
cfe2_ch06_p249_306.indd 251 10/26/10 7:47 PM
252 Chapter 6 arrays and Vectors
6.1.2 accessing array elements
The values stored in an array are called its elements. Each element has a position
number, called an index. To access a value in the values array, you must specify which
index you want to use. That is done with the [] operator:
values[4] = 34.5;
Now the element with index 4 is filled with 34.5. (See Figure 2).
Figure 2
Filling an array element
You can display the contents of the element with index 4 with the following command:
cout << values[4] << endl;
As you can see, the element values[4] can be used like any variable of type double.
In C++, array positions are counted in a way that you may find surprising. If you
look carefully at Figure 2, you will find that the fifth element was filled when we
changed values[4]. In C++, the elements of arrays are numbered starting at 0. That is,
the legal elements for the values array are
values[0], the first element
values[1], the second element
values[2], the third element
values[3], the fourth element
values[4], the fifth element
...
values[9], the tenth element
You will see in Chapter 7 why this numbering scheme was chosen in C++.
You have to be careful about index values. Trying to access a element that does not
exist in the array is a serious error. For example, if values has twenty elements, you are
not allowed to access values[20].
Attempting to access an element whose index is not within the valid index range is
called a bounds error. The compiler does not catch this type of error. Even the run-
ning program generates no error mes sage. If you make a bounds error, you silently
read or overwrite another memory location. As a conse quence, your program may
have random errors, and it can even crash.
Individual elements
in an array values are
accessed by an
integer index i,
using the notation
values[i].
values = [0]
[1]
[2]
[3]
[4]
[5]
[6]
[7]
[8]
[9]
34.5
an array element
can be used like
any variable.
an array index must
be at least zero and
less than the size of
the array.
Like a post office box that is identified by a box
number, an array element is identified by an index.
The most common bounds error is the following:
double values[10];
cout << values[10];
There is no values[10] in an array with ten elements—the legal index values range
from 0 to 9.
To visit all elements of an array, use a variable for the index. Suppose values has ten
elements and the integer variable i takes values 0, 1, 2, and so on, up to 9. Then the
expression values[i] yields each element in turn. For example, this loop displays all
elements.
for (int i = 0; i < 10; i++)
{
cout << values[i] << endl;
}
Note that in the loop condition the index is less than 10 because there is no element
corresponding to val ues[10].
6.1.3 partially Filled arrays
An array cannot change size at run time. This is a problem when you don’t know in
advance how many elements you need. In that situation, you must come up with a
good guess on the maximum number of elements that you need to store. We call this
quantity the capacity. For example, we may decide that we sometimes want to store
more than ten values, but never more than 100:
const int CAPACITY = 100;
double values[CAPACITY];
In a typical program run, only part of the array will be occupied by actual elements.
We call such an array a partially filled array. You must keep a companion variable that
counts how many elements are actually used. In Figure 3 we call the companion vari-
able current_size.
The following loop collects values and fills up the values array.
int current_size = 0;
double input;
while (cin >> input)
{
if (current_size < CAPACITY)
{
values[current_size] = input;
current_size++;
a bounds error,
which occurs if you
supply an invalid
array index, can
corrupt data or cause
your program to
terminate.
With a partially filled
array, you need to
remember how many
elements are filled.
Figure 3
a partially Filled array
values =
24
67
54
32
CAPACITY
.
.
.
Not currently used
current_size
With a partially filled
array, keep a
companion variable
for the current size.
cfe2_ch06_p249_306.indd 252 10/26/10 7:47 PM
6.1 arrays 253
The most common bounds error is the following:
double values[10];
cout << values[10];
There is no values[10] in an array with ten elements—the legal index values range
from 0 to 9.
To visit all elements of an array, use a variable for the index. Suppose values has ten
elements and the integer variable i takes values 0, 1, 2, and so on, up to 9. Then the
expression values[i] yields each element in turn. For example, this loop displays all
elements.
for (int i = 0; i < 10; i++)
{
cout << values[i] << endl;
}
Note that in the loop condition the index is less than 10 because there is no element
corresponding to val ues[10].
6.1.3 partially Filled arrays
An array cannot change size at run time. This is a problem when you don’t know in
advance how many elements you need. In that situation, you must come up with a
good guess on the maximum number of elements that you need to store. We call this
quantity the capacity. For example, we may decide that we sometimes want to store
more than ten values, but never more than 100:
const int CAPACITY = 100;
double values[CAPACITY];
In a typical program run, only part of the array will be occupied by actual elements.
We call such an array a partially filled array. You must keep a companion variable that
counts how many elements are actually used. In Figure 3 we call the companion vari-
able current_size.
The following loop collects values and fills up the values array.
int current_size = 0;
double input;
while (cin >> input)
{
if (current_size < CAPACITY)
{
values[current_size] = input;
current_size++;
a bounds error,
which occurs if you
supply an invalid
array index, can
corrupt data or cause
your program to
terminate.
With a partially filled
array, you need to
remember how many
elements are filled.
Figure 3
a partially Filled array
values =
24
67
54
32
CAPACITY
.
.
.
Not currently used
current_size
With a partially filled
array, keep a
companion variable
for the current size.
cfe2_ch06_p249_306.indd 253 10/26/10 7:47 PM
254 Chapter 6 arrays and Vectors
}
}
At the end of this loop, current_size contains the actual number of elements in the
array. Note that you have to stop accepting inputs if the size of the array reaches the
capacity.
To process the gathered array elements, you again use the companion variable, not
the capacity. This loop prints the partially filled array:
for (int i = 0; i < current_size; i++)
{
cout << values[i] << endl;
}
1. Define an array of integers containing the first five prime numbers.
2. Assume the array primes has been initialized as described in Self Check 1. What is
its contents after executing the following loop?
for (int i = 0; i < 2; i++)
{
primes[4 - i] = primes[i];
}
3. Assume the array primes has been initialized as described in Self Check 1. What is
its contents after executing the following loop?
for (int i = 0; i < 5; i++)
{
primes[i]++;
}
4. Given the definition
const int CAPACITY = 10;
double values[CAPACITY];
write statements to put a zero into the elements of the array values with the
lowest and the highest valid index.
5. Given the array defined in Self Check 4, write a loop to print the elements of the
array values in reverse order, starting with the last element.
6. Define an array called words that can hold ten values of type string.
7. Define an array containing two strings, "Yes", and "No".
Practice it Now you can try these exercises at the end of the chapter: R6.1, R6.2, R6.6, P6.1.
bounds errors
Perhaps the most common error in using arrays is accessing a nonexistent element.
double values[10];
values[10] = 5.4;
// Error—values has 10 elements with subscripts 0 to 9
If your program accesses an array through an out-of-bounds subscript, there is no error mes-
sage. Instead, the pro gram will quietly (or not so quietly) corrupt some memory. Except for
very short programs, in which the problem may go unnoticed, that corruption will make the
s e l F C h e C k
Common error 6.1
program act unpredictably, and it can even cause the program to terminate. These are serious
errors that can be difficult to detect.
use Arrays for sequences of related values
Arrays are intended for storing sequences of values with the same meaning. For example, an
array of test scores makes perfect sense:
int scores[NUMBER_OF_SCORES];
But an array
double personal_data[3];
that holds a person’s age, bank balance, and shoe size in positions 0, 1, and 2 is bad design.
It would be tedious for the programmer to remember which of these data values is stored in
which array location. In this situation, it is far better to use three separate variables.
programming tip 6.1
In november 1988,
robert Morris, a stu-
dent at Cornell Uni versity, launched a
so-called virus program that infected
about 6,000 computers connected to
the Internet across the United states.
tens of thou sands of computer users
were unable to read their e-mail or oth-
erwise use their computers. all major
universities and many high-tech com-
panies were affected. (the Internet was
much smaller then than it is now.)
the particular kind of virus used in
this attack is called a worm. the virus
program crawled from one computer
on the Internet to the next. the worm
would attempt to connect to finger,
a program in the UnIX operating sys-
tem for finding information on a user
who has an account on a particular
com puter on the network. like many
pro grams in UnIX, finger was writ-
ten in the C language. In C, as in C++,
arrays have a fixed size. to store the
user name to be looked up (say, wal-
ters@cs.sjsu.edu), the finger program
allocated an array of 512 characters,
under the assumption that nobody
would ever provide such a long input.
Unfortunately, C, like C++, does not
check that an array index is less than
the length of the array. If you write
into an array using an index that is too
large, you simply overwrite memory
locations that belong to some other
objects. In some versions of the finger
program, the programmer had been
lazy and had not checked whether the
array holding the input characters was
large enough to hold the input. so the
worm program pur posefully filled the
512-character array with 536 bytes.
the excess 24 bytes would overwrite
a return address, which the attacker
knew was stored just after the line buf-
fer. When that function was finished,
it didn’t return to its caller but to code
sup plied by the worm (see Figure 4).
that code ran under the same super-
user privileges as finger, allowing the
worm to gain entry into the remote sys-
tem. had the programmer who wrote
finger been more conscien tious, this
particular attack would not be pos-
sible. In C++, as in C, all pro grammers
must be very careful not to overrun
array boundaries.
one may well speculate what would
possess the virus author to spend
many weeks to plan the antiso cial act
of breaking into thousands of comput-
ers and disabling them. It appears that
the break-in was fully intended by the
author, but the dis abling of the com-
puters was a bug, caused by continu-
ous reinfection. Morris was sentenced
to 3 years pro bation, 400 hours of
community ser vice, and fined $10,000.
In recent years, computer attacks
have intensified and the motives have
become more sinister. Instead of dis-
abling computers, viruses often steal
financial data or use the attacked com-
puters for sending spam e-mail. sadly,
many of these attacks continue to be
possible because of poorly written pro-
grams that are susceptible to buffer
overrun errors.
Return address
Line buffer
(512 bytes)
1 Before the attack
2 After the attack
Return address
Overrun buffer
(536 bytes)
Malicious
code
Figure 4
a “Buffer overrun” attack
Random Fact 6.1 an early Internet Worm
cfe2_ch06_p249_306.indd 254 10/26/10 7:47 PM
6.1 arrays 255
program act unpredictably, and it can even cause the program to terminate. These are serious
errors that can be difficult to detect.
use Arrays for sequences of related values
Arrays are intended for storing sequences of values with the same meaning. For example, an
array of test scores makes perfect sense:
int scores[NUMBER_OF_SCORES];
But an array
double personal_data[3];
that holds a person’s age, bank balance, and shoe size in positions 0, 1, and 2 is bad design.
It would be tedious for the programmer to remember which of these data values is stored in
which array location. In this situation, it is far better to use three separate variables.
programming tip 6.1
In november 1988,
robert Morris, a stu-
dent at Cornell Uni versity, launched a
so-called virus program that infected
about 6,000 computers connected to
the Internet across the United states.
tens of thou sands of computer users
were unable to read their e-mail or oth-
erwise use their computers. all major
universities and many high-tech com-
panies were affected. (the Internet was
much smaller then than it is now.)
the particular kind of virus used in
this attack is called a worm. the virus
program crawled from one computer
on the Internet to the next. the worm
would attempt to connect to finger,
a program in the UnIX operating sys-
tem for finding information on a user
who has an account on a particular
com puter on the network. like many
pro grams in UnIX, finger was writ-
ten in the C language. In C, as in C++,
arrays have a fixed size. to store the
user name to be looked up (say, wal-
ters@cs.sjsu.edu), the finger program
allocated an array of 512 characters,
under the assumption that nobody
would ever provide such a long input.
Unfortunately, C, like C++, does not
check that an array index is less than
the length of the array. If you write
into an array using an index that is too
large, you simply overwrite memory
locations that belong to some other
objects. In some versions of the finger
program, the programmer had been
lazy and had not checked whether the
array holding the input characters was
large enough to hold the input. so the
worm program pur posefully filled the
512-character array with 536 bytes.
the excess 24 bytes would overwrite
a return address, which the attacker
knew was stored just after the line buf-
fer. When that function was finished,
it didn’t return to its caller but to code
sup plied by the worm (see Figure 4).
that code ran under the same super-
user privileges as finger, allowing the
worm to gain entry into the remote sys-
tem. had the programmer who wrote
finger been more conscien tious, this
particular attack would not be pos-
sible. In C++, as in C, all pro grammers
must be very careful not to overrun
array boundaries.
one may well speculate what would
possess the virus author to spend
many weeks to plan the antiso cial act
of breaking into thousands of comput-
ers and disabling them. It appears that
the break-in was fully intended by the
author, but the dis abling of the com-
puters was a bug, caused by continu-
ous reinfection. Morris was sentenced
to 3 years pro bation, 400 hours of
community ser vice, and fined $10,000.
In recent years, computer attacks
have intensified and the motives have
become more sinister. Instead of dis-
abling computers, viruses often steal
financial data or use the attacked com-
puters for sending spam e-mail. sadly,
many of these attacks continue to be
possible because of poorly written pro-
grams that are susceptible to buffer
overrun errors.
Return address
Line buffer
(512 bytes)
1 Before the attack
2 After the attack
Return address
Overrun buffer
(536 bytes)
Malicious
code
Figure 4
a “Buffer overrun” attack
Random Fact 6.1 an early Internet Worm
cfe2_ch06_p249_306.indd 255 10/26/10 7:47 PM
256 Chapter 6 arrays and Vectors
6.2 Common array algorithms
In the following sections, we discuss some of the most common algorithms for
processing sequences of values. We present the algorithms so that you can use them
with fully and partially filled arrays as well as vectors (which we will introduce in
Section 6.7). When we use the expression size of values, you should replace it with a
constant or variable that yields the number of elements in the array (or the expression
values.size() if values is a vector.)
6.2.1 Filling
This loop fills an array with zeroes:
for (int i = 0; i < size of values; i++)
{
values[i] = 0;
}
Next, let us fill an array squares with the numbers 0, 1, 4, 9, 16, and so on. Note that
the element with index 0 contains 02, the element with index 1 contains 12, and so on.
for (int i = 0; i < size of squares; i++)
{
squares[i] = i * i;
}
6.2.2 Copying
Consider two arrays:
int squares[5] = { 0, 1, 4, 9, 16 };
int lucky_numbers[5];
Now suppose you want to copy all values from the first array to the second. The fol-
lowing assignment is an error:
lucky_numbers = squares; // Error
In C++, you cannot assign one array to another. Instead, you must use a loop to copy
all elements:
for (int i = 0; i < 5; i++)
{
lucky_numbers[i] = squares[i];
}
to copy an array,
use a loop to copy
its elements to a
new array.
Figure 5 Copying elements to Copy an array
lucky_numbers = [0]
[1]
[2]
[3]
[4]16
9
4
1
0squares = [0]
[1]
[2]
[3]
[4]16
9
4
1
0
6.2.3 sum and average Value
You have already encountered this algorithm in Section 4.7.1. Here is the code for
computing the sum of all elements in an array:
double total = 0;
for (int i = 0; i < size of values; i++)
{
total = total + values[i];
}
To obtain the average, divide by the number of elements:
double average = total / size of values;
Be sure to check that the size is not zero.
6.2.4 Maximum and Minimum
Use the algorithm from Section 4.7.4 that keeps a variable for
the largest element that you have encoun tered so far. Here is
the implementation for arrays:
double largest = values[0];
for (int i = 1; i < size of values; i++)
{
if (values[i] > largest)
{
largest = values[i];
}
}
Note that the loop starts at 1 because we initialize largest with values[0].
To compute the smallest value, reverse the comparison.
These algorithms require that the array contain at least one element.
6.2.5 element separators
When you display the elements of a collection, you usually want to separate them,
often with commas or vertical lines, like this:
1 | 4 | 9 | 16 | 25
Note that there is one fewer separator than there are numbers.
Print the separator before each element except the initial one
(with index 0):
for (int i = 0; i < size of values; i++)
{
if (i > 0)
{
cout << " | ";
}
cout << values[i];
}
When separating
elements, don’t place
a separator before
the first element.
To print five
elements, you need
four separators.
cfe2_ch06_p249_306.indd 256 10/26/10 7:47 PM
6.2 Common array algorithms 257
6.2.3 sum and average Value
You have already encountered this algorithm in Section 4.7.1. Here is the code for
computing the sum of all elements in an array:
double total = 0;
for (int i = 0; i < size of values; i++)
{
total = total + values[i];
}
To obtain the average, divide by the number of elements:
double average = total / size of values;
Be sure to check that the size is not zero.
6.2.4 Maximum and Minimum
Use the algorithm from Section 4.7.4 that keeps a variable for
the largest element that you have encoun tered so far. Here is
the implementation for arrays:
double largest = values[0];
for (int i = 1; i < size of values; i++)
{
if (values[i] > largest)
{
largest = values[i];
}
}
Note that the loop starts at 1 because we initialize largest with values[0].
To compute the smallest value, reverse the comparison.
These algorithms require that the array contain at least one element.
6.2.5 element separators
When you display the elements of a collection, you usually want to separate them,
often with commas or vertical lines, like this:
1 | 4 | 9 | 16 | 25
Note that there is one fewer separator than there are numbers.
Print the separator before each element except the initial one
(with index 0):
for (int i = 0; i < size of values; i++)
{
if (i > 0)
{
cout << " | ";
}
cout << values[i];
}
When separating
elements, don’t place
a separator before
the first element.
To print five
elements, you need
four separators.
cfe2_ch06_p249_306.indd 257 10/26/10 7:47 PM
258 Chapter 6 arrays and Vectors
6.2.6 linear search
You often need to search for the position of an element so that you can replace or
remove it. Visit all elements until you have found a match or you have come to the
end of the array. Here we search for the position of the first element equal to 100.
int pos = 0;
bool found = false;
while (pos < size of values && !found)
{
if (values[pos] == 100)
{
found = true;
}
else
{
pos++;
}
}
If found is true, then pos is the position of the first match.
6.2.7 removing an element
Consider a partially filled array values whose current size is stored in the variable
current_size. Suppose you want to remove the element with index pos from values. If
the elements are not in any particular order, that task is easy to accomplish. Simply
overwrite the element to be removed with the last element, then decrement the vari-
able tracking the size. (See Figure 6.)
values[pos] = values[current_size - 1];
current_size--;
The situation is more complex if the order of the elements matters. Then you must
move all elements fol lowing the element to be removed to a lower index, then decre-
ment the variable holding the size of the array. (See Figure 7.)
for (int i = pos + 1; i < current_size; i++)
{
values[i - 1] = values[i];
}
current_size--;
a linear search
inspects elements in
sequence until a
match is found.
To search for a
specific element,
visit the elements
and stop when you
encounter the match.
Figure 6
removing an element in an Unordered array
[0]
[1]
[2]
...
[pos]
[current_size - 1]
Decrement after
moving element
current_size
32
54
67.5
29
34.5
80
115
44.5
100
65
Figure 7
removing an element in an ordered array
[0]
[1]
[2]
...
[pos]
[current_size - 1]
1
2
3
4
5
Decrement after
moving elements
32
54
67.5
29
80
115
44.5
100
65
65
6.2.8 Inserting an element
If the order of the elements does not matter, you can simply insert new elements at
the end, incrementing the variable tracking the size. (See Figure 8.) For a partially
filled array:
if (current_size < CAPACITY)
{
current_size++;
values[current_size - 1] = new_element;
}
It is more work to insert an element at a particular position in the middle of a sequence.
First, increase the variable holding the current size. Next, move all elements above
the insertion location to a higher index. Finally, insert the new element. Here is the
code for a partially filled array:
if (current_size < CAPACITY)
{
current_size++;
for (int i = current_size - 1; i > pos; i–)
{
values[i] = values[i – 1];
}
values[pos] = new_element;
}
Note the order of the movement: When you remove an element, you first move the
next element down to a lower index, then the one after that, until you finally get to
the end of the array. When you insert an ele ment, you start at the end of the array,
move that element to a higher index, then move the one before that, and so on until
you finally get to the insertion location (see Figure 9).
6.2.9 swapping elements
You often need to swap elements of an array. For example, the sorting algo rithm in
Special Topic 6.2 on page 263 sorts an array by repeatedly swapping elements.
Before inserting an
element, move
elements to the end
of the array starting
with the last one.
Figure 8
Inserting an element in an Unordered array
[0]
[1]
[2]
…
[current_size – 1]
Incremented before
inserting element
Insert new element here
current_size
32
54
67.5
29
34.5
80
115
44.5
100
Figure 9
Inserting an element in an ordered array
[0]
[1]
[2]
…
[pos]
[current_size – 1]
5
4
3
2
1
Incremented before
moving elements
Insert new element here
32
54
67.5
29
42
34.5
80
115
44.5
100
cfe2_ch06_p249_306.indd 258 10/26/10 7:47 PM
6.2 Common array algorithms 259
6.2.8 Inserting an element
If the order of the elements does not matter, you can simply insert new elements at
the end, incrementing the variable tracking the size. (See Figure 8.) For a partially
filled array:
if (current_size < CAPACITY)
{
current_size++;
values[current_size - 1] = new_element;
}
It is more work to insert an element at a particular position in the middle of a sequence.
First, increase the variable holding the current size. Next, move all elements above
the insertion location to a higher index. Finally, insert the new element. Here is the
code for a partially filled array:
if (current_size < CAPACITY)
{
current_size++;
for (int i = current_size - 1; i > pos; i–)
{
values[i] = values[i – 1];
}
values[pos] = new_element;
}
Note the order of the movement: When you remove an element, you first move the
next element down to a lower index, then the one after that, until you finally get to
the end of the array. When you insert an ele ment, you start at the end of the array,
move that element to a higher index, then move the one before that, and so on until
you finally get to the insertion location (see Figure 9).
6.2.9 swapping elements
You often need to swap elements of an array. For example, the sorting algo rithm in
Special Topic 6.2 on page 263 sorts an array by repeatedly swapping elements.
Before inserting an
element, move
elements to the end
of the array starting
with the last one.
Figure 8
Inserting an element in an Unordered array
[0]
[1]
[2]
…
[current_size – 1]
Incremented before
inserting element
Insert new element here
current_size
32
54
67.5
29
34.5
80
115
44.5
100
Figure 9
Inserting an element in an ordered array
[0]
[1]
[2]
…
[pos]
[current_size – 1]
5
4
3
2
1
Incremented before
moving elements
Insert new element here
32
54
67.5
29
42
34.5
80
115
44.5
100
cfe2_ch06_p249_306.indd 259 10/26/10 7:47 PM
260 Chapter 6 arrays and Vectors
Figure 10
swapping array elements values = [0]
[1]
[2]
[3]
[4]
[i]
[j]
34.5
29
67.5
54
32
1
values =
[i]
[j]
34.5
29
67.5
54
32
2
temp = 54
values =
[i]
[j]
34.5
29
67.5
29
32
3
temp = 54
values =
[i]
[j]
34.5
54
67.5
29
32
4
temp = 54
Values to be swapped
Consider the task of swapping the elements at posi-
tions i and j of an array values. We’d like to set values[i]
to values[j]. But that overwrites the value that is currently
stored in values[i], so we want to save that first:
double temp = values[i];
values[i] = values[j];
Now we can set values[j] to the saved value.
values[j] = temp;
Figure 10 shows the process.
6.2.10 reading Input
If you know how many input values the user will supply, it is simple to place them
into an array:
double values[NUMBER_OF_INPUTS];
for (i = 0; i < NUMBER_OF_INPUTS; i++)
{
cin >> values[i];
}
To swap two elements, you
need a temporary variable.
Use a temporary
variable when
swapping two
elements.
cfe2_ch06_p249_306.indd 260 10/26/10 7:47 PM
6.2 Common array algorithms 261
However, this technique does not work if you need to read an arbitrary number of
inputs. In that case, add the values to an array until the end of the input has been
reached.
double values[CAPACITY];
int current_size = 0;
double input;
while (cin >> input)
{
if (current_size < CAPACITY)
{
values[current_size] = input;
current_size++;
}
}
Now values is a partially filled array, and the companion variable current_size is set to
the number of input values.
This loop discards any inputs that won’t fit in the array. A better approach would
be to copy values to a new larger array when the capacity is reached (see Section 6.2.2).
The following program solves the task that we set ourselves at the beginning of this
chapter, to mark the largest value in an input sequence:
ch06/largest.cpp
1 #include
2
3 using namespace std;
4
5 int main()
6 {
7 const int CAPACITY = 1000;
8 double values[CAPACITY];
9 int current_size = 0;
10
11 cout << "Please enter values, Q to quit:" << endl;
12 double input;
13 while (cin >> input)
14 {
15 if (current_size < CAPACITY)
16 {
17 values[current_size] = input;
18 current_size++;
19 }
20 }
21
22 double largest = values[0];
23 for (int i = 1; i < current_size; i++)
24 {
25 if (values[i] > largest)
26 {
27 largest = values[i];
28 }
29 }
30
31 for (int i = 0; i < current_size; i++)
32 {
33 cout << values[i];
34 if (values[i] == largest)
35 {
cfe2_ch06_p249_306.indd 261 10/26/10 7:47 PM
262 Chapter 6 arrays and Vectors
36 cout << " <== largest value";
37 }
38 cout << endl;
39 }
40
41 return 0;
42 }
Program run
Please enter values, Q to quit:
34.5 80 115 44.5 Q
34.5
80
115 <== largest value
44.5
8. What is the output of the largest.cpp program with the following inputs?
20 10 20 Q
9. Write a loop that counts how many elements in an array are equal to zero.
10. Consider the algorithm to find the largest element in an array. Why don’t we
initialize largest and i with zero, like this?
double largest = 0;
for (int i = 0; i < size of values; i++)
{
if (values[i] > largest)
{
largest = values[i];
}
}
11. When printing separators, we skipped the separator before the initial element.
Rewrite the loop so that the separator is printed after each element, except for
the last element.
12. What is wrong with these statements for printing an array with separators?
cout << values[0];
for (int i = 1; i < size of values; i++)
{
cout << ", " << values[i];
}
13. When searching for a match, we used a while loop, not a for loop. What is wrong
with using this loop instead?
for (pos = 0; pos < size of values && !found; pos++)
{
if (values[pos] == 100)
{
found = true;
}
}
14. When inserting an element into an array, we moved the elements with larger
index values, starting at the end. Why is it wrong to start at the insertion loca-
tion, like this:
s e l F C h e C k
for (int i = pos; i < size of values - 1; i++)
{
values[i + 1] = values[i];
}
Practice it Now you can try these exercises at the end of the chapter: R6.10, R6.13, P6.6, P6.16.
sorting with the C++ library
You often want to sort the elements of an array or vector. Special Topic 6.2 shows you a sort-
ing algorithm that is rel atively simple but not very efficient. Efficient sorting algorithms are
significantly more complex. Fortunately, the C++ library provides an efficient sort function.
To sort an array a with size elements, call
sort(a, a + size);
To sort a vector values, make this call:
sort(values.begin(), values.end());
To fully understand the curious syntax of these calls, you will need to know advanced C++
that is beyond the scope of this book. But don’t hesitate to call the sort function whenever you
need to sort an array or vector.
To use the sort function, include the
A sorting Algorithm
A sorting algorithm rearranges the elements of a sequence so that they are stored in sorted
order. Here is a simple sorting algorithm, called selection sort. Consider sorting the following
array values:
An obvious first step is to find the smallest element. In this case the smallest element is 5,
stored in values[3]. You should move the 5 to the beginning of the array. Of course, there is
already an element stored in values[0], namely 11. Therefore you cannot simply move val-
ues[3] into values[0] without moving the 11 somewhere else. You don’t yet know where the 11
should end up, but you know for certain that it should not be in values[0]. Simply get it out of
the way by swapping it with values[3]:
Now the first element is in the correct place. In the foregoing figure, the darker color indicates
the portion of the array that is already sorted.
Next take the minimum of the remaining entries values[1]…values[4]. That minimum
value, 9, is already in the correct place. You don’t need to do anything in this case, simply
extend the sorted area by one to the right:
Repeat the process. The minimum value of the unsorted region is 11, which needs to be
swapped with the first value of the unsorted region, 17:
special topic 6.1
special topic 6.2
cfe2_ch06_p249_306.indd 262 10/26/10 7:47 PM
6.2 Common array algorithms 263
for (int i = pos; i < size of values - 1; i++)
{
values[i + 1] = values[i];
}
Practice it Now you can try these exercises at the end of the chapter: R6.10, R6.13, P6.6, P6.16.
sorting with the C++ library
You often want to sort the elements of an array or vector. Special Topic 6.2 shows you a sort-
ing algorithm that is rel atively simple but not very efficient. Efficient sorting algorithms are
significantly more complex. Fortunately, the C++ library provides an efficient sort function.
To sort an array a with size elements, call
sort(a, a + size);
To sort a vector values, make this call:
sort(values.begin(), values.end());
To fully understand the curious syntax of these calls, you will need to know advanced C++
that is beyond the scope of this book. But don’t hesitate to call the sort function whenever you
need to sort an array or vector.
To use the sort function, include the
A sorting Algorithm
A sorting algorithm rearranges the elements of a sequence so that they are stored in sorted
order. Here is a simple sorting algorithm, called selection sort. Consider sorting the following
array values:
[0] [1] [2] [3] [4]
11 9 17 5 12
An obvious first step is to find the smallest element. In this case the smallest element is 5,
stored in values[3]. You should move the 5 to the beginning of the array. Of course, there is
already an element stored in values[0], namely 11. Therefore you cannot simply move val-
ues[3] into values[0] without moving the 11 somewhere else. You don’t yet know where the 11
should end up, but you know for certain that it should not be in values[0]. Simply get it out of
the way by swapping it with values[3]:
5 9 17 11 12
Now the first element is in the correct place. In the foregoing figure, the darker color indicates
the portion of the array that is already sorted.
Next take the minimum of the remaining entries values[1]…values[4]. That minimum
value, 9, is already in the correct place. You don’t need to do anything in this case, simply
extend the sorted area by one to the right:
5 9 17 11 12
Repeat the process. The minimum value of the unsorted region is 11, which needs to be
swapped with the first value of the unsorted region, 17:
5 9 11 17 12
special topic 6.1
special topic 6.2
cfe2_ch06_p249_306.indd 263 10/26/10 7:47 PM
264 Chapter 6 arrays and Vectors
Now the unsorted region is only two elements long; keep to the same successful strategy. The
minimum element is 12. Swap it with the first value, 17:
5 9 11 12 17
That leaves you with an unprocessed region of length 1, but of course a region of length 1 is
always sorted. You are done.
Here is the C++ code:
for (int unsorted = 0; unsorted < size - 1; unsorted++)
{
// Find the position of the minimum
int min_pos = unsorted;
for (int i = unsorted + 1; i < size; i++)
{
if (values[i] < values[min_pos]) { min_pos = i; }
}
// Swap the minimum into the sorted area
if (min_pos != unsorted)
{
double temp = values[min_pos];
values[min_pos] = values[unsorted];
values[unsorted] = temp;
}
}
This algorithm is simple to understand, but it is not very efficient. Computer scientists have
studied sorting algorithms extensively and discovered significantly better algorithms. The
sort function of the C++ library provides one such algorithm—see Special Topic 6.1 on page 263.
binary search
When an array is sorted, there is a much faster search algorithm than the linear search of Sec-
tion 6.2.6.
Consider the following sorted array values:
1 5 8 9 12 17 20 32
[0] [1] [2] [3] [4] [5] [6] [7]
We would like to see whether the value 15 is in the array. Let’s narrow our search by finding
whether the value is in the first or second half of the array. The last point in the first half of the
data set, values[3], is 9, which is smaller than the value we are looking for. Hence, we should
look in the second half of the array for a match, that is, in the sequence:
1 5 8 9 12 17 20 32
[0] [1] [2] [3] [4] [5] [6] [7]
Now the last value of the first half of this sequence is 17; hence, the value must be located in
the sequence:
1 5 8 9 12 17 20 32
[0] [1] [2] [3] [4] [5] [6] [7]
The last value of the first half of this very short sequence is 12, which is smaller than the value
that we are searching, so we must look in the second half:
1 5 8 9 12 17 20 32
[0] [1] [2] [3] [4] [5] [6] [7]
special topic 6.3
We still don’t have a match because 15 ≠ 17, and we cannot divide the subsequence further. If
we wanted to insert 15 into the sequence, we would need to insert it just before values[5].
This search process is called a binary search, because we cut the size of the search in half
in each step. That cutting in half works only because we know that the sequence of values is
sorted. Here is an implementation in C++:
bool found = false;
int low = 0;
int high = size - 1;
int pos = 0;
while (low <= high && !found)
{
pos = (low + high) / 2; // Midpoint of the subsequence
if (values[pos] == searched_value) { found = true; }
else if (values[pos] < searched_value) { low = pos + 1; } // Look in second half
else { high = pos - 1; } // Look in first half
}
if (found) { cout << "Found at position " << pos; }
else { cout << "Not found. Insert before position " << pos; }
6.3 arrays and Functions
In this section, we will explore how to write functions that process arrays.
A function that processes the values in an array needs to know the number of valid
elements in the array. For example, here is a sum function that computes the sum of all
elements in an array:
double sum(double values[], int size)
{
double total = 0;
for (int i = 0; i < size; i++)
{
total = total + values[i];
}
return total;
}
Note the special syntax for array parameter variables. When writing an array param-
eter variable, you place an empty [] behind the parameter name. Do not specify the
size of the array inside the brackets.
When you call the function, supply both the name of the array and the size. For
example,
double NUMBER_OF_SCORES = 10;
double scores[NUMBER_OF_SCORES]
= { 32, 54, 67.5, 29, 34.5, 80, 115, 44.5, 100, 65 };
double total_score = sum(scores, NUMBER_OF_SCORES);
You can also pass a smaller size to the function:
double partial_score = sum(scores, 5);
This call computes the sum of the first five elements of the scores array. Remember,
the function has no way of knowing how many elements the array has. It simply
relies on the size that the caller provides.
When passing an
array to a function,
also pass the size of
the array.
cfe2_ch06_p249_306.indd 264 10/26/10 7:47 PM
6.3 arrays and Functions 265
We still don’t have a match because 15 ≠ 17, and we cannot divide the subsequence further. If
we wanted to insert 15 into the sequence, we would need to insert it just before values[5].
This search process is called a binary search, because we cut the size of the search in half
in each step. That cutting in half works only because we know that the sequence of values is
sorted. Here is an implementation in C++:
bool found = false;
int low = 0;
int high = size - 1;
int pos = 0;
while (low <= high && !found)
{
pos = (low + high) / 2; // Midpoint of the subsequence
if (values[pos] == searched_value) { found = true; }
else if (values[pos] < searched_value) { low = pos + 1; } // Look in second half
else { high = pos - 1; } // Look in first half
}
if (found) { cout << "Found at position " << pos; }
else { cout << "Not found. Insert before position " << pos; }
6.3 arrays and Functions
In this section, we will explore how to write functions that process arrays.
A function that processes the values in an array needs to know the number of valid
elements in the array. For example, here is a sum function that computes the sum of all
elements in an array:
double sum(double values[], int size)
{
double total = 0;
for (int i = 0; i < size; i++)
{
total = total + values[i];
}
return total;
}
Note the special syntax for array parameter variables. When writing an array param-
eter variable, you place an empty [] behind the parameter name. Do not specify the
size of the array inside the brackets.
When you call the function, supply both the name of the array and the size. For
example,
double NUMBER_OF_SCORES = 10;
double scores[NUMBER_OF_SCORES]
= { 32, 54, 67.5, 29, 34.5, 80, 115, 44.5, 100, 65 };
double total_score = sum(scores, NUMBER_OF_SCORES);
You can also pass a smaller size to the function:
double partial_score = sum(scores, 5);
This call computes the sum of the first five elements of the scores array. Remember,
the function has no way of knowing how many elements the array has. It simply
relies on the size that the caller provides.
When passing an
array to a function,
also pass the size of
the array.
cfe2_ch06_p249_306.indd 265 10/26/10 7:47 PM
266 Chapter 6 arrays and Vectors
Array parameters are always reference parameters. (You will see the reason in
Chapter 7.) Functions can modify array arguments, and those modifications affect
the array that was passed into the function. For example, the following multiply func-
tion updates all elements in the array:
void multiply(double values[], int size, double factor)
{
for (int i = 0; i < size; i++)
{
values[i] = values[i] * factor;
}
}
You do not use an & symbol to denote the reference parameter in this case.
Although arrays can be function arguments, they cannot be function return types.
If a function com putes multiple values, the caller of the function must provide an
array parameter variable to hold the result.
void squares(int n, int result[])
{
for (int i = 0; i < n; i++)
{
result[i] = i * i;
}
}
When a function changes the size of an array, it should indicate to the caller how
many elements the array has after the call. The easiest way to do this is to return the
new size. Here is an example—a function that adds input values to an array:
int read_inputs(double inputs[], int capacity)
{
int current_size = 0;
double input;
while (cin >> input)
{
if (current_size < capacity)
{
inputs[current_size] = input;
current_size++;
}
}
return current_size;
}
Note that this function also needs to know the capacity of the array. Generally, a
function that adds ele ments to an array needs to know is capacity. You would call this
function like this:
const int MAXIMUM_NUMBER_OF_VALUES = 1000;
double values[MAXIMUM_NUMBER_OF_VALUES];
int current_size = read_inputs(values, MAXIMUM_NUMBER_OF_VALUES);
// values is a partially filled array; the current_size variable specifies its size
Alternatively, you can pass the size as a reference parameter. This is more appropriate
for functions that modify an existing array:
void append_inputs(double inputs[], int capacity, int& current_size)
{
double input;
while (cin >> input)
{
array parameters
are always reference
parameters.
a function’s return
type cannot be
an array.
When a function
modifies the size of
an array, it needs to
tell its caller.
a function that adds
elements to an array
needs to know its
capacity.
cfe2_ch06_p249_306.indd 266 10/26/10 7:47 PM
6.3 arrays and Functions 267
if (current_size < capacity)
{
inputs[current_size] = input;
current_size++;
}
}
}
This function is called as
append_inputs(values, MAXIMUM_NUMBER_OF_VALUES, current_size);
After the call, the current_size variable contains the new size.
The following example program reads values from standard input, doubles them,
and prints the result. The program uses three functions:
• The read_inputs function fills an array with the input values. It returns the number
of elements that were read.
• The multiply function modifies the contents of the array that it receives, demon-
strating that arrays are passed by reference.
• The print function does not modify the contents of the array that it receives.
ch06/functions.cpp
1 #include
2
3 using namespace std;
4
5 /**
6 Reads a sequence of floating-point numbers.
7 @param inputs an array containing the numbers
8 @param capacity the capacity of that array
9 @return the number of inputs stored in the array
10 */
11 int read_inputs(double inputs[], int capacity)
12 {
13 int current_size = 0;
14 cout << "Please enter values, Q to quit:" << endl;
15 bool more = true;
16 while (more)
17 {
18 double input;
19 cin >> input;
20 if (cin.fail())
21 {
22 more = false;
23 }
24 else if (current_size < capacity)
25 {
26 inputs[current_size] = input;
27 current_size++;
28 }
29 }
30 return current_size;
31 }
32
33 /**
34 Multiplies all elements of an array by a factor.
35 @param values a partially filled array
36 @param size the number of elements in values
cfe2_ch06_p249_306.indd 267 10/26/10 7:47 PM
268 Chapter 6 arrays and Vectors
37 @param factor the value with which each element is multiplied
38 */
39 void multiply(double values[], int size, double factor)
40 {
41 for (int i = 0; i < size; i++)
42 {
43 values[i] = values[i] * factor;
44 }
45 }
46
47 /**
48 Prints the elements of a vector, separated by commas.
49 @param values a partially filled array
50 @param size the number of elements in values
51 */
52 void print(double values[], int size)
53 {
54 for (int i = 0; i < size; i++)
55 {
56 if (i > 0) { cout << ", "; }
57 cout << values[i];
58 }
59 cout << endl;
60 }
61
62 int main()
63 {
64 const int CAPACITY = 1000;
65 double values[CAPACITY];
66 int size = read_inputs(values, CAPACITY);
67 multiply(values, size, 2);
68 print(values, size);
69
70 return 0;
71 }
Program run
Please enter values, Q to quit:
12 25 20 Q
24, 50, 40
15. What happens if you call the sum function and you lie about the size? For exam-
ple, calling
double result = sum(values, 1000);
even though values has size 100.
16. How do you call the squares function to compute the first five squares and store
the result in an array numbers?
17. Write a function that returns the first position of an element in an array, or –1 if
the element is not present. Use the linear search algorithm of Section 6.2.6.
18. Rewrite the read_inputs function so that the array size is a reference parameter,
not a return value.
19. Write the header for a function that appends two arrays into another array. Do
not implement the function.
s e l F C h e C k
Practice it Now you can try these exercises at the end of the chapter: R6.14, P6.8, P6.12.
Constant Array Parameters
When a function doesn’t modify an array parameter, it is considered good style to add the
const reserved word, like this:
double sum(const double values[], int size)
The const reserved word helps the reader of the code, making it clear that the function keeps
the array elements unchanged. If the implementation of the function tries to modify the array,
the compiler issues a warning.
6.4 problem solving: adapting algorithms
In Section 6.2, you were introduced to a number of fundamental array algorithms.
These algorithms form the building blocks for many programs that process arrays. In
general, it is a good problem-solving strat egy to have a repertoire of fundamental
algorithms that you can combine and adapt.
Consider this example problem: You are given the quiz scores of a student. You are
to compute the final quiz score, which is the sum of all scores after dropping the low-
est one. For example, if the scores are
8 7 8.5 9.5 7 4 10
then the final score is 50.
We do not have a ready-made algorithm for this situation. Instead, consider which
algorithms may be related. These include:
• Calculating the sum (Section 6.2.3)
• Finding the minimum value (Section 6.2.4)
• Removing an element (Section 6.2.7)
Now we can formulate a plan of attack that combines these algorithms.
Find the minimum.
Remove it from the array.
Calculate the sum.
Let’s try it out with our example. The minimum of
is 4. How do we remove it?
Now we have a problem. The removal algorithm in Section 6.2.7 locates the ele-
ment to be removed by using the position of the element, not the value.
But we have another algorithm for that:
• Linear search (Section 6.2.6)
special topic 6.4
By combining
fundamental
algorithms, you can
solve complex
programming tasks.
cfe2_ch06_p249_306.indd 268 10/26/10 7:48 PM
6.4 problem solving: adapting algorithms 269
Practice it Now you can try these exercises at the end of the chapter: R6.14, P6.8, P6.12.
Constant Array Parameters
When a function doesn’t modify an array parameter, it is considered good style to add the
const reserved word, like this:
double sum(const double values[], int size)
The const reserved word helps the reader of the code, making it clear that the function keeps
the array elements unchanged. If the implementation of the function tries to modify the array,
the compiler issues a warning.
6.4 problem solving: adapting algorithms
In Section 6.2, you were introduced to a number of fundamental array algorithms.
These algorithms form the building blocks for many programs that process arrays. In
general, it is a good problem-solving strat egy to have a repertoire of fundamental
algorithms that you can combine and adapt.
Consider this example problem: You are given the quiz scores of a student. You are
to compute the final quiz score, which is the sum of all scores after dropping the low-
est one. For example, if the scores are
8 7 8.5 9.5 7 4 10
then the final score is 50.
We do not have a ready-made algorithm for this situation. Instead, consider which
algorithms may be related. These include:
• Calculating the sum (Section 6.2.3)
• Finding the minimum value (Section 6.2.4)
• Removing an element (Section 6.2.7)
Now we can formulate a plan of attack that combines these algorithms.
Find the minimum.
Remove it from the array.
Calculate the sum.
Let’s try it out with our example. The minimum of
8
[0]
7
[1]
8.5
[2]
9.5
[3]
7
[4]
4
[5]
10
[6]
is 4. How do we remove it?
Now we have a problem. The removal algorithm in Section 6.2.7 locates the ele-
ment to be removed by using the position of the element, not the value.
But we have another algorithm for that:
• Linear search (Section 6.2.6)
special topic 6.4
By combining
fundamental
algorithms, you can
solve complex
programming tasks.
cfe2_ch06_p249_306.indd 269 10/26/10 7:48 PM
270 Chapter 6 arrays and Vectors
We need to fix our plan of attack:
Find the minimum value.
Find its position.
Remove that position from the array.
Calculate the sum.
Will it work? Let’s continue with our example.
We found a minimum value of 4. Linear search tells us that the value 4 occurs at
position 5.
8
[0]
7
[1]
8.5
[2]
9.5
[3]
7
[4]
4
[5]
10
[6]
We remove it:
8
[0]
7
[1]
8.5
[2]
9.5
[3] [4]
7
[5]
10
Finally, we compute the sum: 8 + 7 + 8.5 + 9.5 + 7 + 10 = 50.
This walkthrough demonstrates that our strategy works.
Can we do better? It seems a bit inefficient to find the minimum and then make
another pass through the array to obtain its position.
We can adapt the algorithm for finding the minimum to yield the position of the
minimum. Here is the original algorithm:
double smallest = values[0];
for (int i = 1; i < size of values; i++)
{
if (values[i] < smallest)
{
smallest = values[i];
}
}
When we find the smallest value, we also want to update the position:
if (values[i] < smallest)
{
smallest = values[i];
smallest_position = i;
}
In fact, then there is no reason to keep track of the smallest value any longer. It is sim-
ply val ues[smallest_position]. With this insight, we can adapt the algorithm as
follows:
int smallest_position = 0;
for (int i = 1; i < size of values; i++)
{
if (values[i] < values[smallest_position])
{
smallest_position = i;
}
}
With this adaptation, our problem is solved with the following strategy:
Find the position of the minimum.
Remove it from the array.
Calculate the sum.
In How To 6.1 on page 271, we develop a C++ program from this strategy.
you should be
familiar with the
implementation of
fundamental
algorithms so that
you can adapt them.
The next section shows you a technique for discovering a new algorithm when
none of the fundamen tal algorithms can be adapted to a task.
20. Section 6.2.7 has two algorithms for removing an element. Which of the two
should be used to solve the task described in this section?
21. It isn’t actually necessary to remove the minimum in order to compute the total
score. Describe an alternative.
22. How can you print the number of positive and negative values in a given array,
using one or more of the algorithms in Section 4.7?
23. How can you print all positive values in an array, separated by commas?
24. Consider the following algorithm for collecting all matches in an array:
int matches_size = 0;
for (int i = 0; i < size of values; i++)
{
if (values[i] fulfills the condition)
{
matches[matches_size] = values[i];
matches_size++;
}
}
How can this algorithm help you with Self Check 23?
Practice it Now you can try these exercises at the end of the chapter: R6.15, R6.16.
step 1 Decompose your task into steps.
You will usually want to break down your task into multiple steps, such as
• Reading the data into an array.
• Processing the data in one or more steps.
• Displaying the results.
In our sample problem, this yields the following pseudocode:
Read inputs.
Compute the final score.
Display the score.
When deciding how to process the data, you should be familiar with the array algorithms in
Section 6.2. Many pro cessing tasks can be solved by combining or adapting one or more of
these algorithms.
s e l F C h e C k
h o W t o 6 . 1 working with Arrays
When you process sequences of values, you usually need to use arrays. (In some very simple
situations, you can process values as you read them in, without storing them.) This How To
walks you through the necessary steps.
Consider the example problem from Section 6.4: You are given the quiz scores of a stu dent.
You are to compute the final quiz score, which is the sum of all scores after dropping the low-
est one. For example, if the scores are
8 7 8.5 9.5 7 5 10
then the final score is 50.
cfe2_ch06_p249_306.indd 270 10/26/10 7:48 PM
6.4 problem solving: adapting algorithms 271
The next section shows you a technique for discovering a new algorithm when
none of the fundamen tal algorithms can be adapted to a task.
20. Section 6.2.7 has two algorithms for removing an element. Which of the two
should be used to solve the task described in this section?
21. It isn’t actually necessary to remove the minimum in order to compute the total
score. Describe an alternative.
22. How can you print the number of positive and negative values in a given array,
using one or more of the algorithms in Section 4.7?
23. How can you print all positive values in an array, separated by commas?
24. Consider the following algorithm for collecting all matches in an array:
int matches_size = 0;
for (int i = 0; i < size of values; i++)
{
if (values[i] fulfills the condition)
{
matches[matches_size] = values[i];
matches_size++;
}
}
How can this algorithm help you with Self Check 23?
Practice it Now you can try these exercises at the end of the chapter: R6.15, R6.16.
step 1 Decompose your task into steps.
You will usually want to break down your task into multiple steps, such as
• Reading the data into an array.
• Processing the data in one or more steps.
• Displaying the results.
In our sample problem, this yields the following pseudocode:
Read inputs.
Compute the final score.
Display the score.
When deciding how to process the data, you should be familiar with the array algorithms in
Section 6.2. Many pro cessing tasks can be solved by combining or adapting one or more of
these algorithms.
s e l F C h e C k
h o W t o 6 . 1 working with Arrays
When you process sequences of values, you usually need to use arrays. (In some very simple
situations, you can process values as you read them in, without storing them.) This How To
walks you through the necessary steps.
Consider the example problem from Section 6.4: You are given the quiz scores of a stu dent.
You are to compute the final quiz score, which is the sum of all scores after dropping the low-
est one. For example, if the scores are
8 7 8.5 9.5 7 5 10
then the final score is 50.
cfe2_ch06_p249_306.indd 271 10/26/10 7:48 PM
272 Chapter 6 arrays and Vectors
The preceding section showed you how to decompose Compute the final score into fundamen-
tal algorithms:
Find the position of the minimum.
Remove it from the array.
Calculate the sum.
step 2 Determine functions, arguments, and return values for each step.
Even though it may be possible to put all steps into the main function, this is rarely a good idea.
The simplest and best approach is to make each nontrivial step into a separate function. In our
example, we will implement four functions:
• read_inputs
• min_position
• remove
• sum
• final_score
For each function that processes an array, you will need to pass the array itself and the array
size. For example,
double sum(double values[], int size)
If the function modifies the size, it needs to tell the caller what the new size is. The function can
return the size, or it can use a reference parameter for the size. The second approach is a better
choice for a function that modifies an existing array.
We use the first approach with the function that reads input values.
int read_inputs(double values[], int capacity) // Returns the size
The remove function modifies the current_size parameter:
void remove(double values[], int& current_size, int pos)
At this point, you should document each function, like this:
/**
Removes an element from an array. The order of the elements is not preserved.
@param values a partially filled array
@param current_size the number of elements in values
(will be reduced by 1 if the position is valid)
@param pos the position of the element to be removed
*/
void remove(double values[], int& current_size, int pos)
step 3 Implement each function, using helper functions when needed.
We won’t show the code for the read_inputs function because you have seen it already. Let us
implement the final_score function. It calls three helper functions, min_position, remove, and
sum:
/**
Removes the smallest value of an array and returns the
sum of the remaining values.
@param values a partially filled array
@param current_size the number of elements in values (will be reduced by 1)
@return the sum of the values, excluding the minimum
*/
double final_score(double values[], int& current_size)
{
int pos = min_position(values, current_size);
remove(values, current_size, pos);
return sum(values, current_size);
cfe2_ch06_p249_306.indd 272 10/26/10 7:48 PM
6.4 problem solving: adapting algorithms 273
}
We discussed the algorithm for min_position in the preceding section:
/**
Gets the position of the minimum value from an array.
@param values a partially filled array
@param size the number of elements in values
@return the position of the smallest element in values
*/
int min_position(double values[], int size)
{
int smallest_position = 0;
for (int i = 1; i < size; i++)
{
if (values[i] < values[smallest_position])
{
smallest_position = i;
}
}
return smallest_position;
}
The remaining helper functions use the algorithms from Section 6.2. You will find the imple-
mentations in the book’s companion code.
step 4 Consider boundary conditions for the functions that you are implementing
Most functions that operate on arrays are a bit intricate, and you have to be careful that you
handle both normal and exceptional situations. What happens with an empty array? An array
that contains a single element? When no match is found? When there are multiple matches?
Consider these boundary conditions and make sure that your functions work correctly.
Here is one example of such a consideration. How do we know that the min_position func-
tion will be called with an array of size at least 1? (Recall that you must have at least one ele-
ment in order to find the minimum.) That func tion is called from the final_score function.
However, the final_score function could conceivably be called with an empty array. We need
to either include a test or add a restriction to the function comment. We will opt for the latter
and change the comment for the values parameter variable of the min_position function to
@param values a partially filled array of size >= 1
Consider another potential problem. What if there are multiple matches? That means that a
student had more than one test with a low score. The final_score function removes only one of
the occurrences of that low score, and that is the desired behavior.
step 5 Assemble and test the complete program.
Now we are ready to combine the individual functions into a complete program. Before doing
this, consider some test cases and their expected output:
test Case expected output Comment
8 7 8.5 9.5 7 5 10 50 See Step 1.
8 7 7 9 24 Only one instance of the low score
should be removed.
8 0 After removing the low score, no
score remains.
(no inputs) error That is not a legal input.
cfe2_ch06_p249_306.indd 273 10/26/10 7:48 PM
274 Chapter 6 arrays and Vectors
This main function completes the solution (see ch06/scores.cpp).
int main()
{
const int CAPACITY = 1000;
double scores[CAPACITY];
int current_size = read_inputs(scores, CAPACITY);
if (current_size == 0)
{
cout << "At least one score is required." << endl;
}
else
{
double score = final_score(scores, current_size);
cout << "Final score: " << score << endl;
}
return 0;
}
6.5 problem solving: discovering algorithms by
Manipulating physical objects
In Section 6.4, you saw how to solve a problem by combining and adapting known
algorithms. But what do you do when none of the standard algorithms is sufficient
for your task? In this section, you will learn a technique for discovering algorithms
by manipulating physical objects.
Consider the following task. You are given an array whose size is an even number,
and you are to switch the first and the second half. For example, if the array contains
the eight numbers
9 13 21 4 11 7 1 3
then you should change it to
9 13 21 411 7 1 3
Many students find it quite challenging to come up with an algorithm. They may
know that a loop is required, and they may realize that elements should be inserted
(Section 6.2.8) or swapped (Section 6.2.9), but they do not have sufficient intuition to
draw diagrams, describe an algorithm, or write down pseudocode.
One useful technique for discovering an algorithm is to manipulate physical
objects. Start by lining up some objects to denote an array. Coins, playing cards, or
small toys are good choices.
W o r k e d e X a M p l e 6 . 1 rolling the dice
This Worked Example shows how to analyze a set of die tosses to see
whether the die is “fair”.
Use a sequence of
coins, playing cards,
or toys to visualize
an array of values.
Available online at www.wiley.com/college/horstmann.
cfe2_ch06_p249_306.indd 274 10/26/10 7:48 PM
www.wiley.com/college/horstmann
6.5 problem solving: discovering algorithms by Manipulating physical objects 275
This main function completes the solution (see ch06/scores.cpp).
int main()
{
const int CAPACITY = 1000;
double scores[CAPACITY];
int current_size = read_inputs(scores, CAPACITY);
if (current_size == 0)
{
cout << "At least one score is required." << endl;
}
else
{
double score = final_score(scores, current_size);
cout << "Final score: " << score << endl;
}
return 0;
}
6.5 problem solving: discovering algorithms by
Manipulating physical objects
In Section 6.4, you saw how to solve a problem by combining and adapting known
algorithms. But what do you do when none of the standard algorithms is sufficient
for your task? In this section, you will learn a technique for discovering algorithms
by manipulating physical objects.
Consider the following task. You are given an array whose size is an even number,
and you are to switch the first and the second half. For example, if the array contains
the eight numbers
then you should change it to
Many students find it quite challenging to come up with an algorithm. They may
know that a loop is required, and they may realize that elements should be inserted
(Section 6.2.8) or swapped (Section 6.2.9), but they do not have sufficient intuition to
draw diagrams, describe an algorithm, or write down pseudocode.
One useful technique for discovering an algorithm is to manipulate physical
objects. Start by lining up some objects to denote an array. Coins, playing cards, or
small toys are good choices.
W o r k e d e X a M p l e 6 . 1 rolling the dice
This Worked Example shows how to analyze a set of die tosses to see
whether the die is “fair”.
Use a sequence of
coins, playing cards,
or toys to visualize
an array of values.
Manipulating physical objects can give you
ideas for discovering algorithms.
Here we arrange eight coins.
Now let’s step back and see what we can do to change the order of the coins.
We can remove a coin (Section 6.2.7):
We can insert a coin (Section 6.2.8):
Or we can swap two coins (Section 6.2.9).
Go ahead—line up some coins and try out these three operations right now so that
you get a feel for them.
Visualizing the
removal of an
array element
Visualizing the
insertion of an
array element
Visualizing the
swapping of
two coins
cfe2_ch06_p249_306.indd 275 10/26/10 7:48 PM
276 Chapter 6 arrays and Vectors
Now how does that help us with our problem, switching the first and the second
half of the array?
Let’s put the first coin into place, by swapping it with the fifth coin. However, as
C++ programmers, we will say that we swap the coins in positions 0 and 4:
Next, we swap the coins in positions 1 and 5:
Two more swaps, and we are done:
Now an algorithm is becoming apparent:
i = 0
j = ... (we’ll think about that in a minute)
while (don’t know yet)
swap elements at positions i and j
i++
j++
Where does the variable j start? When we have eight coins, the coin at position zero is
moved to position 4. In general, it is moved to the middle of the array, or to position
size / 2.
And how many iterations do we make? We need to swap all coins in the first half.
That is, we need to swap size / 2 coins. The pseudocode is
i = 0
j = size / 2
while (i < size / 2)
swap elements at positions i and j
i++
j++
It is a good idea to make a walkthrough of the pseudocode (see Section 4.2). You can
use paper clips to denote the positions of the variables i and j. If the walkthrough is
suc cessful, then we know that there was no “off-by-one” error in the pseudocode.
Self Check 25 asks you to carry out the walkthrough, and Exercise P6.7 asks you to
translate the pseudocode to C++. Exercise R6.17 suggests a different algorithm for
switching the two halves of an array, by repeatedly removing and inserting coins.
Many people find that the manipulation of physical objects is less intimidating
than drawing diagrams or mentally envisioning algorithms. Give it a try when you
need to design a new algorithm!
25. Walk through the algorithm that we developed in this section, using two paper
clips to indicate the positions for i and j. Explain why there are no bounds errors
in the pseudocode.
26. Take out some coins and simulate the following pseudocode, using two paper
clips to indicate the positions for i and j:
i = 0
j = size - 1
while (i < j)
swap elements at positions i and j
i++
j--
What does the algorithm do?
27. Consider the task of rearranging all values in an array so that the even numbers
come first. Other wise, the order doesn’t matter. For example, the array
1 4 14 2 1 3 5 6 23
could be rearranged to
4 2 14 6 1 5 3 23 1
Using coins and paperclips, discover an algorithm that solves this task by
swapping elements, then describe it in pseudocode.
28. Discover an algorithm for the task of Self Check 27 that uses removal and
insertion of elements instead of swapping.
29. Consider the algorithm in Section 4.7.4 that finds the
largest element in a sequence of inputs—not the largest
element in an array. Why is this algorithm better visual-
ized by picking playing cards from a deck rather than
arranging toy soldiers in a sequence?
Practice it Now you can try these exercises at the end of the chapter: R6.17, R6.18, P6.7.
you can use paper
clips as position
markers or counters.
s e l F C h e C k
cfe2_ch06_p249_306.indd 276 10/26/10 7:48 PM
6.5 problem solving: discovering algorithms by Manipulating physical objects 277
And how many iterations do we make? We need to swap all coins in the first half.
That is, we need to swap size / 2 coins. The pseudocode is
i = 0
j = size / 2
while (i < size / 2)
swap elements at positions i and j
i++
j++
It is a good idea to make a walkthrough of the pseudocode (see Section 4.2). You can
use paper clips to denote the positions of the variables i and j. If the walkthrough is
suc cessful, then we know that there was no “off-by-one” error in the pseudocode.
Self Check 25 asks you to carry out the walkthrough, and Exercise P6.7 asks you to
translate the pseudocode to C++. Exercise R6.17 suggests a different algorithm for
switching the two halves of an array, by repeatedly removing and inserting coins.
Many people find that the manipulation of physical objects is less intimidating
than drawing diagrams or mentally envisioning algorithms. Give it a try when you
need to design a new algorithm!
25. Walk through the algorithm that we developed in this section, using two paper
clips to indicate the positions for i and j. Explain why there are no bounds errors
in the pseudocode.
26. Take out some coins and simulate the following pseudocode, using two paper
clips to indicate the positions for i and j:
i = 0
j = size - 1
while (i < j)
swap elements at positions i and j
i++
j--
What does the algorithm do?
27. Consider the task of rearranging all values in an array so that the even numbers
come first. Other wise, the order doesn’t matter. For example, the array
1 4 14 2 1 3 5 6 23
could be rearranged to
4 2 14 6 1 5 3 23 1
Using coins and paperclips, discover an algorithm that solves this task by
swapping elements, then describe it in pseudocode.
28. Discover an algorithm for the task of Self Check 27 that uses removal and
insertion of elements instead of swapping.
29. Consider the algorithm in Section 4.7.4 that finds the
largest element in a sequence of inputs—not the largest
element in an array. Why is this algorithm better visual-
ized by picking playing cards from a deck rather than
arranging toy soldiers in a sequence?
Practice it Now you can try these exercises at the end of the chapter: R6.17, R6.18, P6.7.
you can use paper
clips as position
markers or counters.
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278 Chapter 6 arrays and Vectors
6.6 two-dimensional arrays
It often happens that you want to store collections
of values that have a two-dimensional layout. Such
data sets commonly occur in financial and scientific
applications. An arrangement consisting of rows
and columns of values is called a two-dimensional
array, or a matrix.
Let’s explore how to store the example data
shown in Figure 11: the medal counts of the figure
skating competitions at the 2010 Winter Olympics.
Gold silver Bronze
Canada 1 0 1
China 1 1 0
Germany 0 0 1
Korea 1 0 0
Japan 0 1 1
Russia 0 1 1
United States 1 1 0
Figure 11 Figure skating Medal Counts
6.6.1 defining two-dimensional arrays
C++ uses an array with two subscripts to store a two-dimensional array. For exam-
ple, here is the defini tion of an array with 7 rows and 3 columns, suitable for storing
our medal count data:
const int COUNTRIES = 7;
const int MEDALS = 3;
int counts[COUNTRIES][MEDALS];
You can initialize the array by grouping each row, as follows:
int counts[COUNTRIES][MEDALS] =
{
{ 1, 0, 1 },
{ 1, 1, 0 },
{ 0, 0, 1 },
{ 1, 0, 0 },
{ 0, 1, 1 },
{ 0, 1, 1 },
{ 1, 1, 0 }
};
Just as with one-dimensional arrays, you cannot change the size of a two-dimen-
sional array once it has been defined.
6.6.2 accessing elements
To access a particular element in the two-dimensional array, you need to specify two
subscripts in sepa rate brackets to select the row and column, respectively (see Syntax
6.2 and Figure 12):
int value = counts[3][1];
To access all values in a two-dimensional array, you use two nested loops. For exam-
ple, the following loop prints all elements of counts.
for (int i = 0; i < COUNTRIES; i++)
{
// Process the ith row
for (int j = 0; j < MEDALS; j++)
{
// Process the jth column in the ith row
cout << setw(8) << counts[i][j];
}
cout << endl; // Start a new line at the end of the row
}
syntax 6.2 two-dimensional array definition
int data[4][4] = {
{ 16, 3, 2, 13 },
{ 5, 10, 11, 8 },
{ 9, 6, 7, 12 },
{ 4, 15, 14, 1 },
};
Element type Rows Columns
Name
Optional list of initial values
Individual elements
in a two-dimensional
array are accessed by
using two subscripts,
array[i][j].
Figure 12
accessing an element in a
two-dimensional array
[0]
[1]
[2]
[3]
[4]
[5]
[6]
[0][1][2]
counts[3][1]
Column index
R
o
w
in
d
ex
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6.6 two-dimensional arrays 279
6.6.2 accessing elements
To access a particular element in the two-dimensional array, you need to specify two
subscripts in sepa rate brackets to select the row and column, respectively (see Syntax
6.2 and Figure 12):
int value = counts[3][1];
To access all values in a two-dimensional array, you use two nested loops. For exam-
ple, the following loop prints all elements of counts.
for (int i = 0; i < COUNTRIES; i++)
{
// Process the ith row
for (int j = 0; j < MEDALS; j++)
{
// Process the jth column in the ith row
cout << setw(8) << counts[i][j];
}
cout << endl; // Start a new line at the end of the row
}
syntax 6.2 two-dimensional array definition
int data[4][4] = {
{ 16, 3, 2, 13 },
{ 5, 10, 11, 8 },
{ 9, 6, 7, 12 },
{ 4, 15, 14, 1 },
};
Element type Rows Columns
Name
Optional list of initial values
Individual elements
in a two-dimensional
array are accessed by
using two subscripts,
array[i][j].
Figure 12
accessing an element in a
two-dimensional array
[0]
[1]
[2]
[3]
[4]
[5]
[6]
[0][1][2]
counts[3][1]
Column index
R
o
w
in
d
ex
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280 Chapter 6 arrays and Vectors
6.6.3 Computing row and Column totals
A common task is to compute row or column totals. In our example, the row totals
give us the total num ber of medals won by a particular country.
Finding the right index values is a bit tricky, and it is a good idea to make a quick
sketch. To compute the total of row i, we need to visit the following elements:
[i][0][i][1][i][2]row i
0 MEDALS - 1
As you can see, we need to compute the sum of counts[i][j], where j ranges from 0 to
MEDALS - 1. The fol lowing loop computes the total:
int total = 0;
for (int j = 0; j < MEDALS; j++)
{
total = total + counts[i][j];
}
Computing column totals is similar. Form the sum of counts[i][j], where i ranges
from 0 to COUNTRIES - 1.
[0][j]
[1][j]
[2][j]
[3][j]
[4][j]
[5][j]
[6][j]
column j
COUNTRIES - 1
0
int total = 0;
for (int i = 0; i < COUNTRIES; i++)
{
total = total + counts[i][j];
}
6.6.4 two-dimensional array parameters
When passing a two-dimensional array to a function, you must specify the number of
columns as a con stant with the parameter type. For example, this function computes
the total of a given row:
const int COLUMNS = 3;
int row_total(int table[][COLUMNS], int row)
{
int total = 0;
for (int j = 0; j < COLUMNS; j++)
{
total = total + table[row][j];
}
return total;
}
This function can compute row totals of a two-dimensional array with an arbitrary
number of rows, but the array must have 3 columns. You have to write a different
function if you want to compute row totals of a two-dimensional array with 4
columns.
To understand this limitation, you need to know how the array elements are stored
in memory. Although the array appears to be two-dimensional, the elements are still
stored as a linear sequence. Figure 13 shows how the counts array is stored, row by
row.
For example, to reach
counts[3][1]
the program must first skip past rows 0, 1, and 2 and then locate offset 1 in row 3. The
offset from the start of the array is
3 × number of columns + 1
Now consider the row_total function. The compiler generates code to find the element
table[i][j]
by computing the offset
i * COLUMNS + j
The compiler uses the value that you supplied in the second pair of brackets when
declaring the parame ter:
int row_total(int table[][COLUMNS], int row)
Note that the first pair of brackets should be empty, just as with one-dimensional
arrays.
a two-dimensional
array parameter must
have a fixed number
of columns.
Figure 13 a two-dimensional array is stored as a sequence of rows
counts =
counts[3][1]
row 0 row 1 row 2 row 3
. . .
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6.6 two-dimensional arrays 281
6.6.4 two-dimensional array parameters
When passing a two-dimensional array to a function, you must specify the number of
columns as a con stant with the parameter type. For example, this function computes
the total of a given row:
const int COLUMNS = 3;
int row_total(int table[][COLUMNS], int row)
{
int total = 0;
for (int j = 0; j < COLUMNS; j++)
{
total = total + table[row][j];
}
return total;
}
This function can compute row totals of a two-dimensional array with an arbitrary
number of rows, but the array must have 3 columns. You have to write a different
function if you want to compute row totals of a two-dimensional array with 4
columns.
To understand this limitation, you need to know how the array elements are stored
in memory. Although the array appears to be two-dimensional, the elements are still
stored as a linear sequence. Figure 13 shows how the counts array is stored, row by
row.
For example, to reach
counts[3][1]
the program must first skip past rows 0, 1, and 2 and then locate offset 1 in row 3. The
offset from the start of the array is
3 × number of columns + 1
Now consider the row_total function. The compiler generates code to find the element
table[i][j]
by computing the offset
i * COLUMNS + j
The compiler uses the value that you supplied in the second pair of brackets when
declaring the parame ter:
int row_total(int table[][COLUMNS], int row)
Note that the first pair of brackets should be empty, just as with one-dimensional
arrays.
a two-dimensional
array parameter must
have a fixed number
of columns.
Figure 13 a two-dimensional array is stored as a sequence of rows
counts =
counts[3][1]
row 0 row 1 row 2 row 3
. . .
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282 Chapter 6 arrays and Vectors
The row_total function did not need to know the number of rows of the array. If
the number of rows is required, pass it as a variable, as in this example:
int column_total(int table[][COLUMNS], int rows, int column)
{
int total = 0;
for (int i = 0; i < rows; i++)
{
total = total + table[i][column];
}
return total;
}
Working with two-dimensional arrays is illustrated in the following program. The
program prints out the medal counts and the row totals.
ch06/medals.cpp
1 #include
2 #include
3 #include
4
5 using namespace std;
6
7 const int COLUMNS = 3;
8
9 /**
10 Computes the total of a row in a table.
11 @param table a table with 3 columns
12 @param row the row that needs to be totaled
13 @return the sum of all elements in the given row
14 */
15 double row_total(int table[][COLUMNS], int row)
16 {
17 int total = 0;
18 for (int j = 0; j < COLUMNS; j++)
19 {
20 total = total + table[row][j];
21 }
22 return total;
23 }
24
25 int main()
26 {
27 const int COUNTRIES = 7;
28 const int MEDALS = 3;
29
30 string countries[] =
31 {
32 "Canada",
33 "China",
34 "Germany",
35 "Korea",
36 "Japan",
37 "Russia",
38 "United States"
39 };
40
41 int counts[COUNTRIES][MEDALS] =
42 {
cfe2_ch06_p249_306.indd 282 10/26/10 7:48 PM
6.6 two-dimensional arrays 283
43 { 1, 0, 1 },
44 { 1, 1, 0 },
45 { 0, 0, 1 },
46 { 1, 0, 0 },
47 { 0, 1, 1 },
48 { 0, 1, 1 },
49 { 1, 1, 0 }
50 };
51
52 cout << " Country Gold Silver Bronze Total" << endl;
53
54 // Print countries, counts, and row totals
55 for (int i = 0; i < COUNTRIES; i++)
56 {
57 cout << setw(15) << countries[i];
58 // Process the ith row
59 for (int j = 0; j < MEDALS; j++)
60 {
61 cout << setw(8) << counts[i][j];
62 }
63 int total = row_total(counts, i);
64 cout << setw(8) << total << endl;
65 }
66
67 return 0;
68 }
Program run
Country Gold Silver Bronze Total
Canada 1 0 1 2
China 1 1 0 2
Germany 0 0 1 1
Korea 1 0 0 1
Japan 0 1 1 2
Russia 0 1 1 2
United States 1 1 0 2
30. What results do you get if you total the columns in our sample data?
31. Consider an 8 × 8 array for a board game:
int board[8][8];
Using two nested loops, initialize the board so that zeroes and ones alternate, as
on a checkerboard:
0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
0 1 0 1 0 1 0 1
...
1 0 1 0 1 0 1 0
Hint: Check whether i + j is even.
32. Define a two-dimensional array for representing a tic-tac-toe board. The board
has three rows and columns and contains strings "x", "o", and " ".
33. Write an assignment statement to place an "x" in the upper-right corner of the
tic-tac-toe board.
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284 Chapter 6 arrays and Vectors
34. Which elements are on the diagonal joining the upper-left and the lower-right
corners of the tic-tac-toe board?
Practice it Now you can try these exercises at the end of the chapter: R6.23, P6.19, P6.20.
omitting the Column size of a two-dimensional Array Parameter
When passing a one-dimensional array to a function, you specify the size of the array as a
separate parameter variable:
void print(double values[], int size)
This function can print arrays of any size. However, for two-dimensional arrays you cannot
simply pass the num bers of rows and columns as parameter variables:
void print(double table[][], int rows, int cols) // NO!
The function must know at compile time how many columns the two-dimensional array has.
You must specify the number of columns with the array parameter variable. This number
must be a constant:
const int COLUMNS = 3;
void print(const double table[][COLUMNS], int rows) // OK
This function can print tables with any number of rows, but the column size is fixed.
6.7 Vectors
When you write a program that collects values
from user input, you don’t always know how
many values you will have. Unfortunately, the size
of the array has to be known when the program is
compiled.
In Section 6.1.3, you saw how you can address
this problem with partially filled arrays. The vec-
tor construct, which we discuss in the following
sections, offers a more convenient solution. A vec-
tor collects a sequence of values, just like an array
does, but its size can change.
A vector expands to hold as many elements as needed.
Common error 6.2
W o r k e d e X a M p l e 6 . 2 A world Population table
This Worked Example shows how to print world population data in a table with row and col-
umn headers, and with totals for each of the data columns.
a vector stores
a sequence of
values whose size
can change.
Available online at www.wiley.com/college/horstmann.
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www.wiley.com/college/horstmann
6.7 Vectors 285
34. Which elements are on the diagonal joining the upper-left and the lower-right
corners of the tic-tac-toe board?
Practice it Now you can try these exercises at the end of the chapter: R6.23, P6.19, P6.20.
omitting the Column size of a two-dimensional Array Parameter
When passing a one-dimensional array to a function, you specify the size of the array as a
separate parameter variable:
void print(double values[], int size)
This function can print arrays of any size. However, for two-dimensional arrays you cannot
simply pass the num bers of rows and columns as parameter variables:
void print(double table[][], int rows, int cols) // NO!
The function must know at compile time how many columns the two-dimensional array has.
You must specify the number of columns with the array parameter variable. This number
must be a constant:
const int COLUMNS = 3;
void print(const double table[][COLUMNS], int rows) // OK
This function can print tables with any number of rows, but the column size is fixed.
6.7 Vectors
When you write a program that collects values
from user input, you don’t always know how
many values you will have. Unfortunately, the size
of the array has to be known when the program is
compiled.
In Section 6.1.3, you saw how you can address
this problem with partially filled arrays. The vec-
tor construct, which we discuss in the following
sections, offers a more convenient solution. A vec-
tor collects a sequence of values, just like an array
does, but its size can change.
A vector expands to hold as many elements as needed.
Common error 6.2
W o r k e d e X a M p l e 6 . 2 A world Population table
This Worked Example shows how to print world population data in a table with row and col-
umn headers, and with totals for each of the data columns.
a vector stores
a sequence of
values whose size
can change.
6.7.1
syntax 6.3 defining a Vector
vector
Element type
The index must be
≥ 0 and < values.size().
If you omit the size
and the parentheses, the
vector has initial size 0.
Initial size
Name
Use brackets to access an element.
values[i] = 0;
defining Vectors
When you define a vector, you specify the type of the elements in angle brackets, like
this:
vector
You can optionally specify the initial size. For example, here is a definition of a vector
whose initial size is 10:
vector
If you define a vector without an initial size, it has size 0. While there would be no
point in defining an array of size zero, it is often useful to have vectors with initial
size zero, and then grow them as needed.
In order to use vectors in your program, you need to include the
table 2 defining Vectors
vector
vector
vector
vector
vector
for (int i = 1; i <= 10; i++)
{
numbers.push_back(i);
}
A vector of ten integers, filled with 1, 2,
3, ..., 10.
vector
for (int i = 0; i < numbers.size(); i++)
{
numbers[i] = i + 1;
}
Another way of defining a vector of ten
integers and filling it with
1, 2, 3, ..., 10.
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286 Chapter 6 arrays and Vectors
You access the vector elements as values[i], just as you do with arrays.
The size member function returns the current size of a vector. In a loop that visits
all vector elements, use the size member function like this:
for (int i = 0; i < values.size(); i++)
{
cout << values[i] << endl;
}
6.7.2 Growing and shrinking Vectors
If you need additional elements, you use the push_back function to add an element to
the end of the vector, thereby increasing its size by 1. The push_back function is a mem-
ber function that you must call with the dot notation, like this:
values.push_back(37.5);
After this call, the vector values in Figure 14 has size 3, and values[2] contains the
value 37.5.
It is very common to start with an empty vector and use the push_back function to fill
it. For example,
vector
values.push_back(32); // Now values has size 1 and element 32
values.push_back(54); // Now values has size 2 and elements 32, 54
values.push_back(37.5); // Now values has size 3 and elements 32, 54, 37.5
Another common use for the push_back member function is to fill a vector with input
values.
vector
double input;
while (cin >> input)
{
values.push_back(input);
}
Note how this input loop is much simpler than the one in Section 6.2.10.
Another member function, pop_back, removes the last element of a vector, shrinking
its size by one (see Figure 15):
values.pop_back();
Use the size member
function to obtain
the current size of
a vector.
Use the push_back
member function to
add more elements
to a vector.
Use pop_back to
reduce the size.
Figure 14 adding an element with push_back
1 Before push_back
values =
2
2 After push_back
values =
3
37.5
54
32
Size increased
New element
added at end
32
54
6.7.3 Vectors and Functions
You can use vectors as function arguments in exactly the same way as any other val-
ues. For example, the following function computes the sum of a vector of floating-
point numbers:
double sum(vector
{
double total = 0;
for (int i = 0; i < values.size(); i++)
{
total = total + values[i];
}
return total;
}
This function visits the vector elements, but it does not modify them. If your func-
tion modifies the ele ments, use a reference parameter. The following function multi-
plies all values of a vector with a given fac tor.
void multiply(vector
{
for (int i = 0; i < values.size(); i++)
{
values[i] = values[i] * factor;
}
}
Some programmers use a constant reference (see Special Topic 5.2) for vector param-
eters that are not modified, for example:
double sum(const vector
A function can return a vector. Again, vectors are no different from any other values
in this regard. Sim ply build up the result in the function and return it. In this example,
the squares function returns a vector of squares from 02 up to (n – 1)2:
vector
{
vector
for (int i = 0; i < n; i++)
{
result.push_back(i * i);
}
return result;
}
Vectors can occur as
function arguments
and return values.
Use a reference
parameter to modify
the contents of
a vector.
a function can return
a vector.
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6.7 Vectors 287
6.7.3
Figure 15 removing an element with pop_back
2 After pop_back
values =
2
1 Before pop_back
values =
3
37.5
54
32
This element
to be removed
32
54
Size decreased
Vectors and Functions
You can use vectors as function arguments in exactly the same way as any other val-
ues. For example, the following function computes the sum of a vector of floating-
point numbers:
double sum(vector
{
double total = 0;
for (int i = 0; i < values.size(); i++)
{
total = total + values[i];
}
return total;
}
This function visits the vector elements, but it does not modify them. If your func-
tion modifies the ele ments, use a reference parameter. The following function multi-
plies all values of a vector with a given fac tor.
void multiply(vector
{
for (int i = 0; i < values.size(); i++)
{
values[i] = values[i] * factor;
}
}
Some programmers use a constant reference (see Special Topic 5.2) for vector param-
eters that are not modified, for example:
double sum(const vector
A function can return a vector. Again, vectors are no different from any other values
in this regard. Sim ply build up the result in the function and return it. In this example,
the squares function returns a vector of squares from 02 up to (n – 1)2:
vector
{
vector
for (int i = 0; i < n; i++)
{
result.push_back(i * i);
}
return result;
}
Vectors can occur as
function arguments
and return values.
Use a reference
parameter to modify
the contents of
a vector.
a function can return
a vector.
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288 Chapter 6 arrays and Vectors
As you can see, it is easy to use vectors with functions—there are no special rules to
keep in mind.
6.7.4 Vector algorithms
Most of the algorithms in Section 6.2 apply without change to vectors—simply
replace size of values with values.size(). In this section, we discuss which of the algo-
rithms are different for vectors.
Copying
As discussed in Section 6.2.2, you need an explicit loop to make a copy of an array. It
is much easier to make a copy of a vector. You simply assign it to another vector. Con-
sider this example:
vector
for (int i = 0; i < 5; i++) { squares.push_back(i * i); }
vector
lucky_numbers = squares; // Now lucky_numbers contains the same elements as squares
Finding Matches
Section 6.2.6 shows you how to find the first match, but sometimes you want to have
all matches. This is tedious with arrays, but simple using a vector that collects the
matches. Here we collect all elements that are greater than 100:
vector
for (int i = 0; i < values.size(); i++)
{
if (values[i] > 100)
{
matches.push_back(values[i]);
}
}
removing an element
When you remove an element from a vector, you want to adjust the size of the vector
by calling the pop_back member function. Here is the code for removing an element at
[pos] when the order doesn’t matter.
int last_pos = values.size() – 1;
values[pos] = values[last_pos]; // Replace element at pos with last element
values.pop_back(); // Delete last element
When removing an element from an ordered vector, first move the elements, then
reduce the size:
for (int i = pos + 1; i < values.size(); i++)
{
values[i - 1] = values[i];
}
values.pop_back();
Inserting an element
Inserting an element at the end of a vector requires no special code. Simply use the
push_back member function.
When you insert an element in the middle, you still want to call push_back so that
the size of the vector is increased. Use the following code:
int last_pos = values.size() - 1;
values.push_back(values[last_pos]);
for (int i = last_pos; i > pos; i–)
{
values[i] = values[i – 1];
}
values[pos] = new_element;
35. Define a vector of integers that contains the first five prime numbers (2, 3, 5, 7,
and 11). Use push_back to add the elements.
36. Answer Self Check 35 without using push_back.
37. What is the contents of the vector names after the following statements?
vector
names.push_back(“Ann”);
names.push_back(“Bob”);
names.pop_back();
names.push_back(“Cal”);
38. Suppose you want to store a set of temperature measurements that is taken every
five minutes. Should you use a vector or an array?
39. Suppose you want to store the names of the weekdays. Should you use a vector
or an array of seven strings?
40. Write the header for a function that appends two vectors, yielding a third vector.
Do not implement the function.
41. Consider this partially completed function that appends the elements of one
vector to another.
void append(vector
{
for (int i = 0; i < source.size(); i++)
{
target.push_back(source[i]);
}
}
Specify whether the parameters should be value or reference parameters.
Practice it Now you can try these exercises at the end of the chapter: R6.11, R6.25, P6.26,
P6.27.
Prefer vectors over Arrays
For most programming tasks, vectors are easier to use than arrays. Vectors can grow and
shrink. Even if a vector always stays the same size, it is convenient that a vector remembers its
size. For a beginner, the sole advantage of an array is the initialization syntax. Advanced pro-
grammers sometimes prefer arrays because they are a bit more effi cient. Moreover, you need
to know how to use arrays if you work with older programs.
s e l F C h e C k
programming tip 6.2
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6.7 Vectors 289
When you insert an element in the middle, you still want to call push_back so that
the size of the vector is increased. Use the following code:
int last_pos = values.size() - 1;
values.push_back(values[last_pos]);
for (int i = last_pos; i > pos; i–)
{
values[i] = values[i – 1];
}
values[pos] = new_element;
35. Define a vector of integers that contains the first five prime numbers (2, 3, 5, 7,
and 11). Use push_back to add the elements.
36. Answer Self Check 35 without using push_back.
37. What is the contents of the vector names after the following statements?
vector
names.push_back(“Ann”);
names.push_back(“Bob”);
names.pop_back();
names.push_back(“Cal”);
38. Suppose you want to store a set of temperature measurements that is taken every
five minutes. Should you use a vector or an array?
39. Suppose you want to store the names of the weekdays. Should you use a vector
or an array of seven strings?
40. Write the header for a function that appends two vectors, yielding a third vector.
Do not implement the function.
41. Consider this partially completed function that appends the elements of one
vector to another.
void append(vector
{
for (int i = 0; i < source.size(); i++)
{
target.push_back(source[i]);
}
}
Specify whether the parameters should be value or reference parameters.
Practice it Now you can try these exercises at the end of the chapter: R6.11, R6.25, P6.26,
P6.27.
Prefer vectors over Arrays
For most programming tasks, vectors are easier to use than arrays. Vectors can grow and
shrink. Even if a vector always stays the same size, it is convenient that a vector remembers its
size. For a beginner, the sole advantage of an array is the initialization syntax. Advanced pro-
grammers sometimes prefer arrays because they are a bit more effi cient. Moreover, you need
to know how to use arrays if you work with older programs.
s e l F C h e C k
programming tip 6.2
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290 Chapter 6 arrays and Vectors
use arrays for collecting values.
• Use an array to collect a sequence of values of the same type.
• Individual elements in an array values are accessed by an integer index i,
using the notation values[i].
Before pocket calcu-
lators and personal
computers existed, navigators and
engineers used mechanical adding
machines, slide rules, and tables of
logarithms and trigonometric func-
tions to speed up computations. Unfor-
tunately, the tables—for which values
had to be computed by hand—were
notoriously inaccurate. the mathema-
tician Charles Babbage (1791–1871)
had the insight that if a machine could
be constructed that produced printed
tables automatically, both calcula-
tion and typesetting errors could be
avoided. Babbage set out to develop
a machine for this purpose, which he
called a Dif erence Engine because it
used succes sive differences to com-
pute polynomials. For example, con-
sider the function f (x) = x3. Write down
the values for f (1), f (2), f (3), and so
on. then take the diferences between
successive values:
1
7
8
19
27
37
64
61
125
91
216
repeat the process, taking the differ-
ence of successive values in the sec ond
column, and then repeat once again:
1
7
8 12
19 6
27 18
37 6
64 24
61 6
125 30
91
216
now the differences are all the same.
you can retrieve the function values by
a pattern of additions—you need to
know the values at the fringe of the
pattern and the constant difference.
you can try it out yourself: Write the
highlighted numbers on a sheet of
paper, and fill in the others by adding
the numbers that are in the north and
northwest positions. Replica of Babbage’s Diference Engine
this method was very attractive,
because mechanical addition machines
had been known for some time. they
consisted of cog wheels, with 10 cogs
per wheel, to represent digits, and
mechanisms to handle the carry from
one digit to the next. Mechanical mul-
tiplication machines, on the other
hand, were fragile and unreliable.
Bab bage built a successful prototype
of the difference engine and, with his
own money and government grants,
proceeded to build the table-printing
machine. however, because of funding
problems and the difficulty of building
the machine to the required precision,
it was never completed.
While working on the difference
engine, Babbage conceived of a much
grander vision that he called the Ana
lytical Engine. the difference engine
was designed to carry out a limited set
of computations—it was no smarter
than a pocket calculator is today. But
Babbage realized that such a machine
could be made programmable by stor-
ing programs as well as data. the inter-
nal storage of the analytical engine
was to consist of 1,000 regis ters of 50
decimal digits each. pro grams and con-
stants were to be stored on punched
cards—a technique that was, at that
time, commonly used on looms for
weaving patterned fabrics.
ada augusta, Countess of lovelace
(1815–1852), the only child of lord
Byron, was a friend and sponsor of
Charles Babbage. ada lovelace was
one of the first people to realize the
potential of such a machine, not just
for computing mathematical tables but
for processing data that were not num-
bers. she is considered by many to be
the world’s first programmer.
Random Fact 6.2 the First programmer
C h a p t e r s U M M a r y
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Chapter summary 291
• An array element can be used like any variable.
• An array index must be at least zero and less than the size of the array.
• A bounds error, which occurs if you supply an invalid array index, can corrupt
data or cause your program to terminate.
• With a partially filled array, keep a companion variable for the
cur rent size.
be able to use common array algorithms.
• To copy an array, use a loop to copy its elements to a new array.
• When separating elements, don’t place a separator before the first element.
• A linear search inspects elements in sequence until a match is found.
• Before inserting an element, move elements to the end of the
array starting with the last one.
• Use a temporary variable when swapping two elements.
implement functions that process arrays.
• When passing an array to a function, also pass the size of the array.
• Array parameters are always reference parameters.
• A function’s return type cannot be an array.
• When a function modifies the size of an array, it needs to tell its caller.
• A function that adds elements to an array needs to know its capacity.
be able to combine and adapt algorithms for solving a programming problem.
• By combining fundamental algorithms, you can solve complex programming
tasks.
• You should be familiar with the implementation of fundamental algorithms so
that you can adapt them.
discover algorithms by manipulating physical objects.
• Use a sequence of coins, playing cards, or toys to visualize an array of values.
• You can use paper clips as position markers or counters.
use two-dimensional arrays for data that is arranged in rows and columns.
• Use a two-dimensional array to store tabular data.
• Individual elements in a two-dimensional array are
accessed by using two subscripts, array[i][j].
• A two-dimensional array parameter must have a
fixed number of columns.
Return address
Overrun buffer
(536 bytes)
Malicious
code
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292 Chapter 6 arrays and Vectors
use vectors for managing collections whose size can change.
• A vector stores a sequence of values whose size can change.
• Use the size member function to obtain the current size of a vector.
• Use the push_back member function to add more elements to a vector. Use pop_back
to reduce the size.
• Vectors can occur as function arguments and return values.
• Use a reference parameter to modify the contents of a vector.
• A function can return a vector.
r6.1 Write code that fills an array double values[10] with each set of values below.
a. 1 2 3 4 5 6 7 8 9 10
b. 0 2 4 6 8 10 12 14 16 18
c. 1 4 9 16 25 36 49 64 81 100
d. 0 0 0 0 0 0 0 0 0 0
e. 1 4 9 16 9 7 4 9 11
f. 0 1 0 1 0 1 0 1 0 1
g. 0 1 2 3 4 0 1 2 3 4
r6.2 Consider the following array:
int a[] = { 1, 2, 3, 4, 5, 4, 3, 2, 1, 0 };
What is the value of total after the following loops complete?
a. int total = 0;
for (int i = 0; i < 10; i++) { total = total + a[i]; }
b. int total = 0;
for (int i = 0; i < 10; i = i + 2) { total = total + a[i]; }
c. int total = 0;
for (int i = 1; i < 10; i = i + 2) { total = total + a[i]; }
d. int total = 0;
for (int i = 2; i <= 10; i++) { total = total + a[i]; }
e. int total = 0;
for (int i = 0; i < 10; i = 2 * i) { total = total + a[i]; }
f. int total = 0;
for (int i = 9; i >= 0; i–) { total = total + a[i]; }
g. int total = 0;
for (int i = 9; i >= 0; i = i – 2) { total = total + a[i]; }
h. int total = 0;
for (int i = 0; i < 10; i++) { total = a[i] - total; }
r6.3 Consider the following array:
int a[] = { 1, 2, 3, 4, 5, 4, 3, 2, 1, 0 };
r e V I e W e X e r C I s e s
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review exercises 293
What are the contents of the array a after the following loops complete?
a. for (int i = 1; i < 10; i++) { a[i] = a[i - 1]; }
b. for (int i = 9; i > 0; i–) { a[i] = a[i – 1]; }
c. for (int i = 0; i < 9; i++) { a[i] = a[i + 1]; }
d. for (int i = 8; i >= 0; i–) { a[i] = a[i + 1]; }
e. for (int i = 1; i < 10; i++) { a[i] = a[i] + a[i - 1]; }
f. for (int i = 1; i < 10; i = i + 2) { a[i] = 0; }
g. for (int i = 0; i < 5; i++) { a[i + 5] = a[i]; }
h. for (int i = 1; i < 5; i++) { a[i] = a[9 - i]; }
r6.4 Write a loop that fills an array int values[10] with ten random numbers between 1
and 100. Write code for two nested loops that fill values with ten different random
numbers between 1 and 100.
r6.5 Write C++ code for a loop that simultaneously computes both the maximum and
minimum of an array.
r6.6 What is wrong with the following loop?
int values[10];
for (int i = 1; i <= 10; i++)
{
values[i] = i * i;
}
Explain two ways of fixing the error.
r6.7 What is an index of an array? What are the legal index values? What is a bounds error?
r6.8 Write a program that contains a bounds error. Run the program. What happens on
your computer?
r6.9 Write a loop that reads ten numbers and a second loop that displays them in the
opposite order from which they were entered.
r6.10 Trace the flow of the element separator loop in Section 6.2.5 with the given example.
Show two col umns, one with the value of i and one with the output.
r6.11 Trace the flow of the finding matches loop in Section 6.7.4, where values contains the
elements 110 90 100 120 80. Show two columns, for i and matches.
r6.12 Trace the flow of the linear search loop in Section 6.2.6, where values contains the
elements 80 90 100 120 110. Show two columns, for pos and found. Repeat the trace
when values contains 80 90 100 70.
r6.13 Trace both mechanisms for removing an element described in Section 6.2.7. Use an
array values with elements 110 90 100 120 80, and remove the element at index 2.
r6.14 For the operations on partially filled arrays below, provide the header of a func tion.
a. Sort the elements in decreasing order.
b. Print all elements, separated by a given string.
c. Count how many elements are less than a given value.
d. Remove all elements that are less than a given value.
e. Place all elements that are less than a given value in another array.
Do not implement the functions.
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294 Chapter 6 arrays and Vectors
r6.15 You are given two arrays denoting x- and y-coordinates of a set of points in the
plane. For plotting the point set, we need to know the x- and y-coordinates of the
smallest rectangle containing the points.
y
x
How can you obtain these values from the fundamental algorithms in Section 6.2?
r6.16 Solve the problem described in Section 6.4 by sorting the array first. How do you
need to modify the algorithm for computing the total?
r6.17 Solve the task described in Section 6.5 using an algorithm that removes and inserts
elements instead of switching them. Write the pseudocode for the algorithm, assum-
ing that functions for removal and insertion exist. Act out the algorithm with a
sequence of coins and explain why it is less efficient than the swapping algorithm
developed in Section 6.5.
r6.18 Develop an algorithm for finding the most frequently occurring value in an array of
numbers. Use a sequence of coins. Place paper clips below each coin that count how
many other coins of the same value are in the sequence. Give the pseudocode for an
algorithm that yields the correct answer, and describe how using the coins and paper
clips helped you find the algorithm.
r6.19 Give pseudocode for a function that rotates the elements of an array by one posi tion,
moving the initial element to the end of the array, like this:
3 5 7 11 13 2
2 3 5 7 11 13
r6.20 Give pseudocode for a function that removes all negative values from a partially
filled array, preserving the order of the remaining elements.
r6.21 Suppose values is a sorted partially filled array of integers. Give pseudocode that
describes how a new value can be inserted in its proper position so that the resulting
array stays sorted.
r6.22 A run is a sequence of adjacent repeated values. Give pseudocode for computing the
length of the longest run in an array. For example, the longest run in the array with
elements
1 2 5 5 3 1 2 4 3 2 2 2 2 3 6 5 5 6 3 1
has length 4.
r6.23 Write pseudocode for an algorithm that fills the first and last column as well as the
first and last row of a two-dimensional array of integers with –1.
r6.24 True or false?
a. All elements of an array are of the same type.
b. Arrays cannot contain strings as elements.
c. Two-dimensional arrays always have the same number of rows and columns.
d. Elements of different columns in a two-dimensional array can have
different types.
e. A function cannot return a two-dimensional array.
f. All array parameters are reference parameters.
g. A function cannot change the dimensions of a two-dimensional array that is
passed as a parameter.
r6.25 How do you perform the following tasks with vectors in C++?
a. Test that two vectors contain the same elements in the same order.
b. Copy one vector to another.
c. Fill a vector with zeroes, overwriting all elements in it.
d. Remove all elements from a vector.
r6.26 True or false?
a. All elements of a vector are of the same type.
b. Vector subscripts must be integers.
c. Vectors cannot contain strings as elements.
d. Vectors cannot use strings as subscripts.
e. All vector parameters are reference parameters.
f. A function cannot return a vector.
g. A function cannot change the length of a vector that is a reference parameter.
P6.1 Write a program that initializes an array with ten random integers and then prints
four lines of output, containing
• Every element at an even index.
• Every even element.
• All elements in reverse order.
• Only the first and last element.
P6.2 Write array functions that carry out the following tasks for an array of integers:
a. Swap the first and last element in an array.
b. Shift all elements by one to the right and move the last element into the first
position. For example, 1 4 9 16 25 would be transformed into 25 1 4 9 16.
c. Replace all even elements with 0.
d. Replace each element except the first and last by the larger of its two neighbors.
e. Remove the middle element if the array length is odd, or the middle two
elements if the length is even.
f. Move all even elements to the front, otherwise preserving the order of the
elements.
g. Return the second-largest element in the array.
p r o G r a M M I n G e X e r C I s e s
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programming exercises 295
c. Two-dimensional arrays always have the same number of rows and columns.
d. Elements of different columns in a two-dimensional array can have
different types.
e. A function cannot return a two-dimensional array.
f. All array parameters are reference parameters.
g. A function cannot change the dimensions of a two-dimensional array that is
passed as a parameter.
r6.25 How do you perform the following tasks with vectors in C++?
a. Test that two vectors contain the same elements in the same order.
b. Copy one vector to another.
c. Fill a vector with zeroes, overwriting all elements in it.
d. Remove all elements from a vector.
r6.26 True or false?
a. All elements of a vector are of the same type.
b. Vector subscripts must be integers.
c. Vectors cannot contain strings as elements.
d. Vectors cannot use strings as subscripts.
e. All vector parameters are reference parameters.
f. A function cannot return a vector.
g. A function cannot change the length of a vector that is a reference parameter.
P6.1 Write a program that initializes an array with ten random integers and then prints
four lines of output, containing
• Every element at an even index.
• Every even element.
• All elements in reverse order.
• Only the first and last element.
P6.2 Write array functions that carry out the following tasks for an array of integers:
a. Swap the first and last element in an array.
b. Shift all elements by one to the right and move the last element into the first
position. For example, 1 4 9 16 25 would be transformed into 25 1 4 9 16.
c. Replace all even elements with 0.
d. Replace each element except the first and last by the larger of its two neighbors.
e. Remove the middle element if the array length is odd, or the middle two
elements if the length is even.
f. Move all even elements to the front, otherwise preserving the order of the
elements.
g. Return the second-largest element in the array.
p r o G r a M M I n G e X e r C I s e s
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296 Chapter 6 arrays and Vectors
h. Return true if the array is currently sorted in increasing order.
i. Return true if the array contains two adjacent duplicate values.
j. Return true if the array contains duplicate values (which need not be adjacent).
For each function, provide a test program.
P6.3 Modify the largest.cpp program to mark both the smallest and the largest element.
P6.4 Reimplement How To 6.1 without removing the minimum from the array of scores.
Instead, compute the final score as the difference of the sum and the minimum of the
scores.
P6.5 Write a function void remove_min that removes the minimum value from a partially
filled array without calling other functions.
P6.6 Write a function that computes the alternating sum of all elements in an array. For
example, if alternating_sum is called with an array containing
1 4 9 16 9 7 4 9 11
then it computes
1 – 4 + 9 – 16 + 9 – 7 + 4 – 9 + 11 = –2
P6.7 Write a function that implements the algorithm developed in Section 6.5.
P6.8 Write a function reverse that reverses the sequence of elements in an array. For
example, if reverse is called with an array containing
1 4 9 16 9 7 4 9 11
then the array is changed to
11 9 4 7 9 16 9 4 1
P6.9 Write a function
bool equals(int a[], int a_size, int b[], int b_size)
that checks whether two arrays have the same elements in the same order.
P6.10 Write a function
bool same_set(int a[], int a_size, int b[], int b_size)
that checks whether two vectors have the same elements in some order, ignoring
duplicates. For example, the two arrays
1 4 9 16 9 7 4 9 11
and
11 11 7 9 16 4 1
would be considered identical. You will probably need one or more helper
functions.
P6.11 Write a function
bool same_elements(int a[], int b[], int size)
that checks whether two arrays have the same elements in some order, with the same
multiplicities. For example,
1 4 9 16 9 7 4 9 11
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programming exercises 297
and
11 1 4 9 16 9 7 4 9
would be considered identical, but
1 4 9 16 9 7 4 9 11
and
11 11 7 9 16 4 1 4 9
would not. You will probably need one or more helper functions.
P6.12 Write a function that removes duplicates from an array. For example, if remove_
duplicates is called with an array containing
1 4 9 16 9 7 4 9 11
then the array is changed to
1 4 9 16 7 11
Your function should have a reference parameter for the array size that is updated
when removing the duplicates.
P6.13 A run is a sequence of adjacent repeated values. Write a program that generates a
sequence of 20 random die tosses and prints the die values, marking the runs by
including them in parentheses, like this:
1 2 (5 5) 3 1 2 4 3 (2 2 2 2) 3 6 (5 5) 6 3 1
Use the following pseudocode:
Set a Boolean variable in_run to false.
For each valid index i in the array
If in_run
If values[i] is different from the preceding value
Print ).
in_run = false
If not in_run
If values[i] is the same as the following value
Print (.
in_run = true
Print values[i].
If in_run, print ).
P6.14 Write a program that generates a sequence of 20 random die tosses and that prints the
die values, marking only the longest run, like this:
1 2 5 5 3 1 2 4 3 (2 2 2 2) 3 6 5 5 6 3 1
If there is more than one run of maximum length, mark the first one.
P6.15 Write a program that generates a sequence of 20 random values between 0 and 99,
prints the sequence, sorts it, and prints the sorted sequence. Use the sort function
from the standard C++ library.
P6.16 Write a program that produces ten random permutations of the numbers 1 to 10. To
generate a random permutation, you need to fill an array with the numbers 1 to 10 so
that no two elements have the same contents. You could do it by brute force, by
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298 Chapter 6 arrays and Vectors
generating random values until you have a value that is not yet in the array. But that
is inefficient. Instead, follow this algorithm:
Make a second array and fill it with the numbers 1 to 10.
Repeat 10 times
Pick a random element from the second array.
Remove it and append it to the permutation array.
P6.17 It is a well-researched fact that men in a restroom generally prefer to maximize their
distance from already occupied stalls, by occupying the middle of the longest
sequence of unoccupied places.
For example, consider the situation where all ten stalls are empty.
_ _ _ _ _ _ _ _ _ _
The first visitor will occupy a middle position:
_ _ _ _ X _ _ _ _ _
The next visitor will be in the middle of the empty area at the right.
_ _ _ _ X _ _ X _ _
Given an array of bool values, where true indicates an occupied stall, find the posi-
tion for the next visitor. Your computation should be placed in a function
next_visitor(bool occupied[], int stalls)
P6.18 In this assignment, you will model the game of Bulgarian Solitaire. The game starts
with 45 cards. (They need not be playing cards. Unmarked index cards work just as
well.) Randomly divide them into some number of piles of random size. For exam-
ple, you might start with piles of size 20, 5, 1, 9, and 10. In each round, you take one
card from each pile, forming a new pile with these cards. For example, the sample
starting configuration would be transformed into piles of size 19, 4, 8, 9, and 5. The
solitaire is over when the piles have size 1, 2, 3, 4, 5, 6, 7, 8, and 9, in some order. (It
can be shown that you always end up with such a configuration.)
In your program, produce a random starting configuration and print it. Then keep
applying the soli taire step and print the result. Stop when the solitaire final configura-
tion is reached.
P6.19 Magic squares. An n × n matrix that is filled with the numbers 1, 2, 3, . . ., n2 is a magic
square if the sum of the elements in each row, in each column, and in the two diago-
nals is the same value.
4 15 14 1
9 6 7 12
5 10 11 8
16 3 2 13
Write a program that reads in 16 values from the keyboard and tests whether they
form a magic square when put into a 4 × 4 array. You need to test two features:
1. Does each of the numbers 1, 2, ..., 16 occur in the user input?
2. When the numbers are put into a square, are the sums of the rows, columns,
and diagonals equal to each other?
P6.20 Implement the following algorithm to construct magic n × n squares; it works only if
n is odd.
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programming exercises 299
Set row = n - 1, column = n / 2.
For k = 1 ... n
Place k at [row][column].
Increment row and column.
If the row or column is n, replace it with 0.
If the element at [row][column] has already been filled
Set row and column to their previous value.
Decrement row.
Here is the 5 × 5 square that you get if you follow this method:
17 24 1 8
23 5 7 14
4 6 13 20
10 12 19 21
15
16
22
3
11 18 25 2 9
Write a program whose input is the number n and whose output is the magic square
of order n if n is odd.
P6.21 Write a function
void bar_chart(double values[], int size)
that displays a bar chart of the values in values, using asterisks, like this:
**********************
****************************************
****************************
**************************
**************
You may assume that all values in values are positive. First figure out the maximum
value in values. That value’s bar should be drawn with 40 asterisks. Shorter bars
should use proportionally fewer asterisks.
P6.22 Improve the bar_chart function of Exercise P6.21 to work correctly when values
contains negative values.
P6.23 Improve the bar_chart function of Exercise P6.21 by adding an array of captions for
each bar. The output should look like this:
Egypt **********************
France ****************************************
Japan ****************************
Uruguay **************************
Switzerland **************
P6.24 A theater seating chart is implemented as a two-dimensional array of ticket prices,
like this:
10 10 10 10 10 10 10 10 10 10
10 10 10 10 10 10 10 10 10 10
10 10 10 10 10 10 10 10 10 10
10 10 20 20 20 20 20 20 10 10
10 10 20 20 20 20 20 20 10 10
10 10 20 20 20 20 20 20 10 10
20 20 30 30 40 40 30 30 20 20
20 30 30 40 50 50 40 30 30 20
30 40 50 50 50 50 50 50 40 30
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300 Chapter 6 arrays and Vectors
Write a program that prompts users to pick either a seat or a price. Mark sold seats
by changing the price to 0. When a user specifies a seat, make sure it is available.
When a user specifies a price, find any seat with that price.
P6.25 Write a program that plays tic-tac-toe. The tic-tac-toe game is played on a 3 × 3 grid
as in
O
XO
The game is played by two players, who take turns. The first player marks moves
with a circle, the second with a cross. The player who has formed a horizontal,
vertical, or diagonal sequence of three marks wins. Your program should draw the
game board, ask the user for the coordinates of the next mark, change the players
after every successful move, and pronounce the winner.
P6.26 Write a function
vector
that appends one vector after another. For example, if a is
1 4 9 16
and b is
9 7 4 9 11
then append returns the vector
1 4 9 16 9 7 4 9 11
P6.27 Write a function
vector
that merges two vectors, alternating elements from both vectors. If one vector is
shorter than the other, then alternate as long as you can and then append the remain-
ing elements from the longer vector. For example, if a is
1 4 9 16
and b is
9 7 4 9 11
then merge returns the vector
1 9 4 7 9 4 16 9 11
P6.28 Write a function
vector
that merges two sorted vectors, producing a new sorted vector. Keep an index into
each vector, indicating how much of it has been processed already. Each time,
append the smallest unprocessed element from either vector, then advance the index.
For example, if a is
1 4 9 16
and b is
cfe2_ch06_p249_306.indd 300 10/26/10 7:48 PM
programming exercises 301
4 7 9 9 11
then merge_sorted returns the vector
1 4 4 7 9 9 9 11 16
P6.29 Modify the ch06/image.cpp program in the book’s companion code to generate the
image of a checkerboard.
P6.30 Modify the ch06/animation.cpp program in the book’s companion code to show a
rectangle that travels from the left of the image to the right and then back to the left.
engineering P6.31 Sample values from an experiment often need to be smoothed out. One simple
approach is to replace each value in an array with the average of the value and its two
neighboring values (or one neighboring value if it is at either end of the array).
Implement a function
void smooth(double[] values, int size)
that carries out this operation. You should not create another array in your solution.
engineering P6.32 Sounds can be represented by an array of “sample val-
ues” that describe the intensity of the sound at a point
in time. The sound.cpp program in this book’s compan-
ion code reads a sound file (in WAV format), calls a
function process for processing the sample values, and
saves the sound file. Your task is to implement the
process function by introducing an echo. For each
sound value, add the value from 0.2 seconds ago. Scale
the result so that no value is larger than 32767.
engineering P6.33 You are given a two-dimensional array of values that give the height of a terrain at
different points in a square. Write a function
void flood_map(double heights[10][10], double water_level)
that prints out a flood map, showing which of the points in the terrain would be
flooded if the water level was the given value. In the flood map, print a * for each
flooded point and a space for each point that is not flooded.
Here is a sample map:
* * * * * *
* * * * * * * *
* * * * * *
* * * * * *
* * * * * * * *
* * * * * * * * * *
* * * * *
* * * * * *
* *
* * *
cfe2_ch06_p249_306.indd 301 10/26/10 7:48 PM
302 Chapter 6 arrays and Vectors
Then write a program that reads one hundred terrain height values and shows how
the terrain gets flooded when the water level increases in ten steps from the lowest
point in the terrain to the highest.
engineering P6.34 Modify the ch06/image.cpp program in the book’s companion code to generate the
image of a sine wave.
Draw a line of pixels for every five degrees.
engineering P6.35 Modify the ch06/animation.cpp program to show an animated sine wave. In the ith
frame, shift the sine wave by 5 × i degrees.
engineering P6.36 Write a program that models the movement of an object with mass m that is attached
to an oscillating spring. When a spring is displaced from its equilibrium position by
an amount x, Hooke’s law states that the restoring force is
F = –k x
where k is a constant that depends on the spring. (Use 10 N̸m for this simulation.)
x
F
Unstretched
spring
Start with a given displacement x (say, 0.5 meter). Set the initial velocity v to 0.
Compute the acceleration a from Newton’s law (F = ma) and Hooke’s law, using a
mass of 1 kg. Use a small time interval Δt = 0.01 second. Update the velocity––it
changes by aΔt. Update the displacement––it changes by vΔt.
Every ten iterations, plot the spring displacement as a bar, where 1 pixel represents
1 cm. Modify the program ch06/image.cpp for creating an image.
1. int primes[] = { 2, 3, 5, 7, 11 };
2. 2, 3, 5, 3, 2
3. 3, 4, 6, 8, 12
4. values[0] = 0;
values[CAPACITY – 1] = 0;
5. for (int i = SIZE – 1; i >= 0; i–)
{
cout << values[i] << endl;
}
6. string words[10];
7. string words[] = { "Yes", "No" };
8. 20 <== largest value
10
20 <== largest value
9. int count = 0;
for (int i = 0; i < size; i++)
{
if (values[i] == 0)
{
count++;
}
}
10. If all elements of values are negative, then the result is incorrectly computed as 0.
11. for (int i = 0; i < size; i++)
{
cout << values[i];
if (i < size - 1)
{
cout << " | ";
}
}
Now you know why we set up the loop the other way.
12. If the sequence has no elements, then a random value is printed.
13. If there is a match, then pos is incremented before the loop exits.
14. This loop sets all elements to values[pos].
15. The sum function will add up all the numbers in the values array and the next 900
numbers, yielding a random result. (Actually, there is the chance that the program
doesn’t have the right to access all those numbers, in which case the operating sys tem
will terminate it.)
16. int numbers[5];
squares(5, numbers);
17. int find_first(double values[], int size, double searched_value)
{
for (int pos = 0; pos < size; pos++)
{
if (values[pos] == searched_value)
{
return pos;
a n s W e r s t o s e l F - C h e C k Q U e s t I o n s
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answers to self-Check Questions 303
1. int primes[] = { 2, 3, 5, 7, 11 };
2. 2, 3, 5, 3, 2
3. 3, 4, 6, 8, 12
4. values[0] = 0;
values[CAPACITY - 1] = 0;
5. for (int i = SIZE - 1; i >= 0; i–)
{
cout << values[i] << endl;
}
6. string words[10];
7. string words[] = { "Yes", "No" };
8. 20 <== largest value
10
20 <== largest value
9. int count = 0;
for (int i = 0; i < size; i++)
{
if (values[i] == 0)
{
count++;
}
}
10. If all elements of values are negative, then the result is incorrectly computed as 0.
11. for (int i = 0; i < size; i++)
{
cout << values[i];
if (i < size - 1)
{
cout << " | ";
}
}
Now you know why we set up the loop the other way.
12. If the sequence has no elements, then a random value is printed.
13. If there is a match, then pos is incremented before the loop exits.
14. This loop sets all elements to values[pos].
15. The sum function will add up all the numbers in the values array and the next 900
numbers, yielding a random result. (Actually, there is the chance that the program
doesn’t have the right to access all those numbers, in which case the operating sys tem
will terminate it.)
16. int numbers[5];
squares(5, numbers);
17. int find_first(double values[], int size, double searched_value)
{
for (int pos = 0; pos < size; pos++)
{
if (values[pos] == searched_value)
{
return pos;
a n s W e r s t o s e l F - C h e C k Q U e s t I o n s
cfe2_ch06_p249_306.indd 303 10/26/10 7:48 PM
304 Chapter 6 arrays and Vectors
}
}
return -1;
}
Note that the loop is simpler than that in Section 6.2.6 since we can simply return the
position when a match is found.
18. void read_inputs(double inputs[], int capacity, int& size)
{
size = 0;
double input;
while (cin >> input)
{
if (size < capacity)
{
inputs[size] = input;
size++;
}
}
}
19. int append(double first[], int first_size,
double second[], int second_size,
double target[], int target_capacity)
Note the following:
• You must pass the sizes of the first and second arrays, so that the function knows
how many elements to copy.
• You must pass the capacity of the target, so that the function won’t write past the
end.
• The target array is a parameter variable—functions cannot return arrays.
• The return type is int, so that the function can return the size of the target. (Alter-
natively, you could use a reference parameter int& target_size.)
20. Use the first algorithm. The order of elements does not matter when computing the
sum.
21. Find the minimum value.
Calculate the sum.
Subtract the minimum value.
22. Use the algorithm for counting matches (Section 4.7.2) twice, once for counting the
positive values and once for counting the negative values.
23. You need to modify the algorithm in Section 6.2.5.
bool first = true;
for (int i = 0; i < size of values; i++)
{
if (values[i] > 0))
{
if (first) { first = false; }
else { cout << ", "; }
}
cout << values[i];
}
Note that you can no longer use i > 0 as the criterion for printing a separator.
24. Use the algorithm to collect all positive values in an array, then use the algorithm in
Section 6.2.5 to print the array of matches.
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answers to self-Check Questions 305
25. The paperclip for i assumes positions 0, 1, 2, 3. When i is incremented to 4, the
con dition i < size / 2 becomes false, and the loop ends. Similarly, the paperclip for j
assumes positions 4, 5, 6, 7, which are the valid positions for the second half of the
array.
26. It reverses the elements in the array.
27. Here is one solution. The basic idea is to move all odd elements to the end. Put one
paper clip at the beginning of the array and one at the end. If the element at the first
paper clip is odd, swap it with the one at the other paper clip and move that paper
clip to the left. Otherwise, move the first paper clip to the right. Stop when the two
paper clips meet. Here is the pseudocode:
i = 0
j = size - 1
While (i < j)
If (a[i] is odd)
Swap elements at positions i and j.
j--
Else
i++
28. Here is one solution. The idea is to remove all odd elements and move them to the
end. The trick is to know when to stop. Nothing is gained by moving odd elements
into the area that already contains moved elements, so we want to mark that area
with another paper clip.
i = 0
moved = size
While (i < moved)
If (a[i] is odd)
Remove the element at position i and add it at the end.
moved--
29. When you read inputs, you get to see values one at a time, and you can’t peek ahead.
Picking cards one at a time from a deck of cards simulates this process better than
looking at a sequence of items, all of whom are revealed.
30. You get the total number of gold, silver, and bronze medals in the competition. In
our example, there are four of each.
31. for (int i = 0; i < 8; i++)
{
for (j = 0; j < 8; j++)
{
board[i][j] = (i + j) % 2;
}
}
32. string board[3][3];
33. board[0][2] = "x";
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306 Chapter 6 arrays and Vectors
34. board[0][0], board[1][1], board[2][2]
35. vector
primes.push_back(2);
primes.push_back(3);
primes.push_back(5);
primes.push_back(7);
primes.push_back(11);
36. vector
primes[0] = 2; primes[1] = 3; primes[2] = 5; primes[3] = 7; primes[4] = 11;
37. Ann, Cal
38. The problem doesn’t state how many measurements are taken. If the measurements
go on for many months or years (which could well be the case in a scientific or
industrial application), a vector is the better choice. If you know that the measure-
ments are stored for a fixed period (say, one day), then an array will work equally
well.
39. Because the numbers of weekdays doesn’t change, there is no disadvantage to using
an array.
40. vector
Contrast this with the answer to Self Check 19.
41. target must be a reference parameter, source should be a value parameter.
step 1 Decompose your task into steps.
Overtly, we are given a task that can be solved in one step, namely to visit the die toss values
and increment the counters. However, to produce a runnable program, we also need to pro-
duce an array of test values, and we need to display the outcome. Thus, we have three steps:
• Produce test values.
• Count the occurrences of each value.
• Display the results.
step 2 Determine which algorithm(s) you need for each step.
To produce test values, we could generate random numbers between 1 and 6, but then we
wouldn’t know whether we get the correct answer. Instead, let’s generate test values for which
we know the answer, and fill the array as fol lows:
1 2 3 4 5 1 2 3 4 5 1 2
We now expect the following counts:
1: 3
2: 3
3: 2
4: 2
5: 2
6: 0
We would like to format the output in this way. We haven’t seen this exact output format loop
in this chapter, but it is plausible that a simple loop can generate the desired output.
Finally, let us turn to the task of counting the occurrences of each value. Section 4.7.2 shows
how to count the number of 1s, with a loop
for (int i = 0; i < size; i++)
{
if (values[i] == 1)
{
counters[1]++;
}
}
We could have six loops of this form, but that is not very elegant. After all, when you have a die
value, you know which counter needs to be incremented. It is the counter corresponding to
the value itself. Therefore, the following loop solves our problem.
for (int i = 0; i < size; i++)
{
int value = values[i];
counters[value]++;
}
W o r k e d e X a M p l e 6 . 1 rolling the dice
Your task is to analyze whether a six-sided die is fair by count-
ing how often the values 1, 2, ..., 6 appear. You are given an array
of die toss values, and you should fill an array of counters, where
counters[i] is the number of times the value i was tossed. (Leave
counters[0] unused.)
cfe2_ch06_p249_306.indd 306 10/26/10 7:48 PM
7C h a p t e r
307
p o i n t e r s
to be able to declare, initialize, and
use pointers
to understand the relationship between
arrays and pointers
to be able to convert between string objects and
character pointers
to become familiar with dynamic memory allocation
and deallocation
C h a p t e r G o a l s
C h a p t e r C o n t e n t s
7.1 Defining anD Using Pointers 308
Syntax 7.1: pointer syntax 310
Common Error 7.1: Confusing pointers with the
Data to Which they point 313
Programming Tip 7.1: Use a separate Definition
for each pointer Variable 313
Special Topic 7.1: pointers and references 314
7.2 arrays anD Pointers 314
Special Topic 7.2: Using a pointer to step through
an array 318
Programming Tip 7.2: program Clearly, not
Cleverly 319
Common Error 7.2: returning a pointer to a
local Variable 319
Special Topic 7.3: Constant pointers 320
7.3 C anD C++ strings 320
Special Topic 7.4: Working with C strings 323
7.4 DynamiC memory alloCation 325
Syntax 7.2: Dynamic Memory allocation 325
Common Error 7.3: Dangling pointers 328
Common Error 7.4: Memory leaks 328
7.5 arrays anD VeCtors of
Pointers 329
7.6 Problem solVing: Draw a
PiCtUre 332
How To 7.1: Working with pointers 334
Worked Example 7.1: producing a Mass
Mailing
Random Fact 7.1: embedded systems 336
7.7 strUCtUres anD Pointers
(oPtional) 336
cfe2_ch07_p307_350.indd 307 10/28/10 8:42 PM
308
in the game on the left, the spinner’s pointer moves to an
item. a player follows the pointer and handles the item
to which it points—by taking the ball or following the
instructions written in the space. C++ also has pointers that
can point to different values throughout a program run.
pointers let you work with data whose locations change or
whose size is variable.
7.1 Defining and Using pointers
With a variable, you can access a value at a fixed location. With a pointer, the location
can vary. This capability has many useful applications. Pointers can be used to share
values among different parts of a program. Point ers allow allocation of values on
demand. Furthermore, as you will see in Chapter 10, pointers are neces sary for imple-
menting programs that manipulate objects of multiple related types. In this chapter,
you will learn how to define pointers and access the values to which they point.
7.1.1 Defining pointers
Consider a person who wants a program for making bank deposits and withdrawals,
but who may not always use the same bank account. By using a pointer, it is possible
to switch to a different account with out modifying the code for deposits and
withdrawals.
Let’s start with a variable for storing an account balance:
double harrys_account = 0;
Now suppose that we want to write an algorithm that manipulates a bank account,
but we anticipate that we may sometimes want to use harrys_account, sometimes
another account. Using a pointer gives us that flexibility. A pointer tells you where a
value is located, not what the value is.
Here is the definition of a pointer variable. The pointer variable is initialized with
the location (also called the address) of the variable harrys_account (see Figure 1):
double* account_pointer = &harrys_account; 1
a pointer denotes the
location of a variable
in memory.
Like a pointer that points to different loca tions
on a blackboard, a C++ pointer can point to
different memory locations.
The type double*, or “pointer to double”, denotes the location of a double variable. The
& operator, also called the address operator, yields the location of a variable. Taking
the address of a double variable yields a value of type double*.
Thinking about pointers can be rather abstract, but you can use a simple trick to
make it more tangible. Every variable in a computer program is located in a specific
memory location. You don’t know where each variable is stored, but you can pretend
you do. Let’s pretend that we know that harrys_account is stored in location 20300.
(That is just a made-up value.) As shown in Figure 1, the value of harrys_account is 0,
but the value of &harrys_account is 20300. The value of account_pointer is also 20300. In
our diagrams, we will draw an arrow from a pointer to the location, but of course the
computer doesn’t store arrows, just numbers.
By using a pointer, you can switch to a different account at any time. To access a
different account, simply change the pointer value:
account_pointer = &joint_account;
7.1.2 accessing Variables through pointers
When you have a pointer, you will want to access the variable to which it points. The
* operator is used to read or update the variable to which a pointer points. When used
with pointers, the * operator has no relationship with multiplication. In the C++
standard, this operator is called the indirection operator, but it is also commonly
called the dereferencing operator.
This statement makes an initial deposit into the account to which account_pointer
points (see Figure 2):
*account_pointer = 1000;
In other words, you can use *account_pointer in exactly the same way as harrys_account
or joint_account. Which account is used depends on the value of the pointer. When the
program executes this statement, it fetches the address stored in account_pointer. It
then uses the variable at that address, as shown in Figure 2.
the type T* denotes
a pointer to a
variable of type T.
the & operator
yields the location
of a variable.
the * operator
accesses the variable
to which a pointer
points.
cfe2_ch07_p307_350.indd 308 10/28/10 8:42 PM
7.1 Defining and Using pointers 309
figure 1 pointers and Values in Memory
2
1
harrys_account =
account_pointer = 20300
0
20300
harrys_account =
account_pointer =
joint_account = 0
20312
0
20300
20312
Point to
memory at
given address
Point to
memory at
new address
double* account_pointer = &harrys_account
account_pointer = &joint_account
The type double*, or “pointer to double”, denotes the location of a double variable. The
& operator, also called the address operator, yields the location of a variable. Taking
the address of a double variable yields a value of type double*.
Thinking about pointers can be rather abstract, but you can use a simple trick to
make it more tangible. Every variable in a computer program is located in a specific
memory location. You don’t know where each variable is stored, but you can pretend
you do. Let’s pretend that we know that harrys_account is stored in location 20300.
(That is just a made-up value.) As shown in Figure 1, the value of harrys_account is 0,
but the value of &harrys_account is 20300. The value of account_pointer is also 20300. In
our diagrams, we will draw an arrow from a pointer to the location, but of course the
computer doesn’t store arrows, just numbers.
By using a pointer, you can switch to a different account at any time. To access a
different account, simply change the pointer value:
account_pointer = &joint_account; 2
7.1.2 accessing Variables through pointers
When you have a pointer, you will want to access the variable to which it points. The
* operator is used to read or update the variable to which a pointer points. When used
with pointers, the * operator has no relationship with multiplication. In the C++
standard, this operator is called the indirection operator, but it is also commonly
called the dereferencing operator.
This statement makes an initial deposit into the account to which account_pointer
points (see Figure 2):
*account_pointer = 1000; 1
In other words, you can use *account_pointer in exactly the same way as harrys_account
or joint_account. Which account is used depends on the value of the pointer. When the
program executes this statement, it fetches the address stored in account_pointer. It
then uses the variable at that address, as shown in Figure 2.
the type T* denotes
a pointer to a
variable of type T.
the & operator
yields the location
of a variable.
the * operator
accesses the variable
to which a pointer
points.
cfe2_ch07_p307_350.indd 309 10/28/10 8:42 PM
310 Chapter 7 pointers
figure 2 pointer Variables Can be on either side of an assignment
2
1
joint_account =
account_pointer = 20312
1000
20312
balance = 1000
Update
memory at
given address
Read from
memory
*account_pointer = 1000
balance = *account_pointer
joint_account =
account_pointer = 20312
1000
20312
An expression such as *account_pointer can be on the left or the right of an assign-
ment. When it occurs on the left, then the value on the right is stored in the location to
which the pointer refers. When it occurs on the right, then the value is fetched from
the location and assigned to the variable on the left. For example, the following state-
ment reads the variable to which account_pointer currently points, and places its con-
tents into the balance variable:
balance = *account_pointer; 2
You can have *account_pointer on both sides of an assignment. The following state-
ment withdraws $100:
*account_pointer = *account_pointer - 100;
Table 1 contains additional pointer examples.
syntax 7.1 pointer syntax
double account = 0;
double* ptr = &account;
The & operator yields a memory address.
*ptr = 1000
cout << *ptr;
The * operator accesses the location
to which ptr points.
The type of ptr
is “pointer to double”.
You should always initialize
a pointer variable, either with
a memory address or NULL.
See page 328.
This statement reads from the
location to which ptr points.
This statement
changes account
to 1000.
7.1.3 initializing pointers
With pointers, it is particularly important that you pay attention to proper
initialization.
When you initialize a pointer, be sure that the pointer and the memory address
have the same type. For example, the following initialization would be an error:
int balance = 1000;
double* account_pointer = &balance; // Error!
The address &balance is a pointer to an int value, that is, an expression of type int*. It is
never legal to ini tialize a double* pointer with an int*.
If you define a pointer variable without providing an initial variable, the pointer
contains a random address. Using that random address is an error. In practice, your
program will likely crash or mysteri ously misbehave if you use an uninitialized
pointer:
double* account_pointer;
// Forgot to initialize
*account_pointer = 1000;
// NO! account_pointer contains an unpredictable value
it is an error to use an
uninitialized pointer.
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7.1 Defining and Using pointers 311
7.1.3
table 1 pointer syntax examples
Assume the following declarations:
int m = 10; // Assumed to be at address 20300
int n = 20; // Assumed to be at address 20304
int* p = &m;
expression Value Comment
p 20300 The address of m.
*p 10 The value stored at that address.
&n 20304 The address of n.
p = &n; Set p to the address of n.
*p 20 The value stored at the changed address.
m = *p; Stores 20 into m.
m = p; error m is an int value; p is an int* pointer.
The types are not compatible.
&10 error You can only take the address of a variable.
&p The address of p,
perhaps 20308
This is the location of a pointer variable, not
the location of an integer.
double x = 0;
p = &x;
error p has type int*, &x has type double*.
These types are incompatible.
initializing pointers
With pointers, it is particularly important that you pay attention to proper
initialization.
When you initialize a pointer, be sure that the pointer and the memory address
have the same type. For example, the following initialization would be an error:
int balance = 1000;
double* account_pointer = &balance; // Error!
The address &balance is a pointer to an int value, that is, an expression of type int*. It is
never legal to ini tialize a double* pointer with an int*.
If you define a pointer variable without providing an initial variable, the pointer
contains a random address. Using that random address is an error. In practice, your
program will likely crash or mysteri ously misbehave if you use an uninitialized
pointer:
double* account_pointer;
// Forgot to initialize
*account_pointer = 1000;
// NO! account_pointer contains an unpredictable value
it is an error to use an
uninitialized pointer.
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312 Chapter 7 pointers
There is a special value, NULL, that you should use to indicate a pointer that doesn’t
point anywhere. If you define a pointer variable and are not ready to initialize it quite
yet, set it to NULL.
double* account_pointer = NULL; // Will set later
You can later test whether the pointer is still NULL. If it is, don’t use it.
if (account_pointer != NULL) { cout << *account_pointer; } // OK
Trying to access data through a NULL pointer is illegal, and it will cause your program
to terminate.
The following program demonstrates the behavior of pointers. We execute the
same withdrawal state ment twice, but with different values for account_pointer. Each
time, a different account is modified.
ch07/accounts.cpp
1 #include
2
3 using namespace std;
4
5 int main()
6 {
7 double harrys_account = 0;
8 double joint_account = 2000;
9 double* account_pointer = &harrys_account;
10
11 *account_pointer = 1000; // Initial deposit
12
13 *account_pointer = *account_pointer – 100; // Withdraw $100
14 cout << "Balance: " << *account_pointer << endl; // Print balance
15
16 // Change the pointer value
17 account_pointer = &joint_account;
18
19 // The same statements affect a different account
20 *account_pointer = *account_pointer - 100; // Withdraw $100
21 cout << "Balance: " << *account_pointer << endl; // Print balance
22
23 return 0;
24 }
Program run
Balance: 900
Balance: 1900
1. Consider this set of statements. What is printed?
int a = 1;
int b = 2;
int* p = &a;
cout << *p << endl;
p = &b;
cout << *p << endl;
2. Consider this set of statements. What is printed?
int a = 1;
int b = 2;
the NULL pointer
does not point to
any object.
s e l f C h e C k
int* p = &a;
int* q = &b;
*p = *q;
cout << a << " " << b << endl;
3. Consider this set of statements. What is printed?
int a = 15;
int* p = &a;
int* q = &a;
cout << *p + *q << endl;
4. Consider this set of statements. What is printed?
int a = 15;
int* p = &a;
int* q = &a;
*p = *p + 10;
cout << *q << endl;
5. Consider this set of statements. What is printed?
int a = 15;
int* p = &a;
cout << *p << " " << p << endl;
Practice it Now you can try these exercises at the end of the chapter: R7.1, R7.2, R7.4.
Confusing Pointers with the Data to which they Point
A pointer is a memory address—a number that tells where a value is located in memory. It is a
common error to con fuse the pointer with the variable to which it points:
double* account_pointer = &joint_account;
account_pointer = 1000; // Error
The assignment statement does not set the joint account balance to 1000. Instead, it sets the
pointer to point to mem ory address 1000. The pointer account_pointer only describes where
the joint account variable is, and it almost cer tainly is not located at address 1000. Most com-
pilers will report an error for this assignment.
To actually access the variable, use *account_pointer:
*account_pointer = 1000; // OK
Use a separate Definition for each Pointer Variable
It is legal in C++ to define multiple variables together, like this:
int i = 0, j = 1;
This style is confusing when used with pointers:
double* p, q;
The * associates only with the first variable. That is, p is a double* pointer, and q is a double value.
To avoid any confu sion, it is best to define each pointer variable separately:
double* p;
double* q;
Common error 7.1
programming tip 7.1
cfe2_ch07_p307_350.indd 312 10/28/10 8:42 PM
7.1 Defining and Using pointers 313
int* p = &a;
int* q = &b;
*p = *q;
cout << a << " " << b << endl;
3. Consider this set of statements. What is printed?
int a = 15;
int* p = &a;
int* q = &a;
cout << *p + *q << endl;
4. Consider this set of statements. What is printed?
int a = 15;
int* p = &a;
int* q = &a;
*p = *p + 10;
cout << *q << endl;
5. Consider this set of statements. What is printed?
int a = 15;
int* p = &a;
cout << *p << " " << p << endl;
Practice it Now you can try these exercises at the end of the chapter: R7.1, R7.2, R7.4.
Confusing Pointers with the Data to which they Point
A pointer is a memory address—a number that tells where a value is located in memory. It is a
common error to con fuse the pointer with the variable to which it points:
double* account_pointer = &joint_account;
account_pointer = 1000; // Error
The assignment statement does not set the joint account balance to 1000. Instead, it sets the
pointer to point to mem ory address 1000. The pointer account_pointer only describes where
the joint account variable is, and it almost cer tainly is not located at address 1000. Most com-
pilers will report an error for this assignment.
To actually access the variable, use *account_pointer:
*account_pointer = 1000; // OK
Use a separate Definition for each Pointer Variable
It is legal in C++ to define multiple variables together, like this:
int i = 0, j = 1;
This style is confusing when used with pointers:
double* p, q;
The * associates only with the first variable. That is, p is a double* pointer, and q is a double value.
To avoid any confu sion, it is best to define each pointer variable separately:
double* p;
double* q;
Common error 7.1
programming tip 7.1
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314 Chapter 7 pointers
Pointers and references
In Section 5.9, you saw how reference parameters enable a function to modify variables that
are passed as arguments. Here is an example of a function with a reference parameter:
void withdraw(double& balance, double amount)
{
if (balance >= amount)
{
balance = balance – amount;
}
}
If you call
withdraw(harrys_checking, 1000);
then $1000 is withdrawn from harrys_checking, provided that sufficient funds are available.
You can use pointers to achieve the same effect:
void withdraw(double* balance, double amount)
{
if (*balance >= amount)
{
*balance = *balance – amount;
}
}
However, then you need to call the function with the address of the account:
withdraw(&harrys_checking, 1000);
These solutions are equivalent. Behind the scenes, the compiler translates reference param-
eters into pointers.
7.2 arrays and pointers
Pointers are particularly useful for understanding the peculiarities of arrays. In the
following sections, we describe the relationship between arrays and pointers in C++.
7.2.1 arrays as pointers
Consider this declaration of an array:
int a[10];
As you know, a[3] denotes an array element. The array name without brackets
denotes a pointer to the starting element (see Figure 3).
You can capture that pointer in a variable:
int* p = a; // Now p points to a[0]
You can also use the array name as a pointer in expressions. The statement
cout << *a;
has the same effect as the statement
cout << a[0];
special topic 7.1
the name of an array
variable is a pointer
to the starting
element of the array.
7.2.2 pointer arithmetic
Pointers into arrays support pointer arithmetic. You can add an integer offset to the
pointer to point to another array location. For example, suppose p points to the
beginning of an array:
double a[10];
double* p = a;
Then the expression
p + 3
is a pointer to the array element with index 3, and
*(p + 3)
is that array element.
As you saw in the preceding section, we can use the array name as a pointer. That
is, a + 3 is a pointer to the array element with index 3, and *(a + 3) has exactly the same
meaning as a[3].
In fact, for any integer n, it is true that
a[n] is the same as *(a + n)
This relationship is called the array/pointer duality law.
This law explains why all C++ arrays start with an index of zero. The pointer a (or
a + 0) points to the starting element of the array. That element must therefore be a[0].
To better understand pointer arithmetic, let’s again pretend that we know actual
memory addresses. Suppose the array a starts at address 20300. The array contains ten
values of type double. A double value occupies 8 bytes of memory. Therefore, the array
occupies 80 bytes, from 20300 to 20379. The starting value is located at address 20300,
the next one at address 20308, and so on (see Figure 3). For example, the value of a + 3
is 20300 + 3 × 8 = 20324. (In general, if p is a pointer to a type T, then the address p + n
is obtained by adding n × the size of a T value to the address p.)
Table 2 on page 317 shows pointer arithmetic and the array/pointer duality using
this example.
pointer arithmetic
means adding an
integer offset to an
array pointer,
yielding a pointer
that skips past the
given number of
elements.
the array/pointer
duality law states
that a[n] is identical
to *(a + n), where a
is a pointer into an
array and n is an
integer offset.
cfe2_ch07_p307_350.indd 314 10/28/10 8:42 PM
7.2 arrays and pointers 315
7.2.2
figure 3 pointers into an array
a = 0
1
4
9
16
25
36
49
64
81
a + 3 =
p = 20300
20300
20308
20316
20324
20332
20340
20348
20356
20364
20372
pointer arithmetic
Pointers into arrays support pointer arithmetic. You can add an integer offset to the
pointer to point to another array location. For example, suppose p points to the
beginning of an array:
double a[10];
double* p = a;
Then the expression
p + 3
is a pointer to the array element with index 3, and
*(p + 3)
is that array element.
As you saw in the preceding section, we can use the array name as a pointer. That
is, a + 3 is a pointer to the array element with index 3, and *(a + 3) has exactly the same
meaning as a[3].
In fact, for any integer n, it is true that
a[n] is the same as *(a + n)
This relationship is called the array/pointer duality law.
This law explains why all C++ arrays start with an index of zero. The pointer a (or
a + 0) points to the starting element of the array. That element must therefore be a[0].
To better understand pointer arithmetic, let’s again pretend that we know actual
memory addresses. Suppose the array a starts at address 20300. The array contains ten
values of type double. A double value occupies 8 bytes of memory. Therefore, the array
occupies 80 bytes, from 20300 to 20379. The starting value is located at address 20300,
the next one at address 20308, and so on (see Figure 3). For example, the value of a + 3
is 20300 + 3 × 8 = 20324. (In general, if p is a pointer to a type T, then the address p + n
is obtained by adding n × the size of a T value to the address p.)
Table 2 on page 317 shows pointer arithmetic and the array/pointer duality using
this example.
pointer arithmetic
means adding an
integer offset to an
array pointer,
yielding a pointer
that skips past the
given number of
elements.
the array/pointer
duality law states
that a[n] is identical
to *(a + n), where a
is a pointer into an
array and n is an
integer offset.
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316 Chapter 7 pointers
7.2.3 array parameter Variables are pointers
Once you understand the connection between arrays and pointers, it becomes clear
why array parameter variables are different from other parameter types. As an exam-
ple, consider this function that computes the sum of all values in an array:
double sum(double a[], int size)
{
double total = 0;
for (int i = 0; i < size; i++)
{
total = total + a[i];
}
return total;
}
Here is a call to the function (see Figure 4):
double data[10];
... // Initialize data
double s = sum(data, 10);
The value data is passed to the sum function. It is actually a pointer of type double*,
pointing to the starting element of the array. One would therefore expect that the
function is declared as
double sum(double* a, int size)
However, if you look closely at the function definition, you will see that the parame-
ter variable is declared as an array with empty bounds:
double sum(double a[], int size)
As viewed by the C++ compiler, these parameter declarations are completely equiva-
lent. The [] notation is “syntactic sugar” for declaring a pointer. (Computer scientists
use the term “syntactic sugar” to describe a notation that is easy to read for humans
and that masks a complex implementation detail.) The array notation gives human
When passing an
array to a function,
only the starting
address is passed.
figure 4
passing an array to a Function
data = 0
1
4
9
16
25
36
49
64
81
a + i =
s =
20300
20324
a = 20300
size = 10
total = 5
i = 3
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7.2 arrays and pointers 317
table 2 arrays and pointers
expression Value Comment
a 20300 The starting address of the array, here assumed to be 20300.
*a 0 The value stored at that address. (The array contains values 0, 1, 4, 9, ....)
a + 1 20308 The address of the next double value in the array. A double occupies
8 bytes.
a + 3 20324 The address of the element with index 3, obtained by skipping past
3 × 8 bytes.
*(a + 3) 9 The value stored at address 20324.
a[3] 9 The same as *(a + 3) by array/pointer duality.
*a + 3 3 The sum of *a and 3. Since there are no parentheses, the * refers only to a.
&a[3] 20324 The address of the element with index 3, the same as a + 3.
readers the illusion that an entire array is passed to the function, but in fact the func-
tion receives only the starting address for the array.
Now consider this statement in the body of the function:
total = total + a[i];
The C++ compiler considers a to be a pointer, not an array. The expression a[i] is syn-
tactic sugar for *(a + i). That expression denotes the storage location that is i ele-
ments away from the address stored in the variable a. Figure 4 shows how a[i] is
accessed when i is 3.
You can now understand why it is always necessary to pass the size of the array.
The function receives a single memory address, which tells it where the array starts.
That memory address enables the function to locate the values in the array. But the
function also needs to know where to stop.
For Self Checks 6–9, draw the array and pointer as in Figure 3. Assume a starting
address (20300 will work fine), and assume that each int value occupies 4 bytes.
6. What is the contents of the array a after these statements?
int a[] = { 2, 3, 5 };
int* p = a;
p++;
*p = 0;
7. What is the contents of the array a after these statements?
int a[] = { 2, 3, 5 };
int* p = a + 1;
*(p + 1) = 0;
8. What is the contents of the array a after these statements?
int a[] = { 2, 3, 5 };
int* p = a;
int* q = a + 2;
s e l f C h e C k
cfe2_ch07_p307_350.indd 317 10/28/10 8:42 PM
318 Chapter 7 pointers
p++;
q--;
*p = *q;
9. What do the following statements print?
int a[] = { 2, 3, 5 };
cout << *a + 2 << " ";
cout << *(a + 2) << endl;
10. In Chapter 6, we defined a function
void squares(int n, int result[])
Declare the parameter variable using pointer notation.
Practice it Now you can try these exercises at the end of the chapter: R7.7, R7.8, P7.4.
Using a Pointer to step through an array
Consider again the sum function of Section 7.2.3. Now that you know that the first parameter
variable of the sum function is a pointer, you can implement the function in a slightly different
way. Rather than incrementing an integer index, you can increment a pointer variable to visit
all array elements in turn:
double sum(double* a, int size)
{
double total = 0;
double* p = a; // p starts at the beginning of the array
for (int i = 0; i < size; i++)
{
total = total + *p; // Add the value to which p points
p++; // Advance p to the next array element
}
return total;
}
Initially, the pointer p points to the element a[0]. The increment
p++;
moves it to point to the next element (see Figure 5).
It is a tiny bit more efficient to use and increment a pointer than to access an array element
as a[i]. For this rea son, some programmers routinely use pointers instead of indexes to access
array elements.
special topic 7.2
figure 5 a pointer Variable traversing the elements of an array
p =
a =
Program Clearly, not Cleverly
Some programmers take great pride in minimizing the number of instructions, even if the
resulting code is hard to understand. For example, here is a legal implementation of the sum
function:
double sum(double* a, int size)
{
double total = 0;
while (size-- > 0) // Loop size times
{
total = total + *a++; // Add the value to which a points; increment a
}
return total;
}
This implementation uses two tricks. First, the function parameter variables a and size are
variables, and it is legal to modify them. Moreover, the expressions size– and a++ mean “dec-
rement or increment the variable and return the old value”. In other words, the expression
size– > 0
combines two tasks: to decrement size, and to test whether size was positive before the decre-
ment. Similarly, the expression
*a++
increments the pointer to the next element, and it returns the element to which it pointed
before the increment.
Please do not use this programming style. Your job as a programmer is not to dazzle other
programmers with your cleverness, but to write code that is easy to understand and maintain.
returning a Pointer to a local Variable
Consider this function that tries to return a pointer to an array containing two elements, the
first and the last values of an array:
double* firstlast(double a[], int size)
{
double result[2];
result[0] = a[0];
result[1] = a[size – 1];
return result; // Error!
}
The function returns a pointer to the starting element of the result array. However, that array
is a local variable of the firstlast function. The local variable no longer exists when the func-
tion exits. Its contents will soon be overwritten by other function calls.
You can solve this problem by passing an array to hold the answer:
void firstlast(const double a[], int size, double[] result)
{
result[0] = a[0];
result[1] = a[size – 1];
}
Then it is the responsibility of the caller to allocate an array to hold the result.
programming tip 7.2
Common error 7.2
cfe2_ch07_p307_350.indd 318 10/28/10 8:42 PM
7.2 arrays and pointers 319
Program Clearly, not Cleverly
Some programmers take great pride in minimizing the number of instructions, even if the
resulting code is hard to understand. For example, here is a legal implementation of the sum
function:
double sum(double* a, int size)
{
double total = 0;
while (size– > 0) // Loop size times
{
total = total + *a++; // Add the value to which a points; increment a
}
return total;
}
This implementation uses two tricks. First, the function parameter variables a and size are
variables, and it is legal to modify them. Moreover, the expressions size– and a++ mean “dec-
rement or increment the variable and return the old value”. In other words, the expression
size– > 0
combines two tasks: to decrement size, and to test whether size was positive before the decre-
ment. Similarly, the expression
*a++
increments the pointer to the next element, and it returns the element to which it pointed
before the increment.
Please do not use this programming style. Your job as a programmer is not to dazzle other
programmers with your cleverness, but to write code that is easy to understand and maintain.
returning a Pointer to a local Variable
Consider this function that tries to return a pointer to an array containing two elements, the
first and the last values of an array:
double* firstlast(double a[], int size)
{
double result[2];
result[0] = a[0];
result[1] = a[size – 1];
return result; // Error!
}
The function returns a pointer to the starting element of the result array. However, that array
is a local variable of the firstlast function. The local variable no longer exists when the func-
tion exits. Its contents will soon be overwritten by other function calls.
You can solve this problem by passing an array to hold the answer:
void firstlast(const double a[], int size, double[] result)
{
result[0] = a[0];
result[1] = a[size – 1];
}
Then it is the responsibility of the caller to allocate an array to hold the result.
programming tip 7.2
Common error 7.2
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320 Chapter 7 pointers
Constant Pointers
The following definition specifies a constant pointer:
const double* p = &balance;
You cannot modify the value to which p points. That is, the following statement is illegal:
*p = 0; // Error
Of course, you can read the value:
cout << *p; // OK
A constant array parameter variable is equivalent to a constant pointer. For example, consider
the function
double sum(const double[] values, int size)
Recall from Section 7.2.3 that values is a pointer. The function could have been defined as
double sum(const double* values, int size)
The function can use the pointer values to read the array elements, but it cannot modify them.
7.3 C and C++ strings
C++ has two mechanisms for manipulating strings. The string class supports charac-
ter sequences of arbi trary length and provides convenient operations such as concat-
enation and string comparison. However, C++ also inherits a more primitive level
of string handling from the C language, in which strings are rep resented as arrays of
char values. In this sections, we will discuss the relationships between these types.
7.3.1 the char type
The char type denotes an individual character. Character
literals are delimited by single quotes; for example,
char input = 'y';
Each character is actually encoded as an integer value. (See Appendix D for the
encoding using the ASCII code, which is used on the majority of computers today.)
Note that 'y' is a single character, which is quite different from "y", a string con-
taining the 'y' charac ter.
Table 3 shows typical character literals.
7.3.2 C strings
In the C programming language, strings are always represented as character arrays.
C++ programmers often refer to arrays of char values as “C strings”.
In particular, a literal string, such as "Harry", is not an object of type string. Instead,
it is an array of char values. As with all arrays, a string literal can be assigned to a
pointer variable that points to the initial character in the array:
const char* char_pointer = "Harry"; // Points to 'H'
special topic 7.3
a value of type char
denotes an individual
character. Character
literals are enclosed
in single quotes.
a literal string
(enclosed in double
quotes) is an array of
char values with a
zero terminator.
cfe2_ch07_p307_350.indd 320 10/28/10 8:43 PM
7.3 C and C++ strings 321
figure 6
a Character array [0]
[1]
[2]
[3]
[4]
[5]
'H'
'a'
'r'
'r'
'y'
'\0'
Points to 'H'
char_pointer =
The string is declared as const because you are not supposed to modify a literal string.
(See Special Topic 7.3 on page 320 for more information on constant pointers.)
A C string is terminated by a special character, called a null terminator, denoted
'\0'. For example, the C string "Harry" contains six characters, namely 'H', 'a', 'r',
'r', 'y' and '\0' (See Figure 6.)
The terminator is a character that is encoded as the number zero—this is differ-
ent from the character '0', the character denoting the zero digit. (Under the ASCII
encoding scheme, the character denoting the zero digit is encoded as the number 48.)
Functions that operate on C strings rely on this terminator. Here is a typical exam-
ple, the strlen func tion declared in the
a character array. The function counts the number of characters until it reaches the
terminator.
int strlen(const char s[])
{
int i = 0;
while (s[i] != ‘\0’) { i++; } // Count characters before the null terminator
return i;
}
The call strlen(“Harry”) returns 5.
7.3.3 Character arrays
A literal string such as “Harry” is a constant. You are not allowed to modify its charac-
ters. If you want to modify the characters in a string, define a character array instead.
For example:
char char_array[] = “Harry”; // An array of 6 characters
table 3 Character literals
‘y’ The character y
‘0’ The character for the digit 0. In the
ASCII code, ‘0’ has the value 48.
‘ ‘ The space character
‘\n’ The newline character
‘\t’ The tab character
‘\0’ The null terminator of a string
“y” error: Not a char value
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322 Chapter 7 pointers
The char_array variable is an array of 6 characters, initialized with ‘H’, ‘a’, ‘r’, ‘r’,
‘y’, and a null termi nator. The compiler counts the characters in the string that is used
for initializing the array, including the null terminator.
You can modify the characters in the array:
char_array[0] = ‘L’;
7.3.4 Converting Between C and C++ strings
Before the C++ string class became widely available, direct manipulation of character
arrays was com mon, but also quite challenging (see Special Topic 7.4 on page 323). If
you use functions that receive or return C strings, you need to know how to convert
between C strings and string objects.
For example, the
int atoi(const char s[])
The atoi function converts a character array containing digits into its integer value:
char[] year = “2012”;
int y = atoi(year); // Now y is the integer 2012
This functionality is inexplicably missing from the C++ string class. The c_str mem-
ber function of the string class offers an “escape hatch”. If s is a string, then s.c_str()
yields a char* pointer to the characters in the string. Here is how you use that member
function to call the atoi function:
string year = “2012”;
int y = atoi(year.c_str());
Conversely, converting from a C string to a C++ string is very easy. Simply initialize
a string variable with any value of type char*, such as a string literal or character array.
For example, the definition
string name = “Harry”;
initializes the C++ string object name with the C string “Harry”.
7.3.5 C++ strings and the [] operator
Up to this point, we have always used the substr member function to access individ-
ual characters in a C++ string. For example, if a string variable is defined as
string name = “Harry”;
the expression
name.substr(3, 1)
yields a string of length 1 containing the character at index 3.
You can access individual characters with the [] operator:
name[0] = ‘L’;
Now the string is “Larry”. The [] operator is more convenient than the substr function
if you want to visit a string one character at a time.
Here is a useful example. The following function makes a copy of a string and
changes all characters to uppercase:
Many library
functions use
pointers of
type char*.
the c_str member
function yields a
char* pointer from
a string object.
You can initialize C++
string variables with
C strings.
You can access
characters in a C++
string object with
the [] operator.
/**
Makes an uppercase version of a string.
@param str a string
@return a string with the characters in str converted to uppercase
*/
string uppercase(string str)
{
string result = str; // Make a copy of str
for (int i = 0; i < result.length(); i++)
{
result[i] = toupper(result[i]); // Convert each character to uppercase
}
return result;
}
For example, uppercase("Harry") returns a string with the characters "HARRY".
The toupper function is defined in the
ters to uppercase. (The tolower function does the opposite.)
11. How many char values are stored in the character array “Hello, World!\n”?
12. What is strlen(“Hello, World!\n”)?
13. Allocate a pointer variable that points to the string “Hello”.
14. Consider this statement:
string title = “Agent” + 007;
Does the statement compile? What is its effect?
15. Consider the following statements:
cout << "Enter an integer, Q to quit";
string input;
cin >> input;
if (input == “Q”) { return; }
If the input is not the letter Q, how do you extract the number stored in the string
input?
Practice it Now you can try these exercises at the end of the chapter: R7.15, R7.16, P7.6.
working with C strings
Before the string class became widely available, it was common to work with character arrays
directly.
Table 4 on page 324 shows several commonly used functions for manipulating C strings.
All of these functions are declared in the
Consider the task of concatenating a first name and a last name into a string. The string
class makes this very easy:
string first = “Harry”;
string last = “Smith”;
string name = first + ” ” + last;
Let us implement this task with C strings. Allocate an array of characters for the result:
const int NAME_SIZE = 40;
char name[NAME_SIZE];
s e l f C h e C k
special topic 7.4
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7.3 C and C++ strings 323
/**
Makes an uppercase version of a string.
@param str a string
@return a string with the characters in str converted to uppercase
*/
string uppercase(string str)
{
string result = str; // Make a copy of str
for (int i = 0; i < result.length(); i++)
{
result[i] = toupper(result[i]); // Convert each character to uppercase
}
return result;
}
For example, uppercase("Harry") returns a string with the characters "HARRY".
The toupper function is defined in the
ters to uppercase. (The tolower function does the opposite.)
11. How many char values are stored in the character array “Hello, World!\n”?
12. What is strlen(“Hello, World!\n”)?
13. Allocate a pointer variable that points to the string “Hello”.
14. Consider this statement:
string title = “Agent” + 007;
Does the statement compile? What is its effect?
15. Consider the following statements:
cout << "Enter an integer, Q to quit";
string input;
cin >> input;
if (input == “Q”) { return; }
If the input is not the letter Q, how do you extract the number stored in the string
input?
Practice it Now you can try these exercises at the end of the chapter: R7.15, R7.16, P7.6.
working with C strings
Before the string class became widely available, it was common to work with character arrays
directly.
Table 4 on page 324 shows several commonly used functions for manipulating C strings.
All of these functions are declared in the
Consider the task of concatenating a first name and a last name into a string. The string
class makes this very easy:
string first = “Harry”;
string last = “Smith”;
string name = first + ” ” + last;
Let us implement this task with C strings. Allocate an array of characters for the result:
const int NAME_SIZE = 40;
char name[NAME_SIZE];
s e l f C h e C k
special topic 7.4
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324 Chapter 7 pointers
H a r r y
NAME_SIZE
Last name
and null terminator
length
Space
…
This array can hold strings with a length of at most 39, because one character is required for
the null terminator.
Now copy the first name, using strncpy:
strncpy(name, first, NAME_SIZE – 1);
You must be careful not to overrun the target array. It is unlikely that a first name is longer
than 39 characters, but a hacker could supply a longer input in order to overwrite memory.
Now, if there is still room, add a space and the last name, again being careful not to overrun
the array boundaries.
int length = strlen(name);
if (length < NAME_SIZE - 1)
{
strcat(name, " ");
int n = NAME_SIZE - 2 - length; // Leave room for space, null terminator
if (n > 0)
{
strncat(name, last, n);
}
}
As you can see, the C string code is over three times as long as the code using C++ strings, and
it is not as capable—if the target array is not long enough to hold the result, it is truncated.
table 4 C string Functions
In this table, s and t are character arrays; n is an integer.
Function Description
strlen(s) Returns the length of s.
strcpy(t, s) Copies the characters from s into t.
strncpy(t, s, n) Copies at most n characters from s into t.
strcat(t, s) Appends the characters from s after the end of the
characters in t.
strncat(t, s, n) Appends at most n characters from s after the end of the
characters in t.
strcmp(s, t) Returns 0 if s and t have the same contents, a negative
integer if s comes before t in lexicographic order, a
positive integer otherwise.
7.4 Dynamic Memory allocation
In many programming situations, you do not know beforehand how many values
you need. To solve this problem, you can use dynamic allocation and ask the C++
run-time system to create new values when ever you need them. The run-time system
keeps a large storage area, called the heap, that can allocate val ues and arrays of any
type. When you ask for a
new double
then a storage location of type double is located on the heap, and a pointer to that loca-
tion is returned. More usefully, the expression
new double[n]
allocates an array of size n, and yields a pointer to the starting element. (Here n need
not be a constant.)
You will want to capture that pointer in a variable:
double* account_pointer = new double;
double* account_array = new double[n];
You now use the pointer as described previously in this chapter. If you allocated an
array, the magic of array/pointer duality lets you use the array notation account_
array[i] to access the ith element.
When your program no longer needs memory that you previously allocated with
the new operator, you must return it to the heap, using the delete operator:
delete account_pointer;
However, if you allocated an array, you must use the delete[] operator:
delete[] account_array;
This operator reminds the heap that the pointer points to an array, not a single value.
Use dynamic memory
allocation if you do
not know in advance
how many values
you need.
the new operator
allocates memory
from the heap.
You must reclaim
dynamically allocated
objects with the
delete or delete[]
operator.
syntax 7.2 Dynamic Memory allocation
int* var_ptr = new int;
…
*var_ptr = 1000;
…
delete var_ptr;
The new operator yields a pointer
to a memory block of the given type.
int* array_ptr = new int[size];
…
array_ptr[i] = 1000;
…
delete[] array_ptr;
Remember to use
delete[] when
deallocating the array.
Capture the pointer
in a variable.
Use this form to allocate
an array of the given size
(size need not be a constant).
Use the memory.
Delete the memory
when you are done.
Use the pointer as
if it were an array.
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7.4 Dynamic Memory allocation 325
7.4 Dynamic Memory allocation
In many programming situations, you do not know beforehand how many values
you need. To solve this problem, you can use dynamic allocation and ask the C++
run-time system to create new values when ever you need them. The run-time system
keeps a large storage area, called the heap, that can allocate val ues and arrays of any
type. When you ask for a
new double
then a storage location of type double is located on the heap, and a pointer to that loca-
tion is returned. More usefully, the expression
new double[n]
allocates an array of size n, and yields a pointer to the starting element. (Here n need
not be a constant.)
You will want to capture that pointer in a variable:
double* account_pointer = new double;
double* account_array = new double[n];
You now use the pointer as described previously in this chapter. If you allocated an
array, the magic of array/pointer duality lets you use the array notation account_
array[i] to access the ith element.
When your program no longer needs memory that you previously allocated with
the new operator, you must return it to the heap, using the delete operator:
delete account_pointer;
However, if you allocated an array, you must use the delete[] operator:
delete[] account_array;
This operator reminds the heap that the pointer points to an array, not a single value.
Use dynamic memory
allocation if you do
not know in advance
how many values
you need.
the new operator
allocates memory
from the heap.
You must reclaim
dynamically allocated
objects with the
delete or delete[]
operator.
syntax 7.2 Dynamic Memory allocation
int* var_ptr = new int;
…
*var_ptr = 1000;
…
delete var_ptr;
The new operator yields a pointer
to a memory block of the given type.
int* array_ptr = new int[size];
…
array_ptr[i] = 1000;
…
delete[] array_ptr;
Remember to use
delete[] when
deallocating the array.
Capture the pointer
in a variable.
Use this form to allocate
an array of the given size
(size need not be a constant).
Use the memory.
Delete the memory
when you are done.
Use the pointer as
if it were an array.
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326 Chapter 7 pointers
After you delete a memory block, you can no longer use it. The storage space may
already be used elsewhere.
delete[] account_array;
account_array[0] = 1000; // NO! You no longer own the memory of account_array
Heap arrays have one significant advantage over array variables. If you declare an
array variable, you must specify a fixed array size when you compile the program.
But when you allocate an array on the heap, you can choose the size at run time.
Moreover, if you later need more elements, you are not stuck. You can allocate a
bigger heap array, copy the elements from the smaller array into the bigger array, and
delete the smaller array (see Figure 7):
double* bigger_array = new double[2 * n];
for (int i = 0; i < n; i++)
{
bigger_array[i] = account_array[i];
}
delete[] account_array;
account_array = bigger_array;
n = 2 * n;
This is exactly what a vector does behind the scenes.
Heap allocation is a powerful feature, but
you must be very careful to follow all rules
precisely:
• Every call to new must be matched by
exactly one call to delete.
• Use delete[] to delete arrays.
• Don’t access a memory block after it has
been deleted.
If you don’t follow these rules, your program
can crash or run unpredictably. Table 5 shows
common errors.
Be sure to recycle any heap memory that
your program no longer needs.
figure 7 Growing a Dynamic array
account_array =
bigger_array =
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7.4 Dynamic Memory allocation 327
16.
table 5 Common Memory allocation errors
statements error
int* p;
*p = 5;
delete p;
There is no call to new int.
int* p = new int;
*p = 5;
p = new int;
The first allocated memory block
was never deleted.
int* p = new int[10];
*p = 5;
delete p;
The delete[] operator should have
been used.
int* p = new int[10];
int* q = p;
q[0] = 5;
delete p;
delete q;
The same memory block was
deleted twice.
int n = 4;
int* p = &n;
*p = 5;
delete p;
You can only delete memory
blocks that you obtained from
calling new.
What does this statement sequence print?
int* p = new int;
*p = 3;
cout << *p << endl;
delete p;
17. What does this statement sequence print?
int* p = new int[10];
for (int i = 0; i < 10; i++) { p[i] = i * i; }
cout << *p << " " << *(p + 1) << " " << p[2] << endl;
delete[] p;
18. What is wrong with this sequence of statements?
int* p = new int[10];
p[10] = 5;
delete[] p;
19. Consider this function
int* grow(int a[], int size)
{
int* result = new int[2 * size];
for (int i = 0; i < size; i++) { result[i] = a[i]; }
for (int i = size; i < 2 * size; i++) { result[i] = 0; }
return result;
}
What is the contents of the array to which p points after the following
statements?
s e l f C h e C k
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328 Chapter 7 pointers
int primes[] = { 2, 3, 5, 7, 11 };
int* p = grow(primes, 5);
20. Consider the grow function of Self Check 19. What must its caller remember
to do?
Practice it Now you can try these exercises at the end of the chapter: R7.19, R7.21, P7.10.
Dangling Pointers
A very common pointer error is to use a pointer that points to
memory that has already been deleted. Such a pointer is called a
dangling pointer. Because the freed memory will be reused for
other purposes, you can create real damage by using a dangling
pointer. Con sider this example:
int* values = new int[n];
// Process values
delete[] values;
// Some other work
values[0] = 42;
This code will compile since the compiler does not track whether a pointer points to a valid
memory location. How ever, the program may run with unpredictable results. If the program
calls the new operator anywhere after deleting values, that call may allocate the same memory
again. Now some other part of your program accesses the memory to which values points, and
that program part will malfunction when you overwrite the memory. This can happen even if
you don’t see any call to new—such calls may occur in library functions.
Never use a pointer that has been deleted. Some programmers take the precaution of setting
all deleted pointers to NULL:
delete[] values;
values = NULL;
This is not perfect protection—you might have saved values into another pointer variable—
but it is a reasonable precaution.
memory leaks
Another very common pointer error is to allocate memory on the heap and never deallocate it.
A memory block that is never deallocated is called a memory leak.
If you allocate a few small blocks of memory and forget to deallocate them, this is not a
huge problem. When the program exits, all allocated memory is returned to the operating sys-
tem.
But if your program runs for a long time, or if it allocates lots of
memory (perhaps in a loop), then it can run out of memory. Mem-
ory exhaustion will cause your program to crash. In extreme cases,
the computer may freeze up if your program exhausted all avail able
memory. Avoiding memory leaks is particularly important in pro-
grams that need to run for months or years, without restarting, and in programs that run on
resource-con strained devices such as cell phones.
Even if you write short-lived programs, you should make it a habit to avoid memory leaks.
Make sure that every call to the new operator has a corresponding call to the delete operator.
Common error 7.3
Using a dangling
pointer (a pointer
that points to
memory that has
been deleted) is a
serious programming
error.
Common error 7.4
every call to new
should have a
matching call
to delete.
cfe2_ch07_p307_350.indd 328 10/28/10 8:43 PM
7.5 arrays and Vectors of pointers 329
7.5 arrays and Vectors of pointers
When you have a sequence of pointers, you can place them into an array or vector. An
array and a vector of ten int* pointers are defined as
int* pointer_array[10];
vector
The expression pointer_array[i] or pointer_vector[i] denotes the pointer with index i
in the sequence.
One application of such pointer sequences are two-dimensional arrays in which
each row has a differ ent length, such as the triangular array shown in Figure 8.
In this situation, it would not be very efficient to use a two-dimensional array,
because almost half of the elements would be wasted.
We will develop a program that uses such an array for simulating a Galton board
(Figure 9). A Galton board consists of a pyramidal arrangement of pegs, and a row
of bins at the bottom. Balls are dropped onto the top peg and travel toward the bins.
At each peg, there is a 50 percent chance of moving left or right. The balls in the bins
approximate a bell-curve distribution.
The Galton board can only show the balls in the bins, but we can do better by
keeping a counter for each peg, incrementing it as a ball travels past it.
We will simulate a board with ten rows of pegs. Each row requires an array of
counters.
figure 8 a triangular array
pointer_array =
figure 9
a Galton Board
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330 Chapter 7 pointers
figure 10
Movement of a Ball in
the Galton Board array
row i
row i + 1
column
j
column
j + 1
The following statements initialize the triangular array:
int* counts[10];
for (int i = 0; i < 10; i++)
{
counts[i] = new int[i + 1];
}
Note that the first element counts[0] contains a pointer to an array of length 1, and the
last element counts[9] contains a pointer to an array of length 10.
Before doing the simulation, let us consider how to print out the values. The ele-
ment counts[i] points to an array. The element of index j of that array is
counts[i][j]
This loop prints all elements in the ith row:
for (int j = 0; j <= i; j++)
{
cout << setw(4) << counts[i][j];
}
cout << endl;
Now let’s simulate a falling ball. The movements to the left and right in Figure 9 cor-
respond to move ments to the next row, either straight down or to the right, in
Figure 10. More precisely, if the ball is cur rently at row i and column j, then it will go
to row i + 1 and, with a 50 percent chance, either stay in column j or go to column j + 1.
The program below has the details. In the sample program run, notice how 1,000
balls have hit the top peg, and how the bottommost row of pegs approximates a bell-
curve distribution.
ch07/galton.cpp
1 #include
2 #include
3 #include
4 #include
5
6 using namespace std;
7
8 int main()
9 {
10 srand(time(0));
11
12 int* counts[10];
13
14 // Allocate the rows
15 for (int i = 0; i < 10; i++)
16 {
cfe2_ch07_p307_350.indd 330 10/28/10 8:43 PM
7.5 arrays and Vectors of pointers 331
17 counts[i] = new int[i + 1];
18 for (int j = 0; j <= i; j++)
19 {
20 counts[i][j] = 0;
21 }
22 }
23
24 const int RUNS = 1000;
25
26 // Simulate 1,000 balls
27 for (int run = 0; run < RUNS; run++)
28 {
29 // Add a ball to the top
30 counts[0][0]++;
31 // Have the ball run to the bottom
32 int j = 0;
33 for (int i = 1; i < 10; i++)
34 {
35 int r = rand() % 2;
36 // If r is even, move down, otherwise to the right
37 if (r == 1)
38 {
39 j++;
40 }
41 counts[i][j]++;
42 }
43 }
44
45 // Print all counts
46 for (int i = 0; i < 10; i++)
47 {
48 for (int j = 0; j <= i; j++)
49 {
50 cout << setw(4) << counts[i][j];
51 }
52 cout << endl;
53 }
54
55 // Deallocate the rows
56 for (int i = 0; i < 10; i++)
57 {
58 delete[] counts[i];
59 }
60
61 return 0;
62 }
Program run
1000
480 520
241 500 259
124 345 411 120
68 232 365 271 64
32 164 283 329 161 31
16 88 229 303 254 88 22
9 47 147 277 273 190 44 13
5 24 103 203 288 228 113 33 3
1 18 64 149 239 265 186 61 15 2
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332 Chapter 7 pointers
21. Why didn’t we initialize the triangular array with the following loop?
for (int i = 0; i < 10; i++)
{
counts[i] = new int[i];
}
22. Suppose a program initializes a triangular array as we did in this section, and
then accesses a non-exis tent element with the statement
counts[1][2]++
Will the program compile? If so, what happens at run time?
23. Initialize a triangular 10 × 10 array where the first row has length 10 and the last
row has length 1.
24. What changes would need to be made to the galton.cpp program so that counts is
a vector
25. What changes would need to be made to the galton.cpp program so that each row
is a vector
Practice it Now you can try these exercises at the end of the chapter: R7.22, P7.11, P7.12.
7.6 problem solving: Draw a picture
When designing programs that use pointers, you want to visualize how the pointers
connect the data that you are working with. In most situations, it is essential that you
draw a diagram that shows the connec tions.
Start with the data that will be accessed or modified through the pointers. These
may be account bal ances, counters, character strings, or other items. Focus on what
is being pointed at.
Then draw the variable or variables that contain the pointers. These will be the
front-end to your sys tem. Processing usually starts with a pointer, then locates the
actual data.
Finally, draw the pointers as arrows. If the pointers will vary as the program exe-
cutes, draw a typical arrangement. It can also be useful to draw several diagrams that
show how the pointers change.
Consider the following problem: The media center of a university loans out equip-
ment, such as micro phones, cables, and so on, to faculty and students. We want to
track the name of each item, and the name of the user who checked it out.
In this case, the data that are being accessed are item and user names. We will store
all names in a long array of characters, adding new names to the end as needed. If a
name is already present, we don’t store it twice. This is an efficient way of storing
strings, provided that few strings need to be removed. Here is a section of the charac-
ter array:
Microphone TX-10\0 Smith, Diane\0 Lee, Tim\0 Tape recorder\0 Mini DVI cable\0 …
Next, let us draw the pointers to the item names. they are stored in an array of point-
ers called items. Some items may have the same name. In the example below, we have
two microphones with the same name.
s e l f C h e C k
Draw the data that is
being processed,
then draw the pointer
variables. When
drawing the pointer
arrows, illustrate a
typical situation.
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7.6 problem solving: Draw a picture 333
Microphone TX-10\0 Smith, Diane\0 Lee, Tim\0 Tape recorder\0 Mini DVI cable\0 …
items
Finally, there is a parallel array checked_out_to of pointers to user names. Sometimes,
items can be checked out to the same user. Other items aren’t checked out at all—the
user name pointer is NULL. When you draw a diagram, try to include examples of all
scenarios.
Microphone TX-10\0 Smith, Diane\0 Lee, Tim\0 Tape recorder\0 Mini DVI cable\0 …
items
NULL
checked_out_to
Now that you have a pointer diagram, you can use it to visualize operations on the
data. Suppose we want to print a report of all items that are currently checked out.
This can be achieved by visiting all pointers in both arrays. The pointer items[i] gives
the name of the item, and checked_out_to[i] is either NULL, in which case we do not want
to include this item, or it is the name of the user.
Because programming with pointers is complex, you should always draw dia-
grams whenever you use pointers.
26. Consider the sum function in Section 7.2.3. Draw a picture of its parameter
variables when it is called as sum(data + 3, 7), where data is an array of length 10.
27. One way to reverse all elements in an array is to have pointers to the first and last
location, swap the elements that are being pointed to, and then increment the
first pointer and decrement the second pointer. Repeat until both pointers reach
the middle of the array. Draw a picture of the pointers in this algorithm after a
couple of iterations.
28. One way to test whether two strings have the same contents is to have pointers
to the strings and com pare the characters that are being pointed to. If both are
the ‘\0’ terminator, stop—the strings are identical. If the characters are differ-
ent, stop—the strings are different. Otherwise, increment both pointers. Draw
a picture of the pointers in this algorithm after a couple of iterations. Be sure to
pick strings that match the algorithm state.
29. Draw a picture showing the contents of the counts array in the galton.cpp pro-
gram on page 330 after the fifth iteration of the loop in lines 56–59.
s e l f C h e C k
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334 Chapter 7 pointers
30. Suppose you want to sort the names of the items in the media center. There is
no need to move the strings. All you need to do is rearrange the string pointers.
Draw a picture that shows the result.
Practice it Now you can try these exercises at the end of the chapter: R7.25, R7.26, R7.27.
step 1 Draw a picture.
As described in Section 7.6, it is a good idea to draw a picture that shows the pointers in your
program.
In our example, we need to track the copy accounts: a master account and ten project
accounts. For each user, we need a pointer to the copy account:
users =
master_account =
project_accounts =
.
.
.
.
.
.
step 2 Declare pointer variables.
This step is usually easy. With numerical data, you will have pointers of type int* or double*. If
you manipulate char acter arrays, you use char* pointers. In Chapter 10, you will use pointers
to objects.
How many pointer variables do you have? If you only have one or two, just declare vari-
ables such as
double* account_pointer;
If you have a sequence of pointers, use a vector or array, such as
vector
h o W t o 7 . 1 working with Pointers
You use pointers because you want flexibility in
your program: the ability to change the relationships
between data. This How To walks you through the
decision-making process for using pointers.
We will illustrate the steps with the task of simu-
lating a part of the control logic for a departmental
photocopier. A person making a copy enters a user
number. There are 100 different users, with numbers
0 to 99. Each user is linked to a copy account, either
the master account or one of ten project accounts.
That linkage is maintained by the administrator, and it
can change as users work on different projects. When
copies are made, the appropriate account should be
incremented.
Users identify themselves on the
copier control panel. Using pointers,
the relationships between users and
copy accounts can be flexible.
In our example, the purpose is to manipulate copy counters. Therefore, our data are integers
and the pointers have type int*. We know that we will have exactly 100 pointers, one for each
user. Therefore, we can choose an array
int* users[100];
step 3 Initialize the pointers with variable addresses or heap memory.
Will you allocate variables and then take their addresses with the & operator? That is fine if you
have a fixed amount of data that you want to reach through the pointers. Otherwise, use the
new operator to allocate memory dynami cally. Be sure to deallocate the memory when you are
done, using the delete or delete[] operator.
In our example, there is no need for dynamic allocation, so we will just take addresses of
variables.
int master_account;
int project_accounts[10];
for (int i = 0; i < 100; i++)
{
users[i] = &master_account;
}
// Here we reassign several users to project accounts.
// The following code is a simulation of the actions that would
// occur in an administration interface, which we do not implement.
users[2] = project_accounts + 1
users[3] = project_accounts + 2
users[99] = project_accounts + 2
step 4 Use the * or [] operator to access your data.
When you access a single variable through a pointer, use the * operator to read or write the
variable. For example,
*account_pointer = *account_pointer + 20;
When you have a pointer to an array, use the [] notation. Simply think of the pointer as the
array:
account_array[i] = account_array[i] + 20;
If you have an array or vector of pointers, then you need brackets to get at an individual
pointer. Then supply another * or [] to get at the value. You saw an example in the Galton
board simulator where a count was accessed as
counts[i][j]
Implement your algorithm, keeping these access rules in mind.
In our example, we read the user ID and number of copies. Then we increment the copy
account:
cin >> id >> copies;
*users[id] = *users[id] + copies;
Exercise P7.12 asks you to complete this simulation.
W o r k e D e x a M p l e 7 . 1 Producing a mass mailing
This Worked Example uses pointers to create a template for a mass mailing
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7.6 problem solving: Draw a picture 335
30. Suppose you want to sort the names of the items in the media center. There is
no need to move the strings. All you need to do is rearrange the string pointers.
Draw a picture that shows the result.
Practice it Now you can try these exercises at the end of the chapter: R7.25, R7.26, R7.27.
step 1 Draw a picture.
As described in Section 7.6, it is a good idea to draw a picture that shows the pointers in your
program.
In our example, we need to track the copy accounts: a master account and ten project
accounts. For each user, we need a pointer to the copy account:
step 2 Declare pointer variables.
This step is usually easy. With numerical data, you will have pointers of type int* or double*. If
you manipulate char acter arrays, you use char* pointers. In Chapter 10, you will use pointers
to objects.
How many pointer variables do you have? If you only have one or two, just declare vari-
ables such as
double* account_pointer;
If you have a sequence of pointers, use a vector or array, such as
vector
h o W t o 7 . 1 working with Pointers
You use pointers because you want flexibility in
your program: the ability to change the relationships
between data. This How To walks you through the
decision-making process for using pointers.
We will illustrate the steps with the task of simu-
lating a part of the control logic for a departmental
photocopier. A person making a copy enters a user
number. There are 100 different users, with numbers
0 to 99. Each user is linked to a copy account, either
the master account or one of ten project accounts.
That linkage is maintained by the administrator, and it
can change as users work on different projects. When
copies are made, the appropriate account should be
incremented.
Users identify themselves on the
copier control panel. Using pointers,
the relationships between users and
copy accounts can be flexible.
In our example, the purpose is to manipulate copy counters. Therefore, our data are integers
and the pointers have type int*. We know that we will have exactly 100 pointers, one for each
user. Therefore, we can choose an array
int* users[100];
step 3 Initialize the pointers with variable addresses or heap memory.
Will you allocate variables and then take their addresses with the & operator? That is fine if you
have a fixed amount of data that you want to reach through the pointers. Otherwise, use the
new operator to allocate memory dynami cally. Be sure to deallocate the memory when you are
done, using the delete or delete[] operator.
In our example, there is no need for dynamic allocation, so we will just take addresses of
variables.
int master_account;
int project_accounts[10];
for (int i = 0; i < 100; i++)
{
users[i] = &master_account;
}
// Here we reassign several users to project accounts.
// The following code is a simulation of the actions that would
// occur in an administration interface, which we do not implement.
users[2] = project_accounts + 1
users[3] = project_accounts + 2
users[99] = project_accounts + 2
step 4 Use the * or [] operator to access your data.
When you access a single variable through a pointer, use the * operator to read or write the
variable. For example,
*account_pointer = *account_pointer + 20;
When you have a pointer to an array, use the [] notation. Simply think of the pointer as the
array:
account_array[i] = account_array[i] + 20;
If you have an array or vector of pointers, then you need brackets to get at an individual
pointer. Then supply another * or [] to get at the value. You saw an example in the Galton
board simulator where a count was accessed as
counts[i][j]
Implement your algorithm, keeping these access rules in mind.
In our example, we read the user ID and number of copies. Then we increment the copy
account:
cin >> id >> copies;
*users[id] = *users[id] + copies;
Exercise P7.12 asks you to complete this simulation.
W o r k e D e x a M p l e 7 . 1 Producing a mass mailing
This Worked Example uses pointers to create a template for a mass mailing
Available online at www.wiley.com/college/horstmann.
cfe2_ch07_p307_350.indd 335 10/28/10 8:43 PM
www.wiley.com/college/horstmann
336 Chapter 7 pointers
7.7 structures and pointers (optional)
7.7.1 structures
In C++, you use a structure to aggregate items of arbitrary types into a single value.
For example, a street address is com posed of a house number and a street name. A
structure named StreetAddress can be defined to combine these two values into a sin-
gle entity. In C++, we define a structure with the struct reserved word:
struct StreetAddress
{
int house_number;
string street_name;
};
This definition yields a new type, StreetAddress, that you can use for declaring
variables:
StreetAddress white_house;
The variable white_house has two named parts, called members, house_number and
street_name. You use the “dot notation” to access each member, like this:
white_house.house_number = 1600;
white_house.street_name = “Pennsylvania Avenue”;
an embedded system
is a computer system
that controls a device. the device con
tains a processor and other hardware
and is controlled by a computer pro
gram. Unlike a personal computer,
which has been designed to be flexi
ble and run many different computer
programs, the hardware and software
of an embedded system are tailored
to a specific device. Computer con
trolled devices are becoming increas
ingly common, ranging from wash
ing machines to medical equipment,
cell phones, automobile engines, and
spacecraft.
several challenges are specific to
programming embedded systems.
Most importantly, a much higher stan
dard of quality control applies. Ven
dors are often unconcerned about
bugs in personal computer software,
because they can always make you
install a patch or upgrade to the next
version. But in an embedded system,
that is not an option. Few consumers
would feel comfortable upgrading the
software in their washing machines or
automobile engines. if you ever handed
in a programming assign ment that you
believed to be correct, only to have the
instructor or grader find bugs in it, then
you know how hard it is to write soft
ware that can reliably do its task for
many years without a chance of chang
ing it. Qual ity standards are especially
important in devices whose failure
would destroy property or endanger
human life. Many personal computer
purchas ers buy computers that are fast
and have a lot of storage, because the
investment is paid back over time when
many programs are run on the same
equipment. But the hardware for an
embedded device is not shared––it is
dedicated to one device. a separate pro
cessor, memory, and so on, are built for
every copy of the device. if it is possible
to shave a few pennies off the manufac
turing cost of every unit, the savings
can add up quickly for devices that are
produced in large volumes. thus, the
programmer of an embed dedsystem
has a much larger economic incentive
to conserve resources than the desktop
software programmer. Unfortunately,
try ing to conserve resources usually
makes it harder to write programs that
work correctly.
C and C++ are commonly used
lan guages for developing embedded
sys tems.
The Controller of an Embedded System
Random Fact 7.1 embedded systems
a structure combines
member values into a
single value.
You use the dot
notation to access
members of a
structure.
7.7.2 pointers to structures
It is common to allocate structure values dynamically, using the new operator:
StreetAddress* address_pointer = new StreetAddress;
Suppose you want to set the house number of the structure to which address_pointer
points:
*address_pointer.house_number = 1600; // Error
Unfortunately, that is a syntax error. The dot operator has a higher precedence than
the * operator. That is, the compiler thinks that you mean
*(address_pointer.house_number) = 1600; // Error
However, address_pointer is a pointer, not a structure. You can’t apply the dot (.) oper-
ator to a pointer, and the compiler reports an error. Instead, you must make it clear
that you first want to apply the * operator, then the dot:
(*address_pointer).house_number = 1600; // OK
Because this is such a common situation, the designers of C++ supply an operator to
abbreviate the “fol low pointer and access member” operation. That operator is writ-
ten -> and usually pronounced “arrow”.
address_pointer->house_number = 1600; // OK
7.7.3 structures with pointer Members
A member of a structure can be a pointer. This situation commonly arises when infor-
mation is shared among structure values. Consider this example. In an organization
with multiple offices, each employee has a name and an office location:
struct Employee
{
string name;
StreetAddress* office;
}
Here, we define two employees who both work for the accounting office:
StreetAddress accounting;
accounting.house_number = “1729”;
accounting.street_name = “Park Avenue”;
Employee harry;
harry.name = “Smith, Harry”;
harry.office = &accounting;
Use the -> operator
to access a structure
member through
a pointer.
cfe2_ch07_p307_350.indd 336 10/28/10 8:43 PM
7.7 structures and pointers (optional) 337
7.7.2 pointers to structures
It is common to allocate structure values dynamically, using the new operator:
StreetAddress* address_pointer = new StreetAddress;
Suppose you want to set the house number of the structure to which address_pointer
points:
*address_pointer.house_number = 1600; // Error
Unfortunately, that is a syntax error. The dot operator has a higher precedence than
the * operator. That is, the compiler thinks that you mean
*(address_pointer.house_number) = 1600; // Error
However, address_pointer is a pointer, not a structure. You can’t apply the dot (.) oper-
ator to a pointer, and the compiler reports an error. Instead, you must make it clear
that you first want to apply the * operator, then the dot:
(*address_pointer).house_number = 1600; // OK
Because this is such a common situation, the designers of C++ supply an operator to
abbreviate the “fol low pointer and access member” operation. That operator is writ-
ten -> and usually pronounced “arrow”.
address_pointer->house_number = 1600; // OK
7.7.3
figure 11 a pointer to a structure
address_pointer =
house_number =
StreetAddress
street_name =
1600
Penn. Ave.
structures with pointer Members
A member of a structure can be a pointer. This situation commonly arises when infor-
mation is shared among structure values. Consider this example. In an organization
with multiple offices, each employee has a name and an office location:
struct Employee
{
string name;
StreetAddress* office;
}
Here, we define two employees who both work for the accounting office:
StreetAddress accounting;
accounting.house_number = “1729”;
accounting.street_name = “Park Avenue”;
Employee harry;
harry.name = “Smith, Harry”;
harry.office = &accounting;
Use the -> operator
to access a structure
member through
a pointer.
cfe2_ch07_p307_350.indd 337 10/28/10 8:43 PM
338 Chapter 7 pointers
Employee sally;
sally.name = “Lee, Sally”;
sally.office = &accounting;
Figure 12 shows how these structures are related.
This sharing of information has an important benefit. Suppose the accounting
office moves across the street:
accounting.house_number = 1720;
Now both Harry’s and Sally’s office addresses are automatically updated.
31.
figure 12 two pointers to a shared structure
accounting =
house_number =
StreetAddress
street_name =
1729
Park Ave
harry =
name =
Employee
office =
Smith, Harry
sally =
name =
Employee
office =
Lee, Sally
Declare a variable of type StreetAddress and initialize it with the address of the
State Department: 2201 C Street NW.
32. Declare a pointer variable of type StreetAddress* and initialize it with a structure
containing the address of the State Department: 2201 C Street NW.
33. Define a structure type Date to describe dates such as “July 4” or “December 31”.
34. Using the structure type that you defined in Self Check 33, define a variable
independence_day and initialize it as July 4.
35. How do you print out Harry’s office address?
Practice it Now you can try these exercises at the end of the chapter: R7.23, R7.24, P7.16.
s e l f C h e C k
Define and use pointer variables.
• A pointer denotes the location of a variable in memory.
• The type T* denotes a pointer to a variable of type T.
• The & operator yields the location of a variable.
• The * operator accesses the variable to which a pointer
points.
• It is an error to use an uninitialized pointer.
• The NULL pointer does not point to any object.
Understand the relationship between arrays and pointers in C++.
• The name of an array variable is a pointer to the starting element of the array.
• Pointer arithmetic means adding an integer offset to an array pointer, yielding a
pointer that skips past the given number of elements.
• The array/pointer duality law states that a[n] is identical to *(a + n), where a is a
pointer into an array and n is an integer offset.
• When passing an array to a function, only the starting address is passed.
Use C++ string objects with functions that process character arrays.
• A value of type char denotes an individual character. Character literals are en-
closed in single quotes.
• A literal string (enclosed in double
quotes) is an array of char values
with a zero terminator.
• Many library functions use pointers
of type char*.
• The c_str member function yields a char* pointer from a string object.
• You can initialize C++ string variables with C strings.
• You can access characters in a C++ string object with the [] operator.
allocate and deallocate memory in programs whose memory requirements aren’t
known until run time.
• Use dynamic memory allocation if you do not know in advance how many val ues
you need.
• The new operator allocates memory from the heap.
• You must reclaim dynamically allocated objects
with the delete or delete[] operator.
• Using a dangling pointer (a pointer that points to
mem ory that has been deleted) is a serious
programming error.
• Every call to new should have a matching call to
delete.
C h a p t e r s U M M a r Y
[0]
[1]
[2]
[3]
[4]
[5]
‘H’
‘a’
‘r’
‘r’
‘y’
‘\0’
Points to ‘H’
char_pointer =
cfe2_ch07_p307_350.indd 338 10/28/10 8:43 PM
Chapter summary 339
Define and use pointer variables.
• A pointer denotes the location of a variable in memory.
• The type T* denotes a pointer to a variable of type T.
• The & operator yields the location of a variable.
• The * operator accesses the variable to which a pointer
points.
• It is an error to use an uninitialized pointer.
• The NULL pointer does not point to any object.
Understand the relationship between arrays and pointers in C++.
• The name of an array variable is a pointer to the starting element of the array.
• Pointer arithmetic means adding an integer offset to an array pointer, yielding a
pointer that skips past the given number of elements.
• The array/pointer duality law states that a[n] is identical to *(a + n), where a is a
pointer into an array and n is an integer offset.
• When passing an array to a function, only the starting address is passed.
Use C++ string objects with functions that process character arrays.
• A value of type char denotes an individual character. Character literals are en-
closed in single quotes.
• A literal string (enclosed in double
quotes) is an array of char values
with a zero terminator.
• Many library functions use pointers
of type char*.
• The c_str member function yields a char* pointer from a string object.
• You can initialize C++ string variables with C strings.
• You can access characters in a C++ string object with the [] operator.
allocate and deallocate memory in programs whose memory requirements aren’t
known until run time.
• Use dynamic memory allocation if you do not know in advance how many val ues
you need.
• The new operator allocates memory from the heap.
• You must reclaim dynamically allocated objects
with the delete or delete[] operator.
• Using a dangling pointer (a pointer that points to
mem ory that has been deleted) is a serious
programming error.
• Every call to new should have a matching call to
delete.
C h a p t e r s U M M a r Y
[0]
[1]
[2]
[3]
[4]
[5]
‘H’
‘a’
‘r’
‘r’
‘y’
‘\0’
Points to ‘H’
char_pointer =
cfe2_ch07_p307_350.indd 339 10/28/10 8:43 PM
340 Chapter 7 pointers
work with arrays and vectors of pointers.
Draw diagrams for visualizing pointers and the data to which they point.
• Draw the data that is being processed, then
draw the pointer variables. When drawing the
pointer arrows, illustrate a typical situation.
Use structures to aggregate data items and work with pointers to structures.
• A structure combines member values into a single value.
• You use the dot notation to access members of a structure.
• Use the -> operator to access a structure member through a pointer.
r7.1 Trace the following code. Assume that a and b are stored at 20300 and 20308. Your
trace table should have entries for a, b, and p.
double a = 1000;
double b = 2000;
double* p = &a;
*p = 3000;
p = &b;
a = *p * 2;
r7.2 Trace the following code. Assume that a and b are stored at 20300 and 20308. Your
trace table should have entries for a, b, p, and q.
double a = 1000;
double b = 2000;
double* p = &a;
double* q = &b;
*p = *q;
p = q;
*p = 3000;
r7.3 What does the following code print?
double a = 1000;
double b = 2000;
double* p = &a;
double* q = p;
b = *q;
p = &b;
a = *p + *q;
cout << a << " " << b << endl;
r7.4 Explain the mistakes in the following code. Not all lines contain mistakes. Each line
depends on the lines preceding it.
1 double a = 1000;
2 double* p = a;
3 int* p = &a;
4 double* q;
5 *q = 2000;
Microphone TX-10\0 Smith, Diane\0 Lee, Tim\0 Tape recorder\0 Mini DVI cable\0 …
items
r e V i e W e x e r C i s e s
cfe2_ch07_p307_350.indd 340 10/28/10 8:43 PM
review exercises 341
6 int* r = NULL;
7 *r = 3000;
r7.5 Suppose that a system allows the use of any string as a password, even the empty
string. However, when a user connects to the system for the first time, no password
has been assigned. Describe how you can use a string* variable and the NULL pointer
to distinguish unassigned passwords from empty ones.
r7.6 Given the definitions
double primes[] = { 2, 3, 5, 7, 11, 13 };
double* p = primes + 3;
draw a diagram that explains the meanings of the following expressions:
a. primes[1]
b. primes + 1
c. *(primes + 1)
d. p[1]
e. p + 1
r7.7 Suppose the array primes, defined as
double primes[] = { 2, 3, 5, 7, 11, 13 };
starts at memory location 20300. What are the values of
a. primes
b. *primes
c. primes + 4
d. *(primes + 4)
e. primes[4]
f. &primes[4]
r7.8 Suppose the array primes is defined as
double primes[] = { 2, 3, 5, 7, 11, 13 };
Consider the sum function discussed in Section 7.2.3. What are the values of
a. sum(primes, 6);
b. sum(primes, 4);
c. sum(primes + 2, 4);
d. sum(primes, 0);
e. sum(NULL, 4);
r7.9 Suppose the array primes, defined as
double primes[] = { 2, 3, 5, 7, 11, 13 };
starts at memory location 20300. Trace the function call sum(primes, 4), using the
definition of sum from Special Topic 7.2 on page 318. In your trace table, show the values
for a, size, total, p, and i.
r7.10 Pointers are addresses and have a numerical value. You can print out the value of a
pointer as cout << (unsigned)(p). Write a program to compare p, p + 1, q, and q + 1,
where p is an int* and q is a double*. Explain the results.
r7.11 A pointer variable can contain a pointer to a single variable, a pointer to an array,
NULL, or a random value. Write code that creates and sets four pointer variables a, b, c,
and d to show each of these possibilities.
cfe2_ch07_p307_350.indd 341 10/28/10 8:43 PM
342 Chapter 7 pointers
r7.12 Implement a function firstlast that obtains the first and last values in an array of
integers and stores the result in an array argument.
r7.13 Explain the meanings of the following expressions:
a. "Harry" + 1
b. *("Harry" + 2)
c. "Harry"[3]
d. [4]"Harry"
r7.14 What is the difference between the following two variable definitions?
a. char a[6] = "Hello";
b. char* b = "Hello";
r7.15 What is the difference between the following three variable definitions?
a. char* p = NULL;
b. char* q = "";
c. char r[1] = { '\0' };
r7.16 Consider this program segment:
char a[] = "Mary had a little lamb";
char* p = a;
int count = 0;
while (*p != '\0')
{
count++;
while (*p != ' ' && *p != '\0') { p++; }
while (*p == ' ') { p++; }
}
What is the value of count at the end of the outer while loop?
r7.17 Consider the following code that repeats a C++ string three times.
string a = "Hello";
string b = a + a + a;
Suppose s is a C string, and t is declared as
char t[100];
Write the equivalent code for C strings that stores the threefold repetition of s (or as
much of it as will fit) into t.
r7.18 Which of the following assignments are legal in C++?
void f(int p[])
{
int* q;
const int* r;
int s[10];
p = q; 1
p = r; 2
p = s; 3
q = p; 4
q = r; 5
q = s; 6
r = p; 7
cfe2_ch07_p307_350.indd 342 10/28/10 8:43 PM
review exercises 343
r = q; 8
r = s; 9
s = p; 10
s = q; 11
s = r; 12
}
r7.19 What happens if you forget to delete an object that you obtained from the heap?
What happens if you delete it twice?
r7.20 Write a program that accesses a deleted pointer, an uninitialized pointer, and a NULL
pointer. What happens when you run your program?
r7.21 Find the mistakes in the following code. Not all lines contain mistakes. Each line
depends on the lines preceding it. Watch out for uninitialized pointers, NULL pointers,
pointers to deleted objects, and confusing pointers with objects.
1 int* p = new int;
2 p = 5;
3 *p = *p + 5;
4 string s = "Harry";
5 *s = "Sally";
6 delete &s;
7 int* a = new int[10];
8 *a = 5;
9 a[10] = 5;
10 delete a;
11 int* q;
12 *q = 5;
13 q = p;
14 delete q;
15 delete p;
r7.22 How do you define a triangular two-dimensional array using just vectors, not arrays
or pointers?
r7.23 Rewrite the statements in Section 7.7.3 so that the street address and employee
structures are allocated on the heap.
r7.24 Design a structure type Person that contains the name of a person and pointers to the
person’s father and mother. Write statements that define a structure value for your-
self and your parents, correctly establishing the pointer links. (Use NULL for your
parents’ parents.)
r7.25 Draw a figure showing the result of a call to the maximum function in Exercise P7.4.
r7.26 Draw a figure showing the pointers in the lines array in Exercise P7.13 after reading
a few lines.
r7.27 Section 7.6 described an arrangement where each item had a pointer to the user who
had checked out the item. This makes it difficult to find out the items that a particu-
lar user checked out. To solve this problem, have an array of strings user_names and a
parallel array loaned_items. loaned_items[i] points to an array of char* pointers, each of
which is a name of an item that the ith user checked out. If the ith user didn’t check
out any items, then loaned_items[i] is NULL. Draw a picture of this arrangement.
cfe2_ch07_p307_350.indd 343 10/28/10 8:43 PM
344 Chapter 7 pointers
P7.1 Write a function
void sort2(double* p, double* p)
that receives two pointers and sorts the values to which they point. If you call
sort2(&x, &y)
then x <= y after the call.
P7.2 Write a function
double replace_if_greater(double* p, double x)
that replaces the value to which p points with x if x is greater. Return the old value to
which p pointed.
P7.3 Write a function that computes the average value of an array of floating-point data:
double average(double* a, int size)
In the function, use a pointer variable, not an integer index, to traverse the array
elements.
P7.4 Write a function that returns a pointer to the maximum value of an array of float ing-
point data:
double* maximum(double* a, int size)
If size is 0, return NULL.
P7.5 Write a function that reverses the values of an array of floating-point data:
void reverse(double* a, int size)
In the function, use two pointer variables, not integer indexes, to traverse the array
elements.
P7.6 Implement the strncpy function of the standard library.
P7.7 Implement the standard library function
int strspn(const char s[], const char t[])
that returns the length of the initial portion of s consisting of the characters in t (in
any order).
P7.8 Write a function
void reverse(char s[])
that reverses a character string. For example, "Harry" becomes "yrraH".
P7.9 Using the strncpy and strncat functions, implement a function
void safe_concat(const char a[], const char b[], char result[],
int result_maxlength)
that concatenates the strings a and b to the buffer result. Be sure not to overrun the
buffer. It can hold result_maxlength characters, not counting the '\0' terminator.
(That is, the buffer has result_maxlength + 1 bytes available.)
P7.10 Write a function int* read_data(int& size) that reads data from cin until the user
ter minates input by entering Q. The function should set the size reference parameter
to the number of numeric inputs. Return a pointer to an array on the heap. That
p r o G r a M M i n G e x e r C i s e s
cfe2_ch07_p307_350.indd 344 10/28/10 8:43 PM
programming exercises 345
array should have exactly size elements. Of course, you won’t know at the outset
how many elements the user will enter. Start with an array of 10 elements, and
double the size whenever the array fills up. At the end, allocate an array of the
correct size and copy all inputs into it. Be sure to delete any intermediate arrays.
P7.11 Enhance the Galton board simulation by printing a bar chart of the bottommost
counters. Draw the bars vertically, below the last row of numbers.
1 18 64 149 239 265 186 61 15 2
* * * * * * * *
* * * * * *
* * * * * *
* * * *
* * * *
* * * *
* * * *
* * *
* * *
* *
* *
* *
*
P7.12 Complete the copier simulation of How To 7.1 on page 334. Your program should first
show the main menu:
U)ser A)dministrator Q)uit
For a user, prompt for the ID and the number of copies, increment the appropriate
account, and return to the main menu.
For an administrator, show this menu:
B)alance M)aster P)roject
In the balance option, show the balances of the master account and the ten project
accounts. In the master option, prompt for a user ID and link it to the master
account. In the project option, prompt for user and project IDs. Afterward, return
to the main menu.
P7.13 Write a program that reads lines of text and appends them to a char buffer[1000]. Stop
after reading 1,000 characters. As you read in the text, replace all newline char acters
'\n' with '\0' terminators. Establish an array char* lines[100], so that the pointers in
that array point to the beginnings of the lines in the text. Consider only 100 input
lines if the input has more lines. Then display the lines in reverse order, starting with
the last input line.
P7.14 The program in Exercise P7.13 is limited by the fact that it can only handle inputs of
1,000 characters or 100 lines. Remove this limitation as follows. Concatenate the
input in one long string object. Use the c_str member function to obtain a char* into
the string’s character buffer. Store the beginnings of the lines as a vector
P7.15 Exercise P7.14 demonstrated how to use the string and vector classes to implement
resizable arrays. In this exercise, you should implement that capability manually.
Allocate a buffer of 1,000 characters from the heap (new char[1000]). Whenever the
buffer fills up, allocate a buffer of twice the size, copy the buffer contents, and delete
the old buffer. Do the same for the array of char* point ers—start with a new char*[100]
and keep doubling the size.
cfe2_ch07_p307_350.indd 345 10/28/10 8:43 PM
346 Chapter 7 pointers
P7.16 Modify Exercise P7.13 so that you first print the lines in the order that they were
entered, then print them in sorted order. When you sort the lines, only rearrange the
pointers in the lines array.
P7.17 When you read a long document, there is a good chance that many words occur
multiple times. Instead of storing each word, it may be beneficial to only store
unique words, and to represent the document as a vector of pointers to the unique
words. Write a program that implements this strategy. Read a word at a time from
cin. Keep a vector
allocate memory, copy the word into it, and append a pointer to the new memory. If
the word is already present, then append a pointer to the existing word.
P7.18 Define a structure Student with a name and a vector
structure Course with a name and a vector
Define a function void print_student(Student* s) that prints the name of a student
and the names of all courses that the student takes. Define a function void print_
course(Course* c) that prints the name of a course and the names of all students in
that course.
Define a function void enroll(Student* s, Course* c) that enrolls the given student in
the given course, updating both vectors.
In your main function, define several students and courses, and enroll students in the
courses. Then call print_student for all students and print_course for all courses.
engineering P7.19 Write a program that simulates a device that gathers measurements and processes
them.
A gather function gathers data values (which you should simulate with random
integers between 0 and 100) and places them in an array. When the array is full, the
gather function calls a function new_array to request a new array to fill.
A process function processes a data value (for this exer cise, it simply updates global
variables for computing the maxi mum, minimum, and average). When it has reached
the end of the array, it calls a function next_array to request a new array to process.
Because the gather function may fill arrays faster than the process function processes
them, store the pointers to the filled arrays in another array. The new_array and
next_array functions need to maintain that array of pointers.
In a real device, data gathering and processing happen in parallel. Simulate this by
calling the gather and pro cess functions randomly from main.
engineering P7.20 Write a program that simulates the control
software for a “people mover” system, a set of
driverless trains that move in two concentric
circular tracks. A set of switches allows trains
to switch tracks.
In your program, the outer and inner tracks
should each be divided into ten segments.
Each track segment can contain a train that
moves either clockwise or counterclockwise.
A train moves to an adjacent segment in its track or, if that segment is occupied, to
the adjacent segment in the other track.
Define a Segment structure. Each segment has a pointer to the next and previous
segments in its track, a pointer to the next and previous seg ments in the other track,
and a train indicator that is 0 (empty), +1 (train moving clockwise), or –1 (train
moving counterclockwise). Populate the system with four trains at random seg-
ments, two in each direction. Display the tracks and trains in each step, like this:
+————–>–+
| x x x x x |
| ———-<---- |
| | | |
| ->—<---------- |
| x x x x x |
+-----------------+
The two rectangles indicate the tracks. Each switch that allows a train to switch
between the outer and inner track is indicated by an x. Each train is drawn as a > or <,
indicating its current direction. Your program should show fifty rounds. In each
round, all trains move once.
1. 1
2
2. 2 2
3. 30
4. 25
5. 15 followed by the memory address of the variable a. The address can differ from one
program run to the next.
6. 2 0 5
7. 2 3 0
8. 2 3 5
9. 4 5
10. void squares(int n, int* result)
11. 15:
12. 14. The null terminator is not counted, and '\n' counts as a single character.
13. char* p = "Hello";
14. The statement compiles, but it has a disastrous effect. The type of "Agent" is a char*.
It is legal to add an integer to a char*. The result is a pointer that is seven charac-
ters away from the start of the array. The array only contains 6 characters, so the
result points to some other part of memory. The title string will be constructed
with whatever characters are found there, until a zero is encountered. The result is
unpredictable.
15. int number = atoi(input.c_str());
16. 3
17. 0 1 4
18. The access p[10] is illegal; the returned array has elements with index 0 ... 9.
19. p points to an array of size 10 with elements 2, 3, 5, 7, 11, 0, 0, 0, 0, 0.
20. Call delete[] on the returned pointer after it is no longer needed.
a n s W e r s t o s e l F C h e C k Q U e s t i o n s
cfe2_ch07_p307_350.indd 346 10/28/10 8:43 PM
answers to selfCheck Questions 347
and a train indicator that is 0 (empty), +1 (train moving clockwise), or –1 (train
moving counterclockwise). Populate the system with four trains at random seg-
ments, two in each direction. Display the tracks and trains in each step, like this:
+-------------->–+
| x x x x x |
| ———-<---- |
| | | |
| ->—<---------- |
| x x x x x |
+-----------------+
The two rectangles indicate the tracks. Each switch that allows a train to switch
between the outer and inner track is indicated by an x. Each train is drawn as a > or <,
indicating its current direction. Your program should show fifty rounds. In each
round, all trains move once.
1. 1
2
2. 2 2
3. 30
4. 25
5. 15 followed by the memory address of the variable a. The address can differ from one
program run to the next.
6. 2 0 5
7. 2 3 0
8. 2 3 5
9. 4 5
10. void squares(int n, int* result)
11. 15: 'H''e''l''l''o'','' ''W''o''r''l''d''!''\n''\0'
12. 14. The null terminator is not counted, and '\n' counts as a single character.
13. char* p = "Hello";
14. The statement compiles, but it has a disastrous effect. The type of "Agent" is a char*.
It is legal to add an integer to a char*. The result is a pointer that is seven charac-
ters away from the start of the array. The array only contains 6 characters, so the
result points to some other part of memory. The title string will be constructed
with whatever characters are found there, until a zero is encountered. The result is
unpredictable.
15. int number = atoi(input.c_str());
16. 3
17. 0 1 4
18. The access p[10] is illegal; the returned array has elements with index 0 ... 9.
19. p points to an array of size 10 with elements 2, 3, 5, 7, 11, 0, 0, 0, 0, 0.
20. Call delete[] on the returned pointer after it is no longer needed.
a n s W e r s t o s e l F C h e C k Q U e s t i o n s
cfe2_ch07_p307_350.indd 347 10/28/10 8:43 PM
348 Chapter 7 pointers
21. Then the first row would have had 0 elements and the last row would have had 9
elements.
22. The program will compile. counts[1] is an int* pointer, and it is legal to apply [2].
counts[1][2] will access the memory address counts[1] + 2, which unfortunately
points to some other memory. That memory will be overwritten, and it is not possi-
ble to predict the effect.
23. int* triangular_array[10];
for (int i = 0; i < 10; i++)
{
triangular_array[i] = new int[10 - i];
}
24. Include the
counts(10). The remainder of the program need not change.
25. Change the declaration of counts to vector
loop body to
counts[i] = vector
Drop the deallocation loop.
26.
27.
data = 0
1
4
9
16
25
36
49
64
81
s =
20300
20324
a = 20324
size = 7
data = 81
64
4
9
16
25
36
49
1
0
front =
back =
cfe2_ch07_p307_350.indd 348 10/28/10 8:43 PM
answers to selfCheck Questions 349
28.
29.
30.
31. StreetAddress state_dept;
state_dept.house_number = 2201;
state_dept.street_name = “C Street NW”;
32. StreetAddress* state_dept = new StreetAddress;
state_dept->house_number = 2201;
state_dept->street_name = “C Street NW”;
33. struct Date
{
string month; // It is also OK to define month as an int
int day;
};
34. Date independence_day;
independence_day.month = “July”;
independence_day.day = 4;
35. cout << harry.office->house_number << " " << harry.office->street_name;
Note that you must use arrows because office is a pointer.
c a r g o \0
first =
c a n o l a \0
second =
counts =
×
×
×
×
×
Microphone TX-10\0 Smith, Diane\0 Lee, Tim\0 Tape recorder\0 Mini DVI cable\0 …
items
cfe2_ch07_p307_350.indd 349 10/28/10 8:43 PM
This page intentionally left blank
8C h a p t e r
351
S t r e a m S
to be able to read and write files
to convert between strings and numbers
using string streams
to process command line arguments
to understand the concepts of sequential and random access
C h a p t e r G o a l S
C h a p t e r C o n t e n t S
8.1 Reading and WRiting
text Files 352
Syntax 8.1: Working with File Streams 354
8.2 Reading text input 358
8.3 WRiting text Output 361
8.4 stRing stReams 363
8.5 COmmand line aRguments 365
Random Fact 8.1: encryption algorithms 368
How To 8.1: processing text Files 369
Worked Example 8.1: looking for for Duplicates
8.6 RandOm aCCess and
BinaRy Files 372
Random Fact 8.2: Databases and privacy 377
cfe2_ch08_p351_388.indd 351 10/28/10 8:56 PM
352
In this chapter, you will learn how to read and write files
using the C++ stream library—a very useful skill for
processing real world data. as an application, you will learn
how to encrypt data. (the enigma machine shown at left
is an encryption device used by Germany in World War II.
pioneering British computer scientists broke the code and
were able to intercept encoded messages, which was a
significant help in winning the war.) later in the chapter,
you will learn to process binary files, such as those that
store image data.
8.1 reading and Writing text Files
The C++ input/output library is based on the concept
of streams. An input stream is a source of data, and an
output stream is a destination for data. The most com-
mon sources and destinations for data are the files on
your hard disk.
Data arrive in an input stream just like
items on a conveyor belt, one at a time.
To access a file, you use a file stream. There are three types of file streams: ifstream
(for input), ofstream (for output), and fstream (for both input and output). Include the
In the following sections, you will learn how to process data from files. File pro-
cessing is a very useful skill in many disciplines because it is exceedingly common to
analyze large data sets stored in files.
8.1.1 opening a Stream
To read anything from a file stream, you need to open it. When you open a stream,
you give the name of the file stored on disk. Suppose you want to read data from a file
named input.dat, located in the same directory as the program. Then you use the fol-
lowing function call to open the file:
in_file.open(“input.dat”);
This statement associates the variable in_file with the file named input.dat.
Note that all streams are objects, and you use the dot notation for calling functions
that manipulate them.
To open a file for writing, you use an ofstream variable. To open the same file for
both reading and writ ing, you use an fstream variable.
to read or write files,
you use variables of
type fstream,
ifstream, or
ofstream.
When opening a file
stream, you supply
the name of the
file stored on disk.
File names can contain directory path information, such as
~/homework/input.dat (UNIX)
c:\homework\input.dat (Windows)
When you specify the file name as a string literal, and the name contains backslash
characters (as in a Win dows filename), you must supply each backslash twice:
in_file.open(“c:\\homework\\input.dat”);
Recall that a single backslash inside a string literal is an escape character that is com-
bined with another character to form a special meaning, such as \n for a newline char-
acter. The \\ combination denotes a sin gle backslash.
If you want to pass a name that is stored in a string variable, use the c_str function
to convert the C++ string to a C string:
cout << "Please enter the file name:";
string filename;
cin >> filename;
ifstream in_file;
in_file.open(filename.c_str());
When the program ends, all streams that you have opened will be automatically
closed. You can also manually close a stream with the close member function:
in_file.close();
Manual closing is only necessary if you want to use the stream variable again to pro-
cess another file.
8.1.2 reading from a File
Reading data from a file stream is completely straightforward: You simply use the
same functions that you have always used for reading from cin:
string name;
double value;
in_file >> name >> value;
The fail function tells you whether input has failed. You have already used this func-
tion with cin, to check for errors in console input. File streams behave in the same
way. When you try to read a number from a file, and the next data item is not a prop-
erly formatted number, then the stream fails. After reading data, you should test for
success before processing:
if (!in_file.fail())
{
Process input.
}
Alternatively, you can use the fact that the >> operator returns a “not failed” condi-
tion, allowing you to combine an input statement and a test:
if (in_file >> name >> value)
{
Process input.
}
When you read input from a file, number format errors are not the only reason for
failure. Suppose you have consumed all of the data contained in a file and try to read
more items. A file stream enters the failed state, whereas cin would just wait for more
read from a file
stream with the same
operations that you
use with cin.
cfe2_ch08_p351_388.indd 352 10/28/10 8:56 PM
8.1 reading and Writing text Files 353
File names can contain directory path information, such as
~/homework/input.dat (UNIX)
c:\homework\input.dat (Windows)
When you specify the file name as a string literal, and the name contains backslash
characters (as in a Win dows filename), you must supply each backslash twice:
in_file.open(“c:\\homework\\input.dat”);
Recall that a single backslash inside a string literal is an escape character that is com-
bined with another character to form a special meaning, such as \n for a newline char-
acter. The \\ combination denotes a sin gle backslash.
If you want to pass a name that is stored in a string variable, use the c_str function
to convert the C++ string to a C string:
cout << "Please enter the file name:";
string filename;
cin >> filename;
ifstream in_file;
in_file.open(filename.c_str());
When the program ends, all streams that you have opened will be automatically
closed. You can also manually close a stream with the close member function:
in_file.close();
Manual closing is only necessary if you want to use the stream variable again to pro-
cess another file.
8.1.2 reading from a File
Reading data from a file stream is completely straightforward: You simply use the
same functions that you have always used for reading from cin:
string name;
double value;
in_file >> name >> value;
The fail function tells you whether input has failed. You have already used this func-
tion with cin, to check for errors in console input. File streams behave in the same
way. When you try to read a number from a file, and the next data item is not a prop-
erly formatted number, then the stream fails. After reading data, you should test for
success before processing:
if (!in_file.fail())
{
Process input.
}
Alternatively, you can use the fact that the >> operator returns a “not failed” condi-
tion, allowing you to combine an input statement and a test:
if (in_file >> name >> value)
{
Process input.
}
When you read input from a file, number format errors are not the only reason for
failure. Suppose you have consumed all of the data contained in a file and try to read
more items. A file stream enters the failed state, whereas cin would just wait for more
read from a file
stream with the same
operations that you
use with cin.
cfe2_ch08_p351_388.indd 353 10/28/10 8:56 PM
354 Chapter 8 Streams
Syntax 8.1 Working with File Streams
#include
ifstream in_file;
in_file.open(filename.c_str());
in_file >> name >> value;
ofstream out_file;
out_file.open(“c:\\output.txt”);
out_file << name << " " << value << endl;
Use ifstream for input,
ofstream for output,
fstream for both input
and output.
Include this header
when you use file streams. Call c_str
if the file name is
a C++ string.
Use \\ for
each backslash
in a string literal.
Use the same operations
as with cout.
Use the same operations
as with cin.
user input. Moreover, if you open a file and the name is invalid, or if there is no file of
that name, then the file stream is also in a failed state. It is a good idea to test for failure
immediately after calling open.
8.1.3 Writing to a File
In order to write to a file, you define an ofstream or fstream variable and open it. Then
you send informa tion to the output file, using the same operations that you used with
cout:
ofstream out_file;
out_file.open("output.txt");
out_file << name << " " << value << endl;
8.1.4 a File processing example
Here is a typical example of processing data in a file. The Social Secu rity Administra-
tion publishes lists of the most popular baby names on their web site, http://www.ssa.
gov/OACT/babynames/. If you query the 1,000 most popular names for a given decade, the
browser displays the result on the screen (see Figure 1).
To save the data as text, simply select it and paste the result into a file. This book’s
companion code contains a file called babynames.txt with the data for the 1990s.
Each line in the file contains seven entries:
• The rank (from 1 to 1,000)
• The name, frequency, and percentage of the male name of that rank
• The name, frequency, and percentage of the female name of that rank
For example, the line
10 Joseph 260365 1.2681 Megan 160312 0.8168
Write to a file stream
with the same
operations that you
use with cout.
cfe2_ch08_p351_388.indd 354 10/28/10 8:56 PM
http://www.ssa.gov/OACT/babynames/
http://www.ssa.gov/OACT/babynames/
8.1 reading and Writing text Files 355
shows that the 10th most common boy’s name was Joseph, with 260,365 births, or
1.2681 percent of all births during that period. The 10th most common girl’s name
was Megan. Why are there many more Josephs than Megans? Parents seem to use a
wider set of girl’s names, making each one of them less fre quent.
Let us test that conjecture, by determining
the names given to the top 50 percent of boys
and girls in the list.
To process each line, we first read the rank:
int rank;
in_file >> rank;
We then read a set of three values for the boy’s
name:
string name;
int count;
double percent;
in_file >> name >> count >> percent;
Then we repeat that step for girls. Because the
actions are identical, we supply a helper func-
tion process_name for that purpose. To stop pro-
cessing after reaching 50 percent, we can add up the frequencies and stop when they
reach 50 percent. However, it turns out to be a bit simpler to initialize a total with 50
and subtract the frequencies. We need separate totals for boys and girls. When a total
falls below 0, we stop printing. When both totals fall below 0, we stop reading.
Sellers of personalized items can find
trends in popular names by process-
ing data files from the Social Security
Administration.
Figure 1 Querying Baby names
cfe2_ch08_p351_388.indd 355 10/28/10 8:56 PM
356 Chapter 8 Streams
Note that the in_file parameter variable of the process_name function in the code
below is a reference parameter. Reading or writing modifies a stream variable. The
stream variable monitors how many characters have been read or written so far. Any
read or write operation changes that data. For that reason, you must always make
stream parameter variables reference parameters.
The complete program is shown below. As you can see, reading from a file is just as
easy as reading keyboard input.
Have a look at the program output. Remarkably, only 69 boy names and 153 girl
names account for half of all births. That’s good news for those who are in the busi-
ness of producing personalized doodads. Exercise P8.10 asks you to study how this
distribution has changed over the years.
ch08/babynames.cpp
1 #include
2 #include
3 #include
4
5 using namespace std;
6
7 /**
8 Reads name information, prints the name if total >= 0, and adjusts the total.
9 @param in_file the input stream
10 @param total the total percentage that should still be processed
11 */
12 void process_name(ifstream& in_file, double& total)
13 {
14 string name;
15 int count;
16 double percent;
17 in_file >> name >> count >> percent;
18
19 if (in_file.fail()) { return; } // Check for failure after each input
20 if (total > 0) { cout << name << " "; }
21 total = total - percent;
22 }
23
24 int main()
25 {
26 ifstream in_file;
27 in_file.open("babynames.txt");
28 if (in_file.fail()) { return 0; } // Check for failure after opening
29
30 double boy_total = 50;
31 double girl_total = 50;
32
33 while (boy_total > 0 || girl_total > 0)
34 {
35 int rank;
36 in_file >> rank;
37 if (in_file.fail()) { return 0; }
38
39 cout << rank << " ";
40
41 process_name(in_file, boy_total);
42 process_name(in_file, girl_total);
43
always use a
reference parameter
for a stream.
cfe2_ch08_p351_388.indd 356 10/28/10 8:56 PM
8.1 reading and Writing text Files 357
44 cout << endl;
45 }
46
47 return 0;
48 }
program Run
1 Michael Jessica
2 Christopher Ashley
3 Matthew Emily
4 Joshua Sarah
5 Jacob Samantha
6 Nicholas Amanda
7 Andrew Brittany
8 Daniel Elizabeth
9 Tyler Taylor
10 Joseph Megan
...
68 Dustin Gabrielle
69 Noah Katie
70 Caitlin
71 Lindsey
...
150 Hayley
151 Rebekah
152 Jocelyn
153 Cassidy
1. What happens if you call in_file.open("")?
2. What is wrong with the following code?
ifstream out_file;
out_file.open("output.txt");
out_file << "Hello, World!" << endl;
3. What is wrong with the following function?
double sum(ifstream in)
{
double total = 0;
double input;
while (in >> input) { total = total + input; }
return total;
}
4. How do you modify the babynames.cpp program so that you get the most com-
mon names that make up 10 percent of the population?
5. How do you modify the babynames.cpp program so that the program output is
saved to a file 00?
practice it Now you can try these exercises at the end of the chapter: R8.3, R8.6, P8.1.
s e l F C h e C k
cfe2_ch08_p351_388.indd 357 10/28/10 8:56 PM
358 Chapter 8 Streams
8.2 reading text Input
In the following sections, you will learn how to process text with complex contents
such as that which often occurs in real-life situations.
8.2.1 reading Words
You already know how to read the next word from a stream, using the >> operator.
string word;
in_file >> word;
Here is precisely what happens when that operation is executed. First, any input
characters that are white space are removed from the stream, but they are not added
to the word. White space includes spaces, tab characters, and the newline characters
that separate lines. The first character that is not white space becomes the first charac-
ter in the string word. More characters are added until either another white space char-
acter occurs, or the end of the file has been reached. The white space after the word is
not removed from the stream.
8.2.2 reading Characters
Instead of reading an entire word, you can read one character at a time by calling the
get function:
char ch;
in_file.get(ch);
The get function returns the “not failed” condition. The following loop processes all
characters in a file:
while (in_file.get(ch))
{
Process the character ch.
}
The get function reads white space characters. This is useful if you need to process
characters such as spaces, tabs, or newlines. On the other hand, if you are not inter-
ested in white space, use the >> operator instead.
in_file >> ch; // ch is set to the next non-white space character
If you read a character and you regretted it, you can unget it, so that the next input
operation can read it again. However, you can unget only the last character. This is
called one-character lookahead. You get a chance to look at the next character in the
input stream, and you can make a decision whether you want to consume it or put
it back.
A typical situation for lookahead is to look for numbers:
char ch;
in_file.get(ch);
if (isdigit(ch))
{
in_file.unget(); // Put the digit back so that it is part of the number
int n;
data >> n; // Read integer starting with ch
}
When reading a
string with the >>
operator, the white
space between words
is consumed.
You can get
individual characters
from a stream and
unget the last one.
The isdigit function is one of several useful functions that categorize characters—see
Table 1. All return true or false as to whether the argument passes the test. You must
include the
8.2.3 reading lines
When each line of a file is a data record, it is often best to read entire lines with the
getline function:
string line;
getline(in_file, line);
The next input line (without the newline character) is placed into the string line.
The getline function returns the “not failed” condition. You can use the following
loop to process each line in a file:
while (getline(in_file, line))
{
Process line.
}
Note that getline is not a member function, but an ordinary function that is not called
with the dot nota tion.
You can read a line
of input with the
getline function
and then process
it further.
table 1 Character Functions in
Function accepted Characters
isdigit 0 … 9
isalpha a … z, A … Z
islower a … z
isupper A … Z
isalnum a … z, A … Z, 0 … 9
isspace White space (space, tab, newline, and the rarely
used carriage return, form feed, and vertical tab)
cfe2_ch08_p351_388.indd 358 10/28/10 8:56 PM
8.2 reading text Input 359
If you read a character from a stream
and you don’t like what you get,
you can unget it.
The isdigit function is one of several useful functions that categorize characters—see
Table 1. All return true or false as to whether the argument passes the test. You must
include the
8.2.3 reading lines
When each line of a file is a data record, it is often best to read entire lines with the
getline function:
string line;
getline(in_file, line);
The next input line (without the newline character) is placed into the string line.
The getline function returns the “not failed” condition. You can use the following
loop to process each line in a file:
while (getline(in_file, line))
{
Process line.
}
Note that getline is not a member function, but an ordinary function that is not called
with the dot nota tion.
You can read a line
of input with the
getline function
and then process
it further.
table 1 Character Functions in
Function accepted Characters
isdigit 0 … 9
isalpha a … z, A … Z
islower a … z
isupper A … Z
isalnum a … z, A … Z, 0 … 9
isspace White space (space, tab, newline, and the rarely
used carriage return, form feed, and vertical tab)
cfe2_ch08_p351_388.indd 359 10/28/10 8:56 PM
360 Chapter 8 Streams
Here is a typical example of processing lines in a file. A file with population data
from the CIA World Factbook site (https://www.cia.gov/library/publications/the-
world-factbook/index.html) contains lines such as the follow ing:
China 1330044605
India 1147995898
United States 303824646
…
Because each line is a data record, it is natural to use the getline function for reading
lines into a string variable. To extract the data from that string, you need to find out
where the name ends and the number starts.
Locate the first digit:
int i = 0;
while (!isdigit(line[i])) { i++; }
Then go backward and skip white space:
int j = i – 1;
while (isspace(line[j])) { j–; }
setatSdetinU 3 0 3 8 2 4 6 4 6
i starts here i ends here
j ends here j starts here
length j + 1
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
Finally, extract the country name and population:
string country_name = line.substr(0, j + 1);
string population = line.substr(i);
There is just one problem. The population is stored in a string, not a number. You will
see in Section 8.4 how to extract the population number using streams.
6. Suppose the input stream contains the characters 6,995.0. What is the value of
number and ch after these statements?
int number;
char ch;
in_file >> number;
in_file.get(ch);
7. Suppose the input stream contains the characters Hello, World!. What is the value
of word and ch after these statements?
string word;
char ch;
in_file >> word >> ch;
if (isupper(ch)) { in_file.unget(); in_file >> word; }
8. Your input file contains a sequence of numbers, but sometimes a value is not
available and marked as N/A. How can you read the numbers and skip over the
markers?
s e l F C h e C k
9. What is the effect of the following loop?
char ch;
while (in_file.get(ch) && isspace(ch)) { }
in_file.unget();
10. Why can’t you simply read the population data file with the following loop?
while (in_file >> country_name >> population)
{
Process country name and population.
}
practice it Now you can try these exercises at the end of the chapter: P8.6, P8.7, P8.9.
8.3 Writing text output
You use the >> operator to send strings and numbers to a stream. To write a single
character to a stream, use
out_file.put(ch);
To control how the output is formatted, you use stream manipulators. A manipulator
is a value that affects the behavior of the stream. It is sent to a stream using the <<
operator. The setw manipulator, which you have already used, is a typical example.
The statement
out_file << setw(10);
does not cause any immediate output, but when the next item is written, it is padded
with sufficient spaces so that the output spans ten characters. (If a value does not fit
into the given width, it is not trun cated.)
Occasionally, you need to pad numbers with leading zeroes, for example to print
hours and minutes as 09:01. This is achieved with the setfill manipulator:
out_file << setfill('0') << setw(2) << hours
<< ":" << setw(2) << minutes << setfill(' ');
Now, a zero is used to pad the field. Afterward, the space is restored as the fill
character.
By default, the fill characters appear before the item:
out_file << setw(10) << 123 << endl << setw(10) << 4567;
produces
123
4567
Use the setw
manipulator to
set the width of
the next output.
A manipulator is like a control button on
a sound mixer. It doesn’t produce an output,
but it affects how the output looks.
cfe2_ch08_p351_388.indd 360 10/28/10 8:56 PM
https://www.cia.gov/library/publications/theworld-factbook/index.html
https://www.cia.gov/library/publications/theworld-factbook/index.html
8.3 Writing text output 361
9. What is the effect of the following loop?
char ch;
while (in_file.get(ch) && isspace(ch)) { }
in_file.unget();
10. Why can’t you simply read the population data file with the following loop?
while (in_file >> country_name >> population)
{
Process country name and population.
}
practice it Now you can try these exercises at the end of the chapter: P8.6, P8.7, P8.9.
8.3 Writing text output
You use the >> operator to send strings and numbers to a stream. To write a single
character to a stream, use
out_file.put(ch);
To control how the output is formatted, you use stream manipulators. A manipulator
is a value that affects the behavior of the stream. It is sent to a stream using the <<
operator. The setw manipulator, which you have already used, is a typical example.
The statement
out_file << setw(10);
does not cause any immediate output, but when the next item is written, it is padded
with sufficient spaces so that the output spans ten characters. (If a value does not fit
into the given width, it is not trun cated.)
Occasionally, you need to pad numbers with leading zeroes, for example to print
hours and minutes as 09:01. This is achieved with the setfill manipulator:
out_file << setfill('0') << setw(2) << hours
<< ":" << setw(2) << minutes << setfill(' ');
Now, a zero is used to pad the field. Afterward, the space is restored as the fill
character.
By default, the fill characters appear before the item:
out_file << setw(10) << 123 << endl << setw(10) << 4567;
produces
123
4567
Use the setw
manipulator to
set the width of
the next output.
A manipulator is like a control button on
a sound mixer. It doesn’t produce an output,
but it affects how the output looks.
cfe2_ch08_p351_388.indd 361 10/28/10 8:56 PM
362 Chapter 8 Streams
The numbers line up to the right. That alignment works well for numbers, but not for
strings. Usually, you want strings to line up at the left. You use the left and right
manipulators to set the alignment. The following example uses left alignment for a
string and then switches back to right alignment for a number:
out_file << left << setw(10) << word << right << setw(10) << number;
The default format for floating-point numbers is called general format. That format
displays as many dig its as are specified by the precision (6 by default), switching to
scientific notation for large and small num bers. For example,
out_file << 12.3456789 << " " << 123456789.0;
yields
12.3457 1.23457e+08
The fixed format prints all values with the same number of digits after the decimal
point. In the fixed for mat, the same numbers are displayed as
12.345679 123456789.000000
Use the fixed manipulator to select that format, and the setprecision manipulator to
change the precision.
For example,
out_file << fixed << setprecision(2) << 1.2 << " " << 1.235
yields
1.20 1.24
Use the fixed and
setprecision
manipulators to
format floating-point
numbers with a
fixed number of
digits after the
decimal point.
table 2 Stream manipulators
manipulator purpose example output
setw Sets the field width of
the next item only.
out_file << setw(6) << 123 << endl
<< 123 << endl
<< setw(6) << 12345678;
123
123
12345678
setfill Sets the fill character
for padding a field.
(The default character is
a space.)
out_file << setfill('0') << setw(6)
<< 123;
000123
left Selects left alignment. out_file << left << setw(6) << 123; 123
right Selects right alignment
(default).
out_file << right << setw(6) << 123; 123
fixed Selects fixed format for
floating-point numbers.
double x = 123.4567;
out_file << x << endl << fixed << x;
123.457
123.456700
setprecision Sets the number of
significant digits for
general format, the
number of digits after
the decimal point for
fixed format.
double x = 123.4567;
out_file << fixed << x << endl
<< setprecision(2) << x;
123.456700
123.46
cfe2_ch08_p351_388.indd 362 10/28/10 8:56 PM
8.4 String Streams 363
Table 2 summarizes the stream manipulators. Note that all manipulators set the state
of the stream object for all subsequent operations, with the exception of setw. After
each output operation, the field width is reset to 0. To use any of these manipulators,
include the
11. What is the output of the following statement?
cout << fixed << 123456.0;
12. How do you change the statement so that the result is 123456.00?
13. What is the output of the following statement?
cout << setw(8) << setfill('0') << 123456;
14. What is the output of the following statement?
cout << setw(2) << setfill('0') << 123456;
15. Why doesn’t the following statement sequence line up the numbers?
cout << setw(10) << setprecision(2) << fixed << 1.234 << endl << 56.7;
practice it Now you can try these exercises at the end of the chapter: R8.7, R8.8, P8.18.
8.4 String Streams
In the preceding sections, you saw how file streams read characters from a file and
write characters to a file. The istringstream class reads characters from a string, and
the ostringstream class writes characters to a string. That doesn’t sound so exciting—
we already know how to access and change the characters of a string. However, the
string stream classes have the same interface as the other stream classes. In particu lar,
you can use the familiar >> and << operators to read and write numbers that are con-
tained in strings. For that reason, the istringstream and ostringstream classes are called
adapters—they adapt strings to the stream interface. Include the
when you use string streams.
Here is a typical example. Suppose the string date contains a date such as “January
24, 1973”, and we want to separate it into month, day, and year. First, construct an
istringstream object. Then use the str function to set the stream to the string that you
want to read:
istringstream strm;
strm.str(“January 24, 1973”);
s e l F C h e C k
Like an adapter that converts your power plugs to
international outlets, the string stream adapters
allow you to access strings as streams.
cfe2_ch08_p351_388.indd 363 10/28/10 8:56 PM
364 Chapter 8 Streams
Next, simply use the >> operator to read the month name, the day, the comma separa-
tor, and the year:
string month;
int day;
string comma;
int year;
strm >> month >> day >> comma >> year;
Now month is “January”, day is 24, and year is 1973. Note that this input statement yields
day and year as inte gers. Had we taken the string apart with substr, we would have
obtained only strings, not numbers.
In fact, converting strings that contain digits to their integer values is such a com-
mon operation that it is useful to write a helper function for this purpose:
int string_to_int(string s)
{
istringstream strm;
strm.str(s);
int n = 0;
strm >> n;
return n;
}
For example, string_to_int(“1973”) is the integer 1973.
By writing to a string stream, you can convert integers or floating-point numbers
to strings. First con struct an ostringstream object:
ostringstream strm;
Next, use the << operator to add a number to the stream. The number is converted
into a sequence of characters:
strm << fixed << setprecision(5) << 10.0 / 3;
Now the stream contains the string "3.33333". To obtain that string from the stream,
call the str member function:
string output = strm.str();
You can build up more complex strings in the same way. Here we build a data string
of the month, day, and year:
string month = "January";
int day = 24;
int year = 1973;
ostringstream strm;
strm << month << " " << day << ", " << year;
string output = strm.str();
Now output is the string "January 24, 1973". Note that we converted the integers day
and year into a string. Again, converting an integer into a string is such a common
operation that is useful to have a helper func tion for it:
string int_to_string(int n)
{
ostringstream strm;
strm << n;
return strm.str();
}
For example, int_to_string(1973) is the string "1973".
Use an istringstream
to convert the
numbers inside a
string to integers or
floating-point
numbers.
Use an ostringstream
to convert numeric
values to strings.
16. What is the value of n after these statements?
istrstream strm;
strm.str("123,456");
int n;
int m;
char ch;
strm >> n >> ch >> m;
n = 1000 * n + m;
17. What does string_to_int(“123,456”) return?
18. What does string_to_int(“$123”) return?
19. What is the value of strm.str() after these statements?
ostringstream strm;
strm << setprecision(5) << 10.0 / 3;
20. Write a function that converts a floating-point number to a string. Provide a
parameter variable for the number of digits after the decimal point.
practice it Now you can try these exercises at the end of the chapter: R8.9, P8.10, P8.13.
8.5 Command line arguments
Depending on the operating system and C++ development environment used, there
are different methods of starting a program—for example, by selecting “Run” in the
compilation environment, by clicking on an icon, or by typing the name of the pro-
gram at a prompt in a command shell window. The latter method is called “invoking
the program from the command line”. When you use this method, you must type the
name of the program, of course, but you can also type in additional information that
the program can use. These additional strings are called command line arguments.
For example, if you start a program with the command line
prog -v input.dat
then the program receives two command line arguments: the strings "-v" and "input.
dat". It is entirely up to the program what to do with these strings. It is customary to
interpret strings starting with a hyphen (-) as options and other strings as file names.
To receive command line arguments, you need to define the main function in a
different way. You define two parameter variables: an integer and an array of string
literals of type char*.
int main(int argc, char* argv[])
{
...
}
Here argc is the count of arguments, and argv contains the values of the arguments. In
our example, argc is 3, and argv contains the three strings
argv[0]: "prog"
argv[1]: "-v"
argv[2]: "input.dat"
Note that argv[0] is always the name of the program and that argc is always at least 1.
s e l F C h e C k
programs that start
from the command
line can receive the
name of the program
and the command
line arguments in the
main function.
cfe2_ch08_p351_388.indd 364 10/28/10 8:56 PM
8.5 Command line arguments 365
16. What is the value of n after these statements?
istrstream strm;
strm.str("123,456");
int n;
int m;
char ch;
strm >> n >> ch >> m;
n = 1000 * n + m;
17. What does string_to_int(“123,456”) return?
18. What does string_to_int(“$123”) return?
19. What is the value of strm.str() after these statements?
ostringstream strm;
strm << setprecision(5) << 10.0 / 3;
20. Write a function that converts a floating-point number to a string. Provide a
parameter variable for the number of digits after the decimal point.
practice it Now you can try these exercises at the end of the chapter: R8.9, P8.10, P8.13.
8.5 Command line arguments
Depending on the operating system and C++ development environment used, there
are different methods of starting a program—for example, by selecting “Run” in the
compilation environment, by clicking on an icon, or by typing the name of the pro-
gram at a prompt in a command shell window. The latter method is called “invoking
the program from the command line”. When you use this method, you must type the
name of the program, of course, but you can also type in additional information that
the program can use. These additional strings are called command line arguments.
For example, if you start a program with the command line
prog -v input.dat
then the program receives two command line arguments: the strings "-v" and "input.
dat". It is entirely up to the program what to do with these strings. It is customary to
interpret strings starting with a hyphen (-) as options and other strings as file names.
To receive command line arguments, you need to define the main function in a
different way. You define two parameter variables: an integer and an array of string
literals of type char*.
int main(int argc, char* argv[])
{
...
}
Here argc is the count of arguments, and argv contains the values of the arguments. In
our example, argc is 3, and argv contains the three strings
argv[0]: "prog"
argv[1]: "-v"
argv[2]: "input.dat"
Note that argv[0] is always the name of the program and that argc is always at least 1.
s e l F C h e C k
programs that start
from the command
line can receive the
name of the program
and the command
line arguments in the
main function.
cfe2_ch08_p351_388.indd 365 10/28/10 8:56 PM
366 Chapter 8 Streams
Figure 2 Caesar Cipher
M e e t m e a t t h e
P h h w p h d w w k h
Plain text
Encrypted text # # # #
Let’s write a program that encrypts a file—that is, scrambles it so that it is unread-
able except to those who know the decryption method. Ignoring 2,000 years of prog-
ress in the field of encryption, we will use a method familiar to Julius Caesar, replacing
an A with a D, a B with an E, and so on. That is, each character c is replaced with c + 3
(see Figure 2).
The program takes the following command line
arguments:
• An optional -d flag to indicate decryption instead
of encryption
• The input file name
• The output file name
For example,
caesar input.txt encrypt.txt
encrypts the file input.txt and places the result into
encrypt.txt.
caesar -d encrypt.txt output.txt
decrypts the file encrypt.txt and places the result into
output.txt.
ch08/caesar.cpp
1 #include
2 #include
3 #include
4 #include
5
6 using namespace std;
7
8 /**
9 Encrypts a stream using the Caesar cipher.
10 @param in the stream to read from
11 @param out the stream to write to
12 @param k the encryption key
13 */
14 void encrypt_file(ifstream& in, ofstream& out, int k)
15 {
16 char ch;
17 while (in.get(ch))
18 {
19 out.put(ch + k);
20 }
21 }
22
The emperor Julius Caesar used
a simple scheme to encrypt
messages.
cfe2_ch08_p351_388.indd 366 10/28/10 8:56 PM
8.5 Command line arguments 367
23 int main(int argc, char* argv[])
24 {
25 int key = 3;
26 int file_count = 0; // The number of files specified
27 ifstream in_file;
28 ofstream out_file;
29
30 for (int i = 1; i < argc; i++) // Process all command-line arguments
31 {
32 string arg = argv[i]; // The currently processed argument
33 if (arg == "-d") // The decryption option
34 {
35 key = -3;
36 }
37 else // It is a file name
38 {
39 file_count++;
40 if (file_count == 1) // The first file name
41 {
42 in_file.open(arg.c_str());
43 if (in_file.fail()) // Exit the program if opening failed
44 {
45 cout << "Error opening input file " << arg << endl;
46 return 1;
47 }
48 }
49 else if (file_count == 2) // The second file name
50 {
51 out_file.open(arg.c_str());
52 if (out_file.fail())
53 {
54 cout << "Error opening output file " << arg << endl;
55 return 1;
56 }
57 }
58 }
59 }
60
61 if (file_count != 2) // Exit if the user didn’t specify two files
62 {
63 cout << "Usage: " << argv[0] << " [-d] infile outfile" << endl;
64 return 1;
65 }
66
67 encrypt_file(in_file, out_file, key);
68 return 0;
69 }
21. If the program is invoked with caesar -d encrypt.txt, what is argc, and what are
the elements of argv?
22. Trace the program when it is invoked as described in Self Check 21.
23. Encrypt CAESAR using the Caesar cipher.
24. What does the program do with spaces?
practice it Now you can try these exercises at the end of the chapter: R8.11, P8.5, P8.17.
s e l F C h e C k
cfe2_ch08_p351_388.indd 367 10/28/10 8:56 PM
368 Chapter 8 Streams
the exercises at the
end of this chapter
give a few algorithms to encrypt text.
Don’t actually use any of those meth-
ods to send secret messages to your
lover. any skilled cryptographer can
break these schemes in a very short
time—that is, reconstruct the origi-
nal text without knowing the secret
key word.
In 1978 ron rivest, adi Shamir,
and leonard adleman introduced an
encryption method that is much more
powerful. the method is called RSA
encryption, after the last names of its
inventors. the exact scheme is too
complicated to present here, but it is
not actually difficult to follow. You can
find the details in http://the ory.lcs.
mit.edu/~rivest/rsapaper .
rSa is a remarkable encryption
method. there are two keys: a pub-
lic key and a private key. (See the fig-
ure.) You can print the public key on
your business card (or in your e-mail
signature block) and give it to any-
one. then anyone can send you mes-
sages that only you can decrypt. even
though everyone else knows the pub lic
key, and even if they intercept all the
messages coming to you, they cannot
break the scheme and actually read
the messages. In 1994, hundreds of
researchers, collaborating over the
Internet, cracked an rSa message
encrypted with a 129-digit key. mes-
sages encrypted with a key of 230 dig-
its or more are expected to be secure.
the inventors of the algorithm
obtained a patent for it. a patent is a
deal that society makes with an inven-
tor. For a period of 20 years, the inven-
tor has an exclusive right for its com-
mercialization, may collect royal ties
from others wishing to manufac ture
the invention, and may even stop com-
petitors from using it altogether. In
return, the inventor must publish the
invention, so that others may learn
from it, and must relinquish all claim
to it after the monopoly period ends.
the presumption is that in the absence
of patent law, inventors would be
reluctant to go through the trouble of
inventing, or they would try to cloak
their techniques to prevent others from
copying their devices.
there has been some controversy
about the rSa patent. had there not
been patent protection, would the
inventors have published the method
anyway, thereby giving the benefit to
society without the cost of the 20-year
monopoly? In this case, the answer is
probably yes. the inventors were aca-
demic researchers, who live on sala-
ries rather than sales receipts and are
usually rewarded for their discover-
ies by a boost in their reputation and
careers. Would their followers have
been as active in discovering (and pat-
enting) improvements? there is no way
of knowing, of course. Is an algo rithm
even patentable, or is it a math ematical
fact that belongs to nobody? the pat-
ent office did take the latter attitude
for a long time. the rSa inventors and
many others described their inven-
tions in terms of imaginary electronic
devices, rather than algo rithms, to cir-
cumvent that restriction. nowadays,
the patent office will award software
patents.
there is another interesting aspect
to the rSa story. a programmer, phil
Zimmermann, developed a program
called pGp (for Pretty Good Privacy)
that is based on rSa. anyone can use
the program to encrypt messages, and
decryption is not feasible even with
the most powerful computers. You can
get a copy of a free pGp implementa-
tion from the GnU project (http://www.
gnupg.org). the existence of strong
encryption methods bothers the
United States government to no end.
Criminals and foreign agents can send
communications that the police and
intelligence agencies cannot deci pher.
the government considered charging
Zimmermann with breaching a law
that forbids the unauthorized export
of munitions, arguing that he should
have known that his program would
appear on the Internet. there have
been serious proposals to make it ille-
gal for private citizens to use these
encryption methods, or to keep the
keys secret from law enforcement.
Random Fact 8.1 encryption algorithms
step 1 Understand the processing task.
As always, you need to have a clear understanding of the task before designing a solution. Can
you carry out the task by hand (perhaps with smaller input files)? If not, get more information
about the problem.
The following pseudocode describes our processing task:
While there are more lines to be read
Read a line from each file.
Extract the country name.
population = number following the country name in the first line
area = number following the country name in the second line
If area != 0
density = population / area
Print country name and density.
step 2 Determine which files you need to read and write.
This should be clear from the problem. In our example, there are two input files, the popula-
tion data and the area data, and one output file.
step 3 Choose a method for obtaining the file names.
There are three options:
• Hard-coding the file names (such as "worldpop.txt")
• Asking the user:
cout << "Enter filename: ";
cin >> filename;
in_file.open(filename.c_str());
• Using command-line arguments for the file names
In our example, we use hard-coded file names for simplicity.
step 4 Choose between line, word, and character-based input.
As a rule of thumb, read lines if the input data is grouped by lines. That is the case with tabular
data, as in our example, or when you need to report line numbers.
h o W t o 8 . 1 processing text Files
Processing text files that contain real data can be surprisingly challenging. This How To gives
you step-by-step guidance.
As an example, we will consider this task:
Read two country data files, worldpop.txt and
worldarea.txt (supplied with the book’s com-
panion code). Both files contain the same
countries in the same order. Write a file world_
pop_density.txt that con tains country names
and population densities (people per square
km), with the country names aligned left and
the numbers aligned right:
Afghanistan 50.56
Akrotiri 127.64
Albania 125.91
Algeria 14.18
American Samoa 288.92
. . .
Singapore is one of the most densely
populated countries in the world.
Public-Key Encryption
Meet
me at
the
toga
party
Meet
me at
the
toga
party
Xwya
Txu%
*(Wt
&93ya
=9
Alice
Bob’s public key
encrypts the message;
it has “half the key” Bob
Decrypted
text
Plain
text Encrypted
text
Bob’s private key
knows how to decrypt
the message
cfe2_ch08_p351_388.indd 368 10/28/10 8:56 PM
http://www.gnupg.org
http://www.gnupg.org
http://theory.lcs.mit.edu/~rivest/rsapaper
http://theory.lcs.mit.edu/~rivest/rsapaper
8.5 Command line arguments 369
step 1 Understand the processing task.
As always, you need to have a clear understanding of the task before designing a solution. Can
you carry out the task by hand (perhaps with smaller input files)? If not, get more information
about the problem.
The following pseudocode describes our processing task:
While there are more lines to be read
Read a line from each file.
Extract the country name.
population = number following the country name in the first line
area = number following the country name in the second line
If area != 0
density = population / area
Print country name and density.
step 2 Determine which files you need to read and write.
This should be clear from the problem. In our example, there are two input files, the popula-
tion data and the area data, and one output file.
step 3 Choose a method for obtaining the file names.
There are three options:
• Hard-coding the file names (such as “worldpop.txt”)
• Asking the user:
cout << "Enter filename: ";
cin >> filename;
in_file.open(filename.c_str());
• Using command-line arguments for the file names
In our example, we use hard-coded file names for simplicity.
step 4 Choose between line, word, and character-based input.
As a rule of thumb, read lines if the input data is grouped by lines. That is the case with tabular
data, as in our example, or when you need to report line numbers.
h o W t o 8 . 1 processing text Files
Processing text files that contain real data can be surprisingly challenging. This How To gives
you step-by-step guidance.
As an example, we will consider this task:
Read two country data files, worldpop.txt and
worldarea.txt (supplied with the book’s com-
panion code). Both files contain the same
countries in the same order. Write a file world_
pop_density.txt that con tains country names
and population densities (people per square
km), with the country names aligned left and
the numbers aligned right:
Afghanistan 50.56
Akrotiri 127.64
Albania 125.91
Algeria 14.18
American Samoa 288.92
. . .
Singapore is one of the most densely
populated countries in the world.
cfe2_ch08_p351_388.indd 369 10/28/10 8:56 PM
370 Chapter 8 Streams
When gathering data that can be distributed over several lines, then it makes more sense to
read words. Keep in mind that you lose all white space when you read words.
Reading characters is mostly useful for tasks that require access to individual characters.
Examples include ana lyzing character frequencies, changing tabs to spaces, or encryption.
step 5 With line-oriented input, extract the required data.
It is simple to read a line of input with the getline function. Then you need to get the data
out of that line. You can extract substrings, as described in Section 8.2. Alternatively, you can
turn the line into an istringstream and extract its components with the >> operator. The latter
approach is easier when the number of items on each line is constant. In our example, that is
not the case––country names can consist of more than one string. Therefore, we choose to
extract substrings from each input line.
If you need any of the substrings as numbers, you must convert them (see Section 8.4).
step 6 Place repeatedly occurring tasks into functions.
Processing input files usually has repetitive tasks, such as skipping over white space or extract-
ing numbers from strings. It really pays off to develop a set of functions to handle these tedious
operations.
In our example, we have a common task that calls for a helper function: extracting the coun-
try name and the value that follows. This task can be implemented in a helper function
void read_line(string line, string& country, double& value)
We also need a helper function string_to_double to convert the population and area values
to floating-point numbers. This function is similar to string_to_int that was developed in
Section 8.4.
step 7 If required, use manipulators to format the output.
If you are asked to format your output, use manipulators, as described in Section 8.3. Usually,
you want to switch to fixed format for the output and set the precision. Then use setw before
every value, and use left for aligning strings and right for aligning numbers:
out << setw(40) << left << country << setw(15) << right << density << endl;
Here is the complete program:
ch08/popdensity.cpp
1 #include
2 #include
3 #include
4 #include
5 #include
6 #include
7
8 using namespace std;
9
10 /**
11 Converts a string to a floating-point number, e.g. “3.14” -> 3.14.
12 @param s a string representing a floating-point number
13 @return the equivalent floating-point number
14 */
15 double string_to_double(string s)
16 {
17 istringstream stream;
18 stream.str(s);
19 double x = 0;
cfe2_ch08_p351_388.indd 370 10/28/10 8:56 PM
8.5 Command line arguments 371
20 stream >> x;
21 return x;
22 }
23
24 /**
25 Extracts the country and associated value from an input line.
26 @param line a line containing a country name, followed by a number
27 @param country the string for holding the country name
28 @param value the variable for holding the associated value
29 @return true if a line has been read, false at the end of the stream
30 */
31 void read_line(string line, string& country, double& value)
32 {
33 int i = 0; // Locate the start of the first digit
34 while (!isdigit(line[i])) { i++; }
35 int j = i – 1; // Locate the end of the preceding word
36 while (isspace(line[j])) { j–; }
37
38 country = line.substr(0, j + 1); // Extract the country name
39 value = string_to_double(line.substr(i)); // Extract the number value
40 }
41
42 int main()
43 {
44 ifstream in1;
45 ifstream in2;
46 in1.open(“worldpop.txt”); // Open input files
47 in2.open(“worldarea.txt”);
48
49 ofstream out;
50 out.open(“world_pop_density.txt”); // Open output file
51 out << fixed << setprecision(2);
52
53 string line1;
54 string line2;
55
56 // Read lines from each file
57 while (getline(in1, line1) && getline(in2, line2))
58 {
59 string country;
60 double population;
61 double area;
62
63 // Split the lines into country and associated value
64 read_line(line1, country, population);
65 read_line(line2, country, area);
66
67 // Compute and print the population density
68 double density = 0;
69 if (area != 0) // Protect against division by zero
70 {
71 density = population * 1.0 / area;
72 }
73 out << setw(40) << left << country
74 << setw(15) << right << density << endl;
75 }
76
77 return 0;
78 }
cfe2_ch08_p351_388.indd 371 10/28/10 8:56 PM
372 Chapter 8 Streams
8.6 random access and Binary Files
In the following sections, you will learn how to read and write data at arbitrary posi-
tions in a file, and how to edit image files.
8.6.1 random access
So far, you’ve read from a file an item at a time and written to a file an item at a time,
without skipping forward or backward. That access pattern is called sequential
access. In many applications, we would like to access specific items in a file without
first having to read all preceding items. This access pattern is called random access
(see Figure 3). There is nothing “random” about random access—the term means that
you can read and modify any item stored at any location in the file.
Only file streams support random access; the cin and cout streams, which are
attached to the keyboard and the terminal, do not. Each file stream has two special
positions: the get position and the put position (see Figure 4). These positions deter-
mine where the next character is read or written.
The following function calls move the get and put positions to a given value,
counted from the begin ning of the stream.
strm.seekg(position);
strm.seekp(position);
To determine the current values of the get and put positions (counted from the begin-
ning of the file), use
position = strm.tellg();
position = strm.tellp();
8.6.2 Binary Files
Many files, in particular those containing images and sounds, do not store informa-
tion as text but as binary numbers. The numbers are represented as sequences of
W o r k e D e x a m p l e 8 . 1 looking for for duplicates
This Worked Example processes a file to locate lines that contain repeated words.
You can access any
position in a random
access file by moving
the file pointer
prior to a read or
write operation.
Figure 3 Sequential and random access
Sequential access
Random access
Figure 4 Get and put positions
5 4 9 0 3 . 2
Get
position
Put
position
5
Available online at www.wiley.com/college/horstmann.
cfe2_ch08_p351_388.indd 372 10/28/10 8:56 PM
www.wiley.com/college/horstmann
8.6 random access and Binary Files 373
8.6 random access and Binary Files
In the following sections, you will learn how to read and write data at arbitrary posi-
tions in a file, and how to edit image files.
8.6.1 random access
So far, you’ve read from a file an item at a time and written to a file an item at a time,
without skipping forward or backward. That access pattern is called sequential
access. In many applications, we would like to access specific items in a file without
first having to read all preceding items. This access pattern is called random access
(see Figure 3). There is nothing “random” about random access—the term means that
you can read and modify any item stored at any location in the file.
Only file streams support random access; the cin and cout streams, which are
attached to the keyboard and the terminal, do not. Each file stream has two special
positions: the get position and the put position (see Figure 4). These positions deter-
mine where the next character is read or written.
The following function calls move the get and put positions to a given value,
counted from the begin ning of the stream.
strm.seekg(position);
strm.seekp(position);
To determine the current values of the get and put positions (counted from the begin-
ning of the file), use
position = strm.tellg();
position = strm.tellp();
8.6.2 Binary Files
Many files, in particular those containing images and sounds, do not store informa-
tion as text but as binary numbers. The numbers are represented as sequences of
W o r k e D e x a m p l e 8 . 1 looking for for duplicates
This Worked Example processes a file to locate lines that contain repeated words.
You can access any
position in a random
access file by moving
the file pointer
prior to a read or
write operation.
Figure 3 Sequential and random access
Sequential access
Random access
Figure 4 Get and put positions
5 4 9 0 3 . 2
Get
position
Put
position
5
At a sit-down dinner, food is served
sequentially. At a buffet, you have
“random access” to all food items.
bytes, just as they are in the memory of the computer. (Each byte is a value between 0
and 255.) In binary format, a floating-point number always occupies 8 bytes. We will
study random access with a binary file format for images.
We have to cover a few technical issues about binary files. To open a binary file for
reading and writing, use the following command:
fstream strm;
strm.open(filename, ios::in | ios::out | ios::binary);
You read a byte with the call
int input = strm.get();
This call returns a value between 0 and 255. To read an integer, read four bytes b0, b1,
b2, b3 and combine them to b b b b0 1 2
2
3
3256 256 256+ ⋅ + ⋅ + ⋅ . We will supply a
helper function for this task.
The >> operator cannot be used to read numbers from a binary file.
8.6.3 processing Image Files
In this section, you will learn how to write a program for editing image files in the
BMP format. Unlike the more common GIF, PNG, and JPEG formats, the BMP for-
mat is quite simple because it does not use data compression. As a consequence, BMP
files are huge and you will rarely find them in the wild. How ever, image editors can
convert any image into BMP format.
There are different versions of the BMP format; we will only cover the simplest
and most common one, sometimes called the 24-bit true color format. In this format,
each pixel is represented as a sequence of three bytes, one each for the blue, green, and
red value. For example, the color cyan (a mixture of blue and green) is 255 255 0, red is
0 0 255, and medium gray is 128 128 128.
A BMP file starts with a header that contains various pieces of information. We
only need the follow ing items:
position Item
2 The size of this file in bytes
10 The start of the image data
18 The width of the image in pixels
22 The height of the image in pixels
cfe2_ch08_p351_388.indd 373 10/28/10 8:56 PM
374 Chapter 8 Streams
Figure 5 the Bmp File Format for 24-bit true Color Images
0 2 10 18 22
file size offset width height …
…
…
Padding
Scan line
Scan line
The image is stored as a sequence of pixel rows, starting with the pixels of the bot-
tommost row of the image. Each pixel row contains a sequence of blue/green/red
triplets. The end of the row is padded with additional bytes so that the number of
bytes in the row is divisible by 4. (See Figure 5.) For example, if a row consisted of
merely three pixels, one cyan, one red, and one medium gray one, the row would be
encoded as
255 255 0 0 0 255 128 128 128 x y z
where x y z are padding bytes to bring the row length up to 12, a multiple of 4. It is these
little twists that make working with real-life file formats such a joyful exp erience.
The sample program at the end of this section reads every pixel of a BMP file and
replaces it with its negative, turning white to black, cyan to red, and so on. The result
is a negative image of the kind that old-fashioned film cameras used to produce (see
Figure 6).
Figure 6 an Image and Its negative
cfe2_ch08_p351_388.indd 374 10/28/10 8:56 PM
8.6 random access and Binary Files 375
To try out this program, take one of your favorite images, use an image editor to
convert to BMP format (or use queen-mary.bmp from the code files for this book), then
run the program and view the transformed file in an image editor. Exercises P8.21 and
P8.22 ask you to produce more interesting effects.
ch08/imagemod.cpp
1 #include
2 #include
3 #include
4
5 using namespace std;
6
7 /**
8 Processes a pixel by forming the negative.
9 @param blue the blue value of the pixel
10 @param green the green value of the pixel
11 @param red the red value of the pixel
12 */
13 void process(int& blue, int& green, int& red)
14 {
15 blue = 255 – blue;
16 green = 255 – green;
17 red = 255 – red;
18 }
19
20 /**
21 Gets an integer from a binary stream.
22 @param stream the stream
23 @param offset the offset at which to read the integer
24 @return the integer starting at the given offset
25 */
26 int get_int(fstream& stream, int offset)
27 {
28 stream.seekg(offset);
29 int result = 0;
30 int base = 1;
31 for (int i = 0; i < 4; i++)
32 {
33 result = result + stream.get() * base;
34 base = base * 256;
35 }
36 return result;
37 }
38
39 int main()
40 {
41 cout << "Please enter the file name: ";
42 string filename;
43 cin >> filename;
44
45 fstream stream;
46 // Open as a binary file
47 stream.open(filename.c_str(), ios::in | ios::out | ios::binary);
48
49 int file_size = get_int(stream, 2); // Get the image dimensions
50 int start = get_int(stream, 10);
51 int width = get_int(stream, 18);
52 int height = get_int(stream, 22);
cfe2_ch08_p351_388.indd 375 10/28/10 8:56 PM
376 Chapter 8 Streams
53
54 // Scan lines must occupy multiples of four bytes
55 int scanline_size = width * 3;
56 int padding = 0;
57 if (scanline_size % 4 != 0)
58 {
59 padding = 4 – scanline_size % 4;
60 }
61
62 if (file_size != start + (scanline_size + padding) * height)
63 {
64 cout << "Not a 24-bit true color image file." << endl;
65 return 1;
66 }
67
68 stream.seekg(start); // Go to the start of the pixels
69
70 for (int i = 0; i < height; i++) // For each scan line
71 {
72 for (int j = 0; j < width; j++) // For each pixel
73 {
74 int pos = stream.tellg(); // Go to the start of the pixel
75
76 int blue = stream.get(); // Read the pixel
77 int green = stream.get();
78 int red = stream.get();
79
80 process(blue, green, red); // Process the pixel
81
82 stream.seekp(pos); // Go back to the start of the pixel
83
84 stream.put(blue); // Write the pixel
85 stream.put(green);
86 stream.put(red);
87 }
88
89 stream.seekg(padding, ios::cur); // Skip the padding
90 }
91
92 return 0;
93 }
25. In plain English, what does the following code segment do?
strm.seekp(0);
strm.put(0);
26. How would you modify the imagemod.cpp program to flip the green and blue
values of each pixel for a psychedelic effect?
27. What happens if you run the imagemod.cpp program twice on the same image file?
28. Could we have implemented the image modification program with sequential
access only? If not, why not?
29. Suppose a BMP file stores a 100 × 100 pixel image in BMP format, with the
image data starting at off set 64. What is the total file size?
practice it Now you can try these exercises at the end of the chapter: R8.12, R8.13, P8.21.
s e l F C h e C k
develop programs that read and write files.
• To read or write files, you use variables of type fstream, ifstream, or ofstream.
• When opening a file stream, you supply the name of the file stored on disk.
• Read from a file stream with the same operations that you use with cin.
• Write to a file stream with the same operations that you use with cout.
• Always use a reference parameter for a stream.
most companies use
computers to keep
huge data files of
customer records and other busi-
ness information. Data bases not only
lower the cost of doing business;
they improve the quality of service
that companies can offer. now adays
it is almost unimaginable how time-
consuming it used to be to with draw
money from a bank branch or to make
travel reservations.
today most databases are orga-
nized according to the relational
model. Suppose a company stores your
orders and payments. they will prob-
ably not repeat your name and address
on every order; that would take unnec-
essary space. Instead, they will keep
one file of all their customer names
and identify each customer by a unique
customer number. only that customer
number, not the entire cus tomer infor-
mation, is kept with an order record.
Customers Orders
Cust. #:
11439 59673 11439 DOS for Historians
61013 11439 C++ for Everyone
59897 11439 Big C++
Doe, John
Cust. #:Order #:Name Item
Relational Database Files
to print an invoice, the database
program must issue a query against
both the customer and order files and
pull the necessary information (name,
address, articles ordered) from both.
Frequently, queries involve more than
two files. For example, the company
may have a file of addresses of car
owners and a file of people with good
payment history and may want to find
all of its customers who placed an
order in the last month, drive an expen-
sive car, and pay their bills, so they can
send them another catalog. this kind
of query is, of course, much faster if all
customer files use the same key, which
is why so many orga nizations in the
United States try to collect the Social
Security numbers of their customers.
the Social Security act of 1935
provided that each contributor be
assigned a Social Security number to
track contributions into the Social Secu-
rity Fund. these numbers have a dis-
tinctive format, such as 078-05-1120.
(this particular number was printed on
sample cards that were inserted in wal-
lets. It actually was the Social Security
number of the secre tary of a vice presi-
dent at the wallet manufacturer. When
thousands of peo ple used it as their
own, the number was voided, and the
secretary received a new number.) the
figure at right shows a Social Security
card. although they had not originally
been intended for use as a universal
identi fication number, Social Security
num bers have become just that.
Some people are very concerned
about the fact that just about every
organization wants to store their Social
Security number and other per sonal
information. there is the possi bility
that companies and the government
can merge multiple data bases and
derive information about us that we
may wish they did not have or that
simply may be untrue. an insur ance
company may deny coverage, or
charge a higher premium, if it finds
that you have too many relatives with
a certain disease. You may be denied a
job because of an inaccurate credit or
medical report, and you may not even
know the reason. these are very dis-
turbing developments that have had
a very negative impact for a small but
growing number of people.
In many industrialized countries
(but not currently in the United States),
citizens have a right to control what
information about themselves should
be communicated to others and under
what circumstances.
Social Security Card
Random Fact 8.2 Databases and privacy
C h a p t e r S U m m a r Y
cfe2_ch08_p351_388.indd 376 10/28/10 8:56 PM
Chapter Summary 377
develop programs that read and write files.
• To read or write files, you use variables of type fstream, ifstream, or ofstream.
• When opening a file stream, you supply the name of the file stored on disk.
• Read from a file stream with the same operations that you use with cin.
• Write to a file stream with the same operations that you use with cout.
• Always use a reference parameter for a stream.
most companies use
computers to keep
huge data files of
customer records and other busi-
ness information. Data bases not only
lower the cost of doing business;
they improve the quality of service
that companies can offer. now adays
it is almost unimaginable how time-
consuming it used to be to with draw
money from a bank branch or to make
travel reservations.
today most databases are orga-
nized according to the relational
model. Suppose a company stores your
orders and payments. they will prob-
ably not repeat your name and address
on every order; that would take unnec-
essary space. Instead, they will keep
one file of all their customer names
and identify each customer by a unique
customer number. only that customer
number, not the entire cus tomer infor-
mation, is kept with an order record.
Customers Orders
Cust. #:
11439 59673 11439 DOS for Historians
61013 11439 C++ for Everyone
59897 11439 Big C++
Doe, John
Cust. #:Order #:Name Item
Relational Database Files
to print an invoice, the database
program must issue a query against
both the customer and order files and
pull the necessary information (name,
address, articles ordered) from both.
Frequently, queries involve more than
two files. For example, the company
may have a file of addresses of car
owners and a file of people with good
payment history and may want to find
all of its customers who placed an
order in the last month, drive an expen-
sive car, and pay their bills, so they can
send them another catalog. this kind
of query is, of course, much faster if all
customer files use the same key, which
is why so many orga nizations in the
United States try to collect the Social
Security numbers of their customers.
the Social Security act of 1935
provided that each contributor be
assigned a Social Security number to
track contributions into the Social Secu-
rity Fund. these numbers have a dis-
tinctive format, such as 078-05-1120.
(this particular number was printed on
sample cards that were inserted in wal-
lets. It actually was the Social Security
number of the secre tary of a vice presi-
dent at the wallet manufacturer. When
thousands of peo ple used it as their
own, the number was voided, and the
secretary received a new number.) the
figure at right shows a Social Security
card. although they had not originally
been intended for use as a universal
identi fication number, Social Security
num bers have become just that.
Some people are very concerned
about the fact that just about every
organization wants to store their Social
Security number and other per sonal
information. there is the possi bility
that companies and the government
can merge multiple data bases and
derive information about us that we
may wish they did not have or that
simply may be untrue. an insur ance
company may deny coverage, or
charge a higher premium, if it finds
that you have too many relatives with
a certain disease. You may be denied a
job because of an inaccurate credit or
medical report, and you may not even
know the reason. these are very dis-
turbing developments that have had
a very negative impact for a small but
growing number of people.
In many industrialized countries
(but not currently in the United States),
citizens have a right to control what
information about themselves should
be communicated to others and under
what circumstances.
Social Security Card
Random Fact 8.2 Databases and privacy
C h a p t e r S U m m a r Y
cfe2_ch08_p351_388.indd 377 10/28/10 8:56 PM
378 Chapter 8 Streams
Be able to process text in files.
• When reading a string with the >> operator, the white space
between words is consumed.
• You can get individual characters from a stream and unget the
last one.
• You can read a line of input with the getline function and then
process it further.
Write programs that neatly format their output.
• Use the setw manipulator to set the width of the next output.
• Use the fixed and setprecision manipulators to format floating-point numbers
with a fixed number of digits after the decimal point.
Convert between strings and numbers.
• Use an istringstream to convert the numbers inside a string to
integers or float ing-point numbers.
• Use an ostringstream to convert numeric values to strings.
process the command line arguments of a C++ program.
• Programs that start from the command line can receive the name of the program
and the command line arguments in the main function.
develop programs that read and write binary files.
• You can access any position in a random access file by
moving the file pointer prior to a read or write operation.
R8.1 When do you open a file as an ifstream, as an ofstream, or as an fstream? Could you
simply open all files as an fstream?
R8.2 What happens if you write to a file that you only opened for reading? Try it out if
you don’t know.
R8.3 What happens if you try to open a file for reading that doesn’t exist? What happens if
you try to open a file for writing that doesn’t exist?
R8.4 What happens if you try to open a file for writing, but the file or device is write-
protected (sometimes called read-only)? Try it out with a short test program.
R8.5 How do you open a file whose name contains a backslash, such as temp\output.dat or
c:emp\output.dat?
r e v I e W e x e r C I S e S
R8.6 Why are the in and out parameter variables of the encrypt_file function in Section 8.5
refer ence parameters and not value parameters?
R8.7 Give an output statement to write a date and time in ISO 8601 format, such as
2011-03-01 09:35
Assume that the date and time are given in five integer variables year, month, day,
hour, minute.
R8.8 Give an output statement to write one line of a table containing a product descrip-
tion, quantity, unit price, and total price in dollars and cents. You want the columns
to line up, like this:
R8.9 How can you convert the string “3.14” into the floating-point number 3.14? How
can you convert the floating-point number 3.14 into the string “3.14”?
R8.10 What is a command line? How can a program read its command line?
R8.11 If a program woozle is started with the command
woozle -DNAME=Piglet -I\eeyore -v heff.cpp a.cpp lump.cpp
what is the value of argc, and what are the values of argv[0], argv[1], and so on?
R8.12 What is the difference between sequential access and random access?
R8.13 What is the difference between a text file and a binary file?
R8.14 What are the get and put positions in a file? How do you move them? How do you
tell their current positions?
R8.15 What happens if you try to move the get or put position past the end of a file? What
happens if you try to move the get or put position of cin or cout? Try it out and
report your results.
p8.1 Write a program that carries out the following tasks:
Open a file with the name hello.txt.
Store the message “Hello, World!” in the file.
Close the file.
Open the same file again.
Read the message into a string variable and print it.
p8.2 Write a program that reads a file containing floating-point numbers. Print the
aver age of the numbers in the file. Prompt the user for the file name.
p r o G r a m m I n G e x e r C I S e S
cfe2_ch08_p351_388.indd 378 10/28/10 8:56 PM
programming exercises 379
R8.6 Why are the in and out parameter variables of the encrypt_file function in Section 8.5
refer ence parameters and not value parameters?
R8.7 Give an output statement to write a date and time in ISO 8601 format, such as
2011-03-01 09:35
Assume that the date and time are given in five integer variables year, month, day,
hour, minute.
R8.8 Give an output statement to write one line of a table containing a product descrip-
tion, quantity, unit price, and total price in dollars and cents. You want the columns
to line up, like this:
Item Qty Price Total
Toaster 3 $29.95 $89.85
Hair Dryer 1 $24.95 $24.95
Car Vacuum 2 $19.99 $39.98
R8.9 How can you convert the string “3.14” into the floating-point number 3.14? How
can you convert the floating-point number 3.14 into the string “3.14”?
R8.10 What is a command line? How can a program read its command line?
R8.11 If a program woozle is started with the command
woozle -DNAME=Piglet -I\eeyore -v heff.cpp a.cpp lump.cpp
what is the value of argc, and what are the values of argv[0], argv[1], and so on?
R8.12 What is the difference between sequential access and random access?
R8.13 What is the difference between a text file and a binary file?
R8.14 What are the get and put positions in a file? How do you move them? How do you
tell their current positions?
R8.15 What happens if you try to move the get or put position past the end of a file? What
happens if you try to move the get or put position of cin or cout? Try it out and
report your results.
p8.1 Write a program that carries out the following tasks:
Open a file with the name hello.txt.
Store the message “Hello, World!” in the file.
Close the file.
Open the same file again.
Read the message into a string variable and print it.
p8.2 Write a program that reads a file containing floating-point numbers. Print the
aver age of the numbers in the file. Prompt the user for the file name.
p r o G r a m m I n G e x e r C I S e S
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380 Chapter 8 Streams
p8.3 Repeat Exercise P8.2, but allow the user to specify the file name on the command-
line. If the user doesn’t specify any file name, then prompt the user for the name.
p8.4 Write a program that reads a file containing two columns of floating-point num bers.
Prompt the user for the file name. Print the average of each column.
p8.5 Write a program find that searches all files specified on the command line and prints
out all lines containing a keyword. For example, if you call
find Tim report.txt address.txt homework.cpp
then the program might print
report.txt: discussed the results of my meeting with Tim T
address.txt: Torrey, Tim|11801 Trenton Court|Dallas|TX
address.txt: Walters, Winnie|59 Timothy Circle|Detroit|MI
homework.cpp: Time now;
The keyword is always the first command-line argument.
p8.6 Write a program that checks the spelling of all words in a file. It should read each
word of a file and check whether it is contained in a word list. A word list is avail able
on most UNIX systems (including Linux and Mac OS X) in the file /usr/share/dict/
words. (If you don’t have access to a UNIX system, you can find a copy of the file on
the Internet by searching for /usr/share/dict/words.) The program should print out all
words that it cannot find in the word list. Follow this pseudocode:
Open the dictionary file.
Define a vector of strings called words.
For each word in the dictionary file
Append the word to the words vector.
Open the file to be checked.
For each word in that file
If the word is not contained in the words vector
Print the word.
p8.7 Write a program that reads each line in a file, reverses its characters, and writes the
resulting line to another file. Suppose the user specifies input.txt and output.txt when
prompted for the file names, and input.txt contains the lines
Mary had a little lamb
Its fleece was white as snow
And everywhere that Mary went
the lamb was sure to go.
After the program is finished, output.txt should contain
bmal elttil a dah yraM
wons sa etihw saw eceelf stI
tnew yraM taht erehwyreve dnA
.og ot erus saw bmal ehT
p8.8 Write a program that reads each line in a file, reverses its characters, and writes the
resulting line to the same file. Use the following pseudocode:
While the end of the file has not been reached
pos1 = current get position
Read a line.
cfe2_ch08_p351_388.indd 380 10/28/10 8:56 PM
programming exercises 381
If the line was successfully read
pos2 = current get position
Set put position to pos1.
Write the reversed line.
Set get position to pos2.
p8.9 Write a program that reads each line in a file, reverses its lines, and writes them to
another file. Suppose the user specifies input.txt and output.txt when prompted for
the file names, and input.txt contains the lines
Mary had a little lamb
Its fleece was white as snow
And everywhere that Mary went
The lamb was sure to go.
After the program is finished, output.txt should contain
The lamb was sure to go.
And everywhere that Mary went
Its fleece was white as snow
Mary had a little lamb
p8.10 Get the data for names in prior decades from the Social Security Administration.
Paste the table data in files named babynames80s.txt, etc. Modify the babynames.cpp
program so that it prompts the user for a file name. The numbers in the files have
comma separators, so modify the program to handle them. Can you spot a trend in
the frequencies?
p8.11 Write a program that reads in babynames.txt and produces two files, boynames.txt and
girlnames.txt, separating the data for the boys and girls.
p8.12 Write a program that reads a file in the same format as babynames.txt and prints all
names that are both boy and girl names (such as Alexis or Morgan).
p8.13 Write a program that reads the country data in the file worldpop.txt (included with the
book’s source code). Do not edit the file. Use the following algorithm for pro cessing
each line. Add non-white space characters to the country name. When you encoun-
ter a white space, locate the next non-white space character. If it is not a digit, add a
space and that character to the country name. Otherwise unget it and read the
number. Print the total of all country populations (excepting the entry for
“European Union”).
p8.14 Write a program that asks the user for a file name and displays the number of char-
acters, words, and lines in that file. Then have the program ask for the name of the
next file. When the user enters a file that doesn’t exist (such as the empty string), the
program should exit.
p8.15 Write a program copyfile that copies one file to another. The file names are specified
on the command line. For example,
copyfile report.txt report.sav
p8.16 Write a program that concatenates the contents of several files into one file. For
example,
catfiles chapter1.txt chapter2.txt chapter3.txt book.txt
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382 Chapter 8 Streams
makes a long file book.txt that contains the contents of the files chapter1.txt, chapter2.
txt, and chapter3.txt. The target file is always the last file specified on the command
line.
p8.17 Random monoalphabet cipher. The Caesar cipher, which shifts all letters by a fixed
amount, is far too easy to crack. Here is a better idea. As the key, don’t use numbers
but words. Suppose the key word is FEATHER. Then first remove duplicate letters,
yielding FEATHR, and append the other letters of the alphabet in reverse order:
F E A T H R Z Y X W V U S Q P O N M L K J I G D C B
Now encrypt the letters as follows:
A B C D E F G H I J K L M N O
F E A T H R Z Y X W V U S Q P
P
O
Q
N
R
M
S
L
T
K
U
J
V
I
W
G
X
D
Y
C
Z
B
Write a program that encrypts or decrypts a file using this cipher. For example,
crypt -d -kFEATHER encrypt.txt output.txt
decrypts a file using the keyword FEATHER. It is an error not to supply a keyword.
p8.18 Letter frequencies. If you encrypt a file using the cipher of Exercise P8.17, it will
have all of its letters jumbled up, and will look as if there is no hope of decrypting it
without knowing the keyword. Guessing the keyword seems hopeless too. There
are just too many possible keywords. However, someone who is trained in decryp-
tion will be able to break this cipher in no time at all. The average letter frequencies
of English letters are well known. The most common letter is E, which occurs about
13 percent of the time. Here are the average frequencies of the letters.
Write a program that reads an input file and displays the letter frequencies in that
file. Such a tool will help a code breaker. If the most frequent letters in an encrypted
file are H and K, then there is an excellent chance that they are the encryptions of E
and T.
Show the result in a table such as the one above, and make sure the columns line up.
p8.19 Vigenère cipher. In order to defeat a simple letter frequency analysis, the Vigenère
cipher encodes a letter into one of several cipher letters, depending on its position in
A 8% H 4% O 7% U 3%
B <1% I 7% P 3% V <1%
C 3% J <1% Q <1% W 2%
D 4% K <1% R 8% X <1%
E 13% L 4% S 6% Y 2%
F 3% M 3% T 9% Z <1%
G 2% N 8%
cfe2_ch08_p351_388.indd 382 10/28/10 8:56 PM
programming exercises 383
the input document. Choose a keyword, for example TIGER. Then encode the first
letter of the input text like this:
A B C D E F G H I J K L M N O
T U V W X Y Z A B C D E F G H
P
I
Q
J
R
K
S
L
T
M
U
N
V
O
W
P
X
Q
Y
R
Z
S
The encoded alphabet is just the regular alphabet shifted to start at T, the first letter
of the keyword TIGER. The second letter is encrypted according to the following map.
A B C D E F G H I J K L M N O
I J K L M N O P Q R S T U V W
P
X
Q
Y
R
Z
S
A
T
B
U
C
V
D
W
E
X
F
Y
G
Z
H
The third, fourth, and fifth letters in the input text are encrypted using the alphabet
sequences beginning with characters G, E, and R, and so on. Because the key is only five
letters long, the sixth letter of the input text is encrypted in the same way as the first.
Write a program that encrypts or decrypts an input text according to this cipher.
p8.20 Playfair cipher. Another way of thwarting a simple letter frequency analysis of an
encrypted text is to encrypt pairs of letters together. A simple scheme to do this is the
Playfair cipher. You pick a keyword and remove duplicate letters from it. Then you
fill the keyword, and the remaining letters of the alphabet, into a 5 × 5 square. (Since
there are only 25 squares, I and J are considered the same letter.)
Here is such an arrangement with the keyword PLAYFAIR.
P L A Y F
I R B C D
E G H K M
N O Q S T
U V W X Z
To encrypt a letter pair, say AM, look at the rectangle with corners A and M:
P L A Y F
I R B C D
E G H K M
N O Q S T
U V W X Z
The encoding of this pair is formed by looking at the other two corners of the
rectangle, in this case, FH. If both letters happen to be in the same row or column,
such as GO, simply swap the two letters. Decryption is done in the same way.
Write a program that encrypts or decrypts an input text according to this cipher.
p8.21 Write a program that edits an image file and reduces the blue and green values by 30
percent, giving it a “sunset” effect.
cfe2_ch08_p351_388.indd 383 10/28/10 8:56 PM
384 Chapter 8 Streams
p8.22 Write a program that edits an image file, turning it into grayscale.
Replace each pixel with a pixel that has the same grayness level for the blue, green,
and red component. The grayness level is computed by adding 30 percent of the red
level, 59 percent of the green level, and 11 percent of the blue level. (The color-
sensing cone cells in the human eye differ in their sensitivity for red, green, and
blue light.)
p8.23 Junk mail. Write a program that reads in two files: a template and a database. The
template file contains text and tags. The tags have the form |1| |2| |3|… and need to
be replaced with the first, second, third, … field in the current database record.
A typical database looks like this:
Mr.|Harry|Morgan|1105 Torre Ave.|Cupertino|CA|95014
Dr.|John|Lee|702 Ninth Street Apt. 4|San Jose|CA|95109
Miss|Evelyn|Garcia|1101 S. University Place|Ann Arbor|MI|48105
And here is a typical form letter:
To:
|1| |2| |3|
|4|
|5|, |6| |7|
Dear |1| |3|:
You and the |3| family may be the lucky winners of $10,000,000 in the C++ compiler
clearinghouse sweep stakes! ...
p8.24 Write a program that manipulates three database files. The first file contains the
names and telephone numbers of a group of people. The second file contains the
names and Social Security numbers of a group of people. The third file contains the
Social Security numbers and annual income of a group of people. The groups of
people should overlap but need not be completely identical. Your program should
ask the user for a telephone number and then print the name, Social Security num-
ber, and annual income, if it can determine that information.
p8.25 Write a program that prints out a student grade report. There is a file, classes.txt,
that contains the names of all classes taught at a college, such as
classes.txt
CSC1
CSC2
CSC46
CSC151
MTH121
...
cfe2_ch08_p351_388.indd 384 10/28/10 8:56 PM
programming exercises 385
For each class, there is a file with student ID numbers and grades:
csc2.txt
11234 A-
12547 B
16753 B+
21886 C
...
Write a program that asks for a student ID and prints out a grade report for that
student, by searching all class files. Here is a sample report
Student ID 16753
CSC2 B+
MTH121 C+
CHN1 A
PHY50 A-
engineering p8.26 After the switch in the figure below closes, the voltage (in volts) across the capacitor
is represented by the equation
v t B e t RC( ) = −( )−1 ( )
+
–
v (t)+– C
t = 0
R
B
Suppose the parameters of the electric circuit are B = 12 volts, R = 500 Ω, and
C = 0.25 μF. Consequently
v t e t( ) = − −( )12 1 0 008.
where t has units of μs. Read a file params.txt containing the values for B, R, C, and
the starting and ending values for t. Write a file rc.txt of values for the time t and the
corresponding capacitor voltage v(t), where t goes from the given starting value to
the given ending value in 100 steps. In our example, if t goes from 0 to 1,000 μs, the
twelfth entry in the output file would be:
110 7.02261
engineering p8.27 The figure below shows a plot of the capacitor voltage from the circuit shown in
Exercise P8.26. The capacitor voltage increases from 0 volts to B volts. The “rise
time” is defined as the time required for the capacitor voltage to change from
v1 = 0.05 × B to v2 = 0.95 × B.
0 t (µs)
0
B
cfe2_ch08_p351_388.indd 385 10/28/10 8:56 PM
386 Chapter 8 Streams
1. The stream will be in a failed state because there is no file with an empty name.
2. The stream variable is declared as an input stream. It should have been an ofstream.
3. The stream parameter variable in should be a reference parameter.
4. Initialize boy_total and girl_total with 10.
5. Add the following code to the beginning of main:
ofstream out_file;
out_file.open("output.txt");
Change every cout to out_file.
Add an out_file parameter variable of type ofstream& to process_name.
6. number is 6 because the comma is not recognized as a part of the number.
ch is ','.
7. word is "World!" (including the !) and ch is 'W'.
8. char ch;
double number;
string marker;
in_file >> ch;
in_file.unget();
if (isdigit(ch) || ch == ‘-‘) { cin >> number; } else { cin >> marker; }
9. The loop skips a sequence of white space.
10. Some country names (such as United States) consist of multiple words.
11. 123456.000000, because the default precision is 6.
12. cout << fixed << setprecision(2) << 123456.0;
13. 00123456
14. 123456
15. Because the setw manipulator only affects the first output.
16. 123456
17. 123
18. 0
19. 3.3333. Note that in general format, the precision denotes the total number of dig its,
not the number of digits after the decimal point.
20. string double_to_string(double x, int digits)
{
ostringstream strm;
strm << fixed << setprecision(digits) << x;
return strm.str();
}
21. argc is 3, and argv contains the strings "caesar", "-d", and "encrypt.txt".
a n S W e r S t o S e l F - C h e C k Q U e S t I o n SThe file rc.txt contains a list of values of time t and the corresponding capacitor
voltage v(t). A time in μs and the corresponding voltage in volts are printed on the
same line. For example, the line
110 7.02261
indicates that the capacitor voltage is 7.02261 volts when the time is 110 μs. The time
is increasing in the data file.
Write a program that reads the file rc.txt and uses the data to calculate the rise time.
Approximate B by the voltage in the last line of the file, and find the data points that
are closest to 0.05 × B and 0.95 × B.
engineering p8.28 Suppose a file contains bond energies and bond lengths for covalent bonds in the
following format:
Single, double,
or triple bond
Bond energy
(kJ/mol)
Bond length
(nm)
C|C 370 0.154
C||C 680 0.13
C|||C 890 0.12
C|H 435 0.11
C|N 305 0.15
C|O 360 0.14
C|F 450 0.14
C|Cl 340 0.18
O|H 500 0.10
O|O 220 0.15
O|Si 375 0.16
N|H 430 0.10
N|O 250 0.12
F|F 160 0.14
H|H 435 0.074
Write a program that accepts data from one column and returns the corresponding
data from the other columns in the stored file. If input data matches different rows,
then return all matching row data. For example, a bond length input of 0.12 should
return triple bond C|||C and bond energy 890 kJ̸mol and single bond N|O and bond
energy 250 kJ̸mol.
cfe2_ch08_p351_388.indd 386 10/28/10 8:56 PM
answers to Self-Check Questions 387
1. The stream will be in a failed state because there is no file with an empty name.
2. The stream variable is declared as an input stream. It should have been an ofstream.
3. The stream parameter variable in should be a reference parameter.
4. Initialize boy_total and girl_total with 10.
5. Add the following code to the beginning of main:
ofstream out_file;
out_file.open("output.txt");
Change every cout to out_file.
Add an out_file parameter variable of type ofstream& to process_name.
6. number is 6 because the comma is not recognized as a part of the number.
ch is ','.
7. word is "World!" (including the !) and ch is 'W'.
8. char ch;
double number;
string marker;
in_file >> ch;
in_file.unget();
if (isdigit(ch) || ch == ‘-‘) { cin >> number; } else { cin >> marker; }
9. The loop skips a sequence of white space.
10. Some country names (such as United States) consist of multiple words.
11. 123456.000000, because the default precision is 6.
12. cout << fixed << setprecision(2) << 123456.0;
13. 00123456
14. 123456
15. Because the setw manipulator only affects the first output.
16. 123456
17. 123
18. 0
19. 3.3333. Note that in general format, the precision denotes the total number of dig its,
not the number of digits after the decimal point.
20. string double_to_string(double x, int digits)
{
ostringstream strm;
strm << fixed << setprecision(digits) << x;
return strm.str();
}
21. argc is 3, and argv contains the strings "caesar", "-d", and "encrypt.txt".
a n S W e r S t o S e l F - C h e C k Q U e S t I o n S
cfe2_ch08_p351_388.indd 387 10/28/10 8:56 PM
388 Chapter 8 Streams
22. key file_count i arg
3 0 1 -d
-3 1 2 encrypt.txt
3
Then the program prints the message
Usage: caesar [-d] infile outfile
and exits.
23. FDHVDU
24. It turns them into # characters. The ASCII code for a space is 32, and the # character
has code 35.
25. It replaces the initial byte of a file with 0.
26. Change the process function to swap the values of the green and blue arguments. The
remainder of the program stays unchanged.
27. You get the original image back.
28. We could have read the header values and pixel data sequentially, but to update the
pixels, we had to move backwards.
29. We need 3 × 100 bytes for each scan line. There is no padding since this number is
divisible by 4. The total size = 3 × 100 × 100 + 64 = 30,064 bytes.
step 1 Understand the processing task.
Whenever we find a line containing a repeated word, we are to print it like this:
360:bat?' when suddenly, thump! thump! down she came upon a heap of
2103:'Twinkle, twinkle, twinkle, twinkle--' and went on so long that
A word is only counted as repeated when it is the same as its predecessor. For example, a
line that contains two “the” that are not adjacent would not be reported. The words must be
exactly the same. For example, “Twinkle” and “twinkle” don’t match.
step 2 Determine which files you need to read and write.
We only need to read one file, the one with the words. The result is displayed in the console
window; no output file is required.
step 3 Choose a method for obtaining the file names.
This is a student program with console output; we’ll ask the user through the console.
step 4 Choose between line, word, and character-based input.
We definitely want to use line-based input because we need to count line numbers and print
the entire line if it con tains repeating words.
step 5 With line-oriented input, extract the required data.
When we have an input line, we still need to extract the words. The easiest approach is to use a
string stream, and read words off that stream. We will keep a variable that holds the previous
word.
For each word in the line
If word equals previous word
Found a duplicate.
Else
previous word = word
step 6 Place repeatedly occurring tasks into functions.
In this program, there are no repeated tasks. But let’s take the bigger view. Scanning lines and
printing out the ones that match a particular criterion is a fairly common task. Therefore, let’s
put the checking for repeated words into a separate function,
bool has_repeated_words(string line)
W o r k e D e x a m p l e 8 . 1 looking for for duplicates
Your task is to write a program that reads a file and prints all lines that contain a repeated word
(such as an accidental “the the”), together with their line numbers.
cfe2_ch08_p351_388.indd 388 10/28/10 8:56 PM
9C h a p t e r
389
C l a s s e s
to understand the concept
of encapsulation
to master the separation of interface and
implementation
to be able to implement your own classes
to understand how constructors and member functions act on objects
to discover appropriate classes for solving programming problems
to distribute a program over multiple source files
C h a p t e r G o a l s
C h a p t e r C o n t e n t s
9.1 Object-Oriented
PrOgramming 390
9.2 SPecifying the Public interface
Of a claSS 392
Syntax 9.1: Class Definition 393
Common Error 9.1: Forgetting a semicolon 395
9.3 data memberS 395
9.4 member functiOnS 397
Syntax 9.2: Member Function Definition 400
Programming Tip 9.1: all Data Members should
Be private; Most Member Functions should
Be public 402
Programming Tip 9.2: const Correctness 402
9.5 cOnStructOrS 403
Common Error 9.2: trying to Call a
Constructor 405
Special Topic 9.1: Initializer lists 405
Special Topic 9.2: overloading 406
9.6 PrOblem SOlving: tracing
ObjectS 407
How To 9.1: Implementing a Class 409
Worked Example 9.1: Implementing a Bank
account Class
Random Fact 9.1: electronic Voting Machines 412
9.7 PrOblem SOlving: diScOvering
claSSeS 414
Programming Tip 9.3: Make parallel Vectors into
Vectors of objects 416
9.8 SeParate cOmPilatiOn 417
9.9 POinterS tO ObjectS 422
Special Topic 9.3: Destructors and resource
Management 424
Random Fact 9.2: open source and Free
software 426
cfe2_ch09_p389_440.indd 389 10/28/10 11:42 AM
390
this chapter introduces you to object-oriented program-
ming, an important technique for writing complex
programs. In an object-oriented program, you don’t simply
manipulate numbers and strings, but you work with objects
that are meaningful for your application. objects with
the same behavior (such as the windmills to the left) are
grouped into classes. a programmer provides the desired
behavior by specifying and implementing functions for
these classes. In this chapter, you will learn how to discover,
specify, and implement your own classes, and how to use
them in your programs.
9.1 object-oriented programming
You have learned how to structure your programs by decomposing tasks into func-
tions. This is an excel lent practice, but experience shows that it does not go far
enough. As programs get larger, it becomes increasingly difficult to maintain a large
collection of functions.
To overcome this problem, computer scientists invented object-oriented pro-
gramming, a program ming style in which tasks are solved by collaborating objects.
Each object has its own set of data, together with a set of functions that can act upon
the data. (These functions are called member func tions).
You have already experienced the object-oriented programming style when you
used string objects or streams such as cin and cout. For example, you use the length
and substr member functions to work with string objects. The >> and << operators that
you use with streams are also implemented as member func tions—see Special Topic
9.2 on page 406.
In C++, a programmer doesn’t implement a single object. Instead, the program-
mer provides a class. A class describes a set of objects with the same behavior. For
example, the string class describes the behavior of all strings. The class specifies how a
string stores its characters, which member functions can be used with strings, and
how the member functions are implemented.
a class describes a
set of objects with
the same behavior.
A class describes a set of objects with the same behavior.
For example, a Car class describes all passenger vehicles
that have a certain capacity and shape.
When you develop an object-oriented program, you create your own classes that
describe what is important in your application. For example, in a student database
you might work with Student and Course classes. Of course, then you must supply
member functions for these classes.
When you work with an object, you do not know how it is implemented. You
need not know how a string organizes a character sequence, or how the cin object
reads input from the console. All you need to know is the public interface: the speci-
fications for the member functions that you can invoke. The pro cess of providing a
public interface, while hiding the implementation details, is called encapsulation.
You will want to use encapsulation for your own classes. When you define a class,
you will specify the behavior of the public member functions, but you will hide
the implementation details. Encapsulation benefits the programmers who use your
classes. They can put your classes to work without having to know their implementa-
tions, just as you are able to make use of the string and stream classes without know-
ing their internal details.
Encapsulation is also a benefit for the implementor of a class. When working on a
program that is being developed over a long period of time, it is common for imple-
mentation details to change, usually to make objects more efficient or more capable.
Encapsulation is crucial to enabling these changes. When the implementation is hid-
den, the implementor is free to make improvements. Because the implementation is
hidden, these improvements do not affect the programmers who use the objects.
every class has a
public interface: a
collection of member
functions through
which the objects of
the class can be
manipulated.
encapsulation is the
act of providing a
public interface
and hiding imple-
mentation details.
encapsulation
enables changes in
the implementation
without affecting
users of a class.
A driver of an electric car doesn’t have to learn
new controls even though the car engine is very
different. Neither does the programmer who uses
an object with an improved implemen tation—as
long as the same member functions are used.
cfe2_ch09_p389_440.indd 390 10/28/10 11:42 AM
9.1 object-oriented programming 391
You can drive a car by operating the
steering wheel and pedals, without
know ing how the engine works.
Similarly, you use an object through
its member func tions.
The implementation is hidden.
When you develop an object-oriented program, you create your own classes that
describe what is important in your application. For example, in a student database
you might work with Student and Course classes. Of course, then you must supply
member functions for these classes.
When you work with an object, you do not know how it is implemented. You
need not know how a string organizes a character sequence, or how the cin object
reads input from the console. All you need to know is the public interface: the speci-
fications for the member functions that you can invoke. The pro cess of providing a
public interface, while hiding the implementation details, is called encapsulation.
You will want to use encapsulation for your own classes. When you define a class,
you will specify the behavior of the public member functions, but you will hide
the implementation details. Encapsulation benefits the programmers who use your
classes. They can put your classes to work without having to know their implementa-
tions, just as you are able to make use of the string and stream classes without know-
ing their internal details.
Encapsulation is also a benefit for the implementor of a class. When working on a
program that is being developed over a long period of time, it is common for imple-
mentation details to change, usually to make objects more efficient or more capable.
Encapsulation is crucial to enabling these changes. When the implementation is hid-
den, the implementor is free to make improvements. Because the implementation is
hidden, these improvements do not affect the programmers who use the objects.
every class has a
public interface: a
collection of member
functions through
which the objects of
the class can be
manipulated.
encapsulation is the
act of providing a
public interface
and hiding imple-
mentation details.
encapsulation
enables changes in
the implementation
without affecting
users of a class.
A driver of an electric car doesn’t have to learn
new controls even though the car engine is very
different. Neither does the programmer who uses
an object with an improved implemen tation—as
long as the same member functions are used.
cfe2_ch09_p389_440.indd 391 10/28/10 11:42 AM
392 Chapter 9 Classes
In this chapter, you will learn how to design and implement your own classes in
C++, and how to structure your programs in an object-oriented way, using the prin-
ciple of encapsulation.
1. In C++, is cin an object or a class? Is string an object or a class?
2. When using a string object, you do not know how it stores its characters. How
can you access them?
3. Describe two possible ways in which a string object might store its characters.
4. Suppose the providers of your C++ compiler decide to change the way that a
string object stores its characters, and they update the string member functions
accordingly. Which parts of your code do you need to change when you get the
new compiler?
Practice it Now you can try these exercises at the end of the chapter: R9.1, R9.2.
9.2 specifying the public Interface of a Class
To define a class, we first need to specify
its public interface. The pub lic interface
of a class consists of all member functions
that a user of the class may want to apply
to its objects.
Let’s consider a simple example. We
want to use objects that simu late cash
registers. A cashier who rings up a sale
presses a key to start the sale, then rings
up each item. A display shows the amount
owed as well as the total number of items
purchased.
In our simulation, we want to carry
out the following operations:
• Add the price of an item.
• Get the total amount of all items, and
the count of items purchased.
• Clear the cash register to start a new sale.
The interface is specified in the class definition, sum-
marized in Syntax 9.1 on page 393. We will call our class
CashRegister. (We follow the convention that the name
of a programmer-defined class starts with an uppercase
letter, as does each word within the name. This naming
convention is called camel case because the uppercase
letters in the middle of the name look like the humps of
a camel.)
S e l f c h e c k
Our first example of a class simulates a
cash register.
cfe2_ch09_p389_440.indd 392 10/28/10 11:42 AM
9.2 specifying the public Interface of a Class 393
Here is the C++ syntax for the CashRegister class definition:
class CashRegister
{
public:
void clear();
void add_item(double price);
double get_total() const;
int get_count() const;
private:
data members—see Section 9.3
};
The member functions are declared in the public section of the class. Any part of the
program can call the member functions. The data members are defined in the private
section of the class. Only the member functions of the class can access those data
members; they are hidden from the remainder of the program.
It is legal to declare the private members before the public section, but in this book,
we place the public section first. After all, most programmers reading a class are class
users, not implementors, and they are more interested in the public interface than in
the private implementation.
The member function declarations look similar to the declarations of regular func-
tions. These declara tions do not provide any implementation. You will see in Section
9.4 how to implement the member functions.
There are two kinds of member functions, called mutators and accessors. A muta-
tor is a function that modifies the data members of the object. The CashRegister class
syntax 9.1 Class Definition
Be sure to include this semicolon.
See page 395.
Mark accessors as const.
See page 402.
class CashRegister
{
public:
void clear();
void add_item(double price);
double get_total() const;
int get_count() const;
private:
int item_count;
double total_price;
};
Use CamelCase
for class names.
Data members
should always be private.
See page 402.
Mutator
member functions
Accessor
member functions
Public
section
Private
section
Member functions are declared in the class
and defined outside. See page 400.
cfe2_ch09_p389_440.indd 393 10/28/10 11:42 AM
394 Chapter 9 Classes
figure 1 the Interface of the CashRegister Class
get_total
get_count
CashRegister
Private data
add_item
clear
Mutators
Accessors
has two mutators: clear and add_item. After you call either of these functions, the total
amount and item count are changed.
Accessors just query the object for some information without changing it. The
CashRegister class has two accessors: get_total and get_count. Applying either of these
functions to a CashRegister object simply returns a value and does not modify the
object. In C++, you should use the const reserved word to mark accessor functions
(see Programming Tip 9.2 on page 402), like this:
double get_total() const;
Member functions are invoked using the dot notation that you have already seen with
string and stream functions:
CashRegister register1; // Defines a CashRegister object
register1.clear(); // Invokes a member function
Now we know what a CashRegister object can do, but not how it does it. Of course, to
use CashRegister objects in our programs, we don’t need to know. We simply use the
public interface. Figure 1 shows the interface of the CashRegister class. The mutator
functions are shown with arrows pointing inside the pri vate data to indicate that they
modify the data. The accessor functions are shown with arrows pointing the other
way to indicate that they just read the data.
5. What does the following code segment print?
CashRegister reg;
reg.clear();
reg.add_item(0.95);
reg.add_item(0.95);
cout << reg.get_count() << " " << reg.get_total() << endl;
6. What is wrong with the following code segment?
CashRegister reg;
reg.clear();
reg.add_item(0.95);
cout << reg.get_amount_due() << endl;
7. Declare a member function get_dollars of the CashRegister class that yields the
dollar value of the total amount of the sale.
8. Name two accessor member functions of the string class.
9. Is the get member function of the ifstream class an accessor or a mutator?
Practice it Now you can try these exercises at the end of the chapter: R9.3, R9.7.
a mutator member
function changes the
object on which
it operates.
an accessor member
function does not
change the object on
which it operates.
Use const
with accessors.
S e l f c h e c k
forgetting a Semicolon
Braces { } are common in C++ code, and usually you do not place a semicolon after the closing
brace. However, class definitions always end in };. A common error is to forget that semicolon:
class CashRegister
{
public:
...
private:
...
} // Forgot semicolon
int main()
{
// Many compilers report the error in this line
...
}
This error can be extremely confusing to many compilers. There is syntax, now obsolete but
supported for compati bility with old code, to define class types and variables of that type
simultaneously. Because the compiler doesn’t know that you don’t use that obsolete construc-
tion, it tries to analyze the code wrongly and ultimately reports an error. Unfortunately, it may
report the error several lines away from the line in which you forgot the semicolon.
If the compiler reports bizarre errors in lines that you are sure are correct, check that each
of the preceding class definitions is terminated by a semicolon.
9.3 Data Members
An object stores its data in data members.
These are variables that are declared inside the
class.
When implementing a class, you have to
determine which data each object needs to
store. The object needs to have all the infor-
mation necessary to carry out any member
function call.
Go through all member functions and
consider their data requirements. It is a good
idea to start with the accessor functions. For
example, a CashRegister object must be able
to return the correct value for the get_total
function. That means, it must either store all
entered prices and compute the total in the
func tion call, or it must store the total.
Now apply the same reasoning to the get_
count function. If the cash register stores all
entered prices, it can count them in the get_
count function. Otherwise, you need to have a
variable for the count.
Common error 9.1
Like a wilderness explorer who needs to
carry all items that may be needed, an
object needs to store the data required
for any function calls.
an object holds data
members that are
accessed by member
functions.
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9.3 Data Members 395
forgetting a Semicolon
Braces { } are common in C++ code, and usually you do not place a semicolon after the closing
brace. However, class definitions always end in };. A common error is to forget that semicolon:
class CashRegister
{
public:
...
private:
...
} // Forgot semicolon
int main()
{
// Many compilers report the error in this line
...
}
This error can be extremely confusing to many compilers. There is syntax, now obsolete but
supported for compati bility with old code, to define class types and variables of that type
simultaneously. Because the compiler doesn’t know that you don’t use that obsolete construc-
tion, it tries to analyze the code wrongly and ultimately reports an error. Unfortunately, it may
report the error several lines away from the line in which you forgot the semicolon.
If the compiler reports bizarre errors in lines that you are sure are correct, check that each
of the preceding class definitions is terminated by a semicolon.
9.3 Data Members
An object stores its data in data members.
These are variables that are declared inside the
class.
When implementing a class, you have to
determine which data each object needs to
store. The object needs to have all the infor-
mation necessary to carry out any member
function call.
Go through all member functions and
consider their data requirements. It is a good
idea to start with the accessor functions. For
example, a CashRegister object must be able
to return the correct value for the get_total
function. That means, it must either store all
entered prices and compute the total in the
func tion call, or it must store the total.
Now apply the same reasoning to the get_
count function. If the cash register stores all
entered prices, it can count them in the get_
count function. Otherwise, you need to have a
variable for the count.
Common error 9.1
Like a wilderness explorer who needs to
carry all items that may be needed, an
object needs to store the data required
for any function calls.
an object holds data
members that are
accessed by member
functions.
cfe2_ch09_p389_440.indd 395 10/28/10 11:42 AM
396 Chapter 9 Classes
The add_item function receives a price as an argument, and it must record the price.
If the CashRegister object stores an array of entered prices, then the add_item function
appends the price. On the other hand, if we decide to store just the item total and
count, then the add_item function updates these two variables.
Finally, the clear function must prepare the cash register for the next sale, either by
emptying the array of prices or by setting the total and count to zero.
We have now discovered two different ways of representing the data that the
object needs. Either of them will work, and we have to make a choice. We will choose
the simpler one: variables for the total price and the item count. (Other options are
explored in Exercises P9.3 and P9.4.)
The data members are defined in the private section of the class definition:
class CashRegister
{
public:
// See Section 9.2
private:
int item_count;
double total_price;
};
Every CashRegister object has a separate copy of these data members (see Figure 2).
Because the data members are defined to be private, only the member functions of
the class can access them. Programmers using the CashRegister class cannot access the
data members directly:
int main()
{
...
cout << register1.item_count; // Error—use get_count() instead
...
}
All data access must occur through the public interface. Thus, the data members of an
object are effec tively hidden from the programmer using the class. While it is theo-
retically possible in C++ to leave data members unencapsulated (by placing them into
the public section), this is very uncommon in practice. We will always make all data
members private in this book.
every object has its
own set of
data members.
private data
members can only
be accessed by
member functions
of the same class.
figure 2
Data Members of CashRegister objects
register1 =
register2 =
item_count =
CashRegister
total_price =
1
1.95
item_count =
CashRegister
total_price =
5
17.25
Accessible
only by CashRegister
member functions
cfe2_ch09_p389_440.indd 396 10/28/10 11:42 AM
9.4 Member Functions 397
10.
These clocks have common behavior, but each of them has a different state. Similarly, objects of
a class can have their data members set to different values.
What is the value of register1.item_count, register1.total_price, register2.item_
count, and register2.total_price after these statements?
CashRegister register1;
register1.clear();
register1.add_item(0.90);
register1.add_item(0.95);
CashRegister register2;
register2.clear();
register2.add_item(1.90);
11. What is wrong with this code segment?
CashRegister register2;
register2.clear();
register2.add_item(0.95);
cout << register2.total_price << endl;
12. Consider a class Time that represents a point in time, such as 9 a.m. or 3:30 p.m.
Give two different sets of data members that can be used for implementing the
Time class.
13. Suppose the implementor of the Time class changes from one implementation
strategy to another, keeping the public interface unchanged. What do the pro-
grammers who use the Time class need to do?
14. Consider a class Grade that represents a letter grade, such as A+ or B. Give two dif-
ferent sets of data members that can be used for implementing the Grade class.
Practice it Now you can try these exercises at the end of the chapter: R9.16, R9.17, R9.18.
9.4 Member Functions
The definition of a class declares its member functions. Each member function is
defined separately, after the class definition. The following sections show how to
define member functions.
S e l f c h e c k
cfe2_ch09_p389_440.indd 397 10/28/10 11:42 AM
398 Chapter 9 Classes
9.4.1 Implementing Member Functions
Here is the implementation of the add_item function of the CashRegister class.
void CashRegister::add_item(double price)
{
item_count++;
total_price = total_price + price;
}
The CashRegister:: prefix makes it clear that we are defining the add_item function of
the CashRegister class. In C++ it is perfectly legal to have add_item functions in other
classes as well, and it is important to specify exactly which add_item function we are
defining. (See Syntax 9.2 on page 400.) You use the Class Name::add_item syntax only
when defining the function, not when calling it. When you call the add_item member
function, the call has the form object.add_item(...).
When defining an accessor member function, supply the reserved word const fol-
lowing the closing parenthesis of the parameter list. Here is the get_count member
function:
int CashRegister::get_count() const
{
return item_count;
}
You will find the other member functions with the example program at the end of this
section.
9.4.2 Implicit and explicit parameters
Whenever you refer to a data member, such as item_count or total_price, in a member
function, it denotes the data member of the object on which the member function was
invoked. For example, consider the call
register1.add_item(1.95);
The first statement in the CashRegister::add_item function is
item_count++;
Which item_count is incremented? In this call, it is the item_count of the register1 object.
(See Figure 3.)
Use the ClassName::
prefix when defining
member functions.
When an item is added, it affects the
data members of the cash register
object on which the function is invoked.
When a member function is called on an object, the implicit parameter is a refer-
ence to that object.
You can think of the code of the add_item function like this:
void CashRegister::add_item(double price)
{
implicit parameter.item_count++;
implicit parameter.total_price = implicit parameter.total_price + price;
}
In C++, you do not actually write the implicit parameter in the function definition.
For that reason, the parameter is called “implicit”.
In contrast, parameter variables that are explicitly mentioned in the function defini-
tion, such as the price parameter variable, are called explicit parameters. Every mem-
ber function has exactly one implicit parameter and zero or more explicit parameters.
9.4.3 Calling a Member Function from a
Member Function
When one member function calls another member function on the same object, you
do not use the dot notation. Instead, you simply use the name of the other function.
Here is an example. Suppose we want to implement a member function to add multi-
ple instances of the same item. An easy way to implement this function is to repeat-
edly call the add_item function:
void CashRegister::add_items(int quantity, double price)
{
the implicit
parameter is a
reference to the
object on which a
member function
is applied.
explicit parameters
of a member function
are listed in the
function definition.
When calling another
member function on
the same object,
do not use the
dot notation.
cfe2_ch09_p389_440.indd 398 10/28/10 11:42 AM
9.4 Member Functions 399
figure 3 Implicit and explicit parameters
2 After the member function call register1.add_item(1.95).
1 Before the member function call.
register1 =
item_count =
CashRegister
total_price =
1
1.95
register1 =
item_count =
CashRegister
total_price =
0
0
The implicit parameter
references this object. The explicit parameter
is set to this argument.
When a member function is called on an object, the implicit parameter is a refer-
ence to that object.
You can think of the code of the add_item function like this:
void CashRegister::add_item(double price)
{
implicit parameter.item_count++;
implicit parameter.total_price = implicit parameter.total_price + price;
}
In C++, you do not actually write the implicit parameter in the function definition.
For that reason, the parameter is called “implicit”.
In contrast, parameter variables that are explicitly mentioned in the function defini-
tion, such as the price parameter variable, are called explicit parameters. Every mem-
ber function has exactly one implicit parameter and zero or more explicit parameters.
9.4.3 Calling a Member Function from a
Member Function
When one member function calls another member function on the same object, you
do not use the dot notation. Instead, you simply use the name of the other function.
Here is an example. Suppose we want to implement a member function to add multi-
ple instances of the same item. An easy way to implement this function is to repeat-
edly call the add_item function:
void CashRegister::add_items(int quantity, double price)
{
the implicit
parameter is a
reference to the
object on which a
member function
is applied.
explicit parameters
of a member function
are listed in the
function definition.
When calling another
member function on
the same object,
do not use the
dot notation.
cfe2_ch09_p389_440.indd 399 10/28/10 11:42 AM
400 Chapter 9 Classes
for (int i = 1; i <= quantity; i++)
{
add_item(price);
}
}
Here, the add_item member function is invoked on the implicit parameter.
for (int i = 1; i <= quantity; i++)
{
implicit parameter.add_item(price);
}
That is the object on which the add_items function is invoked. For example, in the call
register1.add_items(6, 0.95);
the add_item function is invoked six times on register1.
ch09/registertest1.cpp
1 #include
2 #include
3
4 using namespace std;
5
6 /**
7 A simulated cash register that tracks the item count and
8 the total amount due.
9 */
10 class CashRegister
11 {
12 public:
13 /**
14 Clears the item count and the total.
15 */
16 void clear();
syntax 9.2 Member Function Definition
void CashRegister::add_item(double price)
{
item_count++;
total_price = total_price + price;
}
int CashRegister::get_count() const
{
return item_count;
}
Use ClassName:: before the
name of the member function. Explicit parameter
Use const
for accessor functions.
See page 402.
Data members
of the implicit
parameter
Data member
of the implicit
parameter
cfe2_ch09_p389_440.indd 400 10/28/10 11:42 AM
9.4 Member Functions 401
17
18 /**
19 Adds an item to this cash register.
20 @param price the price of this item
21 */
22 void add_item(double price);
23
24 /**
25 @return the total amount of the current sale
26 */
27 double get_total() const;
28
29 /**
30 @return the item count of the current sale
31 */
32 int get_count() const;
33
34 private:
35 int item_count;
36 double total_price;
37 };
38
39 void CashRegister::clear()
40 {
41 item_count = 0;
42 total_price = 0;
43 }
44
45 void CashRegister::add_item(double price)
46 {
47 item_count++;
48 total_price = total_price + price;
49 }
50
51 double CashRegister::get_total() const
52 {
53 return total_price;
54 }
55
56 int CashRegister::get_count() const
57 {
58 return item_count;
59 }
60
61 /**
62 Displays the item count and total price of a cash register.
63 @param reg the cash register to display
64 */
65 void display(CashRegister reg)
66 {
67 cout << reg.get_count() << " $" << fixed << setprecision(2)
68 << reg.get_total() << endl;
69 }
70
71 int main()
72 {
73 CashRegister register1;
74 register1.clear();
75 register1.add_item(1.95);
76 display(register1);
cfe2_ch09_p389_440.indd 401 10/28/10 11:42 AM
402 Chapter 9 Classes
77 register1.add_item(0.95);
78 display(register1);
79 register1.add_item(2.50);
80 display(register1);
81 return 0;
82 }
Program run
1 $1.95
2 $2.90
3 $5.40
15. What is wrong with this implementation of the get_total member function?
int get_total()
{
return total_price;
}
16. Implement the add_items member function described in Section 9.4.3 without
calling add_item.
17. Implement a member function get_dollars of the CashRegister class that yields the
amount of the total sale as a dollar value without the cents.
18. Consider the substr member function of the string class. How many parameters
does it have, and what are their types?
19. Consider the length member function of the string class. How many parameters
does it have, and what are their types?
Practice it Now you can try these exercises at the end of the chapter: P9.3, P9.4, P9.6.
all data members Should be Private;
most member functions Should be Public
It is possible to define data members in the public section of a class, but you should not do that
in your own code. Always use encapsulation, with private data members that are manipulated
with member functions.
Generally, member functions should be public. However, sometimes you have a member
function that is only used as a helper function by other member functions. In that case, you
can make the helper function private. Simply declare it in the private section of the class.
const correctness
You should declare all accessor functions in C++ with the const reserved word. (Recall that an
accessor function is a member function that does not modify its implicit parameter.)
For example, suppose you design the following class:
class CashRegister
{
void display(); // Bad style—no const
...
S e l f c h e c k
programming tip 9.1
programming tip 9.2
};
When you compile your code, no error is reported. But now suppose that another program-
mer uses your CashRegis ter class in a function
void display_all(const CashRegister registers[])
{
for (int i = 0; i < NREGISTERS; i++) { registers[i].display(); }
}
That programmer is conscientious and declares the registers parameter variable as const.
But then the call registers[i].dis play() will not compile. Because CashRegister::display is
not tagged as const, the compiler suspects that the call regis ters[i].display() may modify
registers[i]. But the function promised not to modify the registers array.
If you write a program with other team members who are conscientious about const, it is
very important that you do your part as well. You should therefore get into the habit of using
const with all accessor member functions.
9.5 Constructors
A constructor is a member function that initializes the data members of an object.
The constructor is automatically called whenever an object is created. By supplying a
constructor, you can ensure that all data members are properly set before any mem-
ber functions act on an object.
To understand the importance of constructors, consider the following statements:
CashRegister register1;
register1.add_item(1.95);
int count = register1.get_count(); // May not be 1
Here, the programmer forgot to call clear before adding items. Therefore, the data
members of the register1 object were initialized with random values. Constructors
guarantee that an object is always fully initialized when it is defined.
The name of a constructor is identical to the name of its class. You declare con-
structors in the class defi nition, for example:
class CashRegister
{
public:
CashRegister(); // A constructor
...
};
Constructors never return values, but you do not use the void reserved word when
declaring them.
Here is the definition of that constructor:
CashRegister::CashRegister()
{
item_count = 0;
total_price = 0;
}
In the constructor definition, the first CashRegister (before the ::) indicates that we are
about to define a member function of the CashRegister class. The second CashRegister is
the name of that member function.
a constructor is
called automatically
whenever an object
is created.
the name of a
constructor is the
same as the
class name.
cfe2_ch09_p389_440.indd 402 10/28/10 11:42 AM
9.5 Constructors 403
};
When you compile your code, no error is reported. But now suppose that another program-
mer uses your CashRegis ter class in a function
void display_all(const CashRegister registers[])
{
for (int i = 0; i < NREGISTERS; i++) { registers[i].display(); }
}
That programmer is conscientious and declares the registers parameter variable as const.
But then the call registers[i].dis play() will not compile. Because CashRegister::display is
not tagged as const, the compiler suspects that the call regis ters[i].display() may modify
registers[i]. But the function promised not to modify the registers array.
If you write a program with other team members who are conscientious about const, it is
very important that you do your part as well. You should therefore get into the habit of using
const with all accessor member functions.
9.5 Constructors
A constructor is a member function that initializes the data members of an object.
The constructor is automatically called whenever an object is created. By supplying a
constructor, you can ensure that all data members are properly set before any mem-
ber functions act on an object.
To understand the importance of constructors, consider the following statements:
CashRegister register1;
register1.add_item(1.95);
int count = register1.get_count(); // May not be 1
Here, the programmer forgot to call clear before adding items. Therefore, the data
members of the register1 object were initialized with random values. Constructors
guarantee that an object is always fully initialized when it is defined.
The name of a constructor is identical to the name of its class. You declare con-
structors in the class defi nition, for example:
class CashRegister
{
public:
CashRegister(); // A constructor
...
};
Constructors never return values, but you do not use the void reserved word when
declaring them.
Here is the definition of that constructor:
CashRegister::CashRegister()
{
item_count = 0;
total_price = 0;
}
In the constructor definition, the first CashRegister (before the ::) indicates that we are
about to define a member function of the CashRegister class. The second CashRegister is
the name of that member function.
a constructor is
called automatically
whenever an object
is created.
the name of a
constructor is the
same as the
class name.
cfe2_ch09_p389_440.indd 403 10/28/10 11:42 AM
404 Chapter 9 Classes
A constructor is like a set of
assembly instructions for an object.
The constructor that you just saw has no arguments. Such a constructor is called a
default construc tor. It is used whenever you define an object and do not specify any
parameters for the construction. For example, if you define
CashRegister register1;
then the default constructor is called. It sets register1.item_count and register1.total_
price to zero.
Many classes have more than one constructor. This allows you to define objects
in different ways. Consider for example a BankAccount class that has two constructors:
class BankAccount
{
public:
BankAccount(); // Sets balance to 0
BankAccount(double initial_balance); // Sets balance to initial_balance
// Member functions omitted
private:
double balance;
};
Both constructors have the same name as the class, BankAccount. But the default con-
structor has no param eter variables, whereas the second constructor has a double
parameter variable. (This is an example of overloading—see Special Topic 9.2.)
When you construct an object, the compiler chooses the constructor that matches
the arguments that you supply. For example,
BankAccount joes_account;
// Uses default constructor
BankAccount lisas_account(499.95);
// Uses BankAccount(double) constructor
When implementing a constructor, you need to pay particular attention to all data
members that are num bers or pointers. These types are not classes and therefore have
no constructors. If you have a data mem ber that is an object of a class (such as a string
object), then that class has a constructor, and the object will be initialized. For exam-
ple, all string objects are automatically initialized to the empty string.
Consider this class:
class Item
{
a default constructor
has no arguments.
a class can have
multiple
constructors.
the compiler picks
the constructor that
matches the
construction
arguments.
Be sure to initialize
all number and
pointer data
members in a
constructor.
public:
Item();
// Additional member functions omitted
private:
string description;
double price;
};
In the Item constructor, you need to set price to 0, but you need not initialize the
description data member. It is automatically initialized to the empty string.
If you do not supply any constructor for a class, the compiler automatically gener-
ates a default con structor. The default constructor initializes all data members of class
type with their default constructors and leaves the other data members uninitialized.
20. Provide an implementation for the default constructor of the BankAccount class.
21. Provide an implementation for the BankAccount(double) constructor.
22. Provide an implementation for the default constructor of the Item class.
23. Provide an implementation for the default constructor of the CashRegister class
that calls the clear member function.
24. Which constructor is called in each of the following definitions?
a. Item item1;
b. Item item2("Corn flakes");
c. Item item3(3.95);
d. Item item4("Corn flakes", 3.95);
e. Item item5();
Practice it Now you can try these exercises at the end of the chapter: R9.13, R9.14, P9.5, P9.8.
trying to call a constructor
The constructor is invoked only when an object is first created. You cannot invoke it again.
For example, you cannot call the constructor to clear an object:
CashRegister register1;
...
register1.CashRegister(); // Error
It is true that the default constructor sets a new CashRegister object to the cleared state, but you
cannot invoke a con structor on an existing object.
initializer lists
When you construct an object whose data members are themselves objects, those objects are
constructed by their class’s default constructor. However, if a data member belongs to a class
without a default constructor, you need to invoke the data member’s constructor explicitly.
Here is an example.
This Item class has no default constructor:
class Item
{
S e l f c h e c k
Common error 9.2
special topic 9.1
cfe2_ch09_p389_440.indd 404 10/28/10 11:42 AM
9.5 Constructors 405
public:
Item();
// Additional member functions omitted
private:
string description;
double price;
};
In the Item constructor, you need to set price to 0, but you need not initialize the
description data member. It is automatically initialized to the empty string.
If you do not supply any constructor for a class, the compiler automatically gener-
ates a default con structor. The default constructor initializes all data members of class
type with their default constructors and leaves the other data members uninitialized.
20. Provide an implementation for the default constructor of the BankAccount class.
21. Provide an implementation for the BankAccount(double) constructor.
22. Provide an implementation for the default constructor of the Item class.
23. Provide an implementation for the default constructor of the CashRegister class
that calls the clear member function.
24. Which constructor is called in each of the following definitions?
a. Item item1;
b. Item item2("Corn flakes");
c. Item item3(3.95);
d. Item item4("Corn flakes", 3.95);
e. Item item5();
Practice it Now you can try these exercises at the end of the chapter: R9.13, R9.14, P9.5, P9.8.
trying to call a constructor
The constructor is invoked only when an object is first created. You cannot invoke it again.
For example, you cannot call the constructor to clear an object:
CashRegister register1;
...
register1.CashRegister(); // Error
It is true that the default constructor sets a new CashRegister object to the cleared state, but you
cannot invoke a con structor on an existing object.
initializer lists
When you construct an object whose data members are themselves objects, those objects are
constructed by their class’s default constructor. However, if a data member belongs to a class
without a default constructor, you need to invoke the data member’s constructor explicitly.
Here is an example.
This Item class has no default constructor:
class Item
{
S e l f c h e c k
Common error 9.2
special topic 9.1
cfe2_ch09_p389_440.indd 405 10/28/10 11:42 AM
406 Chapter 9 Classes
public:
Item(string item_description, double item_price);
// No other constructors
...
};
This Order class has a data member of type Item:
class Order
{
public:
Order(string customer_name, string item_description, double item_price);
...
private:
Item article;
string customer;
};
The Order constructor must call the Item constructor. That is achieved with an initializer list.
The initializer list is placed before the opening brace of the constructor. The list starts with a
colon and contains names of data members with their construction arguments.
Order(string customer_name, string item_description, double item_price)
: article(item_description, item_price)
{
customer = customer_name;
}
Initializers are separated by commas. The Order constructor can also be written like this:
Order(string customer_name, string item_description, double item_price)
: article(item_description, item_price), customer(customer_name)
{
}
Overloading
When the same function name is used for more than one function, then the name is over-
loaded. In C++ you can overload function names provided the types of the parameter vari-
ables are different. For example, you can define two functions, both called print:
void print(CashRegister r)
void print(Item i)
When the print function is called,
print(x);
the compiler looks at the type of x. If x is a CashRegister object, the first function is called. If x
is an Item object, the second function is called. If x is neither, the compiler generates an error.
It is always possible to avoid overloading by giving each function a unique name, such as
print_register or print_item. However, we have no choice with constructors. C++ demands
that the name of a con structor equal the name of the class. If a class has more than one con-
structor, then that name must be overloaded.
In addition to name overloading, C++ also supports operator overloading. It is possible
to give new meanings to the familiar C++ operators such as +, ==, and <<. This is an advanced
technique that we do not discuss in this book.
special topic 9.2
9.6 problem solving: tracing objects
You have seen how the technique of hand tracing is useful for understanding how a
program works. When your program contains objects, it is useful to adapt the tech-
nique so that you gain a better under standing about object data and encapsulation.
Use an index card or a sticky note for each object. On the front, write the member
functions that the object can execute. On the back, make a table for the values of the
data members.
Here is a card for a CashRegister object:
In a small way, this gives you a feel for encapsulation. An object is manipulated
through its public inter face (on the front of the card), and the data members are hid-
den in the back.
When an object is constructed, fill in the initial values of the data members.
Whenever a mutator member function is executed, cross out the old values and write
the new ones below. Here is what happens after a call to the add_item member function:
If you have more than one object in your program, you will have multiple cards, one
for each object:
Write the member
functions on the
front of a card, and
the data member
values on the back.
Update the values of
the data members
when a mutator
member function
is called.
cfe2_ch09_p389_440.indd 406 10/28/10 11:42 AM
9.6 problem solving: tracing objects 407
9.6 problem solving: tracing objects
You have seen how the technique of hand tracing is useful for understanding how a
program works. When your program contains objects, it is useful to adapt the tech-
nique so that you gain a better under standing about object data and encapsulation.
Use an index card or a sticky note for each object. On the front, write the member
functions that the object can execute. On the back, make a table for the values of the
data members.
Here is a card for a CashRegister object:
CashRegister reg1
clear
add_item(price)
get_total
get_count
item_count total_price
front back
In a small way, this gives you a feel for encapsulation. An object is manipulated
through its public inter face (on the front of the card), and the data members are hid-
den in the back.
When an object is constructed, fill in the initial values of the data members.
item_count total_price
0 0
Whenever a mutator member function is executed, cross out the old values and write
the new ones below. Here is what happens after a call to the add_item member function:
item_count total_price
0 0
1 19.95
If you have more than one object in your program, you will have multiple cards, one
for each object:
Write the member
functions on the
front of a card, and
the data member
values on the back.
Update the values of
the data members
when a mutator
member function
is called.
cfe2_ch09_p389_440.indd 407 10/28/10 11:42 AM
408 Chapter 9 Classes
item_count total_price
0 0
1 19.95
2 14.90
item_count total_price
0 0
1 19.95
These diagrams are also useful when you design a class. Suppose you are asked to
enhance the CashRegister class to compute the sales tax. Add a function get_sales_tax
to the front of the card. Now turn the card over, look over the data members, and ask
yourself whether the object has sufficient information to com pute the answer.
Remember that each object is an autonomous unit. Any data value that can be used in
a computation must be
• A data member.
• A function argument.
• A global constant or variable.
To compute the sales tax, we need to know the tax rate and the total of the taxable
items. (Food items are usually not subject to sales tax.) We don’t have that informa-
tion available. Let us introduce additional data members for the tax rate and the tax-
able total. The tax rate can be set in the constructor (assuming it stays fixed for the
lifetime of the object). When adding an item, we need to be told whether the item is
taxable. If so, we add its price to the taxable total.
For example, consider the following statements.
CashRegister reg2(7.5); // 7.5 percent sales tax
reg2.add_item(3.95, false); // not taxable
reg2.add_item(19.95, true); // taxable
When you record the effect on a card, it looks like this:
taxable_total tax_rate
0 7.5
19.95
item_count total_price
0 0
1 3.95
2 23.90
With this information, it becomes easy to compute the tax. It is taxable_total x tax_rate /
100. Tracing the object helped us understand the need for additional data members.
25. Consider a Car class that simulates fuel consumption in a car. We will assume a
fixed efficiency (in miles per gallon) that is supplied in the constructor. There are
member functions for adding gas, driv ing a given distance, and checking the
amount of gas left in the tank. Make a card for a Car object, choosing suitable
data members and showing their values after the object was constructed.
S e l f c h e c k
26. Trace the following member function calls:
Car my_car(25);
my_car.add_gas(20);
my_car.drive(100);
my_car.drive(200);
my_car.add_gas(5);
27. Suppose you are asked to simulate the odometer
of the car, by adding a mem ber function get_miles_
driven. Add a data member to the object’s card that
is suitable for computing this member function.
28. Trace the member functions of Self Check 26, updating the data member that
you added in Self Check 27.
Practice it Now you can try these exercises at the end of the chapter: R9.20, R9.21, R9.22.
Step 1 Get an informal list of the responsibilities of your objects.
Be careful that you restrict yourself to features that are actually required in the problem. With
real-world items, such as cash registers or bank accounts, there are potentially dozens of fea-
tures that might be worth implementing. However, your job is not to faithfully model the
real world. You need to determine only those responsibilities that you need for solving your
specific problem.
In the case of the menu, you need to
Display the menu.
Get user input.
Now look for hidden responsibilities that aren’t part of the problem description. How do
objects get created? Which mundane activities need to happen, such as clearing the cash regis-
ter at the beginning of each sale?
In the menu example, consider how a menu is produced. The programmer creates an empty
menu object and then adds options “Open new account”, “Help”, and so on. There is a hidden
responsibility:
Add an option.
Step 2 Specify the public interface.
Turn the list in Step 1 into a set of member functions, with specific types for the parameter
variables and the return values. Be sure to mark accessors as const. Many programmers find
h o W t o 9 . 1 implementing a class
A very common task is to implement a class whose objects can carry out a set of specified
actions. This How To walks you through the necessary steps.
As an example, consider a class Menu. An object of this class can display a menu such as
1) Open new account
2) Log into existing account
3) Help
4) Quit
Then the menu waits for the user to supply a value. If the
user does not supply a valid value, the menu is redisplayed,
and the user can try again.
cfe2_ch09_p389_440.indd 408 10/28/10 11:42 AM
9.6 problem solving: tracing objects 409
26. Trace the following member function calls:
Car my_car(25);
my_car.add_gas(20);
my_car.drive(100);
my_car.drive(200);
my_car.add_gas(5);
27. Suppose you are asked to simulate the odometer
of the car, by adding a mem ber function get_miles_
driven. Add a data member to the object’s card that
is suitable for computing this member function.
28. Trace the member functions of Self Check 26, updating the data member that
you added in Self Check 27.
Practice it Now you can try these exercises at the end of the chapter: R9.20, R9.21, R9.22.
Step 1 Get an informal list of the responsibilities of your objects.
Be careful that you restrict yourself to features that are actually required in the problem. With
real-world items, such as cash registers or bank accounts, there are potentially dozens of fea-
tures that might be worth implementing. However, your job is not to faithfully model the
real world. You need to determine only those responsibilities that you need for solving your
specific problem.
In the case of the menu, you need to
Display the menu.
Get user input.
Now look for hidden responsibilities that aren’t part of the problem description. How do
objects get created? Which mundane activities need to happen, such as clearing the cash regis-
ter at the beginning of each sale?
In the menu example, consider how a menu is produced. The programmer creates an empty
menu object and then adds options “Open new account”, “Help”, and so on. There is a hidden
responsibility:
Add an option.
Step 2 Specify the public interface.
Turn the list in Step 1 into a set of member functions, with specific types for the parameter
variables and the return values. Be sure to mark accessors as const. Many programmers find
h o W t o 9 . 1 implementing a class
A very common task is to implement a class whose objects can carry out a set of specified
actions. This How To walks you through the necessary steps.
As an example, consider a class Menu. An object of this class can display a menu such as
1) Open new account
2) Log into existing account
3) Help
4) Quit
Then the menu waits for the user to supply a value. If the
user does not supply a valid value, the menu is redisplayed,
and the user can try again.
cfe2_ch09_p389_440.indd 409 10/28/10 11:42 AM
410 Chapter 9 Classes
this step simpler if they write out member function calls that are applied to a sample object,
like this:
Menu main_menu;
main_menu.add_option("Open new account");
// Add more options
int input = main_menu.get_input();
Now we have a specific list of member functions.
• void add_option(string option)
• int get_input() const
What about displaying the menu? There is no sense in displaying the menu without also ask-
ing the user for input. However, get_input may need to display the menu more than once if the
user provides a bad input. Thus, display is a good candidate for a private member function.
To complete the public interface, you need to specify the constructors. Ask yourself what
information you need in order to construct an object of your class. Sometimes you will want
two constructors: one that sets all data members to a default and one that sets them to user-
supplied values.
In the case of the menu example, we can get by with a single constructor that creates an
empty menu.
Here is the public interface:
class Menu
{
public:
Menu();
void add_option(string option);
int get_input() const
private:
...
};
Step 3 Document the public interface.
Supply a documentation comment for the class, then comment each member function.
/**
A menu that is displayed on a console.
*/
class Menu
{
public:
/**
Constructs a menu with no options.
*/
Menu();
/**
Adds an option to the end of this menu.
@param option the option to add
*/
void add_option(string option);
/**
Displays the menu, with options numbered starting with 1,
and prompts the user for input. Repeats until a valid input
is supplied.
@return the number that the user supplied
*/
int get_input() const;
cfe2_ch09_p389_440.indd 410 10/28/10 11:42 AM
9.6 problem solving: tracing objects 411
private:
...
};
Step 4 Determine data members.
Ask yourself what information an object needs to store to do its job. Remember, the member
functions can be called in any order! The object needs to have enough internal memory to be
able to process every member function using just its data members and the member function
arguments. Go through each member function, perhaps starting with a simple one or an inter-
esting one, and ask yourself what you need to carry out the member function’s task. Make data
members to store the information that the member function needs.
In the menu example, we clearly need to store the menu options so that the menu can
be displayed. How should we store them? As a vector of strings? As one long string? Both
approaches can be made to work. We will use a vec tor here. Exercise P9.7 asks you to imple-
ment the other approach.
class Menu
{
...
private:
...
vector
};
When checking for user input, we need to know the number of menu items. Because we store
them in a vector, the number of menu items is simply obtained as the size of the vector. If you
stored the menu items in one long string, you might want to keep another data member that
stores the menu item count.
Step 5 Implement constructors and member functions.
Implement the constructors and member functions in your class, one at a time, starting with
the easiest ones. For example, here is the implementation of the add_option member function:
void Menu::add_option(string option)
{
options.push_back(option);
}
Here is the get_input member function. This member function is a bit more sophisticated. It
loops until a valid input has been obtained, and it calls the private display member function to
display the menu.
int Menu::get_input() const
{
int input;
do
{
display();
cin >> input;
}
while (input < 1 || input > options.size());
return input;
}
Finally, here is the display member function:
void Menu::display() const
{
for (int i = 0; i < options.size(); i++)
{
cout << i + 1 << ") " << options[i] << endl;
}
cfe2_ch09_p389_440.indd 411 10/28/10 11:42 AM
412 Chapter 9 Classes
}
The Menu constructor is a bit odd. We need to construct a menu with no options. A vector is a
class, and it has a default constructor. That constructor does exactly what we want, namely to
construct an empty vector. Nothing else needs to be done:
Menu::Menu()
{
}
If you find that you have trouble with the implementation of some of your member functions,
you may need to rethink your choice of data members. It is common for a beginner to start out
with a set of data members that cannot accurately describe the state of an object. Don’t hesitate
to go back and rethink your implementation strategy.
Once you have completed the implementation, compile your class and fix any compiler
errors. (See ch09/menu.cpp in your book’s companion code for the completed class.)
Step 6 Test your class.
Write a short tester program and execute it. The tester program should carry out the member
function calls that you found in Step 2.
int main()
{
Menu main_menu;
main_menu.add_option("Open new account");
main_menu.add_option("Log into existing account");
main_menu.add_option("Help");
main_menu.add_option("Quit");
int input = main_menu.get_input();
cout << "Input: " << input << endl;
return 0;
}
Program run
1) Open new account
2) Log into existing account
3) Help
4) Quit
5
1) Open new account
2) Log into existing account
3) Help
4) Quit
3
Input: 3
W o r k e D e x a M p l e 9 . 1 implementing a bank account class
This Worked Example shows how to develop a class that simulates a bank account.
In the 2000 presi-
den tial elections in
the United states, votes were tallied by
a variety of machines. some machines
processed cardboard ballots into which
voters punched holes to indi cate their
choices (see photo below). When voters
were not careful, remains of paper—
the now infamous “chads”—were par-
tially stuck in the punch cards, caus-
ing votes to be mis counted. a manual
recount was neces sary, but it was not
carried out everywhere due to time
constraints and procedural wrangling.
the elec tion was very close, and there
remain doubts in the minds of many
people whether the election outcome
would have been different if the voting
machines had accurately counted the
intent of the voters.
subsequently, voting machine man-
ufacturers have argued that electronic
voting machines would avoid the prob-
lems caused by punch cards or opti-
cally scanned forms. In an elec tronic
voting machine, voters indicate their
preferences by pressing buttons or
touching icons on a computer screen.
typically, each voter is pre sented with
a summary screen for review before
casting the ballot. the process is very
similar to using an automatic bank
teller machine.
It seems plausible that these
machines make it more likely that a
vote is counted in the same way that
the voter intends. however, there has
been significant controversy surround-
ing some types of electronic voting
machines. If a machine simply records
the votes and prints out the totals after
the election has been com pleted, then
how do you know that the machine
worked correctly? Inside the machine
is a computer that executes a program,
and, as you may know from your own
experience, programs can have bugs.
In fact, some electronic voting
machines do have bugs. there have
been isolated cases where machines
reported tallies that were impossible.
When a machine reports far more or far
fewer votes than voters, then it is clear
that it malfunctioned. Unfortu nately, it
is then impossible to find out the actual
votes. over time, one would expect
these bugs to be fixed in the software.
More insidiously, if the results are plau-
sible, nobody may ever investigate.
Punch Card Ballot
Random Fact 9.1 electronic Voting Machines
Touch Screen Voting Machine
Many computer scientists have spo-
ken out on this issue and con firmed
that it is impossible, with today’s
technology, to tell that soft ware is
error free and has not been tampered
with. Many of them recom mend that
electronic voting machines should
employ a voter verifiable audit trail. (a
good source of information is http://
verifiedvoting.org.) typically, a voter-
verifiable machine prints out a ballot.
each voter has a chance to review the
printout, and then deposits it in an old-
fashioned ballot box. If there is a prob-
lem with the electronic equipment, the
printouts can be scanned or counted
by hand.
as this book is written, this con-
cept is strongly resisted both by
man ufacturers of electronic voting
machines and by their customers, the
cities and counties that run elections.
Manufacturers are reluctant to increase
the cost of the machines because
they may not be able to pass the cost
increase on to their custom ers, who
tend to have tight budgets. election
officials fear problems with malfunc-
tioning printers, and some of them
have publicly stated that they actually
prefer equipment that elimi nates both-
ersome recounts.
What do you think? You probably
use an automatic bank teller machine
to get cash from your bank account.
Do you review the paper record that
the machine issues? Do you check your
bank statement? even if you don’t, do
you put your faith in other people who
double-check their bal ances, so that
the bank won’t get away with wide-
spread cheating?
at any rate, is the integrity of
banking equipment more important
or less important than that of voting
machines? Won’t every voting process
have some room for error and fraud
anyway? Is the added cost for equip-
ment, paper, and staff time reason-
able to combat a potentially slight risk
of malfunction and fraud? Computer
sci entists cannot answer these ques-
tions—an informed society must make
these tradeoffs. But, like all profes-
sionals, they have an obligation to
speak out and give accurate testimony
about the capabilities and limitations
of computing equipment.
cfe2_ch09_p389_440.indd 412 10/28/10 11:42 AM
9.6 problem solving: tracing objects 413
int input = main_menu.get_input();
cout << "Input: " << input << endl;
return 0;
}
Program run
1) Open new account
2) Log into existing account
3) Help
4) Quit
5
1) Open new account
2) Log into existing account
3) Help
4) Quit
3
Input: 3
W o r k e D e x a M p l e 9 . 1 implementing a bank account class
This Worked Example shows how to develop a class that simulates a bank account.
In the 2000 presi-
den tial elections in
the United states, votes were tallied by
a variety of machines. some machines
processed cardboard ballots into which
voters punched holes to indi cate their
choices (see photo below). When voters
were not careful, remains of paper—
the now infamous “chads”—were par-
tially stuck in the punch cards, caus-
ing votes to be mis counted. a manual
recount was neces sary, but it was not
carried out everywhere due to time
constraints and procedural wrangling.
the elec tion was very close, and there
remain doubts in the minds of many
people whether the election outcome
would have been different if the voting
machines had accurately counted the
intent of the voters.
subsequently, voting machine man-
ufacturers have argued that electronic
voting machines would avoid the prob-
lems caused by punch cards or opti-
cally scanned forms. In an elec tronic
voting machine, voters indicate their
preferences by pressing buttons or
touching icons on a computer screen.
typically, each voter is pre sented with
a summary screen for review before
casting the ballot. the process is very
similar to using an automatic bank
teller machine.
It seems plausible that these
machines make it more likely that a
vote is counted in the same way that
the voter intends. however, there has
been significant controversy surround-
ing some types of electronic voting
machines. If a machine simply records
the votes and prints out the totals after
the election has been com pleted, then
how do you know that the machine
worked correctly? Inside the machine
is a computer that executes a program,
and, as you may know from your own
experience, programs can have bugs.
In fact, some electronic voting
machines do have bugs. there have
been isolated cases where machines
reported tallies that were impossible.
When a machine reports far more or far
fewer votes than voters, then it is clear
that it malfunctioned. Unfortu nately, it
is then impossible to find out the actual
votes. over time, one would expect
these bugs to be fixed in the software.
More insidiously, if the results are plau-
sible, nobody may ever investigate.
Punch Card Ballot
Random Fact 9.1 electronic Voting Machines
Touch Screen Voting Machine
Many computer scientists have spo-
ken out on this issue and con firmed
that it is impossible, with today’s
technology, to tell that soft ware is
error free and has not been tampered
with. Many of them recom mend that
electronic voting machines should
employ a voter verifiable audit trail. (a
good source of information is http://
verifiedvoting.org.) typically, a voter-
verifiable machine prints out a ballot.
each voter has a chance to review the
printout, and then deposits it in an old-
fashioned ballot box. If there is a prob-
lem with the electronic equipment, the
printouts can be scanned or counted
by hand.
as this book is written, this con-
cept is strongly resisted both by
man ufacturers of electronic voting
machines and by their customers, the
cities and counties that run elections.
Manufacturers are reluctant to increase
the cost of the machines because
they may not be able to pass the cost
increase on to their custom ers, who
tend to have tight budgets. election
officials fear problems with malfunc-
tioning printers, and some of them
have publicly stated that they actually
prefer equipment that elimi nates both-
ersome recounts.
What do you think? You probably
use an automatic bank teller machine
to get cash from your bank account.
Do you review the paper record that
the machine issues? Do you check your
bank statement? even if you don’t, do
you put your faith in other people who
double-check their bal ances, so that
the bank won’t get away with wide-
spread cheating?
at any rate, is the integrity of
banking equipment more important
or less important than that of voting
machines? Won’t every voting process
have some room for error and fraud
anyway? Is the added cost for equip-
ment, paper, and staff time reason-
able to combat a potentially slight risk
of malfunction and fraud? Computer
sci entists cannot answer these ques-
tions—an informed society must make
these tradeoffs. But, like all profes-
sionals, they have an obligation to
speak out and give accurate testimony
about the capabilities and limitations
of computing equipment.
Available online at www.wiley.com/college/horstmann.
cfe2_ch09_p389_440.indd 413 10/28/10 11:42 AM
www.wiley.com/college/horstmann
http://verifiedvoting.org
http://verifiedvoting.org
414 Chapter 9 Classes
9.7 problem solving: Discovering Classes
When you solve a problem using objects and classes, you need to determine the
classes required for the implementation. You may be able to reuse existing classes, or
you may need to implement new ones. One simple approach for discovering classes
and member functions is to look for the nouns and verbs in the problem description.
Often, nouns correspond to classes, and verbs correspond to member functions.
Concepts from the problem domain, be it science, business, or a game, often make
good classes. Exam ples are
• Cannonball
• CashRegister
• Monster
The name for such a class should be a noun that describes the concept. Other
frequently used classes rep resent system services such as files or menus.
What might not be a good class? If you can’t tell from the class name what an
object of the class is sup posed to do, then you are probably not on the right track. For
example, your homework assignment might ask you to write a program that prints
paychecks. Suppose you start by trying to design a class Pay checkProgram. What would
an object of this class do? An object of this class would have to do everything that the
homework needs to do. That doesn’t simplify anything. A better class would be Pay-
check. Then your program can manipulate one or more Paycheck objects.
Another common mistake, particularly by students who are used to writing pro-
grams that consist of functions, is to turn an action into a class. For example, if your
homework assignment is to compute a paycheck, you may consider writing a class
ComputePaycheck. But can you visualize a “ComputePaycheck” object? The fact that
“ComputePaycheck” isn’t a noun tips you off that you are on the wrong track. On
the other hand, a Paycheck class makes intuitive sense. The word “paycheck” is a noun.
You can visualize a paycheck object. You can then think about useful member func-
tions of the Paycheck class, such as compute_taxes, that help you solve the assignment.
When you analyze a problem description, you often find that you need multiple
classes. It is then help ful to consider how these classes are related. One of the funda-
mental relationships between classes is the “aggregation” relationship (which is
informally known as the “has-a” relationship).
to discover classes,
look for nouns in the
problem description.
Concepts from the
problem domain are
good candidates
for classes.
In a class scheduling system, potential
classes from the problem domain include
Class, LectureHall, Instructor, and Student.
The aggregation relationship states that objects of one class contain objects of
another class. Consider a quiz that is made up of questions. Since each quiz has one or
more questions, we say that the class Quiz aggregates the class Question. There is a stan-
dard notation, called a UML (Unified Modeling Language) class diagram, to describe
class relationships. In the UML notation, aggregation is denoted by a line with a dia-
mond-shaped symbol (see Figure 4).
Finding out about aggregation is very helpful for deciding how to implement
classes. For example, when you implement the Quiz class, you will want to store the
questions of a quiz as a data member. Since a quiz can have any number of questions,
you will choose a vector:
class Quiz
{
...
private:
vector
};
In summary, when you analyze a problem description, you will want to carry out
these tasks:
• Find the concepts that you need to implement as classes. Often, these will be
nouns in the problem description.
• Find the responsibilities of the classes. Often, these will be verbs in the problem
description.
• Find relationships between the classes that you have discovered. In this section,
we described the aggregation relationship. In the next chapter, you will learn
about another important relationship between classes, called inheritance.
a class aggregates
another if its objects
contain objects of the
other class.
A car has a motor and tires. In object-oriented design,
this “has-a” relationship is called aggregation.
cfe2_ch09_p389_440.indd 414 10/28/10 11:42 AM
9.7 problem solving: Discovering Classes 415
figure 4 Class Diagram
Quiz Question
The aggregation relationship states that objects of one class contain objects of
another class. Consider a quiz that is made up of questions. Since each quiz has one or
more questions, we say that the class Quiz aggregates the class Question. There is a stan-
dard notation, called a UML (Unified Modeling Language) class diagram, to describe
class relationships. In the UML notation, aggregation is denoted by a line with a dia-
mond-shaped symbol (see Figure 4).
Finding out about aggregation is very helpful for deciding how to implement
classes. For example, when you implement the Quiz class, you will want to store the
questions of a quiz as a data member. Since a quiz can have any number of questions,
you will choose a vector:
class Quiz
{
…
private:
vector
};
In summary, when you analyze a problem description, you will want to carry out
these tasks:
• Find the concepts that you need to implement as classes. Often, these will be
nouns in the problem description.
• Find the responsibilities of the classes. Often, these will be verbs in the problem
description.
• Find relationships between the classes that you have discovered. In this section,
we described the aggregation relationship. In the next chapter, you will learn
about another important relationship between classes, called inheritance.
a class aggregates
another if its objects
contain objects of the
other class.
A car has a motor and tires. In object-oriented design,
this “has-a” relationship is called aggregation.
cfe2_ch09_p389_440.indd 415 10/28/10 11:42 AM
416 Chapter 9 Classes
29. What is the rule of thumb for finding classes?
30. Your job is to write a program that plays chess. Might ChessBoard be an appropri-
ate class? How about MovePiece?
31. In an e-mail system, messages are stored in a mailbox. Draw a UML diagram
that shows the appro priate aggregation relationship.
32. You are implementing a system to manage a library, keeping track of which
books are checked out by whom. Should the Book class aggregate Patron or the
other way around?
33. In a library management system, what would be the relationship between classes
Patron and Author?
Practice it Now you can try these exercises at the end of the chapter: R9.23, R9.24, R9.26.
make Parallel vectors into vectors of Objects
Sometimes, you find yourself using vectors of the same length, each of which stores a part of
what conceptually should be an object. In that situation, it is a good idea to reorganize your
program and use a single vector whose ele ments are objects.
For example, suppose an invoice contains a series of item descriptions and prices. One
solution is to keep two vectors:
vector
vector
Each of the vectors will have the same length, and the ith slice, consisting of descriptions[i]
and prices[i], contains data that needs to be processed together. These vectors are called
parallel vectors (see Figure 5).
Parallel vectors become a headache in larger programs. The programmer must ensure that
the vectors always have the same length and that each slice is filled with values that actually
belong together. Moreover, any function that operates on a slice must get all of the vectors as
arguments, which is tedious to program.
The remedy is simple. Look at the slice and find the concept that
it represents. Then make the concept into a class. In this example,
each slice contains the description and price of an item; turn this
into a class.
class Item
{
public:
…
private:
string description;
double price;
};
S e l f c h e c k
programming tip 9.3
avoid parallel vectors
by changing them
into vectors of
objects.
figure 5 parallel Vectors
[i]
descriptions =
[i]
prices =
A slice
You can now eliminate the parallel vectors and replace them with a single vector:
vector
Each slot in the resulting vector corresponds to a slice in the set of parallel vectors (see
Figure 6).
9.8 separate Compilation
When you write and compile small programs, you can place all your code into a sin-
gle source file. When your programs get larger or you work in a team, that situation
changes. You will want to split your code into separate source files. There are two
reasons why this split becomes necessary. First, it takes time to compile a file, and it
seems silly to wait for the compiler to keep translating code that doesn’t change. If
your code is distributed over several source files, then only those files that you change
need to be recompiled. The second reason becomes apparent when you work with
other programmers in a team. It would be very difficult for multiple programmers to
edit a single source file simultaneously. Therefore, the program code is broken up so
that each programmer is solely responsible for a separate set of files.
If your program is composed of multiple files, some of these files will define data
types or functions that are needed in other files. There must be a path of communica-
tion between the files. In C++, that communication happens through the inclusion of
header files.
A header file contains
• Definitions of classes.
• Definitions of constants.
• Declarations of nonmember functions.
the code of complex
programs is
distributed over
multiple files.
header files contain
the definitions of
classes and
declarations of
nonmember
functions.
cfe2_ch09_p389_440.indd 416 10/28/10 11:42 AM
9.8 separate Compilation 417
figure 6 eliminating parallel Vectors
Parallel vectors
A vector of objects
You can now eliminate the parallel vectors and replace them with a single vector:
vector
Each slot in the resulting vector corresponds to a slice in the set of parallel vectors (see
Figure 6).
9.8 separate Compilation
When you write and compile small programs, you can place all your code into a sin-
gle source file. When your programs get larger or you work in a team, that situation
changes. You will want to split your code into separate source files. There are two
reasons why this split becomes necessary. First, it takes time to compile a file, and it
seems silly to wait for the compiler to keep translating code that doesn’t change. If
your code is distributed over several source files, then only those files that you change
need to be recompiled. The second reason becomes apparent when you work with
other programmers in a team. It would be very difficult for multiple programmers to
edit a single source file simultaneously. Therefore, the program code is broken up so
that each programmer is solely responsible for a separate set of files.
If your program is composed of multiple files, some of these files will define data
types or functions that are needed in other files. There must be a path of communica-
tion between the files. In C++, that communication happens through the inclusion of
header files.
A header file contains
• Definitions of classes.
• Definitions of constants.
• Declarations of nonmember functions.
the code of complex
programs is
distributed over
multiple files.
header files contain
the definitions of
classes and
declarations of
nonmember
functions.
cfe2_ch09_p389_440.indd 417 10/28/10 11:42 AM
418 Chapter 9 Classes
The source file contains
• Definitions of member functions.
• Definitions of nonmember functions.
For the CashRegister class, you create a pair of files, cashregister.h and cashregister.cpp,
that contain the interface and the implementation, respectively.
The header file contains the class definition:
ch09/cashregister.h
1 #ifndef CASHREGISTER_H
2 #define CASHREGISTER_H
3
4 /**
5 A simulated cash register that tracks the item count and
6 the total amount due.
7 */
8 class CashRegister
9 {
10 public:
11 /**
12 Constructs a cash register with cleared item count and total.
13 */
14 CashRegister();
15
16 /**
17 Clears the item count and the total.
18 */
19 void clear();
20
21 /**
22 Adds an item to this cash register.
23 @param price the price of this item
24 */
25 void add_item(double price);
26
27 /**
28 @return the total amount of the current sale
29 */
30 double get_total() const;
31
32 /**
33 @return the item count of the current sale
34 */
35 int get_count() const;
36
37 private:
38 int item_count;
39 double total_price;
40 };
41
42 #endif
You include this header file whenever the definition of the CashRegister class is
required. Because this file is not a standard header file, you must enclose its name in
quotes, not <...>, when you include it, like this:
#include “cashregister.h”
source files contain
function
implementations.
cfe2_ch09_p389_440.indd 418 10/28/10 11:42 AM
9.8 separate Compilation 419
Note the set of directives that bracket the header file:
#ifndef CASHREGISTER_H
#define CASHREGISTER_H
…
#endif
Suppose a file includes two header files: cashregister.h, and another header file that
itself includes cashregister.h. The effect of the directives is to skip the file when it is
encountered the second time. If we did not have that check, the compiler would com-
plain when it saw the definition for the CashRegister class twice. (Sadly, it doesn’t
check whether the definitions are identical.)
The source file for the CashRegister class simply contains the definitions of the
member functions (including constructors).
Note that the source file cashregister.cpp includes its own header file cashregister.h.
The compiler needs to know how the CashRegister class is defined in order to compile
the member functions.
ch09/cashregister.cpp
1 #include “cashregister.h”
2
3 CashRegister::CashRegister()
4 {
5 clear();
6 }
7
8 void CashRegister::clear()
9 {
10 item_count = 0;
11 total_price = 0;
12 }
13
14 void CashRegister::add_item(double price)
15 {
16 item_count++;
17 total_price = total_price + price;
18 }
19
20 double CashRegister::get_total() const
21 {
22 return total_price;
23 }
24
25 int CashRegister::get_count() const
26 {
27 return item_count;
28 }
Note that the function comments are in the header file, because comments are a part
of the interface, not the implementation.
The cashregister.cpp file does not contain a main function. There are many potential
programs that might make use of the CashRegister class. Each of these programs will
need to supply its own main func tion, as well as other functions and classes.
cfe2_ch09_p389_440.indd 419 10/28/10 11:42 AM
420 Chapter 9 Classes
Here is a simple test program that puts the CashRegister class to use. Its source file
includes the cashreg ister.h header file.
ch09/registertest2.cpp
1 #include
2 #include
3 #include “cashregister.h”
4
5 using namespace std;
6
7 /**
8 Displays the item count and total price of a cash register.
9 @param reg the cash register to display
10 */
11 void display(CashRegister reg)
12 {
13 cout << reg.get_count() << " $" << fixed << setprecision(2)
14 << reg.get_total() << endl;
15 }
16
17 int main()
18 {
19 CashRegister register1;
20 register1.clear();
21 register1.add_item(1.95);
22 display(register1);
23 register1.add_item(0.95);
24 display(register1);
25 register1.add_item(2.50);
26 display(register1);
27 return 0;
28 }
To build the complete program, you need to compile both the registertest2.cpp and
cashregister.cpp source files (see Figure 7). The details depend on your compiler. For
example, with the Gnu compiler, you issue the command
g++ -o registertest registertest2.cpp cashregister.cpp
figure 7 Compiling a program from Multiple source Files
registertest2.cppcashregister.cpp
cashregister.h iostreamiomanip
includes
includes
Compiled
into executable
program
cfe2_ch09_p389_440.indd 420 10/28/10 11:42 AM
9.8 separate Compilation 421
You have just seen the simplest and most common case for designing header and
source files. There are a few additional technical details that you should know.
• A header file should include all headers that are necessary for defining the class.
For example, if a class uses the string class, include the
Anytime you include a header from the stan dard library, also include the directive
using namespace std;
item.h
1 #include
2 using namespace std;
3
4 class Item
5 {
6 …
7 private:
8 string description
9 };
• Place shared constants into a header file. For example,
volumes.h
6 const double CAN_VOLUME = 0.355;
• To share a nonmember function, place the function declaration into a header file
and the definition of the function into the corresponding source file.
cube.h
8 double cube_volume(double side_length);
cube.cpp
1 #include “cube.h”
2
3 double cube_volume(double side_length)
4 {
5 double volume = side_length * side_length * side_length;
6 return volume;
7 }
34. Suppose the cash register is enhanced to carry out direct debits from bank
accounts, and a member function debit(BankAccount&) is added to the Cash Register
class. Which header file do you need to include in cashregister.h?
35. In the enhancement described in Self Check 34, what additional file do you need
to include in cashreg ister.cpp?
36. In the enhancement described in Self Check 34, what additional file do you need
to include in the bankaccount.h file?
37. Suppose we want to move the display function from registertest2.cpp to the
cashregister.h and cashregister.cpp files. Explain how those files need to change.
38. Where is the header file located that you include with the #include
directive?
Practice it Now you can try these exercises at the end of the chapter: R9.28, R9.29, P9.20.
S e l f c h e c k
cfe2_ch09_p389_440.indd 421 10/28/10 11:42 AM
422 Chapter 9 Classes
9.9 pointers to objects
The following sections discuss how to work with pointers to objects. As you will see
in the next chapter, pointers to objects are important when you work with multiple
objects from related classes.
9.9.1 Dynamically allocating objects
It is common to allocate objects on the heap. As discussed in Section 7.4, you use the
new operator to obtain memory from the heap. For example, the call
new CashRegister
returns a pointer to a CashRegister object. You can also supply construction
arguments:
new BankAccount(1000)
You usually want to store the pointer that the new operator returns:
CashRegister* register_pointer = new CashRegister;
BankAccount* account_pointer = new BankAccount(1000);
Note that each of these definitions allocates two entities: a pointer variable and an
object on the heap—see Figure 8.
When you no longer need a heap object, be sure to delete it:
delete account_pointer;
9.9.2 the -> operator
Because register_pointer is a pointer to a CashRegister object, the value *register_
pointer denotes the CashRegister object itself. To invoke a member function on that
object, you might call
(*register_pointer).add_item(1.95);
Use the new operator
to obtain an object
that is located on
the heap.
the new operator
returns a pointer to
the allocated object.
When a heap object is
no longer needed,
use the delete
operator to reclaim
its memory.
figure 8 pointers and the objects to Which they point
register_pointer = CashRegister
account_pointer =
BankAccount
The parentheses are necessary because in C++ the dot operator takes precedence over
the * operator. The expression without the parentheses would be a compile-time
error:
*register_pointer.add_item(1.95); // Error—you can’t apply . to a pointer
Because the dot operator has higher precedence than *, the dot would be applied to a
pointer, not an object.
Because calling a member function through a pointer is very common, there is
an operator to abbrevi ate the “follow pointer and call member function” operation.
That operator is written -> and usually pro nounced as “arrow”. Here is how you use
the “arrow” operator:
register_pointer->add_item(1.95);
This call means: When invoking the add_item member function, set the implicit param-
eter to *register_pointer and the explicit parameter to 1.95.
9.9.3 the this pointer
Each member function has a special parameter variable, called this, which is a pointer
to the implicit parameter. For example, consider the CashRegister::add_item function.
If you call
register1.add_item(1.95)
then the this pointer has type CashRegister* and points to the register1 object.
You can use the this pointer inside the definition of a member function. For exam-
ple, you can imple ment the add_item function as
void CashRegister::add_item(double price)
{
this->item_count++;
this->total_price = this->total_price + price;
}
Here, the expression this->item_count refers to the item_count data member of the
implicit parameter (which is register1.item_count in our example). Some program-
mers like to use the this pointer in this fash ion to make it clear that item_count is a data
member and not a variable.
39. Write a statement that dynamically allocates a string object and saves the address
in a pointer variable str_pointer.
40. Write a statement that deallocates the object that was allocated in Self Check 39.
41. Write a statement that dynamically allocates a string object with contents “Hello”
and saves the address in a pointer variable str_pointer.
42. Write a statement that invokes the length member function on the object that
was allocated in Self Check 41 and prints the result.
43. What is the type of this when the string::length member function is called?
Practice it Now you can try these exercises at the end of the chapter: R9.30, P9.21, P9.22.
Use the -> operator
to invoke a member
function through
a pointer.
In a member
function, the this
pointer points to the
implicit parameter.
S e l f c h e c k
cfe2_ch09_p389_440.indd 422 10/28/10 11:42 AM
9.9 pointers to objects 423
The parentheses are necessary because in C++ the dot operator takes precedence over
the * operator. The expression without the parentheses would be a compile-time
error:
*register_pointer.add_item(1.95); // Error—you can’t apply . to a pointer
Because the dot operator has higher precedence than *, the dot would be applied to a
pointer, not an object.
Because calling a member function through a pointer is very common, there is
an operator to abbrevi ate the “follow pointer and call member function” operation.
That operator is written -> and usually pro nounced as “arrow”. Here is how you use
the “arrow” operator:
register_pointer->add_item(1.95);
This call means: When invoking the add_item member function, set the implicit param-
eter to *register_pointer and the explicit parameter to 1.95.
9.9.3 the this pointer
Each member function has a special parameter variable, called this, which is a pointer
to the implicit parameter. For example, consider the CashRegister::add_item function.
If you call
register1.add_item(1.95)
then the this pointer has type CashRegister* and points to the register1 object.
You can use the this pointer inside the definition of a member function. For exam-
ple, you can imple ment the add_item function as
void CashRegister::add_item(double price)
{
this->item_count++;
this->total_price = this->total_price + price;
}
Here, the expression this->item_count refers to the item_count data member of the
implicit parameter (which is register1.item_count in our example). Some program-
mers like to use the this pointer in this fash ion to make it clear that item_count is a data
member and not a variable.
39. Write a statement that dynamically allocates a string object and saves the address
in a pointer variable str_pointer.
40. Write a statement that deallocates the object that was allocated in Self Check 39.
41. Write a statement that dynamically allocates a string object with contents “Hello”
and saves the address in a pointer variable str_pointer.
42. Write a statement that invokes the length member function on the object that
was allocated in Self Check 41 and prints the result.
43. What is the type of this when the string::length member function is called?
Practice it Now you can try these exercises at the end of the chapter: R9.30, P9.21, P9.22.
Use the -> operator
to invoke a member
function through
a pointer.
In a member
function, the this
pointer points to the
implicit parameter.
S e l f c h e c k
cfe2_ch09_p389_440.indd 423 10/28/10 11:42 AM
424 Chapter 9 Classes
destructors and resource management
A destructor is a special member function that is automatically executed under two
circumstances:
• At the end of the block in which an object variable is defined
• When a heap object is deleted
To understand the need for destructors, consider an implementation of a string class, similar to
that of the C++ library. The characters of a string are stored on the heap, and each String object
contains a pointer to the array hold ing its characters.
class String
{
…
private:
char* char_array;
}
The constructor allocates and initializes the character array:
String::String(const char initial_chars[])
{
char_array = new char[strlen(initial_chars) + 1];
strcpy(char_array, initial_chars);
}
It is the job of the destructor to deallocate this memory. The name of the destructor is the ~
character followed by the class name, that is, ~String in our case. A class can have only one
destructor, and the destructor has no arguments.
String::~String()
{
delete[] char_array;
}
When a String object is no longer needed, the destructor is automatically invoked, and the
memory for the characters is properly recycled (see Figure 9):
void fun()
{
String name(“Harry”); // 1 Heap memory is allocated by the constructor
…
} // 2 The destructor is invoked on name, and its heap memory is deallocated
As a rule of thumb, if a constructor calls new, you should supply a destructor that calls delete.
Unfortunately, just supplying a destructor is not enough. Consider this scenario:
String name1(“Harry”);
String name2(“Sally”);
name1 = name2;
The assignment has a very unfortunate effect: The memory for the first String has not been
deallocated. (The destructor is only called at the end of the block in which name1 is defined.)
And the two String objects now share a pointer to the same area of heap memory. Eventually,
that memory location will be deleted twice (see Figure 10).
This problem can be overcome by redefining what it means to assign one object to another.
In this context, the assignment needs to
• Delete the memory of the String object on the left-hand side of the assignment (name1 in
our example).
• Allocate a new memory block to the left-hand side object, and fill it with a copy of the
string on the right-hand side.
special topic 9.3
cfe2_ch09_p389_440.indd 424 10/28/10 11:42 AM
9.9 pointers to objects 425
figure 9 the Destructor Deallocates heap Memory
1
name =
Heap
H a r r y
String
char_array =
2
name =
Heap
H a r r y
String
char_array =
Removed when
object went out
of scope
Deleted
by destructor
In addition, one must also supply a “copy constructor” for making safe copies, for example,
when passing an object as a function argument. The destructor, assignment operator, and copy
constructor are often called the “big 3” operations of memory management in C++. (See
Horstmann & Budd, Big C++, 2nd Ed., Chapter 15, for details.)
figure 10 a Bad assignment
name1 =
Heap
H a r r y
String
char_array =
name2 =
a l l y
String
char_array =
Will be
deleted twice
S
Orphaned
heap memory
cfe2_ch09_p389_440.indd 425 10/28/10 11:43 AM
426 Chapter 9 Classes
understand the concepts of objects and classes.
• A class describes a set of objects with the same behavior.
• Every class has a public interface: a collection of mem ber functions through
which the objects of the class can be manipulated.
• Encapsulation is the act of providing a public interface and
hiding implementation details.
• Encapsulation enables changes in the implementation
without affecting users of a class.
Most companies that
produce software
regard the source code as a trade
secret. after all, if customers or com-
petitors had access to the source code,
they could study it and create similar
programs without paying the original
vendor. For the same reason, custom-
ers dislike secret source code. If a com-
pany goes out of business or decides
to discontinue support for a computer
program, its users are left stranded.
they are unable to fix bugs or adapt
the program to a new operat ing sys-
tem. nowadays, some software pack-
ages are distributed with “open source”
or “free software” licenses. here, the
term “free” doesn’t refer to price, but
to the freedom to inspect and modify
the source code. richard stallman, a
famous computer scientist and win-
ner of a Macarthur “genius” grant,
pioneered the concept of free soft-
ware. he is the inventor of the emacs
text editor and the originator of the
GnU project, which aims to create an
entirely free version of a Unix-compat-
ible operating system. all programs of
the GnU project are licensed under the
General Public License or Gpl. the Gpl
allows you to make as many copies as
you wish, make any modifications to
the source, and redistribute the origi-
nal and modified programs, charging
nothing at all or whatever the market
will bear. In return, you must agree that
your modifications also fall under the
Gpl. You must give out the source code
to any changes that you distrib ute,
and anyone else can distribute them
under the same conditions. the Gpl,
and similar open source licenses, form
a social contract. Users of the software
enjoy the freedom to use and modify
the software, and in return they are
obligated to share any improvements
that they make. Many programs, such
as the linux operating system and
the GnU C++ compiler, are distributed
under the Gpl.
some commercial software ven dors
have attacked the Gpl as “viral” and
“undermining the commercial software
sector”. other companies have a more
nuanced strategy, pro ducing propri-
etary software while also contributing
to open source projects.
Frankly, open source is not a pana-
cea and there is plenty of room for
the commercial software sector. open
source software often lacks the polish
of commercial software because many
of the programmers are volunteers
who are interested in solving their
own problems, not in making a prod-
uct that is easy to use by others. some
product categories are not available at
all as open source software because
the development work is unattractive
when there is little promise of com-
mercial gain. open source software has
been most successful in areas that are
of interest to programmers, such as
the linux operating system, Web serv-
ers, and programming tools.
on the positive side, the open soft-
ware community can be very competi-
tive and creative. It is quite common
to see several competing projects that
take ideas from each other, all rap-
idly becoming more capable. having
many programmers involved, all read-
ing the source code, often means that
bugs tend to get squashed quickly.
eric ray mond describes open source
develop ment in his famous article “the
Cathedral and the Bazaar” (http://
catb.org/~esr/writings/cathedral-
bazaar/cathedral-bazaar/index.html).
he writes “Given enough eyeballs, all
bugs are shallow”.
Richard Stallman, a pioneer of the
free source movement.
Random Fact 9.2 open source and Free software
C h a p t e r s U M M a r Y
cfe2_ch09_p389_440.indd 426 10/28/10 11:43 AM
http://catb.org/~esr/writings/cathedralbazaar/cathedral-bazaar/index.html
http://catb.org/~esr/writings/cathedralbazaar/cathedral-bazaar/index.html
http://catb.org/~esr/writings/cathedralbazaar/cathedral-bazaar/index.html
Chapter summary 427
formulate the public interface of a class in c++.
• A mutator member function changes the object on which it operates.
• An accessor member function does not change the object on which it operates.
Use const with accessors.
choose data members to represent the state of an object.
• An object holds data members that are accessed by member functions.
• Every object has its own set of data members.
• Private data members can only be accessed by
member functions of the same class.
implement member functions of a class.
• Use the ClassName:: prefix when defining member functions.
• The implicit parameter is a reference to the object on which a member function
is applied.
• Explicit parameters of a member function are listed in the function definition.
• When calling another member function on the same object, do not use the dot
notation.
design and implement constructors.
• A constructor is called automatically whenever an object is created.
• The name of a constructor is the same as the class name.
• A default constructor has no arguments.
• A class can have multiple constructors.
• The compiler picks the constructor that matches the construction arguments.
• Be sure to initialize all number and pointer data members in a constructor.
use the technique of object tracing for visualizing object behavior.
• Write the member functions on the front of a card, and the data member values on
the back.
• Update the values of the data members when a mutator member function is called.
discover classes that are needed for solving a programming problem.
• To discover classes, look for nouns in the problem description.
• Concepts from the problem domain are good candi dates for classes.
• A class aggregates another if its objects contain objects
of the other class.
• Avoid parallel vectors by changing them into vectors of
objects.
cfe2_ch09_p389_440.indd 427 10/28/10 11:43 AM
428 Chapter 9 Classes
Separate the interface and implementation of a class in header and source files.
• The code of complex programs is distributed over multiple files.
• Header files contain the definitions of classes and declarations of nonmember
functions.
• Source files contain function implementations.
use pointers to objects and manage dynamically allocated objects.
• Use the new operator to obtain an object that is located on the heap.
• The new operator returns a pointer to the allocated object.
• When a heap object is no longer needed, use the delete operator to reclaim its
memory.
• Use the -> operator to invoke a member function through a pointer.
• In a member function, the this pointer points to the implicit parameter.
r9.1 List all classes in the C++ library that you have encountered in this book up to this
point.
r9.2 Consider a Date class that stores a calendar date such as November 20, 2011. Con-
sider two possible implementations of this class: by storing the day, month, and year,
and by storing the number of days since January 1, 1900. Why might an implemen-
tor prefer the second version? How does the choice affect the user of the class?
r9.3 Write a partial C++ class definition that contains the public interface of the Date class
described in Exercise R9.2. Supply member functions for setting the date to a
particular year, month, and day, for advancing the date by a given number of days,
and for finding the number of days between this date and another. Pay atten tion
to const.
r9.4 What value is returned by the calls reg1.get_count(), reg1.get_total(), reg2.get_
count(), and reg2.get_total() after these statements?
CashRegister reg1;
reg1.clear();
reg1.add_item(0.90);
reg1.add_item(1.95);
CashRegister reg2;
reg2.clear();
reg2.add_item(1.90);
r9.5 Consider the Menu class in How To 9.1 on page 409. What is displayed when the fol-
lowing calls are executed?
Menu simple_menu;
simple_menu.add_option(“Ok”);
simple_menu.add_option(“Cancel”);
int response = simple_menu.get_input();
r9.6 What is the interface of a class? How does it differ from the implementation of
a class?
registertest2.cppcashregister.cpp
cashregister.h iostreamiomanip
includes
includes
r e V I e W e x e r C I s e s
cfe2_ch09_p389_440.indd 428 10/28/10 11:43 AM
review exercises 429
r9.7 What is a member function, and how does it differ from a nonmember function?
r9.8 What is a mutator function? What is an accessor function?
r9.9 What happens if you forget the const in an accessor function? What happens if you
accidentally supply a const in a mutator function?
r9.10 What is an implicit parameter? How does it differ from an explicit parameter?
r9.11 How many implicit parameters can a member function have? How many implicit
parameters can a nonmember function have? How many explicit parameters can a
function have?
r9.12 What is a constructor?
r9.13 What is a default constructor? What is the consequence if a class does not have a
default constructor?
r9.14 How many constructors can a class have? Can you have a class with no construc-
tors? If a class has more than one constructor, which of them gets called?
r9.15 What is encapsulation? Why is it useful?
r9.16 Data members are hidden in the private section of a class, but they aren’t hidden very
well at all. Anyone can read the private section. Explain to what extent the pri vate
reserved word hides the private members of a class.
r9.17 You can read the item_count data member of the CashRegister class with the get_count
accessor function. Should there be a set_count mutator function to change it? Explain
why or why not.
r9.18 Suppose you implement a ChessBoard class. Provide data members to store the pieces
on the board and the player who has the next turn.
r9.19 In a nonmember function, it is easy to differentiate between calls to member func-
tions and calls to nonmember functions. How do you tell them apart? Why is it not
as easy for functions that are called from a member function?
r9.20 Using the object tracing technique described in Section 9.6, trace the program at the
end of Section 9.4.
r9.21 Using the object tracing technique described in Section 9.6, trace the program in
Worked Example 9.1.
r9.22 Design a modification of the BankAccount class in Worked Example 9.1 in which the
first five transactions per month are free and a $1 fee is charged for every additional
transaction. Provide a member function that deducts the fee at the end of a month.
What additional data members do you need? Using the object tracing technique
described in Section 9.6, trace a scenario that shows how the fees are computed over
two months.
r9.23 Consider the following problem description:
Users place coins in a vending machine and select a product by pushing a button. If the inserted coins
are suffi cient to cover the purchase price of the product, the product is dispensed and change is given.
Otherwise, the inserted coins are returned to the user.
What classes should you use to implement it?
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430 Chapter 9 Classes
r9.24 Consider the following problem description:
Employees receive their biweekly paychecks. They are paid their hourly rates for each hour worked;
however, if they worked more than 40 hours per week, they are paid overtime at 150 percent of
their regular wage.
What classes should you use to implement it?
r9.25 Consider the following problem description:
Customers order products from a store. Invoices are generated to list the items and quantities ordered,
payments received, and amounts still due. Products are shipped to the shipping address of the custom-
er, and invoices are sent to the billing address.
What classes should you use to implement it?
r9.26 Suppose a vending machine contains products, and users insert coins into the
vend ing machine to purchase products. Draw a UML diagram showing the aggre-
gation relationships between the classes VendingMachine, Coin, and Product.
r9.27 Suppose an Invoice object contains descriptions of the products ordered and the
billing and shipping address of the customer. Draw a UML diagram showing the
aggregation relationships between the classes Invoice, Address, Customer, and Product.
r9.28 Consider the implementation of a program that plays TicTacToe, with two classes
Player and TicTacToeBoard. Each class implementation is placed in its own C++ source
file and each class interface is placed in its own header file. In addition, a source file
game.cpp contains the code for playing the game and displaying the scores of the
players. Describe the contents of each header file, and determine in which files each
of them is included.
r9.29 What would happen if the display function was moved from registertest2.cpp to
cashregister.h? Try it out if you are not sure.
r9.30 In Exercise P9.22, a MenuItem optionally contains a Menu. Generally, there are two ways
for implementing an “optional” relationship. You can use a pointer that may be NULL.
Or you may use an indicator, such as a bool value, that specifies whether the optional
item is present. Why are pointers required in this case?
P9.1 Implement a class that models a tally counter, a mechani-
cal device that is used to count people—for example, to
find out how many people attend a concert or board a
bus. Whenever the operator pushes a button, the counter
value advances by one. Model this opera tion with a count
member function. A physical counter has a display to
show the current value. In your class, use a get_value
member function instead.
P9.2 Implement a class Rectangle. Provide a constructor to construct a rectangle with a
given width and height, member functions get_perimeter and get_area that compute
the perimeter and area, and a member function void resize(double factor) that resizes
the rectangle by multiplying the width and height by the given factor.
p r o G r a M M I n G e x e r C I s e s
cfe2_ch09_p389_440.indd 430 10/28/10 11:43 AM
programming exercises 431
P9.3 Reimplement the CashRegister class so that it keeps track of the price of each added
item in a vector
Reim plement the clear, add_item, get_total, and get_count member functions. Add a
mem ber function display_all that displays the prices of all items in the current sale.
P9.4 Reimplement the CashRegister class so that it keeps track of the total price as an
inte ger: the total cents of the price. For example, instead of storing 17.29, store the
inte ger 1729. Such an implementation is commonly used because it avoids the
accumulation of roundoff errors. Do not change the public interface of the class.
P9.5 Add a feature to the CashRegister class for computing sales tax. The tax rate should
be supplied when constructing a CashRegister object. Add add_taxable_item and
get_total_tax member functions. (Items added with add_item are not taxable.)
P9.6 After closing time, the store manager would like to know how much business was
transacted during the day. Modify the CashRegister class to enable this functionality.
Supply member functions get_sales_total and get_sales_count to get the total amount
of all sales and the number of sales. Supply a member function reset_sales that resets
any counters and totals so that the next day’s sales start from zero.
P9.7 Reimplement the Menu class so that it stores all menu items in one long string.
Hint: Keep a separate counter for the number of options. When a new option is
added, append the option count, the option, and a newline character.
P9.8 Implement a class StreetAddress. An address has a house number, a street, an optional
apartment number, a city, a state, and a postal code. Supply two construc tors: one
with an apartment number and one without. Supply a print member func tion that
prints the address with the street (and optional apartment number) on one line and
the city, state, and postal code on the next line. Supply a member function comes_be-
fore that tests whether one address comes before another when the addresses are
compared by postal code.
P9.9 Implement a class SodaCan with member functions get_surface_area() and
get_volume(). In the constructor, supply the height and radius of the can.
P9.10 Implement a class Portfolio. This class has two data members, checking
and savings, of the type BankAccount that was developed in Worked
Example 9.1 (ch09/account.cpp in your code files). Implement four
member functions:
deposit(double amount, string account)
withdraw(double amount, string account)
transfer(double amount, string account)
print_balances()
Here the account string is “S” or “C”. For the deposit or withdrawal, it indicates
which account is affected. For a transfer, it indicates the account from which the
money is taken; the money is automatically trans ferred to the other account.
P9.11 Implement a class Student. For the purpose of this exercise, a student has a name and
a total quiz score. Supply an appropriate constructor and functions get_name(),
add_quiz(int score), get_total_score(), and get_average_score(). To compute the latter,
you also need to store the number of quizzes that the student took.
P9.12 Modify the Student class of Exercise P9.11 to compute grade point averages. Mem ber
functions are needed to add a grade and get the current GPA. Specify grades as
cfe2_ch09_p389_440.indd 431 10/28/10 11:43 AM
432 Chapter 9 Classes
elements of a class Grade. Supply a constructor that constructs a grade from a string,
such as “B+”. You will also need a function that translates grades into their numeric
values (for example, “B+” becomes 3.3).
P9.13 Define a class Country that stores the name of the country, its population, and its area.
Using that class, write a program that reads in a set of countries and prints
• The country with the largest area.
• The country with the largest population.
• The country with the largest population density (people per square
kilometer or mile).
P9.14 Design a class Message that models an e-mail message. A message has a recipient, a
sender, and a message text. Support the following member functions:
• A constructor that takes the sender and recipient and sets the time stamp to the
current time
• A member function append that appends a line of text to the message body
• A member function to_string that makes the message into one long string like
this: “From: Harry Hacker\nTo: Rudolf Reindeer\n …”
• A member function print that prints the message text. Hint: Use to_string.
Write a program that uses this class to make a message and print it.
P9.15 Design a class Mailbox that stores e-mail messages, using the Message class of Exercise
P9.14. Implement the following member functions:
void Mailbox::add_message(Message m)
Message Mailbox::get_message(int i) const
void remove_message(int i)
P9.16 Implement a VotingMachine class that can be used for a simple election. Have member
functions to clear the machine state, to vote for a Democrat, to vote for a Republi-
can, and to get the tallies for both parties. (Hint: Use a function in the
to get the current date.)
P9.17 Provide a class for authoring a simple letter. In the constructor, supply the names of
the sender and the recipient:
Letter(string from, string to)
Supply a member function
void add_line(string line)
to add a line of text to the body of the letter. Supply a member function
string get_text()
that returns the entire text of the letter. The text has the form:
Dear recipient name:
blank line
first line of the body
second line of the body
…
last line of the body
blank line
Sincerely,
cfe2_ch09_p389_440.indd 432 10/28/10 11:43 AM
programming exercises 433
blank line
sender name
Also supply a main function that prints this letter.
Dear John:
I am sorry we must part.
I wish you all the best.
Sincerely,
Mary
Construct an object of the Letter class and call add_line twice.
P9.18 Write a class Bug that models a bug moving along a horizontal line. The bug moves
either to the right or left. Initially, the bug moves to the right, but it can turn to
change its direction. In each move, its position changes by one unit in the current
direction. Provide a constructor
Bug(int initial_position)
and member functions
void turn()
void move()
int get_position()
Sample usage:
Bug bugsy(10);
bugsy.move(); // Now the position is 11
bugsy.turn();
bugsy.move(); // Now the position is 10
Your main function should construct a bug, make it move and turn a few times, and
print the actual and expected positions.
P9.19 Implement a class Moth that models a moth flying in a straight line. The moth has a
position, the distance from a fixed origin. When the moth moves toward a point of
light, its new position is halfway between its old position and the position of the
light source. Supply a constructor
Moth(double initial_position)
and member functions
void move_to_light(double light_position)
double get_position()
Your main function should construct a moth, move it toward a couple of light
sources, and check that the moth’s position is as expected.
P9.20 Implement classes Person and StreetAddress. Each person has a street address. Pro vide
display functions in each class for displaying their contents. Distribute your code
over three source files, one for each class, and one containing the main function.
Construct two Person objects and display them.
P9.21 Modify the Person class in Exercise P9.20 so that it contains a pointer to the street
address. Construct and display two Person objects that share the same StreetAddress
object.
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434 Chapter 9 Classes
P9.22 Reimplement the Menu class from How To 9.1 to support submenus similar to the
ones in a graphical user interface.
A menu contains a sequence of menu items. Each menu item has a name and, option-
ally, a submenu.
Implement classes Menu and MenuItem. Supply a function to display a menu and get
user input. Simply num ber the displayed items and have the user enter the number
of the selected item. When an item with a sub menu is selected, display the submenu.
Otherwise simply print a message with the name of the selected item.
Note that there is a circular dependency between the two classes. To break it, first
provide a declaration
class Menu;
Then define the MenuItem class and finally define the Menu class and the main function.
engineering P9.23 Define a class ComboLock that works like the combination lock in a
gym locker, as shown here. The lock is constructed with a combina-
tion—three numbers between 0 and 39. The reset function resets
the dial so that it points to 0. The turn_left and turn_right functions
turn the dial by a given number of ticks to the left or right. The open
func tion attempts to open the lock. The lock opens if the user first
turned it right to the first number in the combination, then left to
the second, and then right to the third.
class ComboLock
{
public:
ComboLock(int secret1, int secret2, int secret3);
void reset();
void turn_left(int ticks);
void turn_right(int ticks);
bool open() const;
cfe2_ch09_p389_440.indd 434 10/28/10 11:43 AM
programming exercises 435
private:
…
};
engineering P9.24 Implement a class Car with the following properties. A car has a certain fuel effi-
ciency (measured in miles̸gallon) and a certain amount of fuel in the gas tank. The
efficiency is specified in the constructor, and the initial fuel level is 0. Supply a
member function drive that simulates driving the car for a certain distance, reducing
the fuel level in the gas tank, and member functions get_gas_level, to return the
cur rent fuel level, and add_gas, to tank up. Sample usage:
Car my_hybrid(50); // 50 miles per gallon
my_hybrid.add_gas(20); // Tank 20 gallons
my_hybrid.drive(100); // Drive 100 miles
cout << my_hybrid.get_gas_level() << endl; // Print fuel remaining
engineering P9.25 Write a program that prints all real solutions to the quadratic equation
ax2 + bx + c = 0. Read in a, b, c and use the quadratic formula. You may assume that
a ≠ 0. If the discriminant b2 – 4ac is negative, display a message stating that there are
no real solutions.
Implement a class QuadraticEquation whose constructor receives the coefficients a, b, c
of the quadratic equation. Supply member functions get_solution1 and get_solution2
that get the solutions, using the qua dratic formula, or 0 if no solution exists. The
get_solution1 function should return the smaller of the two solutions.
Supply a function
bool has_solutions() const
that returns false if the discriminant is negative.
engineering P9.26 Design a class Cannonball to model a cannonball that is fired into the air. A ball has
• An x- and a y-position.
• An x- and a y-velocity.
Supply the following member functions:
• A constructor with an x-position (the y-position is initially 0)
• A member function move(double sec) that moves the ball to the next position
(First compute the dis tance traveled in sec seconds, using the current velocities,
then update the x- and y-positions; then update the y-velocity by taking into
account the gravitational acceleration of –9.81 m/sec2; the x-velocity is
unchanged.) (See Exercise P4.29 for additional details.)
• A member function shoot whose parameters are the angle a and initial velocity
v (Compute the x-velocity as v cos a and the y-velocity as v sin a; then keep
calling move with a time interval of 0.1 sec onds until the y-position is ≤ 0; display
the (x, y) position after every move.)
Use this class in a program that prompts the user for the starting angle and the initial
velocity. Then call shoot.
engineering P9.27 The colored bands on the top-most resistor shown in the photo below indicate a
resistance of 6.2 kΩ ±5%. The resistor tolerance of ±5% indicates the acceptable
variation in the resistance. A 6.2 kΩ ±5% resistor could have a resistance as small as
5.89 kΩ or as large as 6.51 kΩ. We say that 6.2 kΩ is the nominal value of the resis-
tance and that the actual value of the resistance can be any value between 5.89 kΩ and
6.51 kΩ.
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436 Chapter 9 Classes
Write a C++ program that represents a resistor as a class. Provide a single construc-
tor that accepts values for the nominal resistance and tolerance and then determines
the actual value randomly. The class should provide public member functions to get
the nominal resistance, tolerance, and the actual resistance.
Write a main function for the C++ program that demonstrates that the class works
properly by displaying actual resistances for ten 330 Ω ±10% resistors.
engineering P9.28 In the Resistor class from Exercise P9.27, supply a
method that returns a description of the “color bands”
for the resistance and tolerance. A resistor has four color
bands:
• The first band is the first significant digit of the
resistance value.
• The second band is the second significant digit of the resistance value.
• The third band is the decimal multiplier.
• The fourth band indicates the tolerance.
First band
Second band Multiplier
Tolerance
Color DIgit Multiplier tolerance
Black 0 ×100 —
Brown 1 ×101 ±1%
Red 2 ×102 ±2%
Orange 3 ×103 —
Yellow 4 ×104 —
Green 5 ×105 ±0.5%
Blue 6 ×106 ±0.25%
Violet 7 ×107 ±0.1%
Gray 8 ×108 ±0.05%
White 9 ×109 —
Gold — ×10–1 ±5%
Silver — ×10–2 ±10%
None — — ±20%
For example (using the values from the table as a key), a resistor with red, violet,
green, and gold bands (left to right) will have 2 as the first digit, 7 as the second digit,
a multiplier of 105, and a tolerance of ±5%, for a resistance of 2,700 kΩ, plus or
minus 5%.
engineering P9.29 The figure below shows a frequently used electric circuit called a “voltage divider”.
The input to the circuit is the voltage vi. The output is the voltage vo. The output of a
voltage divider is proportional to the input, and the constant of proportionality is
called the “gain” of the circuit. The voltage divider is represented by the equation
where G is the gain and R1 and R2 are the resistances of the two resistors that com-
prise the voltage divider.
Manufacturing variations cause the actual resistance values to deviate from the
nominal values, as described in Exercise P9.27. In turn, variations in the resistance
values cause variations in the values of the gain of the voltage divider. We calculate
the nominal value of the gain using the nominal resistance values and the actual
value of the gain using actual resistance values.
Write a C++ program that contains two classes, VoltageDivider and Resistor. The
Resistor class is described in Exercise P9.27. The VoltageDivider class should have two
data members that are objects of the Resistor class. Provide a single constructor that
accepts two Resistor objects, nominal values for their resistances, and the resistor
tolerance. The class should provide public member functions to get the nominal and
actual values of the voltage divider’s gain.
Write a main function for the program that demonstrates that the class works prop-
erly by displaying nominal and actual gain for ten voltage dividers each consisting of
5% resistors having nominal values R1 = 250 Ω and R2 = 750 Ω.
1. cin is an object, string is a class.
2. Through the substr member function and the [] operator.
3. As a vector
4. None. The member functions will have the same effect, and your code could not
have manipulated string objects in any other way.
5. 2 1.90
6. There is no member function named get_amount_due.
7. int get_dollars() const;
8. length, substr
a n s W e r s t o s e l F – C h e C k Q U e s t I o n s
cfe2_ch09_p389_440.indd 436 10/28/10 11:43 AM
answers to self-Check Questions 437
For example (using the values from the table as a key), a resistor with red, violet,
green, and gold bands (left to right) will have 2 as the first digit, 7 as the second digit,
a multiplier of 105, and a tolerance of ±5%, for a resistance of 2,700 kΩ, plus or
minus 5%.
engineering P9.29 The figure below shows a frequently used electric circuit called a “voltage divider”.
The input to the circuit is the voltage vi. The output is the voltage vo. The output of a
voltage divider is proportional to the input, and the constant of proportionality is
called the “gain” of the circuit. The voltage divider is represented by the equation
G
v
v
R
R R
o
i
= =
+
2
1 2
where G is the gain and R1 and R2 are the resistances of the two resistors that com-
prise the voltage divider.
+
–
vi
R1
vo
+
–
R2
Manufacturing variations cause the actual resistance values to deviate from the
nominal values, as described in Exercise P9.27. In turn, variations in the resistance
values cause variations in the values of the gain of the voltage divider. We calculate
the nominal value of the gain using the nominal resistance values and the actual
value of the gain using actual resistance values.
Write a C++ program that contains two classes, VoltageDivider and Resistor. The
Resistor class is described in Exercise P9.27. The VoltageDivider class should have two
data members that are objects of the Resistor class. Provide a single constructor that
accepts two Resistor objects, nominal values for their resistances, and the resistor
tolerance. The class should provide public member functions to get the nominal and
actual values of the voltage divider’s gain.
Write a main function for the program that demonstrates that the class works prop-
erly by displaying nominal and actual gain for ten voltage dividers each consisting of
5% resistors having nominal values R1 = 250 Ω and R2 = 750 Ω.
1. cin is an object, string is a class.
2. Through the substr member function and the [] operator.
3. As a vector
4. None. The member functions will have the same effect, and your code could not
have manipulated string objects in any other way.
5. 2 1.90
6. There is no member function named get_amount_due.
7. int get_dollars() const;
8. length, substr
a n s W e r s t o s e l F – C h e C k Q U e s t I o n s
cfe2_ch09_p389_440.indd 437 10/28/10 11:43 AM
438 Chapter 9 Classes
9. A mutator. Getting a character removes it from the stream, thereby modifying it.
Not convinced? Consider what happens if you call the get function twice. You will
usually get two different characters. But if you call an accessor twice on an object
(without a mutation between the two calls), you are sure to get the same result.
10. 2, 1.85, 1, 1.90
11. The code tries to access a private data member.
12. (1) int hours; // Between 1 and 12
int minutes; // Between 0 and 59
bool pm; // True for p.m., false for a.m.
(2) int hours; // Military time, between 0 and 23
int minutes; // Between 0 and 59
(3) int total_minutes // Between 0 and 60 * 24 – 1
13. They need not change their programs at all since the public interface has not
changed. They need to recompile with the new version of the Time class.
14. (1) string letter_grade; // “A+”, “B”
(2) double number_grade; // 4.3, 3.0
15. (1) The CashRegister:: is missing. (2) The const is missing.
16. void CashRegister::add_items(int quantity, double price)
{
item_count = item_count + quantity;
total_price = total_price + quantity * price;
}
17. int CashRegister::get_dollars() const
{
int dollars = total_price; // Truncates cents
return dollars;
}
18. Three parameters: two explicit parameters of type int, and one implicit parameter of
type string.
19. One parameter: the implicit parameter of type string. The function has no explicit
parameters.
20. BankAccount::BankAccount() { balance = 0; }
21. BankAccount::BankAccount(double initial_balance)
{
balance = initial_balance;
}
22. Item::Item()
{
price = 0;
}
Note that the description data member need not be initialized.
23. CashRegister::CashRegister()
{
clear();
}
24. (a) Default constructor. (b) Item(string) or Item(const char[])
(c) Item(double) (d) Item(string, double) (e) Does not define an object.
cfe2_ch09_p389_440.indd 438 10/28/10 11:43 AM
answers to self-Check Questions 439
25.
Car my_car
Car(mpg)
add_gas(amount)
drive(distance)
get_gas_left
gas_left miles_per_gallon
0 25
front back
26.
gas_left miles_per_gallon
0
20
16
8
13
25
27.
gas_left miles_per_gallon
0 25
total_miles
0
28.
gas_left miles_per_gallon
0
20
16
8
13
25
total_miles
0
100
200
29. Look for nouns in the problem description.
30. Yes (ChessBoard) and no (MovePiece).
31.
Mailbox Message
32. Typically, a library system wants to track which books a patron has checked out, so
it makes more sense to have Patron aggregate Book. However, there is not always one
true answer in design. If you feel strongly that it is important to identify the patron
cfe2_ch09_p389_440.indd 439 10/28/10 11:43 AM
440 Chapter 9 Classes
who had checked out a particular book (perhaps to notify the patron to return it
because it was requested by someone else), then you can argue that the aggregation
should go the other way around.
33. There would be no relationship.
34. The header file that defines the BankAccount class, probably named bankaccount.h.
35. None. The cashregister.cpp file includes cashregister.h, which includes bankaccount.h.
36. None. The bank account need not know anything about cash registers.
37. Add the following line to cashregister.h:
void display(CashRegister reg);
Add the implementation of the display function to cashregister.cpp.
38. The answer depends on your system. On my system, the file is located at /usr/in-
clude/c++/4.2/iostream.
39. string* str_pointer = new string;
40. delete str_pointer;
41. string* str_pointer = new string(“Hello”);
or
string* str_pointer = new string;
*str_pointer = “Hello”;
42. cout << str_pointer->length();
43. string*, or more accurately, const string* because length is an accessor function.
Step 1 Get an informal list of the responsibilities of your objects.
The following responsibilities are mentioned in the problem statement:
Deposit funds.
Withdraw funds.
Add interest.
There is a hidden responsibility as well. We need to be able to find out how much money is in
the account.
Get balance.
Step 2 Specify the public interface.
We need to specify parameter variables and determine which member functions are accessors.
To deposit or withdraw money, one needs to know the amount of the deposit or withdrawal:
void deposit(double amount);
void withdraw(double amount);
To add interest, one needs to know the interest rate that is to be applied:
void add_interest(double rate);
Clearly, all these member functions are mutators because they change the balance.
Finally, we have
double get_balance() const;
This function is an accessor because inquiring about the balance does not change it.
Now we move on to constructors. A default constructor makes an account with a zero
balance. It can also be use ful to supply a constructor with an initial balance.
Here is the complete public interface:
class BankAccount
{
public:
BankAccount();
BankAccount(double initial_balance);
void deposit(double amount);
void withdraw(double amount);
void add_interest(double rate);
double get_balance() const;
private:
…
};
W o r k e D e x a M p l e 9 . 1 implementing a bank account class
Write a class that simulates a bank account. Customers can deposit and withdraw funds. If
sufficient funds are not available for withdrawal, a $10 overdraft penalty is charged. At the end
of the month, interest is added to the account. The interest rate can vary every month.
cfe2_ch09_p389_440.indd 440 10/28/10 11:43 AM
10C h a p t e r
441
I n h e r I ta n C e
to understand the concepts of
inheritance and polymorphism
to learn how to inherit and override
member functions
to be able to implement constructors for derived classes
to be able to design and use virtual functions
C h a p t e r G o a l s
C h a p t e r C o n t e n t s
10.1 InherItance hIerarchIes 442
10.2 ImplementIng DerIveD
classes 446
Syntax 10.1: Derived-Class Definition 448
Common Error 10.1: private Inheritance 449
Common Error 10.2: replicating Base-Class
Members 450
Programming Tip 10.1: Use a single Class for
Variation in Values, Inheritance for Variation
in Behavior 450
Special Topic 10.1: Calling the Base-Class
Constructor 451
Syntax 10.2: Constructor with Base-Class
Initializer 451
10.3 OverrIDIng member
FunctIOns 451
Common Error 10.3: Forgetting the Base-
Class name 455
10.4 vIrtual FunctIOns anD
pOlymOrphIsm 455
Programming Tip 10.2: Don’t Use type tags 462
Common Error 10.4: slicing an object 462
Special Topic 10.2: Virtual self-Calls 463
How To 10.1: Developing an Inheritance
hierarchy 464
Worked Example 10.1: Implementing an
employee hierarchy for payroll processing
Random Fact 10.1: the limits of
Computation 469
cfe2_ch10_p441_480.indd 441 10/27/10 12:12 PM
442
objects from related classes usually share common
behavior. For example, shovels, rakes, and clippers all
perform gardening tasks. In this chapter, you will learn
how the notion of inheritance expresses the relationship
between specialized and general classes. By using
inheritance, you will be able to share code between classes
and provide services that can be used by multiple classes.
10.1 Inheritance hierarchies
In object-oriented design, inheritance is a relationship between a more general class
(called the base class) and a more specialized class (called the derived class). The
derived class inherits data and behavior from the base class. For example, consider the
relationships between different kinds of vehicles depicted in Figure 1.
Cars share the common traits of all vehicles, such as the ability to transport people
from one place to another. We say that the class Car inherits from the class Vehicle. In
this relationship, the Vehicle class is the base class and the Car class is the derived class.
Informally, the inheritance relationship is called the is-a relationship. Contrast this
relationship with the has-a relationship that we discussed in Section 9.7. Every car is a
vehicle. Every vehicle has an engine.
The inheritance relationship is very powerful because it allows us to reuse algo-
rithms with objects of different classes. Suppose we have an algorithm that manipu-
lates a Vehicle object. Because a car is a special kind of vehicle, we can supply a Car
object to such an algorithm, and it will work correctly. This is an example of the sub-
stitution principle that states that you can always use a derived-class object when a
base-class object is expected.
a derived class
inherits data and
behavior from a
base class.
You can always use a
derived-class object
in place of a
base-class object.
Figure 1 an Inheritance hierarchy of Vehicle Classes
Vehicle
Motorcycle Car Truck
Sedan SUV
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10.1 Inheritance hierarchies 443
Figure 2 the Inheritance hierarchy of stream Classes
istringstream ifstream iostream
fstream
istream ostream
ofstream ostringstream
The inheritance relationship can give rise to hierarchies where classes get ever
more specialized, as shown in Figure 1. The C++ stream classes, shown in Figure 2,
are another example of such a hierarchy. Figure 2 uses the UML notation for inheri-
tance where the base and derived class are joined with an arrow that points to the
base class.
As you can see, an ifstream (an input stream that reads from a file) is a special case of
an istream (an input stream that reads data from any source). If you have an ifstream, it
can be the argument for a function that expects an istream.
void process_input(istream& in) // Can call with an ifstream object
Why provide a function that processes istream objects instead of ifstream objects?
That function is more useful because it can handle any kind of input stream (such as
an istringstream, which is convenient for testing). This again is the substitution prin-
ciple at work.
In this chapter, we will consider a simple
hierarchy of classes. Most likely, you have
taken computer-graded quizzes. A quiz
consists of ques tions, and there are different
kinds of questions:
• Fill-in-the-blank
• Choice (single or multiple)
• Numeric (where an approximate answer is
ok; e.g., 1.33 when the actual answer is 4/3)
• Free response
Figure 3 shows an inheritance hierarchy for these question types.
We will develop a simple but flexible quiz-
taking program to illustrate inheritance.
cfe2_ch10_p441_480.indd 443 10/27/10 12:12 PM
444 Chapter 10 Inheritance
Figure 3 Inheritance hierarchy of Question types
Choice
Question
FillIn
Question
Numeric
Question
FreeResponse
Question
MultiChoice
Question
Question
At the root of this hierarchy is the Question type. A question can dis play its text,
and it can check whether a given response is a correct answer:
class Question
{
public:
Question();
void set_text(string question_text);
void set_answer(string correct_response);
bool check_answer(string response) const;
void display() const;
private:
string text;
string answer;
};
How the text is displayed depends on the question type. Later in this chapter, you
will see some varia tions, but the base class simply sends the question text to cout.
How the response is checked also depends on the question type. As already men-
tioned, a numeric question might accept approximate answers (see Exercise P10.1).
In Exercise P10.3, you will see another way of checking the response. But in the base
class, we will simply require that the response match the correct answer exactly.
In the following sections, you will see how to form derived classes that inherit the
member functions and data members of this base class.
Here is the implementation of the Question class and a simple test program. Note
that the Question class constructor needs to do no work because the text and answer
fields are automatically set to the empty string. The boolalpha stream manipulator in
the main function causes Boolean values to be displayed as true and false instead of the
default 1 and 0.
ch10/quiz1/test.cpp
1 #include
2 #include
3 #include
cfe2_ch10_p441_480.indd 444 10/27/10 12:12 PM
10.1 Inheritance hierarchies 445
4
5 using namespace std;
6
7 class Question
8 {
9 public:
10 /**
11 Constructs a question with empty text and answer.
12 */
13 Question();
14
15 /**
16 @param question_text the text of this question
17 */
18 void set_text(string question_text);
19
20 /**
21 @param correct_response the answer for this question
22 */
23 void set_answer(string correct_response);
24
25 /**
26 @param response the response to check
27 @return true if the response was correct, false otherwise
28 */
29 bool check_answer(string response) const;
30
31 /**
32 Displays this question.
33 */
34 void display() const;
35
36 private:
37 string text;
38 string answer;
39 };
40
41 Question::Question()
42 {
43 }
44
45 void Question::set_text(string question_text)
46 {
47 text = question_text;
48 }
49
50 void Question::set_answer(string correct_response)
51 {
52 answer = correct_response;
53 }
54
55 bool Question::check_answer(string response) const
56 {
57 return response == answer;
58 }
59
60 void Question::display() const
61 {
62 cout << text << endl;
63 }
cfe2_ch10_p441_480.indd 445 10/27/10 12:12 PM
446 Chapter 10 Inheritance
64
65 int main()
66 {
67 string response;
68 cout << boolalpha; // Show Boolean values as true, false
69
70 Question q1;
71 q1.set_text("Who was the inventor of C++?");
72 q1.set_answer("Bjarne Stroustrup");
73
74 q1.display();
75 cout << "Your answer: ";
76 getline(cin, response);
77 cout << q1.check_answer(response) << endl;
78
79 return 0;
80 }
program run
Who was the inventor of C++?
Your answer: Bjarne Stroustrup
true
1. Consider classes Manager and Employee. Which should be the base class and which
should be the derived class?
2. What are the inheritance relationships between classes BankAccount, Checking
Account, and SavingsAccount?
3. Consider the function do_something(istream& stream). List all stream classes from
Figure 2 whose objects can be passed to this function.
4. Consider the function do_something(Car& c). List all vehicle classes from Figure 1
whose objects cannot be passed to this function.
5. Should a class Quiz inherit from the class Question? Why or why not?
practice It Now you can try these exercises at the end of the chapter: R10.1, R10.2, R10.5.
10.2 Implementing Derived Classes
In C++, you form a derived class from a base class by specifying what makes the
derived class different. You define the member functions that are new to the derived
class. The derived class inherits all member functions from the base class, but you can
change the implementation if the inherited behavior is not appropriate.
The derived class automatically inherits all data members from the base class. You
only define the added data members.
Here is the syntax for the definition of a derived class:
class ChoiceQuestion : public Question
{
public:
New and changed member functions
private:
Additional data members
};
s e l F c h e c k
The : symbol denotes inheritance. The reserved word public is required for a techni-
cal reason (see Common Error 10.1 on page 449).
A ChoiceQuestion object differs from a Question object in three ways:
• Its objects store the various choices for the answer.
• There is a member function for adding another choice.
• The display function of the ChoiceQuestion class shows these choices so that the
respondent can choose one of them.
When the ChoiceQuestion class inherits from the Question class, it needs only to spell
out these three differ ences:
class ChoiceQuestion : public Question
{
public:
ChoiceQuestion();
void add_choice(string choice, bool correct);
void display() const;
private:
vector
};
Figure 4 shows the layout of a ChoiceQuestion object. It inherits the text and answer data
members from the Question base object, and it adds an additional data member: the
choices vector.
The add_choice function is specific to the ChoiceQuestion class. You can only apply it
to ChoiceQuestion objects, not general Question objects. However, the display function
is a redefinition of a function that exists in the base class, to take into account the spe-
cial needs of the derived class. We say that the derived class overrides this function.
You will see how in Section 10.3
In the ChoiceQuestion class definition you specify only new member functions and
data members. All other member functions and data members of the Question class are
automatically inherited by the Question class. For example, each ChoiceQuestion object
still has text and answer data members, and set_text, set_answer, and check_answer mem-
ber functions.
a derived class can
override a base-class
function by
providing a new
implementation.
the derived class
inherits all data
members and all
functions that it does
not override.
Figure 4 Data layout of a Derived-Class object
text =
ChoiceQuestion
answer =
choices =
Question portion
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10.2 Implementing Derived Classes 447
Like the manufacturer of a
stretch limo, who starts with a
regular car and modifies it, a
programmer makes a derived
class by modifying another class.
The : symbol denotes inheritance. The reserved word public is required for a techni-
cal reason (see Common Error 10.1 on page 449).
A ChoiceQuestion object differs from a Question object in three ways:
• Its objects store the various choices for the answer.
• There is a member function for adding another choice.
• The display function of the ChoiceQuestion class shows these choices so that the
respondent can choose one of them.
When the ChoiceQuestion class inherits from the Question class, it needs only to spell
out these three differ ences:
class ChoiceQuestion : public Question
{
public:
ChoiceQuestion();
void add_choice(string choice, bool correct);
void display() const;
private:
vector
};
Figure 4 shows the layout of a ChoiceQuestion object. It inherits the text and answer data
members from the Question base object, and it adds an additional data member: the
choices vector.
The add_choice function is specific to the ChoiceQuestion class. You can only apply it
to ChoiceQuestion objects, not general Question objects. However, the display function
is a redefinition of a function that exists in the base class, to take into account the spe-
cial needs of the derived class. We say that the derived class overrides this function.
You will see how in Section 10.3
In the ChoiceQuestion class definition you specify only new member functions and
data members. All other member functions and data members of the Question class are
automatically inherited by the Question class. For example, each ChoiceQuestion object
still has text and answer data members, and set_text, set_answer, and check_answer mem-
ber functions.
a derived class can
override a base-class
function by
providing a new
implementation.
the derived class
inherits all data
members and all
functions that it does
not override.
Figure 4 Data layout of a Derived-Class object
text =
ChoiceQuestion
answer =
choices =
Question portion
cfe2_ch10_p441_480.indd 447 10/27/10 12:12 PM
448 Chapter 10 Inheritance
syntax 10.1 Derived-Class Definition
class ChoiceQuestion : public Question
{
public:
ChoiceQuestion();
void add_choice(string choice, bool correct);
void display() const;
private:
vector
};
Derived class Base class
The : symbol
denotes inheritance.
Declare functions
that are added
to the derived class.
Define data members
that are added to
the derived class.
Declare functions
that the derived
class overrides.
Always place
public after the :.
See page 449.
You can call the inherited member functions on a derived-class object:
choice_question.set_answer(“2”);
However, the inherited data members are inaccessible. Because these members are
private data of the base class, only the base class has access to them. The derived class
has no more access rights than any other class.
In particular, the ChoiceQuestion member functions cannot directly access the answer
member. These member functions must use the public interface of the Question class
to access its private data, just like every other function.
To illustrate this point, let us implement the add_choice member function. The
function has two param eters: the choice to be added (which is appended to the vector
of choices), and a Boolean value to indicate whether this choice is correct. If it is true,
set the answer to the current choice number. (We use an ostringstream to convert the
number to a string—see Section 8.4 for details.)
void ChoiceQuestion::add_choice(string choice, bool correct)
{
choices.push_back(choice);
if (correct)
{
// Convert choices.size() to string
ostringstream stream;
stream << choices.size();
string num_str = stream.str();
// Set num_str as the answer
...
}
}
You can’t just access the answer member in the base class. Fortunately, the Question
class has a set_answer member function. You can call that member function. On which
object? The question that you are cur rently modifying—that is, the implicit parame-
ter of the ChoiceQuestion::add_choice function. As you saw in Chapter 9, if you invoke
a member function on the implicit parameter, you don’t specify the parameter but
just write the member function name:
set_answer(num_str);
The compiler interprets this call as
implicit parameter.set_answer(num_str);
6. Suppose q is an object of the class Question and cq an object of the class Choice
Question. Which of the following calls are legal?
a. q.set_answer(response)
b. cq.set_answer(response)
c. q.add_option(option, true)
d. cq.add_option(option, true)
7. Define a class Manager that inherits from the class Employee and adds a data mem-
ber bonus for storing a salary bonus. Omit the constructor declaration.
8. Suppose the class Employee is defined as follows:
class Employee
{
public:
Employee();
void set_name(string new_name);
void set_base_salary(double new_salary);
string get_name();
double get_salary() const;
private:
string name;
double base_salary;
}
Which data members does the Manager class from Self Check 7 have?
9. Define a class Manager that inherits from the class Employee and overrides the
get_salary function.
10. Which member functions does the Manager class from Self Check 9 inherit?
practice It Now you can try these exercises at the end of the chapter: P10.2, P10.6.
private Inheritance
It is a common error to forget the reserved word public that must follow the colon after the
derived-class name.
class ChoiceQuestion : Question // Error
{
...
};
The class definition will compile. The ChoiceQuestion still inherits from Question, but it inherits
privately. That is, only the member functions of ChoiceQuestion get to call member functions of
Question. Whenever another function invokes a Question member function on a ChoiceQuestion
object, the compiler will flag this as an error:
int main()
{
s e l F c h e c k
Common error 10.1
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10.2 Implementing Derived Classes 449
a member function on the implicit parameter, you don’t specify the parameter but
just write the member function name:
set_answer(num_str);
The compiler interprets this call as
implicit parameter.set_answer(num_str);
6. Suppose q is an object of the class Question and cq an object of the class Choice
Question. Which of the following calls are legal?
a. q.set_answer(response)
b. cq.set_answer(response)
c. q.add_option(option, true)
d. cq.add_option(option, true)
7. Define a class Manager that inherits from the class Employee and adds a data mem-
ber bonus for storing a salary bonus. Omit the constructor declaration.
8. Suppose the class Employee is defined as follows:
class Employee
{
public:
Employee();
void set_name(string new_name);
void set_base_salary(double new_salary);
string get_name();
double get_salary() const;
private:
string name;
double base_salary;
}
Which data members does the Manager class from Self Check 7 have?
9. Define a class Manager that inherits from the class Employee and overrides the
get_salary function.
10. Which member functions does the Manager class from Self Check 9 inherit?
practice It Now you can try these exercises at the end of the chapter: P10.2, P10.6.
private Inheritance
It is a common error to forget the reserved word public that must follow the colon after the
derived-class name.
class ChoiceQuestion : Question // Error
{
...
};
The class definition will compile. The ChoiceQuestion still inherits from Question, but it inherits
privately. That is, only the member functions of ChoiceQuestion get to call member functions of
Question. Whenever another function invokes a Question member function on a ChoiceQuestion
object, the compiler will flag this as an error:
int main()
{
s e l F c h e c k
Common error 10.1
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450 Chapter 10 Inheritance
ChoiceQuestion q;
...
cout << q.check_answer(response); // Error
}
This private inheritance is rarely useful. In fact, it violates the spirit of using inheritance in the
first place—namely, to create objects that are usable just like the base-class objects. You should
always use public inheritance and remember to supply the public reserved word in the defini-
tion of the derived class.
replicating base-class members
A derived class has no access to the data members of the base class.
ChoiceQuestion::ChoiceQuestion(string question_text)
{
text = question_text; // Error—tries to access private base-class member
}
When faced with a compiler error, beginners commonly “solve” this issue by adding another
data member with the same name to the derived class:
class ChoiceQuestion : public Question
{
...
private:
vector
string text; // Don’t!
}
Sure, now the constructor compiles, but it doesn’t set the correct text! Such a ChoiceQuestion
object has two data members, both named text. The constructor sets one of them, and the
display member function displays the other.
Instead of uselessly replicating a base-class data member, you need to call a member func-
tion that updates the base-class member, such as the set_text function in our example.
text =
ChoiceQuestion
answer =
choices =
text =
Question portion
use a single class for variation in values,
Inheritance for variation in behavior
The purpose of inheritance is to model objects with different behavior. When students first
learn about inheritance, they have a tendency to overuse it, by creating multiple classes even
though the variation could be expressed with a simple data member.
Consider a program that tracks the fuel efficiency of a fleet of cars by logging the distance
traveled and the refuel ing amounts. Some cars in the fleet are hybrids. Should you create a
derived class HybridCar? Not in this application. Hybrids don’t behave any differently than
Common error 10.2
programming tip 10.1
other cars when it comes to driving and refueling. They just have a better fuel efficiency. A
single Car class with a data member
double miles_per_gallon;
is entirely sufficient.
However, if you write a program that shows how to repair different kinds of vehicles, then
it makes sense to have a separate class HybridCar. When it comes to repairs, hybrid cars behave
differently from other cars.
calling the base-class constructor
Consider the process of constructing a derived-class object. A
derived-class constructor can only initialize the data members of the
derived class. But the base-class data members also need to be initial-
ized. Unless you specify otherwise, the base-class data members are
initialized with the default constructor of the base class.
In order to specify another constructor, you use an initializer list,
as described in Special Topic 9.1. Specify the name of the base class
and the construction arguments in the initializer list. For example, suppose the Question base
class had a constructor for set ting the question text. Here is how a derived-class constructor
could call that base-class constructor:
ChoiceQuestion::ChoiceQuestion(string question_text)
: Question(question_text)
{
}
The derived-class constructor calls the base-class constructor before
executing the code inside the { }.
In our example program, we used the default constructor of the
base class. However, if a base class has no default constructor, you
must use the initializer list syntax.
10.3 overriding Member Functions
The derived class inherits the member functions from the base class. If you are not
satisfied with the behavior of the inherited member function, you can override it by
specifying a new implementation in the derived class.
special topic 10.1
Unless specified
otherwise, the
base-class data
members are
initialized with the
default constructor.
the constructor of a
derived class can
supply arguments
to a base-class
constructor.
syntax 10.2 Constructor with Base-Class Initializer
ChoiceQuestion::ChoiceQuestion(string question_text)
: Question(question_text)
{
}
The base-class
constructor
is called first. If you omit the base-class
constructor call, the default
constructor is invoked.
This block can contain
additional statements.
cfe2_ch10_p441_480.indd 450 10/27/10 12:12 PM
10.3 overriding Member Functions 451
other cars when it comes to driving and refueling. They just have a better fuel efficiency. A
single Car class with a data member
double miles_per_gallon;
is entirely sufficient.
However, if you write a program that shows how to repair different kinds of vehicles, then
it makes sense to have a separate class HybridCar. When it comes to repairs, hybrid cars behave
differently from other cars.
calling the base-class constructor
Consider the process of constructing a derived-class object. A
derived-class constructor can only initialize the data members of the
derived class. But the base-class data members also need to be initial-
ized. Unless you specify otherwise, the base-class data members are
initialized with the default constructor of the base class.
In order to specify another constructor, you use an initializer list,
as described in Special Topic 9.1. Specify the name of the base class
and the construction arguments in the initializer list. For example, suppose the Question base
class had a constructor for set ting the question text. Here is how a derived-class constructor
could call that base-class constructor:
ChoiceQuestion::ChoiceQuestion(string question_text)
: Question(question_text)
{
}
The derived-class constructor calls the base-class constructor before
executing the code inside the { }.
In our example program, we used the default constructor of the
base class. However, if a base class has no default constructor, you
must use the initializer list syntax.
10.3 overriding Member Functions
The derived class inherits the member functions from the base class. If you are not
satisfied with the behavior of the inherited member function, you can override it by
specifying a new implementation in the derived class.
special topic 10.1
Unless specified
otherwise, the
base-class data
members are
initialized with the
default constructor.
the constructor of a
derived class can
supply arguments
to a base-class
constructor.
syntax 10.2 Constructor with Base-Class Initializer
ChoiceQuestion::ChoiceQuestion(string question_text)
: Question(question_text)
{
}
The base-class
constructor
is called first. If you omit the base-class
constructor call, the default
constructor is invoked.
This block can contain
additional statements.
cfe2_ch10_p441_480.indd 451 10/27/10 12:12 PM
452 Chapter 10 Inheritance
Consider the display function of the ChoiceQuestion class. It needs to override the
base-class display function in order to show the choices for the answer. Specifically,
the derived-class function needs to
• Display the question text.
• Display the answer choices.
The second part is easy because the answer choices are a data member of the derived
class.
void ChoiceQuestion::display() const
{
// Display the question text
…
// Display the answer choices
for (int i = 0; i < choices.size(); i++)
{
cout << i + 1 << ": " << choices[i] << endl;
}
}
But how do you get the question text? You can’t access the text member of the base
class directly because it is private.
Instead, you can call the display function of the base class.
void ChoiceQuestion::display() const
{
// Display the question text
display(); // Invokes implicit parameter.display()
// Display the answer choices
...
}
However, this won’t quite work. Because the implicit parameter of
ChoiceQuestion::display is of type ChoiceQuestion, and there is a function named display
in the ChoiceQuestion class, that function will be called—but that is just the function
you are currently writing! The function would call itself over and over.
To display the question text, you must be more specific about which function
named display you want to call. You want Question::display:
void ChoiceQuestion::display() const
{
// Display the question text
Question::display(); // OK
// Display the answer choices
...
}
When you override a function, you usually want to extend the functionality of the
base-class version. Therefore, you often need to invoke the base-class version before
extending it. To invoke it, you need to use the BaseClass::function notation. However,
you have no obligation to call the base-class function. Occasionally, a derived class
overrides a base-class function and specifies an entirely different functional ity.
Here is the complete program that displays a plain Question object and a Choice
Question object. (The defi nition of the Question class, which you have already seen,
is placed into question.h, and the implementa tion is in question.cpp.) This example
shows how you can use inheritance to form a more specialized class from a base class.
a derived class can
inherit a function
from the base class,
or it can override it
by providing another
implementation.
Use
BaseClass::function
notation to explicitly
call a base-class
function.
cfe2_ch10_p441_480.indd 452 10/27/10 12:12 PM
10.3 overriding Member Functions 453
ch10/quiz2/test.cpp
1 #include
2 #include
3 #include
4 #include “question.h”
5
6 class ChoiceQuestion : public Question
7 {
8 public:
9 /**
10 Constructs a choice question with no choices.
11 */
12 ChoiceQuestion();
13
14 /**
15 Adds an answer choice to this question.
16 @param choice the choice to add
17 @param correct true if this is the correct choice, false otherwise
18 */
19 void add_choice(string choice, bool correct);
20
21 void display() const;
22 private:
23 vector
24 };
25
26 ChoiceQuestion::ChoiceQuestion()
27 {
28 }
29
30 void ChoiceQuestion::add_choice(string choice, bool correct)
31 {
32 choices.push_back(choice);
33 if (correct)
34 {
35 // Convert choices.size() to string
36 ostringstream stream;
37 stream << choices.size();
38 string num_str = stream.str();
39 set_answer(num_str);
40 }
41 }
42
43 void ChoiceQuestion::display() const
44 {
45 // Display the question text
46 Question::display();
47 // Display the answer choices
48 for (int i = 0; i < choices.size(); i++)
49 {
50 cout << i + 1 << ": " << choices[i] << endl;
51 }
52 }
53
54 int main()
55
56 string response;
57 cout << boolalpha;
58
cfe2_ch10_p441_480.indd 453 10/27/10 12:12 PM
454 Chapter 10 Inheritance
59 // Ask a basic question
60
61 Question q1;
62 q1.set_text("Who was the inventor of C++?");
63 q1.set_answer("Bjarne Stroustrup");
64
65 q1.display();
66 cout << "Your answer: ";
67 getline(cin, response);
68 cout << q1.check_answer(response) << endl;
69
70 // Ask a choice question
71
72 ChoiceQuestion q2;
73 q2.set_text("In which country was the inventor of C++ born?");
74 q2.add_choice("Australia", false);
75 q2.add_choice("Denmark", true);
76 q2.add_choice("Korea", false);
77 q2.add_choice("United States", false);
78
79 q2.display();
80 cout << "Your answer: ";
81 getline(cin, response);
82 cout << q2.check_answer(response) << endl;
83
84 return 0;
85 }
program run
Who was the inventor of C++?
Your answer: Bjarne Stroustrup
true
In which country was the inventor of C++ born?
1: Australia
2: Denmark
3: Korea
4: United States
Your answer: 2
true
11. What is wrong with the following implementation of the display function?
void ChoiceQuestion::display() const
{
cout << text << endl;
for (int i = 0; i < choices.size(); i++)
{
cout << i + 1 << ": " << choices[i] << endl;
}
}
12. What is wrong with the following implementation of the display function?
void ChoiceQuestion::display() const
{
this>display();
for (int i = 0; i < choices.size(); i++)
{
s e l F c h e c k
cout << i + 1 << ": " << choices[i] << endl;
}
}
13. Look again at the implementation of the add_choice function that calls
the set_answer function of the base class. Why don’t you need to call
Question:: set_answer?
14. In the Manager class of Self Check 9, override the get_name function so that manag-
ers have a * before their name (such as *Lin, Sally).
15. In the Manager class of Self Check 9, override the get_salary function so that it
returns the sum of the salary and the bonus.
practice It Now you can try these exercises at the end of the chapter: R10.12, P10.1.
Forgetting the base-class name
A common error in extending the functionality of a base-class function is to forget the base-
class name. For example, to compute the salary of a manager, get the salary of the underlying
Employee object and add a bonus:
double Manager::get_salary() const
{
double base_salary = get_salary();
// Error—should be Employee::get_salary()
return base_salary + bonus;
}
Here get_salary() refers to the get_salary function applied to the implicit parameter of the
member function. The implicit parameter is of type Manager, and there is a Manager::get_salary
function, so that function is called. Of course, that is a recursive call to the function that we are
writing. Instead, you must specify which get_salary function you want to call. In this case, you
need to call Employee::get_salary explicitly.
Whenever you call a base-class function from a derived-class function with the same name,
be sure to give the full name of the function, including the base-class name.
10.4 Virtual Functions and polymorphism
In the preceding sections you saw one important use of inheritance: to form a more
specialized class from a base class. In the following sections you will see an even more
powerful application of inheritance: to work with objects whose type and behavior
can vary at run time. This variation of behavior is achieved with virtual functions.
When you invoke a virtual function on an object, the C++ run-time system deter-
mines which actual member function to call, depending on the class to which the
object belongs.
In the following sections, you will see why you need to use pointers to access
objects whose class can vary at run-time, and how a virtual function selects the mem-
ber function that is appropriate for a given object.
Common error 10.3
cfe2_ch10_p441_480.indd 454 10/27/10 12:12 PM
10.4 Virtual Functions and polymorphism 455
cout << i + 1 << ": " << choices[i] << endl;
}
}
13. Look again at the implementation of the add_choice function that calls
the set_answer function of the base class. Why don’t you need to call
Question:: set_answer?
14. In the Manager class of Self Check 9, override the get_name function so that manag-
ers have a * before their name (such as *Lin, Sally).
15. In the Manager class of Self Check 9, override the get_salary function so that it
returns the sum of the salary and the bonus.
practice It Now you can try these exercises at the end of the chapter: R10.12, P10.1.
Forgetting the base-class name
A common error in extending the functionality of a base-class function is to forget the base-
class name. For example, to compute the salary of a manager, get the salary of the underlying
Employee object and add a bonus:
double Manager::get_salary() const
{
double base_salary = get_salary();
// Error—should be Employee::get_salary()
return base_salary + bonus;
}
Here get_salary() refers to the get_salary function applied to the implicit parameter of the
member function. The implicit parameter is of type Manager, and there is a Manager::get_salary
function, so that function is called. Of course, that is a recursive call to the function that we are
writing. Instead, you must specify which get_salary function you want to call. In this case, you
need to call Employee::get_salary explicitly.
Whenever you call a base-class function from a derived-class function with the same name,
be sure to give the full name of the function, including the base-class name.
10.4 Virtual Functions and polymorphism
In the preceding sections you saw one important use of inheritance: to form a more
specialized class from a base class. In the following sections you will see an even more
powerful application of inheritance: to work with objects whose type and behavior
can vary at run time. This variation of behavior is achieved with virtual functions.
When you invoke a virtual function on an object, the C++ run-time system deter-
mines which actual member function to call, depending on the class to which the
object belongs.
In the following sections, you will see why you need to use pointers to access
objects whose class can vary at run-time, and how a virtual function selects the mem-
ber function that is appropriate for a given object.
Common error 10.3
cfe2_ch10_p441_480.indd 455 10/27/10 12:12 PM
456 Chapter 10 Inheritance
10.4.1 the slicing problem
In this section, we will discuss a problem that commonly arises when you work with
a collection of objects that belong to different classes in a class hierarchy.
If you look into the main function of quiz2/test.cpp, you will find that there was
some repetitive code to display each question and check the responses. It would be
nicer if all questions were collected in an array and one could use a loop to present
them to the user:
const int QUIZZES = 2;
Question quiz[QUIZZES];
quiz[0].set_text("Who was the inventor of C++?");
quiz[0].set_answer("Bjarne Stroustrup");
ChoiceQuestion cq;
cq.set_text("In which country was the inventor of C++ born?");
cq.add_choice("Australia", false);
...
quiz[1] = cq;
for (int i = 0; i < QUIZZES; i++)
{
quiz[i].display();
cout << "Your answer: ";
getline(cin, response);
cout << quiz[i].check_answer(response) << endl;
}
The array quiz holds objects of type Question. The compiler realizes that a Choice
Question is a special case of a Question. Thus it permits the assignment from a choice
question to a question:
quiz[1] = cq;
However, a ChoiceQuestion object has three data members, whereas a Question object has
just two. There is no room to store the derived-class data. That data simply gets sliced
away when you assign a derived-class object to a base-class variable (see Figure 5).
If you run the resulting program, the options are not displayed:
Who was the inventor of C++?
Your answer: Bjarne Stroustrup
true
In which country was the inventor of C++ born?
Your answer:
This problem is very typical of code that needs to manipulate objects from a mixture
of classes in an inheritance hierarchy. Derived-class objects are usually bigger than
base-class objects, and objects of dif ferent derived classes have different sizes. An
array of objects cannot deal with this variation in sizes.
When converting a
derived-class object
to a base class, the
derived-class data is
sliced away.
Figure 5 slicing away Derived-Class Data
text =
ChoiceQuestion
answer =
choices =
text =
Question
answer =
Instead, you need to store the actual objects elsewhere and collect their locations
in an array by storing pointers. We will discuss the use of pointers in the next section.
10.4.2 pointers to Base and Derived Classes
To access objects from different classes in a class hierarchy, use pointers. Pointers to
the various objects all have the same size—namely, the size of a memory address—
even though the objects themselves may have different sizes.
Here is the code to set up the array of pointers (see Figure 6):
Question* quiz[2];
quiz[0] = new Question;
quiz[0]>set_text(“Who was the inventor of C++?”);
quiz[0]>set_answer(“Bjarne Stroustrup”);
ChoiceQuestion* cq_pointer = new ChoiceQuestion;
cq_pointer>set_text(“In which country was the inventor of C++ born?”);
cq_pointer>add_choice(“Australia”, false);
…
quiz[1] = cq_pointer;
As the highlighted code shows, you simply define the array to hold pointers, allocate
all objects by calling new, and use the > operator instead of the dot operator.
Note that the last assignment assigns a derived-class pointer of type ChoiceQuestion*
to a base-class pointer of type Question*. This is perfectly legal. A pointer is the start-
ing address of an object. Because every ChoiceQuestion is a special case of a Question,
the starting address of a ChoiceQuestion object is, in par ticular, the starting address of a
Question object. The reverse assignment—from a base-class pointer to a derived-class
pointer—is an error.
The code to present all questions is
for (int i = 0; i < QUIZZES; i++)
{
quiz[i]>display();
cout << "Your answer: ";
getline(cin, response);
cout << quiz[i]>check_answer(response) << endl;
}
Again, note the use of the > operator because quiz[i] is a pointer.
a derived-class
pointer can be
converted to a
base-class pointer.
Figure 6 an array of pointers Can store objects from Different Classes
quiz =
text =
Question
answer =
text =
ChoiceQuestion
answer =
choices =
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10.4 Virtual Functions and polymorphism 457
Instead, you need to store the actual objects elsewhere and collect their locations
in an array by storing pointers. We will discuss the use of pointers in the next section.
10.4.2 pointers to Base and Derived Classes
To access objects from different classes in a class hierarchy, use pointers. Pointers to
the various objects all have the same size—namely, the size of a memory address—
even though the objects themselves may have different sizes.
Here is the code to set up the array of pointers (see Figure 6):
Question* quiz[2];
quiz[0] = new Question;
quiz[0]>set_text(“Who was the inventor of C++?”);
quiz[0]>set_answer(“Bjarne Stroustrup”);
ChoiceQuestion* cq_pointer = new ChoiceQuestion;
cq_pointer>set_text(“In which country was the inventor of C++ born?”);
cq_pointer>add_choice(“Australia”, false);
…
quiz[1] = cq_pointer;
As the highlighted code shows, you simply define the array to hold pointers, allocate
all objects by calling new, and use the > operator instead of the dot operator.
Note that the last assignment assigns a derived-class pointer of type ChoiceQuestion*
to a base-class pointer of type Question*. This is perfectly legal. A pointer is the start-
ing address of an object. Because every ChoiceQuestion is a special case of a Question,
the starting address of a ChoiceQuestion object is, in par ticular, the starting address of a
Question object. The reverse assignment—from a base-class pointer to a derived-class
pointer—is an error.
The code to present all questions is
for (int i = 0; i < QUIZZES; i++)
{
quiz[i]>display();
cout << "Your answer: ";
getline(cin, response);
cout << quiz[i]>check_answer(response) << endl;
}
Again, note the use of the > operator because quiz[i] is a pointer.
a derived-class
pointer can be
converted to a
base-class pointer.
Figure 6 an array of pointers Can store objects from Different Classes
quiz =
text =
Question
answer =
text =
ChoiceQuestion
answer =
choices =
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458 Chapter 10 Inheritance
10.4.3 Virtual Functions
When you collect objects of different classes in a class hierarchy, and then invoke a
member function, you want the appropriate member function to be applied. For
example, when you call the display member function on a Question* pointer that hap-
pens to point to a ChoiceQuestion, you want the choices to be dis played.
For reasons of efficiency, this is not the default in C++. By default, a call
quiz[i]>display();
always calls Question::display because the type of quiz[i] is Question*.
However, in this case you really want to determine the actual type of the object
to which quiz[i] points, which can be either a Question or a ChoiceQuestion object, and
then call the appropriate function. In C++, you must alert the compiler that the func-
tion call needs to be preceded by the appropriate function selection, which can be a
different one for every iteration in the loop. You use the virtual reserved word for
this purpose:
class Question
{
public:
Question();
void set_text(string question_text);
void set_answer(string correct_response);
virtual bool check_answer(string response) const;
virtual void display() const;
private:
…
};
The virtual reserved word must be used in the base class. All functions with the same
name and parame ter variable types in derived classes are then automatically virtual.
However, it is considered good taste to supply the virtual reserved word for the
derived-class functions as well.
class ChoiceQuestion : public Question
{
public:
ChoiceQuestion();
void add_choice(string choice, bool correct);
virtual void display() const;
private:
…
};
You do not supply the reserved word virtual in the function definition:
void Question::display() const // No virtual reserved word
{
cout << text << endl;
}
Whenever a virtual function is called, the compiler determines the type of the implicit
parameter in the particular call at run time. The appropriate function for that object is
then called. For example, when the display function is declared virtual, the call
quiz[i]>display();
always calls the function belonging to the actual type of the object to which display[i]
points—either Question::display or ChoiceQuestion::display.
When a virtual
function is called, the
version belonging to
the actual type of the
implicit parameter
is invoked.
10.4.4 polymorphism
The quiz array collects a mixture of both kinds of questions. Such a collection is called
polymorphic (liter ally, “of multiple shapes”). Objects in a polymorphic collection
have some commonality but are not nec essarily of the same type. Inheritance is used
to express this commonality, and virtual functions enable variations in behavior.
Virtual functions give programs a great deal of flexibility. The question presen-
tation loop describes only the general mechanism: “Display the question, get a
response, and check it”. Each object knows on its own how to carry out the specific
tasks: “Display the question” and “Check a response”.
Using virtual functions makes programs easily extensible. Suppose we want
to have a new kind of question for calculations, where we are willing to accept an
approximate answer. All we need to do is to define a new class NumericQuestion, with
its own check_answer function. Then we can populate the quiz array with a mixture of
plain questions, choice questions, and numeric questions. The code that presents the
questions need not be changed at all! The calls to the virtual functions automatically
select the correct member functions of the newly defined classes.
Here is the final version of the quiz program, using pointers and virtual functions.
When you run the program, you will find that the appropriate versions of the virtual
functions are called. (The files question.cpp and choicequestion.cpp are included in your
book’s companion code.)
ch10/quiz3/question.h
1 #ifndef QUESTION_H
2 #define QUESTION_H
3
4 #include
5
6 using namespace std;
7
8 class Question
9 {
10 public:
11 /**
12 Constructs a question with empty question and answer.
13 */
14 Question();
15
polymorphism
(literally, “having
multiple shapes”)
describes objects
that share a set of
tasks and execute
them in different
ways.
cfe2_ch10_p441_480.indd 458 10/27/10 12:12 PM
10.4 Virtual Functions and polymorphism 459
10.4.4
In the same way that vehicles can
differ in their method of locomotion,
polymorphic objects carry out tasks
in different ways.
polymorphism
The quiz array collects a mixture of both kinds of questions. Such a collection is called
polymorphic (liter ally, “of multiple shapes”). Objects in a polymorphic collection
have some commonality but are not nec essarily of the same type. Inheritance is used
to express this commonality, and virtual functions enable variations in behavior.
Virtual functions give programs a great deal of flexibility. The question presen-
tation loop describes only the general mechanism: “Display the question, get a
response, and check it”. Each object knows on its own how to carry out the specific
tasks: “Display the question” and “Check a response”.
Using virtual functions makes programs easily extensible. Suppose we want
to have a new kind of question for calculations, where we are willing to accept an
approximate answer. All we need to do is to define a new class NumericQuestion, with
its own check_answer function. Then we can populate the quiz array with a mixture of
plain questions, choice questions, and numeric questions. The code that presents the
questions need not be changed at all! The calls to the virtual functions automatically
select the correct member functions of the newly defined classes.
Here is the final version of the quiz program, using pointers and virtual functions.
When you run the program, you will find that the appropriate versions of the virtual
functions are called. (The files question.cpp and choicequestion.cpp are included in your
book’s companion code.)
ch10/quiz3/question.h
1 #ifndef QUESTION_H
2 #define QUESTION_H
3
4 #include
5
6 using namespace std;
7
8 class Question
9 {
10 public:
11 /**
12 Constructs a question with empty question and answer.
13 */
14 Question();
15
polymorphism
(literally, “having
multiple shapes”)
describes objects
that share a set of
tasks and execute
them in different
ways.
cfe2_ch10_p441_480.indd 459 10/27/10 12:12 PM
460 Chapter 10 Inheritance
16 /**
17 @param question_text the text of this question
18 */
19 void set_text(string question_text);
20
21 /**
22 @param correct_response the answer for this question
23 */
24 void set_answer(string correct_response);
25
26 /**
27 @param response the response to check
28 @return true if the response was correct, false otherwise
29 */
30 virtual bool check_answer(string response) const;
31
32 /**
33 Displays this question.
34 */
35 virtual void display() const;
36 private:
37 string text;
38 string answer;
39 };
40
41 #endif
ch10/quiz3/choicequestion.h
1 #ifndef CHOICEQUESTION_H
2 #define CHOICEQUESTION_H
3
4 #include
5 #include “question.h”
6
7 class ChoiceQuestion : public Question
8 {
9 public:
10 /**
11 Constructs a choice question with no choices.
12 */
13 ChoiceQuestion();
14
15 /**
16 Adds an answer choice to this question.
17 @param choice the choice to add
18 @param correct true if this is the correct choice, false otherwise
19 */
20 void add_choice(string choice, bool correct);
21
22 virtual void display() const;
23 private:
24 vector
25 };
26
27 #endif
cfe2_ch10_p441_480.indd 460 10/27/10 12:12 PM
10.4 Virtual Functions and polymorphism 461
ch10/quiz3/test.cpp
1 #include
2 #include “question.h”
3 #include “choicequestion.h”
4
5 int main()
6 {
7 string response;
8 cout << boolalpha;
9
10 // Make a quiz with two questions
11 const int QUIZZES = 2;
12 Question* quiz[QUIZZES];
13 quiz[0] = new Question;
14 quiz[0]>set_text(“Who was the inventor of C++?”);
15 quiz[0]>set_answer(“Bjarne Stroustrup”);
16
17 ChoiceQuestion* cq_pointer = new ChoiceQuestion;
18 cq_pointer>set_text(
19 “In which country was the inventor of C++ born?”);
20 cq_pointer>add_choice(“Australia”, false);
21 cq_pointer>add_choice(“Denmark”, true);
22 cq_pointer>add_choice(“Korea”, false);
23 cq_pointer>add_choice(“United States”, false);
24 quiz[1] = cq_pointer;
25
26 // Check answers for all questions
27 for (int i = 0; i < QUIZZES; i++)
28 {
29 quiz[i]>display();
30 cout << "Your answer: ";
31 getline(cin, response);
32 cout << quiz[i]>check_answer(response) << endl;
33 }
34
35 return 0;
36 }
16. Why did the test program introduce the variable cq_pointer instead of directly
calling quiz[1]>add_choice(“Australia”, false)?
17. Which of the following statements are legal?
a. ChoiceQuestion q;
Question a = q;
b. ChoiceQuestion b = a;
c. a.add_choice(“Yes”, true);
d. b.add_choice(“No”, false);
18. Which of the following statements are legal?
a. Question* p = new ChoiceQuestion;
b. ChoiceQuestion* q = p;
c. p>add_choice(“Yes”, true);
d. q>add_choice(“No”, false);
s e l F c h e c k
cfe2_ch10_p441_480.indd 461 10/27/10 12:12 PM
462 Chapter 10 Inheritance
19. What is displayed as the result of the following statements?
ChoiceQuestion* p = new ChoiceQuestion;
p>set_text(“What is the answer?”);
p>add_choice(“42”, true);
p>add_choice(“Something else”, false);
Question q = *p;
q.display();
20. Suppose check_answer was not declared virtual in question.h. How would the
behavior of test.cpp change?
practice It Now you can try these exercises at the end of the chapter: R10.14, R10.15, P10.11,
P10.12.
Don’t use type tags
Some programmers build inheritance hierarchies in which each object has a tag that indicates
its type, commonly a string. They then query that string:
if (q>get_type() == “Question”)
{
// Do something
}
else if (q>get_type() == “ChoiceQuestion”)
{
// Do something else
}
This is a poor strategy. If a new class is added, then all these queries need to be revised. In con-
trast, consider the addi tion of a class NumericQuestion to our quiz program. Nothing needs to
change in that program because it uses virtual functions, not type tags.
Whenever you find yourself adding a type tag to a hierarchy of classes, reconsider and use
virtual functions instead.
slicing an Object
In C++ it is legal to copy a derived-class object into a base-class variable. However, any
derived-class information is lost in the process. For example, when a Manager object is assigned
to a variable of type Employee, the result is only the employee portion of the manager data:
Manager m;
…
Employee e = m; // Holds only the Employee base data of m
Any information that is particular to managers is sliced off, because it would not fit into a vari-
able of type Employee. To avoid slicing, you can use pointers.
The slicing problem commonly occurs when a function has a polymorphic parameter (that
is, a parameter that can belong to a base class or a derived class). In that case, the parameter
variable must be a pointer or a reference. Consider this example:
void ask(Question q) // Error
{
programming tip 10.2
Common error 10.4
q.display();
cout << "Your answer: ";
getline(cin, response);
cout << q.check_answer(response) << endl;
}
If you call this function with a ChoiceQuestion object, then the parameter variable q is initialized
with a copy of that object. But q is a Question object; the derived-class information is sliced
away. The simplest remedy is to use a refer ence:
void ask(const Question& q)
Now only the address is passed to the function. A reference is really a pointer in disguise. No
slicing occurs, and vir tual functions work correctly.
virtual self-calls
Suppose we add the following function to the Question class:
void Question::ask() const
{
display();
cout << "Your answer: ";
getline(cin, response);
cout << check_answer(response) << endl;
}
Now consider the call
ChoiceQuestion cq;
cq.set_text("In which country was the inventor of C++ born?");
...
cq.ask();
Which display and check_answer function will the ask function call? If you look inside the code
of the Question::ask function, you can see that these functions are executed on the implicit
parameter:
void Question::ask() const
{
implicit parameter.display();
cout << "Your answer: ";
getline(cin, response);
cout << implicit parameter.check_answer(response) << endl;
}
The implicit parameter in our call is cq, an object of type ChoiceQuestion. Because the display
and check_answer func tions are virtual, the ChoiceQuestion versions of the functions are called
automatically. This happens even though the ask function is defined in the Question class,
which has no knowledge of the ChoiceQuestion class.
As you can see, virtual functions are a very powerful mechanism. The Question class sup-
plies an ask function that specifies the common nature of asking a question, namely to dis-
play it and check the response. How the displaying and checking are carried out is left to the
derived classes.
special topic 10.2
cfe2_ch10_p441_480.indd 462 10/27/10 12:12 PM
10.4 Virtual Functions and polymorphism 463
q.display();
cout << "Your answer: ";
getline(cin, response);
cout << q.check_answer(response) << endl;
}
If you call this function with a ChoiceQuestion object, then the parameter variable q is initialized
with a copy of that object. But q is a Question object; the derived-class information is sliced
away. The simplest remedy is to use a refer ence:
void ask(const Question& q)
Now only the address is passed to the function. A reference is really a pointer in disguise. No
slicing occurs, and vir tual functions work correctly.
virtual self-calls
Suppose we add the following function to the Question class:
void Question::ask() const
{
display();
cout << "Your answer: ";
getline(cin, response);
cout << check_answer(response) << endl;
}
Now consider the call
ChoiceQuestion cq;
cq.set_text("In which country was the inventor of C++ born?");
...
cq.ask();
Which display and check_answer function will the ask function call? If you look inside the code
of the Question::ask function, you can see that these functions are executed on the implicit
parameter:
void Question::ask() const
{
implicit parameter.display();
cout << "Your answer: ";
getline(cin, response);
cout << implicit parameter.check_answer(response) << endl;
}
The implicit parameter in our call is cq, an object of type ChoiceQuestion. Because the display
and check_answer func tions are virtual, the ChoiceQuestion versions of the functions are called
automatically. This happens even though the ask function is defined in the Question class,
which has no knowledge of the ChoiceQuestion class.
As you can see, virtual functions are a very powerful mechanism. The Question class sup-
plies an ask function that specifies the common nature of asking a question, namely to dis-
play it and check the response. How the displaying and checking are carried out is left to the
derived classes.
special topic 10.2
cfe2_ch10_p441_480.indd 463 10/27/10 12:12 PM
464 Chapter 10 Inheritance
step 1 List the classes that are part of the hierarchy.
In our case, the problem description yields two classes: SavingsAccount and Checking Account. To
express the common ality between them, we will introduce a class BankAccount.
step 2 Organize the classes into an inheritance hierarchy.
Draw a UML diagram that shows base and derived classes. Here is the diagram for our
example:
Savings
Account
Checking
Account
BankAccount
step 3 Determine the common responsibilities.
In Step 2, you will have identified a class at the base of the hierarchy. That class needs to have
sufficient responsibili ties to carry out the tasks at hand. To find out what those tasks are, write
pseudocode for processing the objects:
For each user command
If it is a deposit or withdrawal
Deposit or withdraw the amount from the specified account.
Print the balance.
If it is month end processing
For each account
Call month end processing.
Print the balance.
h o W t o 1 0 . 1 Developing an Inheritance hierarchy
When you work with a set of classes, some of which are more general and others more spe-
cialized, you want to organize them into an inheritance hierarchy. This enables you to pro cess
objects of different classes in a uniform way.
As an example, we will consider a bank that offers its customers the following account
types:
• A savings account that earns interest. The interest compounds monthly and is computed on the min imum
monthly balance.
• A checking account that has no interest, gives you three free withdrawals per month, and charges a $1
transaction fee for each additional withdrawal.
The program will manage a set of accounts of both types, and it should be structured so that
other account types can be added without affecting the main processing loop. Supply a menu
D)eposit W)ithdraw M)onth end Q)uit
For deposits and withdrawals, query the account number and amount. Print the balance of the
account after each transaction.
In the “Month end” command, accumulate interest or clear the transaction counter,
depending on the type of the bank account. Then print the balance of all accounts.
cfe2_ch10_p441_480.indd 464 10/27/10 12:12 PM
10.4 Virtual Functions and polymorphism 465
step 1 List the classes that are part of the hierarchy.
In our case, the problem description yields two classes: SavingsAccount and Checking Account. To
express the common ality between them, we will introduce a class BankAccount.
step 2 Organize the classes into an inheritance hierarchy.
Draw a UML diagram that shows base and derived classes. Here is the diagram for our
example:
step 3 Determine the common responsibilities.
In Step 2, you will have identified a class at the base of the hierarchy. That class needs to have
sufficient responsibili ties to carry out the tasks at hand. To find out what those tasks are, write
pseudocode for processing the objects:
For each user command
If it is a deposit or withdrawal
Deposit or withdraw the amount from the specified account.
Print the balance.
If it is month end processing
For each account
Call month end processing.
Print the balance.
h o W t o 1 0 . 1 Developing an Inheritance hierarchy
When you work with a set of classes, some of which are more general and others more spe-
cialized, you want to organize them into an inheritance hierarchy. This enables you to pro cess
objects of different classes in a uniform way.
As an example, we will consider a bank that offers its customers the following account
types:
• A savings account that earns interest. The interest compounds monthly and is computed on the min imum
monthly balance.
• A checking account that has no interest, gives you three free withdrawals per month, and charges a $1
transaction fee for each additional withdrawal.
The program will manage a set of accounts of both types, and it should be structured so that
other account types can be added without affecting the main processing loop. Supply a menu
D)eposit W)ithdraw M)onth end Q)uit
For deposits and withdrawals, query the account number and amount. Print the balance of the
account after each transaction.
In the “Month end” command, accumulate interest or clear the transaction counter,
depending on the type of the bank account. Then print the balance of all accounts.
From the pseudocode, we obtain the following list of common responsibilities that every bank
account must carry out:
Deposit money.
Withdraw money.
Get the balance.
Carry out month end processing.
step 4 Decide which functions are overridden in derived classes.
For each derived class and each of the common responsibilities, decide whether the behavior
can be inherited or whether it needs to be overridden. Declare any functions that are overrid-
den as virtual in the root of the hierarchy.
Getting the balance is common to all account types. Withdrawing and end of month pro-
cessing are different for the derived classes, so they need to be declared virtual. Because it is
entirely possible that some future account type will levy a fee for deposits, it seems prudent to
declare the deposit member function virtual as well.
class BankAccount
{
public:
virtual void deposit(double amount);
virtual void withdraw(double amount);
virtual void month_end();
double get_balance() const;
private:
...
};
step 5 Define the public interface of each derived class.
Typically, derived classes have responsibilities other than those of the base class. List those,
as well as the member functions that need to be overridden. You also need to specify how the
objects of the derived classes should be con structed.
In this example, we need a way of setting the interest rate for the savings account. In addi-
tion, we need to specify constructors and overridden functions.
class SavingsAccount : public BankAccount
{
public:
/**
Constructs a savings account with a zero balance.
*/
SavingsAccount();
/**
Sets the interest rate for this account.
@param rate the monthly interest rate in percent
*/
void set_interest_rate(double rate);
virtual void withdraw(double amount);
virtual void month_end();
private:
...
};
class CheckingAccount : public BankAccount
{
cfe2_ch10_p441_480.indd 465 10/27/10 12:12 PM
466 Chapter 10 Inheritance
public:
/**
Constructs a checking account with a zero balance.
*/
CheckingAccount();
virtual void withdraw(double amount);
virtual void month_end();
private:
...
};
step 6 Identify data members.
List the data members for each class. If you find a data member that is common to all classes,
be sure to place it in the base of the hierarchy.
All accounts have a balance. We store that value in the BankAccount base class:
class BankAccount
{
...
private:
double balance;
};
The SavingsAccount class needs to store the interest rate. It also needs to store the minimum
monthly balance, which must be updated by all withdrawals:
class SavingsAccount : public BankAccount
{
...
private:
double interest_rate;
double min_balance;
};
The CheckingAccount class needs to count the withdrawals, so that the charge can be applied
after the free withdrawal limit is reached:
class CheckingAccount : public BankAccount
{
...
private:
int withdrawals;
};
step 7 Implement constructors and member functions.
The member functions of the BankAccount class update or return the balance:
BankAccount::BankAccount()
{
balance = 0;
}
void BankAccount::deposit(double amount)
{
balance = balance + amount;
}
void BankAccount::withdraw(double amount)
{
balance = balance amount;
}
cfe2_ch10_p441_480.indd 466 10/27/10 12:12 PM
10.4 Virtual Functions and polymorphism 467
double BankAccount::get_balance() const
{
return balance;
}
At the level of the BankAccount base class, we can say nothing about end of month processing.
We choose to make that function do nothing:
void BankAccount::month_end()
{
}
In the withdraw member function of the SavingsAccount class, the minimum balance is updated.
Note the call to the base-class member function:
void SavingsAccount::withdraw(double amount)
{
BankAccount::withdraw(amount);
double balance = get_balance();
if (balance < min_balance)
{
min_balance = balance;
}
}
In the month_end member function of the SavingsAccount class, the interest is deposited into the
account. We must call the deposit member function since we have no direct access to the bal
ance data member. The minimum balance is reset for the next month:
void SavingsAccount::month_end()
{
double interest = min_balance * interest_rate / 100;
deposit(interest);
min_balance = get_balance();
}
The withdraw function of the CheckingAccount class needs to check the withdrawal count. If there
have been too many withdrawals, a charge is applied. Again, note how the function invokes
the base-class function, using the BankAc count:: syntax:
void CheckingAccount::withdraw(double amount)
{
const int FREE_WITHDRAWALS = 3;
const int WITHDRAWAL_FEE = 1;
BankAccount::withdraw(amount);
withdrawals++;
if (withdrawals > FREE_WITHDRAWALS)
{
BankAccount::withdraw(WITHDRAWAL_FEE);
}
}
End of month processing for a checking account simply resets the withdrawal count:
void CheckingAccount::month_end()
{
withdrawals = 0;
}
step 8 Allocate objects on the heap and process them.
For polymorphism (that is, variation of behavior) to work in C++, you need to call virtual
functions through point ers. The easiest strategy is to allocate all polymorphic objects on the
heap, using the new operator.
cfe2_ch10_p441_480.indd 467 10/27/10 12:12 PM
468 Chapter 10 Inheritance
In our sample program, we allocate 5 checking accounts and 5 savings accounts and store
their addresses in an array of bank account pointers. Then we accept user commands and exe-
cute deposits, withdrawals, and monthly processing.
int main()
{
cout << fixed << setprecision(2);
// Create accounts
const int ACCOUNTS_SIZE = 10;
BankAccount* accounts[ACCOUNTS_SIZE];
for (int i = 0; i < ACCOUNTS_SIZE / 2; i++)
{
accounts[i] = new CheckingAccount;
}
for (int i = ACCOUNTS_SIZE / 2; i < ACCOUNTS_SIZE; i++)
{
SavingsAccount* account = new SavingsAccount;
account>set_interest_rate(0.75);
accounts[i] = account;
}
// Execute commands
bool more = true;
while (more)
{
cout << "D)eposit W)ithdraw M)onth end Q)uit: ";
string input;
cin >> input;
if (input == “D” || input == “W”) // Deposit or withdrawal
{
cout << "Enter account number and amount: ";
int num;
double amount;
cin >> num >> amount;
if (input == “D”) { accounts[num]>deposit(amount); }
else { accounts[num]>withdraw(amount); }
cout << "Balance: " << accounts[num]>get_balance() << endl;
}
else if (input == "M") // Month end processing
{
for (int n = 0; n < ACCOUNTS_SIZE(); n++)
{
accounts[n]>month_end();
cout << n << " " << accounts[n]>get_balance() << endl;
}
}
else if (input == "Q")
{
more = false;
}
}
return 0;
}
See ch10/accounts.cpp for the complete program.
W o r k e D e x a M p l e 1 0 . 1 Implementing an employee
hierarchy for payroll processing
This Worked Example shows how to implement payroll
pro cessing that works for different kinds of employees.
have you ever won-
dered how your
instr uc tor or grader makes sure your
programming homework is correct? In
all likelihood, they look at your solu-
tion and perhaps run it with some test
inputs. But usually they have a correct
solution available. that suggests that
there might be an easier way. perhaps
they could feed your program and
their correct program into a “program
comparator”, a computer program that
analyzes both programs and deter-
mines whether they both compute the
same results. of course, your solution
and the program that is known to be
correct need not be identical—what
matters is that they produce the same
output when given the same input.
how could such a program compar-
ator work? Well, the C++ compiler
knows how to read a program and
make sense of the classes, functions,
and statements. so it seems plausible
that someone could, with some effort,
write a program that reads two C++
programs, analyzes what they do, and
determines whether they solve the
same task. of course, such a program
would be very attractive to instruc tors,
because it could automate the grad-
ing process. thus, even though no
such program exists today, it might be
tempting to try to develop one and sell
it to universities around the world.
however, before you start raising
venture capital for such an effort, you
should know that theoretical com puter
scientists have proven that it is impos-
sible to develop such a pro gram, no
matter how hard you try.
there are quite a few of these
unsolvable problems. the first one,
called the halting problem, was dis-
covered by the British researcher alan
turing in 1936. Because his research
occurred before the first actual com-
puter was constructed, turing had to
devise a theoretical device, the Turing
machine, to explain how computers
could work. the turing machine con-
sists of a long magnetic tape, a read/
write head, and a program that has
numbered instructions of the form: “If
the current symbol under the head is
x, then replace it with y, move the head
one unit left or right, and con tinue
with instruction n” (see figure on the
next page). Interestingly enough, with
just these instructions, you can pro-
gram just as much as with C++, even
though it is incredibly tedious to do so.
theoretical computer scientists like
turing machines because they can be
described using nothing more than the
laws of mathematics.
Alan Turing
expressed in terms of C++, the halt-
ing problem states: “It is impossi ble
to write a program with two inputs,
namely the source code of an arbi-
trary C++ program P and a string I,
that decides whether the program P,
when executed with the input I, will
halt without getting into an infinite
loop”. of course, for some kinds of
programs and inputs, it is possible to
decide whether the programs halt with
the given input. the halting prob lem
asserts that it is impossible to come
up with a single decision-mak ing algo-
rithm that works with all pro grams and
inputs. note that you can’t simply run
the program P on the input I to settle
this question. If the program runs for
1,000 days, you don’t know that the
program is in an infinite loop. Maybe
you just have to wait another day for
it to stop.
such a “halt checker”, if it could be
written, might also be useful for grad-
ing homework. an instructor could use
it to screen student submissions to see
if they get into an infinite loop with a
particular input, and then not check
them any further. however, as turing
demonstrated, such a program cannot
be written. his argument is ingenious
and quite simple.
suppose a “halt checker” program
existed. let’s call it H. From H, we will
develop another program, the “killer”
program K. K does the following com-
putation. Its input is a string contain-
ing the source code for a program R.
It then applies the halting checker on
the input program R and the input
string r. that is, it checks whether the
program R halts if its input is its own
source code. It sounds bizarre to feed
a program to itself, but it isn’t impos-
sible. For example, the C++ compiler is
written in C++, and you can use it to
compile itself. or, as a simpler exam-
ple, you can use a word count pro gram
to count the words in its own source
code. (continued)
Random Fact 10.1 the limits of Computation
cfe2_ch10_p441_480.indd 468 10/27/10 12:12 PM
10.4 Virtual Functions and polymorphism 469
W o r k e D e x a M p l e 1 0 . 1 Implementing an employee
hierarchy for payroll processing
This Worked Example shows how to implement payroll
pro cessing that works for different kinds of employees.
have you ever won-
dered how your
instr uc tor or grader makes sure your
programming homework is correct? In
all likelihood, they look at your solu-
tion and perhaps run it with some test
inputs. But usually they have a correct
solution available. that suggests that
there might be an easier way. perhaps
they could feed your program and
their correct program into a “program
comparator”, a computer program that
analyzes both programs and deter-
mines whether they both compute the
same results. of course, your solution
and the program that is known to be
correct need not be identical—what
matters is that they produce the same
output when given the same input.
how could such a program compar-
ator work? Well, the C++ compiler
knows how to read a program and
make sense of the classes, functions,
and statements. so it seems plausible
that someone could, with some effort,
write a program that reads two C++
programs, analyzes what they do, and
determines whether they solve the
same task. of course, such a program
would be very attractive to instruc tors,
because it could automate the grad-
ing process. thus, even though no
such program exists today, it might be
tempting to try to develop one and sell
it to universities around the world.
however, before you start raising
venture capital for such an effort, you
should know that theoretical com puter
scientists have proven that it is impos-
sible to develop such a pro gram, no
matter how hard you try.
there are quite a few of these
unsolvable problems. the first one,
called the halting problem, was dis-
covered by the British researcher alan
turing in 1936. Because his research
occurred before the first actual com-
puter was constructed, turing had to
devise a theoretical device, the Turing
machine, to explain how computers
could work. the turing machine con-
sists of a long magnetic tape, a read/
write head, and a program that has
numbered instructions of the form: “If
the current symbol under the head is
x, then replace it with y, move the head
one unit left or right, and con tinue
with instruction n” (see figure on the
next page). Interestingly enough, with
just these instructions, you can pro-
gram just as much as with C++, even
though it is incredibly tedious to do so.
theoretical computer scientists like
turing machines because they can be
described using nothing more than the
laws of mathematics.
Alan Turing
expressed in terms of C++, the halt-
ing problem states: “It is impossi ble
to write a program with two inputs,
namely the source code of an arbi-
trary C++ program P and a string I,
that decides whether the program P,
when executed with the input I, will
halt without getting into an infinite
loop”. of course, for some kinds of
programs and inputs, it is possible to
decide whether the programs halt with
the given input. the halting prob lem
asserts that it is impossible to come
up with a single decision-mak ing algo-
rithm that works with all pro grams and
inputs. note that you can’t simply run
the program P on the input I to settle
this question. If the program runs for
1,000 days, you don’t know that the
program is in an infinite loop. Maybe
you just have to wait another day for
it to stop.
such a “halt checker”, if it could be
written, might also be useful for grad-
ing homework. an instructor could use
it to screen student submissions to see
if they get into an infinite loop with a
particular input, and then not check
them any further. however, as turing
demonstrated, such a program cannot
be written. his argument is ingenious
and quite simple.
suppose a “halt checker” program
existed. let’s call it H. From H, we will
develop another program, the “killer”
program K. K does the following com-
putation. Its input is a string contain-
ing the source code for a program R.
It then applies the halting checker on
the input program R and the input
string r. that is, it checks whether the
program R halts if its input is its own
source code. It sounds bizarre to feed
a program to itself, but it isn’t impos-
sible. For example, the C++ compiler is
written in C++, and you can use it to
compile itself. or, as a simpler exam-
ple, you can use a word count pro gram
to count the words in its own source
code. (continued)
Random Fact 10.1 the limits of Computation
Available online at www.wiley.com/college/horstmann.
cfe2_ch10_p441_480.indd 469 10/27/10 12:12 PM
www.wiley.com/college/horstmann
470 Chapter 10 Inheritance
explain the notions of inheritance, base class, and derived class.
• A derived class inherits data and behavior from a
base class.
• You can always use a derived-class object in place
of a base-class object.
When K gets the answer from H
that R halts when applied to itself, it is
programmed to enter an infinite loop.
otherwise K exits. In C++, the pro gram
might look like this:
int main()
{
string r = read program input;
HaltChecker checker;
if (checker.check(r, r))
{
while (true) { }
// Infinite loop
}
else
{
return 0;
}
}
now ask yourself: What does the halt
checker answer when asked if K halts
when given K as the input? Maybe it
finds out that K gets into an infinite
loop with such an input. But wait, that
can’t be right. that would mean that
checker.check(r,r) returns false when
r is the program code of K. as you can
plainly see, in that case, the main func-
tion returns, so K didn’t get into an
infinite loop. that shows that K must
halt when analyzing itself, so checker.
check(r, r) should return true. But
then the main function doesn’t termi-
nate—it goes into an infi nite loop. that
shows that it is logi cally impossible to
implement a program that can check
whether every program halts on a par-
ticular input.
It is sobering to know that there are
limits to computing. there are prob-
lems that no computer program, no
matter how ingenious, can answer.
theoretical computer scientists are
working on other research involving
the nature of computation. one impor-
tant question that remains unsettled
to this day deals with problems that
in practice are very time-consuming to
solve. It may be that these problems
are intrinsically hard, in which case it
would be pointless to try to look for
better algorithms. such theoretical
research can have important practi-
cal applications. For example, right
now, nobody knows whether the most
com mon encryption schemes used
today could be broken by discover-
ing a new algorithm (see random Fact
8.1 for more information on encryp-
tion algo rithms). knowing that no fast
algo rithms exist for breaking a par-
ticular code could make us feel more
com fortable about the security of
encryp tion.
C h a p t e r s U M M a r Y
Implement derived classes in c++.
• A derived class can override a base-class function
by providing a new implementation.
• The derived class inherits all data members and all
functions that it does not override.
• Unless specified otherwise, the base-class data
members are initialized with the default construc tor.
• The constructor of a derived class can supply arguments to a base-class
constructor.
Describe how a derived class can override functions from a base class.
• A derived class can inherit a function from the base class, or it can override it by
providing another implementation.
• Use BaseClass::function notation to explicitly call a base-class function.
Describe virtual functions and polymorphism.
• When converting a derived-class object to a base class, the derived-class data is
sliced away.
• A derived-class pointer can be converted to a base-class pointer.
• When a virtual function is called, the version belonging
to the actual type of the implicit parameter is invoked.
• Polymorphism (literally, “having multiple shapes”)
describes objects that share a set of tasks and execute
them in different ways.
r10.1 Identify the base class and the derived class in each of the following pairs of classes.
a. Employee, Manager
b. Polygon, Triangle
c. GraduateStudent, Student
d. Person, Student
e. Employee, Professor
f. BankAccount, CheckingAccount
g. Vehicle, Car
h. Vehicle, Minivan
i. Car, Minivan
j. Truck, Vehicle
r e V I e W e x e r C I s e s
Instruction
number
If tape
symbol is
Replace
with
Then move
head
Then go to
instruction
1
1
2
2
2
3
3
3
4
4
0
1
0
1
2
0
1
2
1
2
2
1
0
1
0
0
1
2
1
0
right
left
right
right
left
left
left
right
right
left
2
4
2
2
3
3
3
1
5
4
Program
Control unit
Read/write head
Tape
A Turing Machine
cfe2_ch10_p441_480.indd 470 10/27/10 12:12 PM
review exercises 471
Implement derived classes in c++.
• A derived class can override a base-class function
by providing a new implementation.
• The derived class inherits all data members and all
functions that it does not override.
• Unless specified otherwise, the base-class data
members are initialized with the default construc tor.
• The constructor of a derived class can supply arguments to a base-class
constructor.
Describe how a derived class can override functions from a base class.
• A derived class can inherit a function from the base class, or it can override it by
providing another implementation.
• Use BaseClass::function notation to explicitly call a base-class function.
Describe virtual functions and polymorphism.
• When converting a derived-class object to a base class, the derived-class data is
sliced away.
• A derived-class pointer can be converted to a base-class pointer.
• When a virtual function is called, the version belonging
to the actual type of the implicit parameter is invoked.
• Polymorphism (literally, “having multiple shapes”)
describes objects that share a set of tasks and execute
them in different ways.
r10.1 Identify the base class and the derived class in each of the following pairs of classes.
a. Employee, Manager
b. Polygon, Triangle
c. GraduateStudent, Student
d. Person, Student
e. Employee, Professor
f. BankAccount, CheckingAccount
g. Vehicle, Car
h. Vehicle, Minivan
i. Car, Minivan
j. Truck, Vehicle
r e V I e W e x e r C I s e s
cfe2_ch10_p441_480.indd 471 10/27/10 12:12 PM
472 Chapter 10 Inheritance
r10.2 An object-oriented traffic simulation system has the following classes:
Vehicle PickupTruck
Car SportUtilityVehicle
Truck Minivan
Sedan Bicycle
Coupe Motorcycle
Draw a UML diagram that shows the inheritance relationships between these
classes.
r10.3 What inheritance relationships would you establish among the following classes?
Student Professor
Employee Secretary
Person Janitor
DepartmentChair
r10.4 Draw a UML diagram that shows the inheritance and aggregation relationships
between the classes:
Person Instructor
Student Lecture
Course Lab
r10.5 Consider a program for managing inventory in a small appliance store. Why isn’t it
useful to have a base class SmallAppliance and derived classes Toaster, CarVacuum,
Trave lIron, and so on?
r10.6 Which data members does the CheckingAccount class in How To 10.1 on page 464 inherit
from its base class? Which data members does it add?
r10.7 Which functions does the SavingsAccount class in How To 10.1 on page 464 inherit from
its base class? Which functions does it override? Which functions does it add?
r10.8 Design an inheritance hierarchy for geometric shapes: rectangles, squares, and
cir cles. Draw a UML diagram. Provide a virtual function to compute the area of a
shape. Provide appropriate constructors for each class. Write the class definitions
but do not provide implementations of the member functions.
r10.9 Continue Exercise R10.8 by writing a main function that executes the following steps:
Fill a vector of shape pointers with a rectangle, a square, and a circle.
Print the area of each shape.
Deallocate all heap objects.
r10.10 Can you convert a base-class object into a derived-class object? A derived-class
object into a base-class object? A base-class pointer into a derived-class pointer? A
derived-class pointer into a base-class pointer? If so, give examples. If not, explain
why not.
r10.11 Consider a function process_file(ostream& str). Objects from which of the classes in
Figure 2 can be passed as parameters to this function?
cfe2_ch10_p441_480.indd 472 10/27/10 12:12 PM
review exercises 473
r10.12 What does the following program print?
class B
{
public:
void print(int n) const;
};
void B::print(int n) const
{
cout << n << endl;
}
class D : public B
{
public:
void print(int n) const;
};
void D::print(int n) const
{
if (n <= 1) { B::print(n); }
else if (n % 2 == 0) { print(n / 2); }
else { print(3 * n + 1); }
}
int main()
{
D d;
d.print(3);
return 0;
}
Determine the answer by hand, not by compiling and running the program.
r10.13 Suppose the class D inherits from B. Which of the following assignments are legal?
B b;
D d;
B* pb;
D* pd;
a. b = d;
b. d = b;
c. pd = pb;
d. pb = pd;
e. d = pd;
f. b = *pd;
g. *pd = *pb;
r10.14 Suppose the class Sub is derived from the class Sandwich. Which of the following
assignments are legal?
Sandwich* x = new Sandwich();
Sub* y = new Sub();
a. x = y;
b. y = x;
c. y = new Sandwich();
cfe2_ch10_p441_480.indd 473 10/27/10 12:12 PM
474 Chapter 10 Inheritance
d. x = new Sub();
e. *x = *y;
f. *y = *x;
r10.15 What does the program print? Explain your answers by tracing the flow of each call.
class B
{
public:
B();
virtual void p() const;
void q() const;
};
B::B() {}
void B::p() const { cout << "B::p\n"; }
void B::q() const { cout << "B::q\n"; }
class D : public B
{
public:
D();
virtual void p() const;
void q() const;
};
D::D() {}
void D::p() const { cout << "D::p\n"; }
void D::q() const { cout << "D::q\n"; }
int main()
{
B b;
D d;
B* pb = new B;
B* pd = new D;
D* pd2 = new D;
b.p(); b.q();
d.p(); d.q();
pb>p(); pb>q();
pd>p(); pd>q();
pd2>p(); pd2>q();
return 0;
}
r10.16 Using the Employee class from Worked Example 10.1 (ch10/salaries.cpp in your
companion code), form a subclass Volunteer of Employee and provide a constructor
Volun teer(string name) that sets the salary to 0.
r10.17 In the accounts.cpp program of How To 10.1, would it be reasonable to make the
get_balance function virtual? Explain your reasoning.
r10.18 What is the effect of declaring the display member function virtual only in the
ChoiceQuestion class?
p10.1 Add a class NumericQuestion to the question hierarchy of Section 10.1. If the response
and the expected answer differ by no more than 0.01, then accept it as correct.
p10.2 Add a class FillInQuestion to the question hierarchy of Section 10.1. Such a question
is constructed with a string that contains the answer, surrounded by _ _, for exam ple,
“The inventor of C++ was _Bjarne Stroustrup_”. The question should be displayed as
The inventor of C++ was _____
Provide a main function that demonstrates your class.
p10.3 Modify the check_answer member function of the Question class so that it does not take
into account different spaces or upper/lowercase characters. For example, the
response ” bjarne stroustrup” should match an answer of “Bjarne Stroustrup”.
p10.4 Add a class MultiChoiceQuestion to the question hierarchy of Section 10.1 that allows
multiple correct choices. The respondent should provide any one of the correct
choices. The answer string should contain all of the correct choices, separated by
spaces.
p10.5 Add a class ChooseAllCorrect to the question hierarchy of Section 10.1 that allows
multiple correct choices. The respondent should provide all correct choices, sepa-
rated by spaces.
p10.6 Add a member function add_text to the Question base class and provide a different
implementation of ChoiceQuestion that calls add_text rather than storing a vector of
choices.
p10.7 Change the CheckingAccount class in How To 10.1 so that a $1 fee is levied for depos its
or withdrawals in excess of three free monthly transactions. Place the code for
computing the fee into a private member function that you call from the deposit and
withdraw member functions.
p10.8 Derive a class Programmer from Employee. Supply a constructor Programmer(string name,
double salary) that calls the base-class constructor. Supply a function get_name that
returns the name in the format “Hacker, Harry (Programmer)”.
p10.9 Implement a base class Person. Derive classes Student and Instructor from Person.
A person has a name and a birthday. A student has a major, and an instructor has a
sal ary. Write the class definitions, the constructors, and the member functions
display for all classes.
p10.10 Derive a class Manager from Employee. Add a data member named department of type
string. Supply a function display that displays the manager’s name, department, and
salary. Derive a class Executive from Manager. Supply a function display that displays
the string Executive, followed by the information stored in the Manager base object.
p10.11 Implement a base class Account and derived classes Savings and Checking. In the base
class, supply member functions deposit and withdraw. Provide a function daily_
interest that computes and adds the daily interest. For calculations, assume that
every month has 30 days. Checking accounts yield interest of 3 percent annu ally on
balances over $1,000. Savings accounts yield interest of 6 percent annually on the
entire balance. Write a test program that makes a month’s worth of depos its and
withdrawals and calculates the interest every day.
p r o G r a M M I n G e x e r C I s e s
cfe2_ch10_p441_480.indd 474 10/27/10 12:12 PM
programming exercises 475
p10.1 Add a class NumericQuestion to the question hierarchy of Section 10.1. If the response
and the expected answer differ by no more than 0.01, then accept it as correct.
p10.2 Add a class FillInQuestion to the question hierarchy of Section 10.1. Such a question
is constructed with a string that contains the answer, surrounded by _ _, for exam ple,
“The inventor of C++ was _Bjarne Stroustrup_”. The question should be displayed as
The inventor of C++ was _____
Provide a main function that demonstrates your class.
p10.3 Modify the check_answer member function of the Question class so that it does not take
into account different spaces or upper/lowercase characters. For example, the
response ” bjarne stroustrup” should match an answer of “Bjarne Stroustrup”.
p10.4 Add a class MultiChoiceQuestion to the question hierarchy of Section 10.1 that allows
multiple correct choices. The respondent should provide any one of the correct
choices. The answer string should contain all of the correct choices, separated by
spaces.
p10.5 Add a class ChooseAllCorrect to the question hierarchy of Section 10.1 that allows
multiple correct choices. The respondent should provide all correct choices, sepa-
rated by spaces.
p10.6 Add a member function add_text to the Question base class and provide a different
implementation of ChoiceQuestion that calls add_text rather than storing a vector of
choices.
p10.7 Change the CheckingAccount class in How To 10.1 so that a $1 fee is levied for depos its
or withdrawals in excess of three free monthly transactions. Place the code for
computing the fee into a private member function that you call from the deposit and
withdraw member functions.
p10.8 Derive a class Programmer from Employee. Supply a constructor Programmer(string name,
double salary) that calls the base-class constructor. Supply a function get_name that
returns the name in the format “Hacker, Harry (Programmer)”.
p10.9 Implement a base class Person. Derive classes Student and Instructor from Person.
A person has a name and a birthday. A student has a major, and an instructor has a
sal ary. Write the class definitions, the constructors, and the member functions
display for all classes.
p10.10 Derive a class Manager from Employee. Add a data member named department of type
string. Supply a function display that displays the manager’s name, department, and
salary. Derive a class Executive from Manager. Supply a function display that displays
the string Executive, followed by the information stored in the Manager base object.
p10.11 Implement a base class Account and derived classes Savings and Checking. In the base
class, supply member functions deposit and withdraw. Provide a function daily_
interest that computes and adds the daily interest. For calculations, assume that
every month has 30 days. Checking accounts yield interest of 3 percent annu ally on
balances over $1,000. Savings accounts yield interest of 6 percent annually on the
entire balance. Write a test program that makes a month’s worth of depos its and
withdrawals and calculates the interest every day.
p r o G r a M M I n G e x e r C I s e s
cfe2_ch10_p441_480.indd 475 10/27/10 12:12 PM
476 Chapter 10 Inheritance
p10.12 Implement a base class Appointment and derived classes Onetime, Daily, Weekly, and
Monthly. An appointment has a description (for example, “see the dentist”) and a date
and time. Write a virtual function occurs_on(int year, int month, int day) that checks
whether the appointment occurs on that date. For example, for a monthly appoint-
ment, you must check whether the day of the month matches. Then fill a vector of
Appointment* with a mixture of appointments. Have the user enter a date and print out
all appointments that happen on that date.
p10.13 Improve the appointment book program of Exercise P10.12. Give the user the
option to add new appointments. The user must specify the type of the appoint ment,
the description, and the date and time.
p10.14 Improve the appointment book program of Exercises P10.12 and P10.13 by letting
the user save the appointment data to a file and reload the data from a file. The sav ing
part is straightforward: Make a virtual function save. Save out the type, descrip tion,
date, and time. The loading part is not so easy. You must first determine the type of
the appointment to be loaded, create an object of that type with its default construc-
tor, and then call a virtual load function to load the remainder.
p10.15 Use polymorphism to carry out image manipulations such as those described in
Exercises P8.10 and P8.11. Design a base class Effect and derived classes Sunset and
Grayscale. Use a virtual function process. The file processing part of your program
should repeatedly call process on an Effect* pointer, without having to know which
effect is applied.
engineering p10.16 In this problem, you will model a circuit consisting of an
arbitrary configuration of resistors. Provide a base class
Circuit with a member function get_resistance. Pro vide a
derived class Resistor representing a single resistor. Pro-
vide derived classes Serial and Parallel, each of which
contains a vector
of circuits, each of which can be a single resistor or another
circuit. Simi larly, a Parallel circuit models a set of circuits
in parallel. For example, the following circuit is a Parallel
circuit containing a single resistor and one Serial cir cuit.
A Serial circuit
Use Ohm’s law to compute the combined resistance.
engineering p10.17 Part (a) of the figure below shows a symbolic representation of an electric circuit
called an amplifier. The input to the amplifier is the voltage vi and the output is the
voltage vo. The output of an amplifier is proportional to the input. The constant of
proportionality is called the “gain” of the amplifier.
Parts (b), (c), and (d) show schematics of three specific types of amplifier: the
inverting amplifier, noninverting amplifier, and voltage divider amplifier. Each of
these three amplifiers consists of two resistors and an op amp. The value of the gain
of each amplifier depends on the values of its resistances. In particular, the gain, g, of
cfe2_ch10_p441_480.indd 476 10/27/10 12:12 PM
programming exercises 477
the inverting amplifier is given by g
R
R
= − 2
1
. Similarly the gains of the noninverting
amplifier and voltage divider amplifier are given by g
R
R
= +1 2
1
and g
R
R R
=
+
2
1 2
,
respectively.
–
+
–
+
–
+
R2
R1
R1
vi R2
R1 R2
vo
vovi
vi
vo
vovi
(a) Amplifier (b) Inverting amplifier
(c) Noninverting amplifier (d) Voltage divider amplifier
Write a C++ program that represents the amplifier as a base class and represents the
inverting, noninverting, and voltage divider amplifiers as derived classes. Give the
base class two virtual functions, get_gain and a get_description function that returns a
string identifying the amplifier. Each derived class should have a constructor with
two arguments, the resistances of the amplifier.
The derived classes need to override the get_gain and get_description functions of the
base class.
Write a main function for the C++ program that demonstrates that the derived classes
all work properly for sample values of the resistances.
engineering p10.18 Resonant circuits are used to select a signal (e.g., a radio station or TV channel)
from among other competing signals. Resonant circuits are characterized by the
frequency response shown in the figure below. The resonant frequency response is
completely described by three parameters: the resonant frequency, ωo, the band-
width, B, and the gain at the resonant frequency, k.
Frequency (rad/s, log scale)
k
ω o
B0.707k
cfe2_ch10_p441_480.indd 477 10/27/10 12:12 PM
478 Chapter 10 Inheritance
Two simple resonant circuits are shown in the figure below. The circuit in (a) is
called a parallel resonant circuit. The circuit in (b) is called a series resonant circuit.
Both resonant circuits consist of a resistor having resistance R, a capacitor having
capacitance C, and an inductor having inductance L.
R L C C
L
R
(a) Parallel resonant circuit (b) Series resonant circuit
These circuits are designed by determining values of R, C, and L that cause the
resonant frequency response to be described by specified values of ωo, B, and k. The
design equations for the parallel resonant circuit are:
R k C
BR
L
C
= = =, , and
o
1 1
2ω
Similarly, the design equations for the series resonant circuit are:
R
k
L
R
B
C
L
= = =
1 1
2
, , and
oω
Write a C++ program that represents ResonantCircuit as a base class and represents
the SeriesResonantCircuit and ParallelResonantCircuit as derived classes. Give the base
class three private data members representing the parameters ωo, B, and k of the
resonant frequency response. The base class should provide public member func-
tions to get and set each of these members. The base class should also provide a
display function that prints a description of the resonant frequency response.
Each derived class should provide a function that designs the corresponding reso-
nant circuit. The derived classes should also override the display function of the base
class to print descriptions of both the frequency response (the values of ωo, B, and k)
and the circuit (the values of R, C, and L).
All classes should provide appropriate constructors.
Write a main function for the C++ program that demonstrates that the derived classes
all work properly.
1. Because every manager is an employee but not the other way around, the Manager
class is more specialized. It is the derived class, and Employee is the base class.
2. CheckingAccount and SavingsAccount both inherit from the more general class
BankAc count.
3. istream, istringstream, ifstream, iostream, fstream
4. Vehicle, Truck, Motorcycle
5. It shouldn’t. A quiz isn’t a question; it has questions.
6. a, b, d
7. class Manager : public Employee
{
private:
double bonus;
}
8. name, base_salary, and bonus
9. class Manager : public Employee
{
public:
double get_salary() const;
}
10. get_name, set_name, set_base_salary
11. The function is not allowed to access the text member from the base class.
12. The type of the this pointer is ChoiceQuestion*. Therefore, the display function of
ChoiceQuestion is selected, and the function calls itself.
13. Because there is no ambiguity. The derived class doesn’t have a set_answer
function.
14. string Manager::get_name() const
{
return “*” + Employee::get_name();
}
15. double Manager::get_salary() const
{
return Employee::get_salary() + bonus;
}
16. The type of quiz[1] is Question*, and the Question class has no member function called
add_choice.
17. a and d are legal.
18. a and d are legal.
19. Only the question text, not the choices. The choices are sliced away when *p is
assigned to the object q.
20. It wouldn’t change. The function is never overridden in the classes used in our test
program.
a n s W e r s t o s e l F – C h e C k Q U e s t I o n s
cfe2_ch10_p441_480.indd 478 10/27/10 12:12 PM
answers to self-Check Questions 479
1. Because every manager is an employee but not the other way around, the Manager
class is more specialized. It is the derived class, and Employee is the base class.
2. CheckingAccount and SavingsAccount both inherit from the more general class
BankAc count.
3. istream, istringstream, ifstream, iostream, fstream
4. Vehicle, Truck, Motorcycle
5. It shouldn’t. A quiz isn’t a question; it has questions.
6. a, b, d
7. class Manager : public Employee
{
private:
double bonus;
}
8. name, base_salary, and bonus
9. class Manager : public Employee
{
public:
double get_salary() const;
}
10. get_name, set_name, set_base_salary
11. The function is not allowed to access the text member from the base class.
12. The type of the this pointer is ChoiceQuestion*. Therefore, the display function of
ChoiceQuestion is selected, and the function calls itself.
13. Because there is no ambiguity. The derived class doesn’t have a set_answer
function.
14. string Manager::get_name() const
{
return “*” + Employee::get_name();
}
15. double Manager::get_salary() const
{
return Employee::get_salary() + bonus;
}
16. The type of quiz[1] is Question*, and the Question class has no member function called
add_choice.
17. a and d are legal.
18. a and d are legal.
19. Only the question text, not the choices. The choices are sliced away when *p is
assigned to the object q.
20. It wouldn’t change. The function is never overridden in the classes used in our test
program.
a n s W e r s t o s e l F – C h e C k Q U e s t I o n s
cfe2_ch10_p441_480.indd 479 10/27/10 12:12 PM
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AA P P E N D I X
481
C + + L A N g u A g E
C o D I N g g u I D E L I N E s
Introduction
This coding style guide is a simplified version of one that has been used with good
success both in industrial practice and for college courses. It lays down rules that you
must follow for your programming assignments.
A style guide is a set of mandatory requirements for layout and formatting. Uni
form style makes it easier for you to read code from your instructor and classmates.
You will really appreciate the consistency if you do a team project. It is also easier for
your instructor and your grader to grasp the essence of your programs quickly.
A style guide makes you a more productive programmer because it reduces gra
tuitous choice. If you don’t have to make choices about trivial matters, you can spend
your energy on the solution of real problems.
In these guidelines a number of constructs are plainly outlawed. That doesn’t
mean that programmers using them are evil or incompetent. It does mean that the
constructs are of marginal utility and can be expressed just as well or even better with
other language constructs.
If you have already programmed in C or C++, you may be initially uncomfort able
about giving up some fond habits. However, it is a sign of professionalism to set aside
personal preferences in minor matters and to compromise for the benefit of your
group.
These guidelines are necessarily somewhat long and dull. They also mention fea
tures that you may not yet have seen in class. Here are the most important highlights:
• Tabs are set every three spaces.
• Variable and function names are lowercase.
• Constant names are uppercase. Class names start with an uppercase letter.
• There are spaces after reserved words and between binary operators.
• Braces must line up.
• No magic numbers may be used.
• Every function must have a comment.
• At most 30 lines of code may be used per function.
• No goto, continue, or break is allowed.
• At most two global variables may be used per file.
A note to the instructor: Of course, many programmers and organizations have strong
feelings about coding style. If this style guide is incompatible with your own prefer
ences or with local custom, please feel free to modify it. For that purpose, this coding
style guide is available in electronic form on the companion web site for this book.
cfe2_appA_481_488.indd 481 10/28/10 2:41 PM
482 Appendix A C++ Language Coding guidelines
source Files
Each program is a collection of one or more files or modules. The executable pro gram
is obtained by compiling and linking these files. Organize the material in each file as
follows:
• Header comments
• #include statements
• Constants
• Classes
• Functions
It is common to start each file with a comment block. Here is a typical format:
/**
@file invoice.cpp
@author Jenny Koo
@date 2012-01-24
@version 3.14
*/
You may also want to include a copyright notice, such as
/* Copyright 2012 Jenny Koo */
A valid copyright notice consists of
• the copyright symbol or the word “Copyright” or the abbreviation “Copr.”
• the year of first publication of the work
• the name of the owner of the copyright
(Note: To save space, this header comment has been omitted from the programs in
this book as well as the programs on disk so that the actual line numbers match those
that are printed in the book.)
Next, list all included header files.
#include
#include “question.h”
Do not embed absolute path names, such as
#include “c:\me\my_homework\widgets.h” // Don’t !!!
After the header files, list constants that are needed throughout the program file.
const int GRID_SIZE = 20;
const double CLOCK_RADIUS = 5;
Then supply the definitions of all classes.
class Product
{
…
};
Order the class definitions so that a class is defined before it is used in another class.
Finally, list all functions, including member functions of classes and nonmember
functions. Order the nonmember functions so that a function is defined before it is
called. As a consequence, the main function will be the last function in your file.
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Appendix A C++ Language Coding guidelines 483
Functions
Supply a comment of the following form for every function.
/**
Explanation.
@param parameter variable1 explanation
@param parameter variable2 explanation
…
@return explanation
*/
The introductory explanation is required for all functions except main. It should start
with an uppercase letter and end with a period. Some documentation tools extract
the first sentence of the explanation into a summary table. Thus, if you provide an
explanation that consists of multiple sentences, formulate the explanation such that
the first sentence is a concise explanation of the function’s purpose.
Omit the @param comment if the function has no parameter variables. Omit the
@return comment for void functions. Here is a typical example:
/**
Converts calendar date into Julian day. This algorithm is from Press
et al., Numerical Recipes in C, 2nd ed., Cambridge University Press, 1992.
@param year the year of the date to be converted
@param month the month of the date to be converted
@param day the day of the date to be converted
@return the Julian day number that begins at noon of the given
calendar date
*/
long dat2jul(int year, int month, int day)
{
…
}
Parameter variable names must be explicit, especially if they are integers or Boolean.
Employee remove(int d, double s); // Huh?
Employee remove(int department, double severance_pay); // OK
Of course, for very generic functions, short names may be very appropriate.
Do not write void functions that return exactly one answer through a reference.
Instead, make the result into a return value.
void find(vector
bool find(vector
Of course, if the function computes more than one value, some or all results can be
returned through reference parameters.
Functions must have at most 30 lines of code. (Comments, blank lines, and lines
containing only braces are not included in this count.) Functions that consist of
one long if/else/else statement sequence may be longer, provided each branch is 10
lines or less. This rule forces you to break up complex computations into separate
func tions.
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484 Appendix A C++ Language Coding guidelines
Local Variables
Do not define all local variables at the beginning of a block. Define each variable just
before it is used for the first time.
Every variable must be either explicitly initialized when defined or set in the
immediately following statement (for example, through a >> instruction).
int pennies = 0;
or
int pennies;
cin >> pennies;
Move variables to the innermost block in which they are needed:
while (…)
{
double xnew = (xold + a / xold) / 2;
…
}
Do not define two variables in one statement:
int dimes = 0, nickels = 0; // Don’t
When defining a pointer variable, place the * with the type, not the variable:
Link* p; // OK
not
Link *p; // Don’t
Constants
In C++, do not use #define to define constants:
#define CLOCK_RADIUS 5 // Don’t
Use const instead:
const double CLOCK_RADIUS = 5; // The radius of the clock face
You may not use magic numbers in your code. (A magic number is an integer con
stant embedded in code without a constant definition.) Any number except 0, 1, or 2
is considered magic:
if (p.get_x() < 10) // Don’t
Use a const variable instead:
const double WINDOW_XMAX = 10;
if (p.get_x() < WINDOW_XMAX) // OK
Even the most reasonable cosmic constant is going to change one day. You think
there are 365 days per year? Your customers on Mars are going to be pretty unhappy
about your silly prejudice.
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Appendix A C++ Language Coding guidelines 485
Make a constant
const int DAYS_PER_YEAR = 365;
so that you can easily produce a Martian version without trying to find all the 365’s,
364’s, 366’s, 367’s, and so on in your code.
Classes
Lay out the items of a class as follows:
class ClassName
{
public:
constructors
mutators
accessors
private:
data
};
All data fields of classes must be private.
Control Flow
The for statement
Use for loops only when a variable runs from somewhere to somewhere else with
some constant increment/decrement.
for (i = 0; i < a.size(); i++)
{
cout << a[i] << endl;
}
Do not use the for loop for weird constructs such as
for (xnew = a / 2; count < ITERATIONS; cout << xnew) // Don’t
{
xold = xnew;
xnew = xold + a / xold;
count++;
}
Make such a loop into a while loop, so the sequence of instructions is much clearer.
xnew = a / 2;
while (count < ITERATIONS) // OK
{
xold = xnew;
xnew = xold + a / xold;
count++;
cout << xnew;
}
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486 Appendix A C++ Language Coding guidelines
Nonlinear Control Flow
Don’t use the switch statement. Use if/else instead.
Do not use the break, continue, or goto statement. Use a bool variable to control the
execution flow.
Lexical Issues
Naming Conventions
The following rules specify when to use upper and lowercase letters in identifier
names.
1. All variable and function names and all data fields of classes are in lowercase,
sometimes with an underscore in the middle. For example, first_player.
2. All constants are in uppercase, with an occasional underscore. For example,
CLOCK_RADIUS.
3. All class names start with uppercase and are followed by lowercase letters, with
an occasional uppercase letter in the middle. For example, BankTeller.
Names must be reasonably long and descriptive. Use first_player instead of fp. No
drppng f vwls. Local variables that are fairly routine can be short (ch, i) as long as they
are really just boring holders for an input character, a loop counter, and so on. Also,
do not use ctr, c, cntr, cnt, c2 for five counter variables in your function. Surely each of
these variables has a specific purpose and can be named to remind the reader of it (for
example, ccurrent, cnext, cprevious, cnew, cresult).
Indentation and White space
Use tab stops every three columns. Save your file so that it contains no tabs at all.
That means you will need to change the tab stop setting in your editor! In the edi tor,
make sure to select “3 spaces per tab stop” and “save all tabs as spaces”. Every pro
gramming editor has these settings. If yours doesn’t, don’t use tabs at all but type the
correct number of spaces to achieve indentation.
Use blank lines freely to separate logically distinct parts of a function.
Use a blank space around every binary operator:
x1 = (-b - sqrt(b * b - 4 * a * c)) / (2 * a); // Good
x1=(-b-sqrt(b*b-4*a*c))/(2*a); // Bad
Leave a blank space after (and not before) each comma, semicolon, and reserved
word, but not after a function name.
if (x == 0) ...
f(a, b[i]);
Every line must fit in 80 columns. If you must break a statement, add an indentation
level for the continuation:
a[n] = ...................................................
+ .................;
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Appendix A C++ Language Coding guidelines 487
Braces
Opening and closing braces must line up, either horizontally or vertically.
while (i < n) { cout << a[i] << endl; i++; } // OK
while (i < n)
{
cout << a[i] << endl;
i++;
} // OK
Some programmers don’t line up vertical braces but place the { behind the while:
while (i < n) { // Don’t
cout << a[i] << endl;
i++;
}
This style saves a line, but it is difficult to match the braces.
Always use braces with if, while, do, and for statements, even if the body is only a
single statement.
if (floor > 13)
{ // OK
floor––;
}
if (floor > 13)
floor––; // Don’t
unstable Layout
Some programmers take great pride in lining up certain columns in their code:
class Employee
{
…
private:
string name;
int age;
double hourly_wage;
};
This is undeniably neat, and we recommend it if your editor does it for you, but don’t
do it manually. The layout is not stable under change. A data type that is longer than
the preallotted number of columns requires that you move all entries around.
Some programmers like to format multiline comments so that every line starts
with **:
/* This is a comment
** that extends over
** three source lines
*/
Again, this is neat if your editor has a command to add and remove the asterisks, and
if you know that all programmers who will maintain your code also have such an
editor. Otherwise, it can be a powerful method of discouraging programmers from
editing the comment. If you have to choose between pretty comments and comments
that reflect the current facts of the program, facts win over beauty.
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BA P P E N D I X
489
R E s E R v E D W o R D
s u m m A R y
Reserved Word Description Reference Location
bool The Boolean type Section 3.7
break Break out of a loop or switch Special Topic 3.3, 4.2
case A label in a switch statement Special Topic 3.3
char The character type Section 7.3
class Definition of a class Section 9.2
const Definition of a constant value, reference, member
function, or pointer
Section 2.1.5, Special
Topic 5.2, Special Topic 6.4,
Section 9.2
default The default case of a switch statement Special Topic 3.3
delete Return a memory block to the heap Section 7.4
do A loop that is executed at least once Section 4.4
double The double-precision, floating-point type Section 2.1.2
else The alternative clause in an if statement Section 3.1
false The false Boolean value Section 3.7
float The single-precision, floating-point type Special Topic 2.1
for A loop that is intended to initialize, test, and
update a variable
Section 4.3
if The conditional branch statement Section 3.1
int The integer type Section 2.1
long A modifier for the int and double types that
indicates that the type may have more bytes
Special Topic 2.1
namespace A name space for disambiguating names Section 1.5
new Allocate a memory block from the heap Section 7.4
private Features of a class that can only be accessed by
this class and its friends
Section 9.2
cfe2_appB_489_490.indd 489 10/28/10 2:42 PM
490 Appendix B Reserved Word summary
The following reserved words are not covered in this book:
Reserved Word Description Reference Location
public Features of a class that can be accessed by
all functions
Section 9.2
return Returns a value from a function Section 5.4
short A modifier for the int type that indicates that the
type may have fewer bytes
Special Topic 2.1
static_cast Convert from one type to another Special Topic 2.3
struct A construct for aggregating items of arbitrary
types into a single value
Section 7.7
switch A statement that selects among multiple branches,
depending upon the value of an expression
Special Topic 3.3
this The pointer to the implicit parameter of a
member function
Section 9.9.3
true The true value of the Boolean type Section 3.7
unsigned A modifier for the int and char types that indicates
that values of the type cannot be negative
Special Topic 2.1
using Importing a name space Section 1.5
virtual A member function with dynamic dispatch Section 10.4
void The empty type of a function or pointer Section 5.5
while A loop statement that is controlled by a condition Section 4.1
asm
auto
catch
const_cast
continue
dynamic_cast
enum
explicit
export
extern
friend
goto
inline
mutable
operator
protected
register
reinterpret_cast
signed
sizeof
static
template
throw
try
typedef
typeid
type_info
typename
union
volatile
wchar_t
cfe2_appB_489_490.indd 490 10/28/10 2:42 PM
CA P P E N D I X
491
O P E r At O r
S u m m A r y
The operators are listed in groups of decreasing precedence in the table below. The
horizontal lines in the table indicate a change in operator prece dence. For example,
z = x – y; means z = (x – y); because = has a lower precedence than – .
The prefix unary operators and the assignment operators associate right-to-left.
All other operators associate left-to-right. For example, x – y – z means (x – y) – z
because – associates left-to-right, but x = y = z means x = (y = z) because = associates
right-to-left.
Operator Description reference Location
:: Scope resolution Section 9.4.1
. Access member Section 2.5.4,
Section 7.7.2
-> Dereference and access member Section 9.9
[] Vector or array subscript Section 6.1
() Function call Section 5.1
++ Increment Section 2.2.2
— Decrement Section 2.2.2
! Boolean not Section 3.8
~ Bitwise not Appendix G
+ (unary) Positive Section 2.2.1
– (unary) Negative Section 2.2.1
* (unary) Pointer dereferencing Section 7.1
& (unary) Address of variable Section 5.9,
Section 7.1
new Heap allocation Section 7.4
delete Heap recycling Section 7.4
sizeof Size of variable or type Appendix F
(type) Cast not covered
cfe2_appC_491_492.indd 491 10/28/10 2:43 PM
492 Appendix C Operator Summary
Operator Description reference Location
.* Access pointer to member not covered
->* Dereference and access pointer to member not covered
* Multiplication Section 2.2.1
/ Division or integer division Section 2.2.1,
Section 2.2.3
% Integer remainder Section 2.2.3
+ Addition Section 2.2.1
– Subtraction Section 2.2.1
<< Output Section 1.5, Section
2.3.2, Appendix G
>> Input Section 2.3.1,
Appendix G
< Less than Section 3.2
<= Less than or equal Section 3.2
> Greater than Section 3.2
>= Greater than or equal Section 3.2
== Equal Section 3.2
!= Not equal Section 3.2
& Bitwise and Appendix G
^ Bitwise xor Appendix G
| Bitwise or Appendix G
&& Boolean and Section 3.7
|| Boolean or Section 3.7
? : Selection Special Topic 3.1
= Assignment Section 2.1.4
+= -= *=
/= %= &=
|= ̂ = >>=
<<=
Combined operator and assignment Special Topic 2.4
, Sequencing of expressions not covered
cfe2_appC_491_492.indd 492 10/28/10 2:43 PM
DA P P E N D I X
493
C h A r A C t E r
C o D E s
These escape sequences can occur in strings (for example, "\n") and characters
(for example, '\'').
Escape sequence Description
\n Newline
\r Carriage return
\t Tab
\v Vertical tab
\b Backspace
\f Form feed
\a Alert
\\ Backslash
\" Double quote
\' Single quote
\? Question mark
\xh1h2 Code specified in hexadecimal
\o1o2o3 Code specified in octal
cfe2_appD_493_494.indd 493 10/28/10 2:44 PM
494 Appendix D Character Codes
table 1 AsCII Code table
Dec.
Code
hex
Code
Char-
acter
Dec.
Code
hex
Code
Char-
acter
Dec.
Code
hex
Code
Char-
acter
Dec.
Code
hex
Code
Char-
acter
0 00 \0 32 20 Space 64 40 @ 96 60 ‘
1 01 33 21 ! 65 41 A 97 61 a
2 02 34 22 " 66 42 B 98 62 b
3 03 35 23 # 67 43 C 99 63 c
4 04 36 24 $ 68 44 D 100 64 d
5 05 37 25 % 69 45 E 101 65 e
6 06 38 26 & 70 46 F 102 66 f
7 07 \a 39 27 ' 71 47 G 103 67 g
8 08 \b 40 28 ( 72 48 H 104 68 h
9 09 \t 41 29 ) 73 49 I 105 69 i
10 0A \n 42 2A * 74 4A J 106 6A j
11 0B \v 43 2B + 75 4B K 107 6B k
12 0C \f 44 2C , 76 4C L 108 6C l
13 0D \r 45 2D - 77 4D M 109 6D m
14 0E 46 2E . 78 4E N 110 6E n
15 0F 47 2F / 79 4F O 111 6F o
16 10 48 30 0 80 50 P 112 70 p
17 11 49 31 1 81 51 Q 113 71 q
18 12 50 32 2 82 52 R 114 72 r
19 13 51 33 3 83 53 S 115 73 s
20 14 52 34 4 84 54 T 116 74 t
21 15 53 35 5 85 55 U 117 75 u
22 16 54 36 6 86 56 V 118 76 v
23 17 55 37 7 87 57 W 119 77 w
24 18 56 38 8 88 58 X 120 78 x
25 19 57 39 9 89 59 Y 121 79 y
26 1A 58 3A : 90 5A Z 122 7A z
27 1B 59 3B ; 91 5B [ 123 7B {
28 1C 60 3C < 92 5C \ 124 7C |
29 1D 61 3D = 93 5D ] 125 7D }
30 1E 62 3E > 94 5E ^ 126 7E ~
31 1F 63 3F ? 95 5F _ 127 7F
cfe2_appD_493_494.indd 494 10/28/10 2:44 PM
EA P P E N D I X
495
C + + L I b r A r y
S u m m A r y
Standard Code Libraries
• double sqrt(double x)
Function: Square root, x
• double pow(double x, double y)
Function: Power, xy. If x > 0, y can be any value. If x is 0, y must be > 0.
If x < 0, y must be an integer.
• double sin(double x)
Function: Sine, sin x (x in radians)
• double cos(double x)
Function: Cosine, cos x (x in radians)
• double tan(double x)
Function: Tangent, tan x (x in radians)
• double log10(double x)
Function: Decimal log, log10 (x), x > 0
• double fabs(double x)
Function: Absolute value, | x|
• int abs(int x)
Function: Absolute value, |x|
• void exit(int n)
Function: Exits the program with status code n.
• int rand()
Function: Random integer
• void srand(int n)
Function: Sets the seed of the random number generator to n.
• bool isalpha(char c)
Function: Tests whether c is a letter.
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496 Appendix E C++ Library Summary
• char isalnum(char c)
Function: Test whether c is a letter or a number.
• bool isdigit(char c)
Function: Tests whether c is a digit.
• bool isspace(char c)
Function: Tests whether c is white space.
• bool islower(char c)
Function: Tests whether c is lowercase.
• bool isupper(char c)
Function: Tests whether c is uppercase.
• char tolower(char c)
Function: Returns the lowercase of c.
• char toupper(char c)
Function: Returns the uppercase of c.
• time_t time(time_t* p)
Function: Returns the number of seconds since January 1, 1970, 00:00:00 GMT.
If p is not NULL, the return value is also stored in the location to which p points.
• istream& getline(istream& in, string s)
Function: Gets the next input line from the input stream in and stores it in the
string s.
Class string
• int string::length() const
Member function: Returns the length of the string.
• string string::substr(int i) const
Member function: Returns the substring from index i to the end of the string.
• string string::substr(int i, int n) const
Member function: Returns the substring of length n starting at index i.
• const char* string::c_str() const
Member function: Returns a char array with the characters in this string.
• boolalpha
Manipulator: Causes Boolean values to be displayed as true and false instead of
the default 1 and 0.
• fixed
Manipulator: Selects fixed floating-point format, with trailing zeroes.
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Appendix E C++ Library Summary 497
• left, right
Manipulator: Left- or right-justifies values if they are shorter than the field width.
• scientific
Manipulator: Selects scientific floating-point format, such as 1.729000e+03.
• setfill(char c)
Manipulator: Sets the fill character to the character c.
• setprecision(int n)
Manipulator: Sets the precision of floating-point values to n digits after the
deci mal point in fixed and scientific formats.
• setw(int n)
Manipulator: Sets the width of the next field.
Class istream
• bool istream::fail() const
Function: True if input has failed.
• istream& istream::get(char& c)
Function: Gets the next character and places it into c.
• istream& istream::unget()
Function: Puts the last character read back into the stream, to be read again in the
next input operation; only one character can be put back at a time.
• istream& istream::seekg(long p)
Function: Moves the get position to position p.
• long istream::tellg()
Function: Returns the get position.
Class ostream
• ostream& ostream::seekp(long p)
Function: Moves the put position to position p.
• long ostream::tellp()
Function: Returns the put position.
Class ifstream
• void ifstream::open(const char n[])
Function: Opens a file with name n for reading.
Class ofstream
• void ofstream::open(const char n[])
Function: Opens a file with name n for writing.
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498 Appendix E C++ Library Summary
Class fstream
• void fstream::open(const char n[])
Function: Opens a file with name n for reading and writing.
Class fstreambase
• void fstreambase::close()
Function: Closes the file stream.
Notes: • fstreambase is the common base class of ifstream, ofstream, and fstream.
• To open a binary file both for input and output, use f.open(n, ios::in |
ios::out | ios::binary)
Class istringstream
• istringstream::istringstream(string s)
Constructs a string stream that reads from the string s.
Class ostringstream
• string ostringstream::str() const
Function: Returns the string that was collected by the string stream.
Notes: • Call istrstream(s.c_str()) to construct an istrstream.
• Call s = string(out.str()) to get a string object that contains the characters
col lected by the ostrstream out.
Class vector
• int vector
Function: Returns the number of elements in the container.
• vector
Function: Constructs a vector with n elements.
• void vector
Function: Inserts x after the last element.
• void vector
Function: Removes (but does not return) the last element.
• T& vector
Function: Accesses the element at index n.
• T& vector
Function: Accesses the element at index n, checking that the index is in range.
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499
G l o s s a r y
Accessor function A function that accesses an object but does not change it.
Address A value that specifies the location of a variable in memory.
Aggregation relationship The “has-a” relationship between classes.
Algorithm An unambiguous, executable, and terminating specification to solve a
problem.
ANSI/ISO C++ Standard The standard for the C++ language that was developed
by the American National Standards Institute and the International Standards
Organization.
Argument A parameter value in a function call, or one of the values combined by an
operator.
Array A collection of values of the same type, each of which can be accessed by an
integer index.
Arrow operator The -> operator. p->m is the same as (*p).m.
ASCII code The American Standard Code for Information Interchange, which
associates code values between 0 and 127 to letters, digits, punctuation marks, and
control characters.
Assignment Placing a new value into a variable.
Base class A class from which another class is derived.
Binary file A file in which values are stored in their binary representation and cannot
be read as text.
Binary search A fast algorithm for finding a value in a sorted array. It narrows the
search down to half of the array in every step.
Bit Binary digit; the smallest unit of information, having two possible values, 0 and 1.
A data element consisting of n bits has 2n possible values.
Black Box A device with a given specification but unknown implementation.
Block A group of statements bracketed by {}.
Boolean operator See Logical operator.
Boolean type A type with two values, true and false.
Boundary test case A test case involving values that are at the outer boundary of
the set of legal values. For example, if a function is expected to work for all nonnega-
tive integers, then 0 is a boundary test case.
Bounds error Trying to access an array element that is outside the legal range.
cfe2_gloss_p499_506.indd 499 10/28/10 5:36 PM
500 Glossary
break statement A statement that terminates a loop or switch statement.
Byte A number between 0 and 255 (eight bits). Essentially all currently manufactured
computers use a byte as the smallest unit of storage in memory.
Capacity The number of values that a data structure such as an array can potentially
hold, in contrast to the size (the number of elements it currently holds).
Case-sensitive Distinguishing upper- and lowercase characters.
Cast Converting a value from one type to a different type. For example, the cast
from a floating-point number x to an integer is expressed in C++ by the static cast
notation, static_cast
Character A single letter, digit, or symbol.
Class A programmer-defined data type.
Command line The line you type when you start a program in a command window.
It consists of the program name and the command line arguments.
Command line arguments Additional strings of information provided at the com-
mand line that the program can use.
Comment An explanation to make the human reader understand a section of a
program; ignored by the compiler.
Compiler A program that translates code in a high-level language such as C++ to
machine instructions.
Compile-time error See Syntax error.
Concatenation Placing one string after another.
Constant A value that cannot be changed by the program. In C++, constants are
marked with the reserved word const.
Constructor A function that initializes a newly allocated object.
CPU (Central Processing Unit) The part of a computer that executes the machine
instructions.
Dangling pointer A pointer that does not point to a valid location.
Data member A variable that is present in every object of a class.
Debugger A program that lets a user run another program one or a few steps at a
time, stop execution, and inspect the variables in order to analyze the program for
bugs.
Declaration A statement that announces the existence of a variable, function, or
class but does not define it.
Default constructor A constructor that can be invoked with no parameters.
Definition A statement or series of statements that fully describes a variable, a func-
tion and its implementation, a type, or a class and its properties.
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Glossary 501
delete operator The operator that recycles memory to the heap.
Derived class A class that modifies a base class by adding data members, adding
member functions, or redefining member functions.
Directory A structure on a disk that can hold files or other directories; also called a
folder.
Dot notation The notation object.function(parameters) used to invoke a member
function on an object.
Element A storage location in an array.
Encapsulation The hiding of implementation details.
Escape character A character in text that is not taken literally but has a special
meaning when combined with the character or characters that follow it. The \ charac-
ter is an escape character in C++ strings.
Exception A condition that prevents a program from continuing normally.
Executable file The file that contains a program’s machine instructions.
Explicit parameter A parameter of a member function other than the object on
which the function is invoked.
Expression A syntactical construct that is made up of constants, variables, and/or
function calls, and the operators combining them.
Extension The last part of a file name, which specifies the file type. For example, the
extension .cpp denotes a C++ file.
Failed stream state The state of a stream after an invalid operation has been
attempted, such as reading a number when the next stream position yielded a non-
digit, or reading after the end of file was reached.
File A sequence of bytes that is stored on disk.
File pointer The position within a file of the next byte to be read or written. It can
be moved so as to access any byte in the file.
Floating-point number A number with a fractional part.
Folder Directory.
Function A sequence of statements that can be invoked multiple times, with
different values for its parameter variables.
Global variable A variable whose scope is not restricted to a single function.
Header file A file that informs the compiler of features that are available in another
module or library.
Heap A reservoir of storage from which memory can be allocated when a program
runs.
cfe2_gloss_p499_506.indd 501 10/28/10 5:36 PM
502 Glossary
IDE (Integrated Development Environment) A programming environment that
includes an editor, compiler, and debugger.
Implicit parameter The object on which a member function is called. For example,
in the call x.f(y), the object x is the implicit parameter of f.
#include directive An instruction to the preprocessor to include a header file.
Index The position of an element in an array.
Inheritance The “is-a” relationship between a general base class and a specialized
derived class.
Initialization Setting a variable to a well-defined value when it is created.
Integer A number without a fractional part.
Integer division Taking the quotient of two integers and discarding the remainder.
In C++, the / symbol denotes integer division if both arguments are integers. For
example, 11 / 4 is 2, not 2.75.
Interface The set of functions that can be applied to objects of a given type.
Lexicographic ordering Ordering strings in the same order as in a dictionary, by
skipping all matching characters and comparing the first nonmatching characters of
both strings. For example, “orbit” comes before “orchid” in the lexicographic order-
ing. Note that in C++, unlike a dictionary, the ordering is case-sensitive: Z comes
before a.
Library A set of precompiled functions that can be included in programs.
Linker The program that combines object and library files into an executable file.
Local variable A variable whose scope is a single block.
Logic error An error in a syntactically correct program that causes it to act differ-
ently from its specification.
Logical operator An operator that can be applied to Boolean values. C++ has three
logical operators: &&, ||, and !.
Loop A sequence of instructions that is executed repeatedly.
Loop and a half A loop whose termination decision is neither at the beginning nor
at the end.
Machine code Instructions that can be executed directly by the CPU.
Magic number A number that appears in a program without explanation.
main function The function that is called first when a program executes.
Member function A function that is defined by a class and operates on objects of
that class.
Memory The circuitry that stores code and data in a computer.
cfe2_gloss_p499_506.indd 502 10/28/10 5:36 PM
Glossary 503
Memory leak Memory that is dynamically allocated but never returned to the heap
manager. A succession of memory leaks can cause the heap manager to run out of
memory.
Modulus operator The % operator that yields the remainder of an integer division.
Mutator function A member function that changes the state of an object.
Nested block A block that is contained inside another block.
Nested loop A loop that is contained in another loop.
new operator The operator that allocates new memory from the heap.
Newline The ’\n’ character, which indicates the end of a line.
Object-oriented programming A programming style in which tasks are solved by
collaborating objects.
Off-by-one error A common programming error in which a value is one larger or
smaller than it should be.
Opening a file Preparing a file for reading or writing.
Operator A symbol denoting a mathematical or logical operation, such as + or &&.
Operator precedence The rule that governs which operator is evaluated first. For
example, in C++ the && operator has a higher precedence than the || operator. Hence
a || b && c is interpreted as a || (b && c).
Overloading Giving more than one meaning to a function name or operator.
Overriding Redefining a function from a base class in a derived class.
Parallel arrays Arrays of the same length, in which corresponding elements are
logically related.
Parameter variable A variable in a function that is initialized with the argument
value when the function is called.
Pointer A value that denotes the memory location of an object.
Polymorphism Selecting a function among several functions with the same name,
by comparing the actual types of the parameters.
Prompt A string that prompts the program user to provide input.
Prototype The declaration of a function, including its parameter types and return
type.
Pseudocode A mixture of English and C++ used when developing the code for a
program.
Pseudorandom numbers A number that appears to be random but is generated by
a formula.
Public interface The features of a class that are accessible to all clients.
cfe2_gloss_p499_506.indd 503 10/28/10 5:36 PM
504 Glossary
Random access The ability to access any value directly without having to read the
values preceding it.
Recursive function A function that can call itself with simpler values. It must
han dle the simplest values without calling itself.
Reference parameter A parameter that is bound to a variable supplied in the call.
Changes made to the parameter within the function affect the variable outside the
function.
Relational operator An operator that compares two values, yielding a Boolean
result.
Reserved word A word that has a special meaning in a programming language and
therefore cannot be used as a name by the programmer.
Return value The value returned by a function through a return statement.
Roundoff error An error introduced by the fact that the computer can store only a
finite number of digits of a floating-point number.
Run-time error See Logic error.
Scope The part of a program in which a variable is defined.
Selection sort A sorting algorithm in which the smallest element is repeatedly
found and removed until no elements remain.
Sentinel A value in input that is not to be used as an actual input value but to signal
the end of input.
Sequential access Accessing values one after another without skipping over any of
them.
Slicing an object Copying an object of a derived class into a variable of the base
class, thereby losing the derived-class data.
Source file A file containing instructions in a programming language.
Stepwise refinement Solving a problem by breaking it into smaller problems and
then further decomposing those smaller problems.
Stream An abstraction for a sequence of bytes from which data can be read or to
which data can be written.
String A sequence of characters.
Structure A construct for aggregating items of arbitrary types into a single value.
Stub A function with no or minimal functionality.
Substitution principle The rule that states that you can use a derived-class object
whenever a base-class object is expected.
Syntax Rules that define how to form instructions in a particular programming
language.
cfe2_gloss_p499_506.indd 504 10/28/10 5:36 PM
Glossary 505
Syntax error An instruction that does not follow the programming language rules
and is rejected by the compiler.
Tab character The ’\t’ character, which advances the next character on the line to
the next one of a set of fixed screen positions known as tab stops.
Test coverage The instructions of a program that are executed when a set of test
cases are run.
Text file A file in which values are stored in their text representation.
Trace message A message that is printed during a program run for debugging
purposes.
Type A named set of values and the operations that can be carried out with them.
Unary operator An operator with one argument.
Unicode A standard code that assigns values consisting of two bytes to characters
used in scripts around the world.
Uninitialized variable A variable that has not been set to a particular value. It is
filled with whatever “random” bytes happen to be present in the memory location
that the variable occupies.
Unit test A test of a function by itself, isolated from the remainder of the program.
Value parameter A function parameter whose value is copied into a parameter vari-
able of a function. If a variable is passed as a value parameter, changes made to the
parameter variable inside the function do not affect the original variable outside the
program.
Variable A storage location that can hold different values.
Vector The standard C++ template for a dynamically-growing array.
Virtual function A function that can be redefined in a derived class. The actual
function called depends on the type of the object on which it is invoked at run time.
void A reserved word indicating no type or an unknown type.
Walkthrough Simulating a program or a part of a program by hand to test for correct
behavior.
White space A sequence consisting of space, tab, and/or newline characters.
cfe2_gloss_p499_506.indd 505 10/28/10 5:36 PM
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507
I n d e x
Symbols
( ) (parentheses)
order of arithmetic operations, 40
unbalanced, 42
& (ampersand)
address operator, 309
reference parameter indicator, 222, 266
-> (arrow operator), 337, 422–423
* (asterisk)
indirection operator, 309
multiplication operator, 40, 105t
\ (backslash)
displaying as a literal, 14
escape sequence, 14, 493t
&& (Boolean AND)
confusing with || (OR), 107
description, 105t
inverting, 108
short-circuit evaluation, 108
! (Boolean NOT)
description, 105t
inverting conditions, 108
|| (Boolean OR)
confusing with && (AND), 107
description, 105t
inverting, 108
short-circuit evaluation, 108
= (equal sign)
vs. ==, equality relational operator, 83
assignment operator, 34–35
>> (extraction operator)
assigning values to variables, 48
reading input from a screen, 48
reading numbers from binary files, 373
reading strings from the console, 58
<< (insertion operator)
displaying to the screen, 12
ending displayed output, 14
formatting printed output, 51t
starting a new line, 15
syntax, 13
- (minus sign)
subtraction operator, 40
unary negative operator, 105t
unary operator, spaces in, 46
+ (plus sign)
addition operator, 40, 105t
concatenation operator, 57
unary positive operator, 105t
"..." (quotation marks)
displaying as literals, 14, 493t
string indicators, 13
/ (slash)
division operator, 40, 105t
: (colon), inheritance indicator, 447
. (dot), dot operator, 337
/*...*/, multiline comments, C++ code, 36
/**...*/, multiline comments, Java style, 199
{ } (braces)
coding guidelines, 487
in if statements, 79–80, 98
? : (question mark, colon), selection
operator, 81
\" (backslash, double quote), double quote
escape sequence, 493t
\? (backslash, question mark), question
mark escape sequence, 493t
\' (backslash, quote), single quote escape
sequence, 493t
== (equal signs), equality operator
vs. = (equal sign), assignment operator, 83
Boolean operations, 105t
description, 83t
equality testing, 83
!= (exclamation, equal), not equal operator,
83t, 105t
> (greater than), relational operator, 83t, 105t
>= (greater than or equal), relational opera-
tor, 83t, 105t
– (hyphen), in option names, 365
< (less than), relational operator, 83t, 105t
<= (less than or equal), relational operator,
83t, 105t
- (minus sign), subtraction operator, 105t
-- (minus signs), decrement operator, 40,
105t
% (percent sign), modulus operator, 41, 105t
++ (plus signs), increment operator, 40, 105t
// (slashes), single-line comment
indicator, 36
Page references followed by t indicate material in tables. Functions and classes from the C++
library are included by name. For functions and classes created for application examples, refer
to the program listed under the main heading “applications.”
cfe2_index_p507_528.indd 507 10/28/10 4:30 PM
508 Index
_ (underscore), in variable names, 33
[ ] operator
accessing array elements, 252
accessing individual string characters,
322–323
; (semicolon)
in class definitions, 395
ending C++ statements, 13
in if statements, 80
omitting, 14
\0, null terminator, 321
A
\a (alert escape sequence), 493t
abs function, 43t, 495
absolute path names, embedding, 482
absolute values, calculating, 43t
accessor functions, declaring, 402–403
accessors, definition, 393–394
account.cpp (ch05), 223–224
accounts.cpp (ch07), 312
adapters, string streams, 363
adapting algorithms, 269–274
addition. See also summing
+ (arithmetic operator), 40, 105t
Adleman, Leonard, 368
advisory messages. See error messages
aggregation relationships, 414–415
airport luggage handling system, 89
alert escape sequence (\a), 493t
algorithms. See also array algorithms; loops,
common algorithms
adapting, 269–274
defining characteristics, 19
describing with pseudocode, 18, 20–21
designing, 17–21
discovering, 274–277
encryption, 368
executable, 19
terminating, 19
unambiguous, 19
aligning
file stream output, 361
if statements, 81
Altair 8800, 230
ambiguity, algorithms, 19
ampersand (&)
address operator, 309
reference parameter indicator, 222, 266
Analytical Engine, 290
AND operator (&&)
confusing with || (OR), 107
description, 105t
inverting, 108
short-circuit evaluation, 108
ANSI (American National Standards
Institute), 7
Apple II, 230
Apple Macintosh, 231
applications
baby-name program, 354–357
bank account withdrawal simulation,
220–225
bell-curve distribution, simulating,
329–332
cash register simulation, 400–402,
418–421
check-printing program, 211–215
coffee-making simulation, 210–211
converting numbers to words, 211–215
elevator floor simulation, 76–78, 110–111
file encryption, 366–367
Galton board simulation, 329–332
income tax simulation, 94–96
input validation with if statements,
109–111
investment program, 17–18, 132–136
Monte Carlo method, simulation,
170–172
printing a table, 165–168
quiz scoring program, 269–274
quiz-taking program, 443–446, 452–455,
459–462
Richter scale program, 90–92
triangle pattern program, 227–229
vending machine simulation, 54–56
world population density, 369–371
applications, volume calculations
bottles, 30–51
cans, 30–51
cubes, 196–198, 204
pyramids, 204–205
soft drink containers, 30–51
arguments, command line, 365–367
arguments, functions
definition, 195
modifying, 220–225
passing, 199–201
arithmetic. See also number types
combining with assignment, 47
decrementing variables, 40
incrementing variables, 40
missing
rounding floating-point numbers, 41
roundoff errors, 45–46
unbalanced parentheses, 42
cfe2_index_p507_528.indd 508 10/28/10 4:30 PM
Index 509
arithmetic, with loops
averaging values, 157–158
comparing adjacent values, 160–161
counting items, 158
finding the first item, 159
maximum/minimum values, finding,
159–160
summing values, 157–158
arithmetic conversions
casts, 46
floating-point to integer, 41
arithmetic operators
absolute values, calculating, 43t
addition, 40
cosines, calculating, 43t
decimal log values, calculating, 43t
mixing with logical, 107
multiplication, 40
powers, calculating, 41–43
roots, calculating, 41–43
sines, calculating, 43t
spacing for readability, 46
subtraction, 40
summary of, 40
tangents, calculating, 43t
arithmetic operators, division
/ (slash), division operator, 40
floating-point numbers, 40–41, 43–44
integers, intentional, 40–41
integers, unintentional, 43–44
modulus, calculating, 41
with remainder, 40–41
without remainder, 40
array algorithms
adapting, 269–274
averaging elements, 257
copying arrays, 256
discovering, 274–277
displaying elements, with separators, 257
filling arrays, 256
finding largest/smallest elements, 257
inserting elements, 259
reading input, 260–263
removing elements, 258
searching, binary, 264–265
searching, linear, 258
selection sort, 263–264
summing elements, 256
swapping elements, 259–260
array elements
accessing, 252–253
averaging, 257
definition, 252
displaying, with separators, 257
finding largest/smallest, 257
index number, 252
inserting, 259
removing, 258
searching, binary, 264–265
searching, linear, 258
summing, 256
swapping, 259–260
array/pointer duality law, 315
arrays. See also vectors
bounds errors, 252–253, 254–255
capacity, 253–254
of char values. See C strings
companion variables, 253–254
copying, 256
current size, determining, 253–254
defining, 250–251
definition, 250
filling, 256
largest input, marking, 261–263
parameter variables, 316–318
parameters, 266
partially filled, 253–254
purpose of, 255
reading input, 260–263
selection sort, 263–264
sorting, 263–264
syntax, 251
vs. vectors, 289
arrays, processing with functions
&, reference parameter indicator, 266
array parameters, 266
array size, increasing, 266
array size, passing, 265, 317
array size, returning, 266
capacity, determining, 266–267
constant array parameters, 269
multiplying elements, 267–268
printing elements, 267–268
reading input, 267–268
arrays, two-dimensional
accessing elements, 279
column count, specifying, 284
defining, 278–279
definition, 278
parameters, 281–284
passing to functions, 281–284
row and column totals, computing, 280
syntax, 279
arrays, using with pointers
array parameter variables, 316–318
array/pointer duality law, 315
declaring arrays as pointers, 314
passing arrays to functions, 316–318
cfe2_index_p507_528.indd 509 10/28/10 4:30 PM
510 Index
arrays, using with pointers (continued)
pointer arithmetic, 315
returning a pointer to a local variable, 319
simulating a Galton board, 329–332
stepping through arrays, 318
arrow operator (->), 337, 422–423
artificial intelligence, 112–113
ASCII codes, 494t
assignment
= (equal sign), assignment operator,
34–35
combining with arithmetic, 47
asterisk (*)
indirection operator, 309
multiplication operator, 40, 105t
asymmetric loop bounds vs. symmetric, 147
at function, 498
atoi function, 322
Augusta, Ada, 290
autonomous vehicles, 113
averaging
array elements, 257
values, with loops, 157–158
B
\b (backspace escape sequence), 493t
Babbage, Charles, 290
baby-name program, 354–357
babynames.cpp (ch08), 356–357
backing up data, 10–11
backslash (\)
displaying as a literal, 14
escape sequence, 14, 493t
backslash, double quote (\”), double quote
escape sequence, 493t
backslash, question mark (\?), question
mark escape sequence, 493t
backslash, quote (\’), single quote escape
sequence, 493t
bank account withdrawal simulation,
220–225
base classes
constructing, 451
constructor syntax, 451
definition, 442
forgetting the name of, 455
inheriting derived classes from, 451–455
overriding with derived classes, 448
bell-curve distribution, simulating, 329–332
binary files, reading/writing, 372–373
binary search, 264–265
black box, functions as, 195
blank lines, coding guidelines, 486
blank spaces, coding guidelines, 486
BMP images, reading/writing, 373–376
bool, reserved word, 489t
bool data type, 103–104.
boolalpha manipulator, 444–446, 496
Boole, George, 103
Boolean operations
operator, summary of, 104
operator precedence, 105t
truth tables, 104
variables, 103–104
Boolean operations, && (AND)
confusing with || (OR), 107
description, 105t
inverting, 108
short-circuit evaluation, 108
Boolean operations, ! (NOT)
description, 105t
inverting conditions, 108
Boolean operations, || (OR)
confusing with && (AND), 107
description, 105t
inverting, 108
short-circuit evaluation, 108
Boolean variables, 103–104
to control a loop, 152
bottle volume program, 30–51
boundary conditions, testing, 102
bounds errors, arrays, 252–253, 254–255
braces ({ })
coding guidelines, 487
in if statements, 79–80, 98
break, reserved word, 489t
break statements
coding guidelines, 486
loop termination, 153
missing, 93
breaking complex tasks into simple steps
recursive thinking, 229–230
stepwise refinement, 210–215
breaking out of loops, 153
breaking statements, coding guidelines, 486
buffer overrun attacks, 255
bugs, historical first, 138
C
C++ programs
case sensitivity, 8
ending statements, 13–14
history of, 6
inventor of, 6
main function, 12
namespaces, 12
naming conflicts, avoiding, 12
cfe2_index_p507_528.indd 510 10/28/10 4:30 PM
Index 511
style guide. See coding guidelines
syntax, 12
C strings
\0, null terminator, 321
atoi function, 322
converting characters to integers, 322
converting to/from C++, 322
definition, 320–321
functions for, 323–324
length, determining, 321
manipulating directly, 323–324
strlen function, 321
terminating, 321
caesar.cpp (ch08), 366–367
calling
constructors, 405
functions, 194, 226–230
member functions from member
functions, 399–402
member functions from pointers,
422–423
can volume program, 30–51
carriage return escape sequence (\r), 493t
case
converting, 322–323, 496
determining, 496
case, reserved word, 93, 489t
case sensitivity
C++ programs, 8
coding guidelines, 486
definition, 16
file systems, 9
if statement comparisons, 86
integrated development environment, 8
misspelling words, 16
variables, 33
cash register simulation, 400–402, 418–421
cashregister.cpp (ch09), 419
cashregister.h (ch09), 418
casts, 46
“The Cathedral and the Bazaar,” 426
chads, 412
char, reserved word, 489t
char* pointer, 322
char type, 320
character arrays, strings, 321–322
character codes
ASCII codes, 494t
escape sequences, 493t
characters. See also strings
converting to integers, 322
definition, 56
extracting from strings, 58–60, 496
reading from text files, 358–359
check-printing program, 211–215
chips, definition, 3
cin (>>)
assigning values to variables, 48
reading input from a screen, 48
class, reserved word, 392–394, 489t
classes. See also public interface
aggregation relationships, 414–415
coding guidelines, 485
definition, 390
discovering, 414–417
“has-a” relationships, 414–415, 442
“is-a” relationships, 442
naming conventions, 392
classes, defining
constructors, implementing, 411–412
data members, determining, 411
member functions, implementing,
411–412
object responsibilities, listing, 409
order of definitions, 482
process overview, 409–412
public interface, documenting, 410–411
public interface, specifying, 409–410
syntax, 393
testing your results, 412
clear function, 153
cloning PCs, 230
close function, 498
code libraries,
isalnum function, 496
isalpha function, 495
isdigit function, 496
islower function, 496
isspace function, 496
isupper function, 496
tolower function, 496
toupper function, 496
code libraries,
cos function, 495
fabs function, 495
log10 function, 495
pow function, 495
sin function, 495
sqrt function, 495
tan function, 495
code libraries,
abs function, 495
exit function, 495
rand function, 495
srand function, 495
cfe2_index_p507_528.indd 511 10/28/10 4:30 PM
512 Index
code libraries,
time function, 496
code libraries,
close function, 498
fstream class, 498
fstreambase class, 498
ifstream class, 497
ofstream class, 497
open function, 497–498
code libraries,
boolalpha manipulator, 496
fixed manipulator, 496
left manipulator, 497
right manipulator, 497
scientific manipulator, 497
setfill manipulator, 497
setprecision manipulator, 497
setw manipulator, 497
code libraries,
fail function, 497
get function, 497
istream class, 497
ostream class, 497
seekg function, 497
seekp function, 497
setw manipulator, 497
tellg function, 497
tellp function, 497
unget function, 497
code libraries,
c_str function, 496
getline function, 496
length function, 496
string class, 496
substr function, 496
code libraries,
istringstream class, 498
istringstream function, 498
ostringstream class, 498
str function, 498
code libraries,
at function, 498
operator function, 498
pop_back function, 498
push_back function, 498
size function, 498
vector function, 498
vector
coding guidelines
classes, 485
constants, 484–485
copyright notices, 482
local variables, 484
magic numbers, 484
pointer variables, 484
prefer clarity to cleverness, 318
coding guidelines, control flow
break statements, 486
continue statements, 486
goto statements, 486
if/else statements, 486
nonlinear, 486
for statements, 485
switch statements, 486
while loops, 485
coding guidelines, functions
comment blocks, 483
introductory explanation, 483
maximum size, 483
@param comments, 483
parameter variable names, 483
@return comments, 483
coding guidelines, lexical issues
{ } (braces), 487
blank lines, 486
blank spaces, 486
breaking statements, 486
case sensitivity, 486
indentation, 486
line width, 486
multiline comments, 487
naming conventions, 486
tab stops, 486
unstable layout, 487
white space, 486
coding guidelines, source files
comment blocks, 482
copyright notices, 482
embedding absolute path names, 482
order of class definitions, 482
organizing source file content, 482
coffee-making simulation, 210–211
colon (:), inheritance indicator, 447
column width (data field), formatting,
49–51t
columns, two-dimensional arrays
count, specifying, 284
summing, 280
command line arguments, 365–367
comments
/*…*/ (multiline), 36
/**…*/ (multiline), 199
// (single-line), 36
definition, 35
functions, 199
comments, coding guidelines
in functions, 483
multiline, 487
cfe2_index_p507_528.indd 512 10/28/10 4:30 PM
Index 513
in source files, 482
companion variables, arrays, 253–254
comparisons
adjacent values, with loops, 160–161
arithmetic. See arithmetic operators
logical. See Boolean operations
true/false. See Boolean operations
comparisons, if statements
case sensitivity, 86
floating-point numbers, 86
lexicographic order of strings, 86–87
numbers, 82–85
numbers vs. letters, 87
punctuation marks, 87
relational operators, 82–83t
roundoff errors, 86
spaces, 87
strings, 82–85, 86–87
syntax, 83
compiler, integrated development
environment, 8
compiling source code
connecting separate files, 417–421
header files, 417–421
separate compilation, 417–421
with zero warning messages, 85
computer controlled devices. See embedded
systems
computer programs. See also C++
definition, 2
hardware independence, 6
high level languages, 6
machine instructions, 6
computers (electronic). See also personal
computers
Analytical Engine, 290
artificial intelligence, 112–113
CYC project, 112–113
Difference Engine, 290
ENIAC, 5
expert-systems, 112
Fifth-Generation Project, 112
first programmer, 290
history of, 5, 290, 469–470
language translation, 112
limits of computation, 469–470
Turing machines, 469–470
unsolvable problems, 469–470
computers (human), 5
concatenating strings, 57
connecting personal computers. See
networks
console windows, integrated development
environment, 7
const, reserved word
constant array parameters, 269
constant references, 225
declaring accessor functions, 402–403
defining constants, 35
description, 489t
member functions, 402–403
constant array parameters, 269
constant references, 225
constants. See also variables
coding guidelines, 484–485
definition, 35
sharing, 421
constructors
base classes, 451
calling, 405
copy, 425
data pointers, initializing, 404–405
default, arguments, 404
definition, 403
for existing objects, 405
explicit declaration, 405–406
implementing, 411–412
initializer lists, 405–406
member functions, 403–406
multiple per class, 404
naming, 403
number pointers, initializing, 404–405
overloading, 404, 406
return values, 403
safe copies, 425
void reserved word, 403
continue statements, coding guidelines, 486
control flow, coding guidelines
break statements, 486
continue statements, 486
goto statements, 486
if/else statements, 486
nonlinear, 486
for statements, 485
switch statements, 486
while loops, 485
control flow, implementing. See also
Boolean operations; loops
break statement, 93, 153
for, reserved word, 489t
for statement, syntax, 144. See also for
loops
switch, reserved word, 490t
switch statements, 93. See also if
statements
while, reserved word, 490t
while statements, syntax, 133. See also
while loops
cfe2_index_p507_528.indd 513 10/28/10 4:30 PM
514 Index
converting
case, 322–323, 496
characters to integers, 322
numbers to words, 211–215
strings between C and C++, 322
copy constructors, 425
copy protection schemes, 172
copying
arrays, 256
vectors, 288
cos function, 43t, 495
cosines, calculating, 43t
counting
items with loops, 158
loop iterations, 147–148
cout (<<)
displaying to the screen, 12
ending displayed output, 14
formatting printed output, 51t
starting a new line, 15
syntax, 13
coverage, testing cases, 102
CPU (central processing unit), 3
c_str function, 496
cube.cpp (ch05), 198, 204
cube.cpp (ch09), 421
cube.h (ch09), 421
cubes, volume calculation, 196–198, 204
currency values, formatting, 49
CYC project, 112–113
D
dangling else problem, 98
dangling pointers, 328
DARPA, autonomous vehicle challenge,
113
data members
declaring, 393
definition, 395
description, 395–397
determining, 411
initializing. See constructors
public vs. private, 396, 402
data pointers, initializing, 404–405
data types. See also number types; strings;
variables
ASCII code, 61
bool, 103–104. See also Boolean
operations
changing. See casts
international alphabets, 61
Unicode, 61
databases, privacy issues, 377
De Morgan, Augustus, 108
De Morgan’s Law, 108–109
debugging. See also error messages; errors;
exceptions; testing code
confusing = (assignment operator), with
== (equality operator), 85
confusing && (Boolean AND) with ||
(Boolean OR), 107
the first bug, 138
with flowcharts, 99–101
by hand, 52–53
if statements, 97–98
infinite loops, 136–137
missing header files, 45
roundoff errors, 45–46
spelling errors, 16
with test cases, 102
unbalanced parentheses, 42
undefined variables, 37
uninitialized variables, 37
unintentional integer division, 43–44
while loops, 139–142
decimal log values, calculating, 43t
decisions. See control flow
declaring
accessor functions, 402–403
arrays as pointers, 314
data members, 393
function parameter variables, 196
functions, 196–198, 203
member functions, 393
parameter variables, 196
decrementing
— (minus signs), decrement operator, 40,
105t
loop iterations, 144
variables, 40
default, reserved word, 93, 489t
defining
arrays, one-dimensional, 250–251
arrays, two-dimensional, 278–279
classes. See classes, defining
constants, 35
derived classes, 446–449
member functions, 398
pointers, 308–309
variables, individual, 30–32
variables, multiple, 313
vectors, 285–286
delete, reserved word, 489t
delete[] operator, 325
delete operator, 325
Denver luggage handling system, 89
cfe2_index_p507_528.indd 514 10/28/10 4:30 PM
Index 515
dereferencing operator. See indirection
operator (*)
derived classes
: (colon), denoting inheritance, 447
defining, 446–449
definition, 442
inheriting from base classes, 451–455
overriding base classes, 448
overriding member functions, 451–455
private inheritance, 449–450
public reserved word, 447, 449–450
replicating base-class members, 450
syntax, 448
dice.cpp (ch04), 170
die tosses, simulating, 169–170
Difference Engine, 290
directories, 9–10. See also folders
discovering classes, 414–417
division
/ (slash), division operator, 40
floating-point numbers, 40–41, 43–44
integers, intentional, 40–41
integers, unintentional, 43–44
modulus, calculating, 41
with remainder, 40–41
without remainder, 40
do, reserved word, 489t
do loops
description, 148–149
flowcharts, 149
input validation, 148–149
purpose of, 148
dongles, 172
DOS (Disk Operating System), 231
dot (.), dot operator, 337
dot notation
accessing structure members, 336
invoking the length function, 58
member functions, 399–400
double, reserved word, 489t
double number type, 38t
doublinv.cpp (ch04), 135
duplicate names. See namespaces;
overloading
dynamic memory allocation, 325–328
E
earthquakes, 90
Eckert, J. Presper, 5
editors, integrated development
environment, 7–8
electronic voting machines, 412–413
elevator floor simulation, 76–78, 110–111
elevator1.cpp (ch03), 78
elevator2.cpp (ch03), 110–111
else, reserved word, 489t
else statements, dangling else problem, 98
embedded systems, 336
embedding absolute path names, 482
encapsulation, 391
encryption
algorithms, 368
patent for, 368
PGP (Pretty Good Privacy), 368
private keys, 368
public keys, 368
RSA, 368
endl (end of line marker), 13
ENIAC (electronic numerical integrator
and computer), 5
equal sign (=)
vs. ==, equality relational operator, 83
assignment operator, 34–35
equal signs (==), equality operator
vs. = (equal sign), assignment operator,
83
Boolean operations, 105t
description, 83t
equality testing, 83
error conditions, testing, 102
error messages. See also debugging; errors;
specific messages
advisory, 85
compiling with zero warnings, 85
fatal, 85
missing main function, 16
severity of, 85
spelling errors, 16
undefined symbol, 16
warnings, 85
errors. See also debugging; exceptions;
specific errors
compile-time, 15–16
logic, 16
run-time, 15–16
syntax, 15–16
escape sequences, summary of, 493t. See also
specific sequences
examples. See applications
exceptions, 16
exclamation, equal (!=), not equal operator,
83t, 105t
executable files, 9
execution, algorithms, 19
exit function, 495
expert systems, 112
explicit parameters, member functions,
398–399
cfe2_index_p507_528.indd 515 10/28/10 4:30 PM
516 Index
exponentiation. See powers, calculating
extraction operator (>>)
assigning values to variables, 48
reading input from a screen, 48
reading numbers from binary files, 373
reading strings from the console, 58
F
\f (form feed escape sequence), 493t
fabs function, 495
fail function
description, 497
file stream input, 353–354
if statements, 110
failure state, clearing, 153
false, reserved word, 103–106, 489t
“fence post errors,” 148
Fifth-Generation Project, 112
file encryption application, 366–367
file extensions, 9
file streams. See also string streams
definition, 352
file processing, example, 354–357
opening, 352–353
reference parameters, 356
syntax, 354
file streams, input
definition, 352
fail function, 353–354
failure indicator, 353–354
file name, specifying, 353
fstream variable, 352
ifstream variable, 352
opening, 352–353
file streams, input from text files
character-by-character, 358–359
get function, 358–359
getline function, 359–361
isdigit function, 359
line-by-line, 359–361
one-character lookahead, 358–359
overview, 353–354
process overview, 369–370
unget function, 358–359
white spaces, removing, 358
word-by-word, 358
file streams, output
alignment, setting, 361
definition, 352
file name, specifying, 353
fixed format, 361
fixed manipulator, 361
floating-point numbers, default format,
361
formatting, 361
fstream variable, 352
general format, 361
left manipulator, 361
manipulators, 361
opening, 352–353
ostream variable, 352
padding with leading zeros, 361
precision, 361
right manipulator, 361
setfill manipulator, 361
setprecision manipulator, 361
setw manipulator, 361
single characters, 361
to text files, 354, 369–370
width, setting, 361
file systems
case sensitivity, 9
directories, 9–10
extensions, 9
files, 9–10
folders, 9–10
files
extensions, 9
in folder hierarchies, 10
integrated development environment, 9
reading input from, 154
filling arrays
inserting elements, 259
with loops, 256
reading data into, 260–263
filling vectors, 286, 288
finding
largest array input, 261–263
largest/smallest array elements, 257
vector elements, 288
first programmer, 290
fixed format, file stream output, 361
fixed manipulator, 49, 51t, 361, 496
flags. See Boolean variables
float, reserved word, 489t
float number type, 38t
floating-point numbers
converting to integer, 41
definition, 33
division, 40–41, 43–44
double number type, 38t
file stream output, default format, 361
float number type, 38t
if statement comparisons, 86
Pentium bug, 47
rounding, 41
roundoff errors, 45–46
flow control. See control flow
cfe2_index_p507_528.indd 516 10/28/10 4:30 PM
Index 517
flowcharts. See also storyboards
common symbols, 99
debugging with, 99–101
do loops, 149
if statements, 77
loops, 149
for loops, 145
mapping multiple choices, 99
reusing tasks, 100–101
“spaghetti code,” avoiding, 100–101
while loops, 133, 149
folders, 9–10. See also directories
for, reserved word, 489t
for loops
counting iterations, 147–148
description, 142–143
“fence post errors,” 148
flowchart, 145
incrementing/decrementing, 144
off-by-one errors, 148
purpose of, 147
symmetric bounds vs. asymmetric, 147
for statements
coding guidelines, 485
syntax, 144
form feed escape sequence (\f), 493t
formatting output. See output, formatting
free software, 426
freeing memory, 325, 328
fstream class, 498
fstream variable, 352
fstreambase class, 498
functions. See also member functions;
specific functions
actual parameters. See arguments
as a black box, 195
body of, 196
calling, 194, 226–230
coding guidelines, 483
comments, 199
declaring, 196–198, 203
definition, 12, 194
dot notation, 58
execution flow, 194
explicit parameters, 398–399
formal parameters. See parameter
variables
hand-tracing, 216–217
implementing, 196–198
implicit parameter, 398–399
input to. See arguments
invoking, 58
manual walkthroughs, 216–217
naming, 196
nonmember, sharing, 421
output from. See return values
processing arrays. See arrays, processing
with functions
prototypes. See declaring, functions
recommended length, 216
recursive, 226–230
reusing, 208–210
for string manipulation, 323–324
stubs, 217
syntax, 197
temporary replacements for, 217
vectors as arguments, 287–288
functions, arguments
definition, 195
modifying. See reference parameters
passing, 199–201
functions, parameter variables
declaring, 196
definition, 196
modifying, 201
passing, 199–201
functions, reference parameters
&, parameter indicator, 222
constant references, 225
definition, 220
modifying arguments, 220–225
vs. return values, 225
value parameters, 222, 225
functions, return values
definition, 195
ensuring, 203
missing, 203
omitting, 206–207
vs. reference parameters, 225
return statement, 196, 202
type, specifying, 196
void type, 206–207
functions, variable scope
definition, 218
global, 219–220
inside functions. See local variables
local, 219–220
outside functions. See global variables
reusing names, 218–219
shadowing, 219
functions.cpp (ch06), 267–268
G
Galton board simulation, 329–332
galton.cpp (ch07), 330–331
general format, file stream output, 361
get function, 358–359, 497
cfe2_index_p507_528.indd 517 10/28/10 4:30 PM
518 Index
getline function, 359–361, 496
global variables, 219–220
goto statements, coding guidelines, 486
GPL (General Public License), 426
greater than (>), relational operator, 83t, 105t
greater than or equal (>=), relational
operator, 83t, 105t
guidelines for coding. See coding guidelines
H
hand-tracing
functions, 216–217
if statements, 97–98
while loops, 139–142
hanging loops, 136–137
hard disk storage, 3–4
hardware, definition, 2
hardware independence, 6
“has-a” relationships, 414–415
header files
compiling source code, 417–421
contents, 417
definition, 417
missing, 45
using namespace directive, 421
the heap, 325
hello.cpp (ch01), 11
hexadecimal code (\xh1h2), 493t
high level languages, 6
history of
C++ programs, 6
computers, 5, 290, 469–470
electronic computers, 5
IBM PC, 230
operating systems, 230–231
personal computers, 230–231
user interface, 231
Hoff, Marcian E., 230–231
hyphen (-), in option names, 365
I
IBM PC, history of, 230
if, reserved word, 489t
if statements
? : (question mark, colon), selection
operator, 81
{ } (braces) in, 79–80, 98
; (semicolons) in, 80
aligning, 81
debugging, 97–98
description, 76
duplicate code in branches, 82
fail function, 110
flowchart, 77
hand-tracing, 97–98
within if statements. See if statements,
nested branches
implementing, 87–89
indenting, 81
input validation, 109–111
multiple alternatives, 90–93. See also
switch statements
syntax, 78
tab characters, 81
if statements, comparisons
case sensitivity, 86
floating-point numbers, 86
lexicographic order of strings, 86–87
numbers, 82–85
numbers vs. letters, 87
punctuation marks, 87
relational operators, 82–83t
roundoff errors, 86
spaces, 87
strings, 82–85, 86–87
syntax, 83
if statements, nested branches
dangling else problem, 98
description, 94–96
if/else statements, coding guidelines, 486
ifstream class, 497
ifstream variable, 352
imagemod.cpp (ch08), 375–376
implicit parameters, member functions,
398–399
income tax simulation, 94–96
incrementing
++ (plus signs), increment operator, 40,
105t
loop iterations, 144
variables, 40
indentation
coding guidelines, 486
if statements, 81
indirection operator (*), 309
infinite loops, 136–137
inheritance
polymorphism, 459
substitution principle, 442
value variations vs. behavior variations,
450–451
inheritance, base classes
constructing, 451
constructor syntax, 451
definition, 442
forgetting the name of, 455
inheritance, derived classes
: (colon), denoting inheritance, 447
cfe2_index_p507_528.indd 518 10/28/10 4:30 PM
Index 519
defining, 446–449
definition, 442
inheriting from base classes, 451–455
overriding base classes, 448
overriding member functions, 451–455
private inheritance, 449–450
public reserved word, 447, 449–450
replicating base-class members, 450
syntax, 448
inheritance, hierarchies
developing, 464–468
overview, 442–446
inheritance, virtual functions
base classes, pointers to, 457
definition, 455
derived classes, pointers to, 457
overview, 458–459
slicing problem, 456–457, 462–463
vs. type tags, 462
virtual reserved word, 458
virtual self-calls, 463
initializer lists, 405–406
initializing
data members. See constructors
pointers, 311–313
strings, 57
variables, 31
input. See also file streams; string streams
definition, 4
redirecting, 154
input, processing with loops
Boolean sentinels, 152
clear function, 153
failure state, clearing, 153
from a file, 154
numeric sentinels, 151
sentinel values, 150–152
input statement, assigning values to
variables, 48
input validation
do loops, 148–149
if statements, 109–111
inserting
array elements, 259
vector elements, 288
insertion operator (<<)
displaying to the screen, 12
ending displayed output, 14
formatting printed output, 51t
starting a new line, 15
syntax, 13
int, reserved word, 489t
int number type
description, 38t
precision, 39
range, 39
range overflow, 39
value truncation, 39
integers. See also int number type
converting from floating-point, 41
converting to strings, 364
definition, 31–33
division, intentional, 40–41
division, unintentional, 43–44
integrated development environment
case sensitivity, 8
compiler, 8
console windows, 7
definition, 7
editors, 7–8
executable files, 9
files, 9
libraries, 9
linker, 9
source code, 9
storing programs, 9–10
Intel 8086 processor, 230
interconnecting personal computers. See
networks
intname.cpp (ch05), 213–215
inventor of C++ programs, 6
inverting Boolean conditions, 108
investment program, 17–18, 132–136
invtable.cpp (ch04), 145
“is-a” relationship, 442
isalnum function, 496
isalpha function, 495
isdigit function, 359, 496
islower function, 496
ISO (International Organization for
Standardization), 7
isspace function, 496
istream class, 497
istringstream class, 363–365, 498
istringstream function, 498
isupper function, 496
item.h (ch09), 421
L
language translation, 112
largest.cpp (ch06), 261–262
leading zeros, padding with, 361
left manipulator, 361, 497
Lenat, Douglas, 112
cfe2_index_p507_528.indd 519 10/28/10 4:30 PM
520 Index
length
C strings, determining, 321
strings, determining, 58–60
length function, 58–60, 496
less than (<), relational operator, 83t, 105t
less than or equal (<=), relational operator,
83t, 105t
lexical issues, coding guidelines. See coding
guidelines, lexical issues
libraries, integrated development
environment, 9. See also code libraries
licenses, software, 426
limits of computation, 469–470
line width, coding guidelines, 486
linear search, 258
linker, integrated development
environment, 9
literals, empty string, 57
local variables, 219–220, 484
log10 function, 43t, 495
long, reserved word, 489t
long long number type, 38–39t
long number type, 39, 39t
loop-and-a-half problem, 153
loops. See also do loops; for loops; while
loops
break statement, 153
breaking out of, 153
count controlled. See for loops
definite. See for loops
definition, 132
event controlled. See while loops
flowcharts, 149
hanging, 136–137
indefinite. See while loops
infinite, 136–137
loop-and-a-half problem, 153
within loops. See nested loops
nested, 165–168
off-by-one errors, 137–138
post-test. See do loops
pre-test. See for loops; while loops
termination conditions, setting, 137
writing, 162–164
loops, common algorithms
averaging values, 157–158
comparing adjacent values, 160–161
counting items, 158
finding the first item, 159
maximum/minimum values, finding,
159–160
summing values, 157–158
loops, processing input
Boolean sentinels, 152
clear function, 153
end-of-sequence indicator. See sentinel
values
failure state, clearing, 153
from a file, 154
numeric sentinels, 151
sentinel values, 150–152
luggage handling system, 89
M
machine instructions, 6
magic numbers, 39, 484
main function, 12
manipulators
file stream output, 361
formatting printed output, 49–51t
manual walkthroughs, functions, 216–217
mathematical operations. See arithmetic
matrices. See arrays, two-dimensional
Mauchly, John, 5
maximum/minimum values, finding with
loops, 159–160
medals.cpp (ch06), 282–283
member functions
accessors, 393–394
const reserved word, 402–403
constructors, 403–406
declaring, 393
defining, 398
definition, 390
definition syntax, 400
dot notation, 58, 399–400
explicit parameters, 398–399
implementing, 411–412
implicit parameters, 398–399
mutators, 393–394
overriding with derived classes, 451–455
public vs. private, 402
this pointer, 423
types of, 393–394
member functions, calling
with dot notation, 58
from member functions, 399–402
from pointers, 422–423
memory
deconstructors, 424–425
definition, 3–4
delete[] operator, 325
delete operator, 325
dynamic allocation, 325–328
freeing, 325, 328, 424–425
the heap, 325
new operator, 325
memory leaks, 328
cfe2_index_p507_528.indd 520 10/28/10 4:30 PM
Index 521
minimum/maximum values, finding with
loops, 159–160
minus sign (-)
subtraction operator, 40, 105t
unary negative operator, 105t
unary operator, spaces in, 46
minus signs (--), decrement operator, 40,
105t
missing
break statements, 93
function return values, 203
header files, 45
main function, 16
modifying
parameter variables, 201
strings, 321–322
modulus, calculating, 41. See also
remainders
Monte Carlo method, simulation, 170–172
montecarlo.cpp (ch04), 171
Morris, Robert, 255
multiplication, 40
multiplying array elements, 267–268
mutators, member functions, 393–394
N
\n (newline character), 14, 493t
namespace, reserved word, 489t
namespaces
avoiding naming conflicts, 12
definition, 12
specifying, 12, 421
naming conflicts. See namespaces;
overloading
naming conventions
camel case, 392
classes, 392
coding guidelines, 486
constructors, 403
functions, 196
naming conventions, variables
descriptive names, 38
syntax, 33
nested branches
dangling else problem, 98
description, 94–96
nested loops, 165–168
networks, definition, 4
new, reserved word, 489t
new operator, 325
newline character (\n), 493t
Nicely, Thomas, 47
nonlinear coding guidelines, 486
NOT operator (!)
description, 105t
inverting conditions, 108
NULL pointers, 312
null terminator (\0), 321
number pointers, initializing, 404–405
number types. See also data types
double, 38t
float, 38t
int, 38t
long, 39
long long, 38–39t
short, 38–39t
summary of, 38, 38t
unsigned, 38–39t
unsigned short, 38t
whole numbers, with fractions. See
floating-point numbers
whole numbers, without fractions. See
integers
numbers
converting to words, 211–215
if statement comparisons, 82–85
vs. letters, if statement comparisons, 87
numeric sentinels, 151
O
\o1o2o3 (code specified in octal), 493t
object-oriented programming, 390–391
objects
dynamic allocation, 422
responsibilities, listing, 409
vectors of, 416–417
objects, pointers to
->, arrow operator, 422–423
calling member functions, 422–423
dynamic object allocation, 422
this pointer, 423
octal code (\o1o2o3), 493t
off-by-one errors, 137–138, 148
ofstream class, 497
omitting function return values, 206–207
one-character lookahead, 358–359
open function
fstream class, 498
ifstream class, 497
ofstream class, 497
open source software, 426
opening file stream input/output, 352–353,
497, 498
operating systems, history of, 230–231
operator function, 498
operators
order of precedence, 105t, 491t–492t
cfe2_index_p507_528.indd 521 10/28/10 4:30 PM
522 Index
operators (continued)
overloading, 406
OR operator (||)
confusing with && (AND), 107
description, 105t
inverting, 108
short-circuit evaluation, 108
order of class definitions, coding guidelines,
482
ostream class, 497
ostream variable, 352
ostringstream class, 363–365, 498
output. See also file streams; string streams
definition, 4
displaying to the screen, 12, 15
ending displayed output, 14
redirecting, 154
starting a new line, 15
syntax, 13
output, formatting
column width, 49–51t
currency values, 49
fixed manipulator, 49, 51t
floating-point numbers, file stream
default format, 361
with manipulators, 49–51t
printed output, 49–51t
setprecision manipulator, 49, 51t
setw manipulator, 49, 51t
output, formatting file streams
fixed format, 361
floating-point numbers, default
format, 361
formatting, 361
general format, 361
output statements
displaying values on screens, 12–13
formatting printed output, 51t
overloading
constructors, 404, 406
operators, 406
P
padding with leading zeros, 361
parallel vectors, 416–417
@param comments
coding guidelines, 483
function documentation, 199
parameter variables, functions
declaring, 196
definition, 196
modifying, 201
names, coding guidelines, 483
passing, 199–201
parentheses (())
order of arithmetic operations, 40
unbalanced, 42
partially filled arrays, 253–254
passing
arguments to functions, 199–201
arrays to functions, 316–318
parameter variables, 199–201
PCs. See personal computers
Pentium floating-point bug, 47
percent sign (%), modulus operator, 41, 105t
peripheral devices, 4
personal computers. See also computers
Altair 8800, 230
Apple II, 230
Apple Macintosh, 231
chips, 3
cloning, 230
components of, 3–5
CPU (central processing unit), 3
DOS, 231
first kit, 230
growth of, 230–231
hard disk storage, 3–4
history of, 230–231
IBM PC, 230
input, 4
Intel 8086 processor, 230
interconnecting. See networks
memory storage, 3–4
operating systems, 230–231
output, 4
peripheral devices, 4
primary storage, 3–4
schematic design, 4
secondary storage, 3–4
storage, 3–4
transistors, 3
user interface, 231
VisiCalc spreadsheets, 230–231
PGP (Pretty Good Privacy), 368
piracy, software, 172
planning program actions. See flowcharts;
storyboards
plus sign (+)
addition operator, 40, 105t
concatenation operator, 57
unary positive operator, 105t
plus signs (++), increment operator, 40, 105t
pointer members, using structures with
pointers, 337–338
pointer variables, coding guidelines, 484
cfe2_index_p507_528.indd 522 10/28/10 4:30 PM
Index 523
pointers
&, address operator, 309
to base classes from virtual functions, 457
dangling, 328
defining, 308–309
defining multiple variables, 313
definition, 308
to derived classes from virtual
functions, 457
drawing pictures of, 332–335
initializing, 311–313
NULL, 312
simulating a Galton board, 329–332
syntax, 310
translating from reference
parameters, 314
uninitialized, 311–312
vs. variables pointed to, 313
as vectors, 329–332
pointers, accessing variables with
*, indirection operator, 309
overview, 309–311
pointer syntax, 310
process description, 334–335
pointers, to objects
->, arrow operator, 422–423
calling member functions, 422–423
dynamic object allocation, 422
this pointer, 423
pointers, using with arrays
array parameter variables, 316–318
array/pointer duality law, 315
declaring arrays as pointers, 314
passing arrays to functions, 316–318
pointer arithmetic, 315
returning a pointer to a local variable, 319
simulating a Galton board, 329–332
stepping through arrays, 318
pointers, using with structures
->, arrow operator, 337
., dot operator, 337
allocating structure values, 337
dot notation, 336
pointer members, 337–338
pointing to structures, 337
structures, definition, 336
pop_back function, 286, 288, 498
popdensity.cpp (ch08), 370–371
pow function, 41–42, 495
powers, calculating, 41–43
powtable.cpp (ch04), 166
precision
file stream output, 361
int number type, 39
Pretty Good Privacy (PGP), 368
primary storage, 3–4
printing
array elements, 267–268
formatting printed output, 49–51t
tables, rows and columns, 165–168
privacy issues, databases, 377
private
data members, 396, 402
inheritance, 449–450
keys, 368
member functions, 402
private, reserved word, 489t
problem solving techniques
discovering classes, 414–417
drawing pictures of pointers, 332–334
flowcharts, 99–101
hand calculations, 52–53
reusable functions, 208–210
stepwise refinement, 210–215
storyboards, 154–157
test cases, 102
thinking recursively, 229–230
problem solving techniques, hand tracing
code simulation, 139–142
objects, 407–409
problem solving techniques, with
algorithms
adapting algorithms, 269–274
designing algorithms, 17–21
discovering algorithms, 274–277
problem statements, converting to C++
programs, 54–56
programmers, first, 290
programming, 2. See also software develop-
ment process
programming environment. See integrated
development environment
programming languages. See also C++
hardware independence, 6
high level, 6
machine instructions, 6
programs, starting from the command line,
365–367
prompt
definition, 48
reading from, 48
pseudocode
converting to C++ programs, 54–56
definition, 18
designing algorithms, 18, 20–21
pseudorandom numbers, 169
public
data members, 396, 402
cfe2_index_p507_528.indd 523 10/28/10 4:30 PM
524 Index
public (continued)
keys, 368
member functions, 402
public, reserved word, 490t
public interface
class definition syntax, 392–395
definition, 391
documenting, 410–411
specifying, 392–395, 409–410
public reserved word, derived classes, 447,
449–450
punctuation marks, if statement
comparisons, 87
push_back function, 286, 288–289, 498
pyramids, volume calculation, 204–205
Q
question mark, colon (?:)
selection operator, 81
quiz scoring program, 269–274
quiz1/test.cpp (ch10), 444–446
quiz2/test.cpp (ch10), 453–454
quiz3/choicequestion.h (ch10), 460
quiz3/question.h (ch10), 459–460
quiz3/test.cpp (ch10), 461
quiz-taking program, 443–446, 452–455,
459–462
quotation marks (“…”)
displaying as literals, 14, 493t
string indicators, 13
R
\r (carriage return escape sequence), 493t
random access files, reading/writing, 372
random numbers
between bounds, 169–170
generating, 168–169
Monte Carlo method, 170–172
pseudorandom numbers, 169
rand function, 168–169
seed numbers, 169
simulating die tosses, 169–170
random.cpp (ch04), 168–169
range, number types, 38t
range overflow, int number type, 39
Raymond, Eric, 426
reading
input into arrays, 260–263, 267–268
from a screen prompt, 48
strings from the console, 57–58. See also
string streams
reading files. See also file streams
binary, 372–373
BMP images, 373–376
random access, 372
reading files, text
character-by-character, 358–359
get function, 358–359
getline function, 359–361
isdigit function, 359
line-by-line, 359–361
one-character lookahead, 358–359
overview, 353–354
process overview, 369–370
unget function, 358–359
white spaces, removing, 358
word-by-word, 358
recursive functions, 226–230
recursive thinking, 229–230
reference parameters, functions
&, parameter indicator, 222
constant references, 225
definition, 220
modifying arguments, 220–225
vs. return values, 225
translating pointers from, 314
value parameters, 222, 225
registertest1.cpp (ch09), 400–402
registertest2.cpp (ch09), 420
relational comparisons, 105t
relational operators
combining, 107
defined, 82–83t
examples, 84t
multiple, 107
remainders, 40–41. See also modulus
removing
array elements, 258
vector elements, 286–287, 288
repeating statement execution. See loops
reserved characters, displaying as literals, 14
reserved words. See also specific words
summary of, 489t–490t
as variables, 33
resizing vectors, 286
return, reserved word, 490t
@return comments, coding guidelines, 483
return statement, 196, 202
return values, constructors, 403
return values, functions
definition, 195
ensuring, 203
missing, 203
omitting, 206–207
vs. reference parameters, 225
return statement, 196, 202
type, specifying, 196
void type, 206–207
cfe2_index_p507_528.indd 524 10/28/10 4:30 PM
Index 525
reusing
functions, 208–210
variable names, 218–219
Richter scale, 90t
Richter scale program, 90–92
right manipulator, 361, 497
Rivest, Ron, 368
roots, calculating, 41–43
rounding floating-point numbers, 41
roundoff errors, 45–46, 86
rows in two-dimensional arrays,
summing, 280
RSA encryption, 368
S
safe copies, 425
sample programs. See applications
scientific manipulator, 497
searching arrays
binary, 264–265
linear, 258
secondary storage, 3–4
security
copy protection schemes, 172
dongles, 172
software piracy, 172
seed numbers for random numbers, 169
seekg function, 497
seekp function, 497
selection sort, 263–264
semicolon (;)
in class definitions, 395
in else statements, 80
ending C++ statements, 13
in if statements, 80
omitting, 14
sentinel values, 150–152
sentinel.cpp (ch04), 151
separate compilation, 417–421
setfill manipulator, 361, 497
setprecision manipulator, 49, 51t, 361, 497
setw manipulator, 49, 51t, 361, 497
shadowing variables, 219
Shamir, Adi, 368
short, reserved word, 490t
short number type, 38–39t
short-circuit evaluation, 108
simulating die tosses, 169–170
simulation programs, 168–172
sin function, 43t, 495
sines, calculating, 43t
size function, 498
size of arrays
determining, 253–254, 266–267
increasing, 266
passing to functions, 265, 317
returning from functions, 266
size of vectors, modifying, 286
slash (/)
division operator, 40, 105t
slashes (//), single-line comment
indicator, 36
slicing problem, 456–457, 462–463
software
definition, 2
free, 426
licenses, 426
open source, 426
software development process
estimating time for, 103
planning actions. See flowcharts;
storyboards
scheduling, 103
software piracy, 172
sort function, 263
sorting arrays, 263–264
source code
converting to executable program, 8–9
definition, 9
integrated development environment, 9
storing, 9
source files
coding guidelines, 482
content organization guidelines, 482
spaces
after function names, 46
between arithmetic operators, 46
in expressions, 46
if statement comparisons, 87
spelling errors, messages, 16
sqrt function, 41–42, 495
srand function, 169
Stallman, Richard, 426
standards organizations, 7
static_cast, reserved word, 46, 490t
stepwise refinement, 210–215
storage, definition, 3–4
storing
programs, 9–10
source code, 9
values. See constants; variables
storyboards, 154–157. See also flowcharts
str function, 363–365, 498
string class, 320, 496
string data types, variable vs. literal, 57
string streams. See also file streams
adapters, 363
cfe2_index_p507_528.indd 525 10/28/10 4:30 PM
526 Index
string streams (continued)
converting integers to strings, 364
converting strings to integers, 364
interface, adapting strings to, 363
istringstream class, 363–365
ostringstream class, 363–365
reading individual characters, 363
str function, 363–365
strings
+ (plus sign), concatenation operator, 57
case conversion, 322–323
char type, 320
character arrays, 321–322
comparing, lexicographic order, 86–87
concatenating, 57
converting to integers, 364
definition, 13, 56
empty literals, 57
enclosing in quotation marks, 13
if statement comparisons, 82–85, 86–87
initializing, 57
length, determining, 58–60
length function, 58–60
modifying, 321–322
reading from the console, 57–58
string class, 320
tolower function, 323
toupper function, 323
strings, C
\0, null terminator, 321
atoi function, 322
converting characters to integers, 322
converting to/from C++, 322
definition, 320–321
functions for, 323–324
length, determining, 321
manipulating directly, 323–324
strlen function, 321
terminating, 321
strings, characters
definition, 56
reading, 57–58, 363–365
strings, substrings
[] operator, 322–323
extracting individual characters, 322–323
extracting strings, 58–60
substr member function, 58–60
strlen function, 321
Stroustrup, Bjarne, 6
struct, reserved word, 336–338 , 490t
structures, using with pointers
->, arrow operator, 337
., dot operator, 337
allocating structure values, 337
dot notation, 336
pointer members, 337–338
pointers to structures, 337
structures, definition, 336
stubs for functions, 217
substitution principle, 442
substr function, 58–60, 496
substrings, extracting, 58–60
subtraction, 40
summing arrays
columns, 280
elements, 256
rows, 280
summing values with loops, 157–158
swapping array elements, 259–260
switch, reserved word, 490t
switch statements, 93, 486
symmetric loop bounds vs. asymmetric, 147
syntactic sugar, 316–318
T
\t (tab escape sequence), 493t
tab stops
coding guidelines, 486
indenting if statements, 81
tan function, 43t, 495
tangents, calculating, 43t
tax.cpp (ch03), 95–96
tellg function, 497
tellp function, 497
terminating
\0, null terminator, 321
algorithms, 19
C strings, 321
loops, setting conditions for, 137
loops with break statements, 153
testing code. See also debugging
boundary conditions, 102
coverage, 102
designing test cases, 102
error conditions, 102
new classes, 412
text files, reading. See reading files, text
this, reserved word, 490t
this pointer, 423
Thrun, Sebastian, 113
time function, 496
tolower function, 323, 496
toupper function, 323, 496
transistors, 3
triangle pattern program, 227–229
triangle.cpp (ch05), 227–228
cfe2_index_p507_528.indd 526 10/28/10 4:30 PM
Index 527
troubleshooting. See debugging; error mes-
sages; errors; exceptions
true, reserved word, 103–106, 490t
truncation, int number type, 39
truth tables, Boolean operations, 104
Turing, Alan, 469–470
Turing machines, 469–470
type tags vs. virtual functions, 462
U
unbalanced parentheses, 42
“undefined symbol” message, 16
undefined variables, 37
underscore (_), in variable names, 33
unget function, 358–359, 497
uninitialized pointers, 311–312
uninitialized variables, 37
unsigned, reserved word, 490t
unsigned number type, 38–39t
unsigned short number type, 38t
unsolvable computer problems, 469–470
unstable layout, coding guidelines, 487
user interface, history of, 231
using, reserved word, 12, 490t
using namespace directive, 421
V
\v (vertical tab escape sequence), 493t
value parameters, 222, 225
variable scope, functions
definition, 218
global, 219–220
local, 219–220
reusing names, 218–219
shadowing, 219
variables
_ (underscore), in names, 33
Boolean operations, 103–104
case sensitivity, 33
decrementing, 40
defining, 30–32
defining multiple, 313
definition, 30
incrementing, 40
initializing, 31
vs. pointers to variables, 313. See also
pointers
reserved words, 33
unchanging. See constants
undefined, 37
uninitialized, 37
variables, assigning values to
= (equal sign), assignment operator,
34–35
assignment statement, 34–35
initially, 31, 34
input statement, 48
replacing an existing value, 34–35
variables, naming conventions
descriptive names, 38
reusing names, 218–219
syntax, 33
variables, scope
definition, 218
global, 219–220
inside functions. See local variables
local, 219–220
outside functions. See global variables
reusing names, 218–219
shadowing, 219
variables, syntax
assigning a value, 34, 48
defining a variable, 31
input statement, 48
vector function, 498
vectors. See also arrays
vs. arrays, 289
common algorithms, 288–289
copying, 288
defining, 285–286
definition, 284
filling, 286
finding elements, 288
as function arguments, 287–288
inserting elements, 288
of objects, 416–417
parallel, 416–417
pointers as, 329–332
pop_back function, 286, 288
push_back function, 286, 288–289
removing elements, 286–287, 288
resizing, 286
returned by functions, 287–288
syntax, 285
vector
vertical tab escape sequence (\v), 493t
virtual, reserved word, 458, 490t
virtual functions
base classes, pointers to, 457
definition, 455
derived classes, pointers to, 457
overview, 458–459
slicing problem, 456–457, 462–463
vs. type tags, 462
virtual reserved word, 458
virtual self-calls, 463
virtual self-calls, 463
cfe2_index_p507_528.indd 527 10/28/10 4:30 PM
528 Index
viruses, 255
VisiCalc spreadsheets, 230–231
void, reserved word, 403, 490t
void type, function return values, 206–207
volume calculations
bottles, 30–51
cans, 30–51
cubes, 196–198, 204
pyramids, 204–205
soft drink containers, 30–51
volume1.cpp (ch02), 36
volume2.cpp (ch02), 50
volumes.h (ch09), 421
voting machines, electronic, 412–413
W
while, reserved word, 490t
while loops
coding guidelines, 485
debugging, 139–142
description, 132–136
flowcharts, 133, 149
hand-tracing, 139–142
while statements, syntax, 133
white space
coding guidelines, 486
removing, 358
whole numbers
with fractions. See floating-point
numbers
without fractions. See integers
width
code lines, guidelines, 486
column (data field), formatting, 49–51t
file stream output, setting, 361
Wilkes, Maurice, 138
world population density program, 369–371
worms, 255
writing files
binary, 372–373
BMP images, 373–376
random access, 372
text, 352–357. See also file streams,
output
writing strings. See string streams
X
\xh1h2 (code specified in hexadecimal), 493t
Z
Zimmermann, Phil, 368
cfe2_index_p507_528.indd 528 10/28/10 4:30 PM
529
C r e d i t s
text Credits
Exercises P2.25, P5.32, P5.34: Adapted from MATLAB: An Introduction with Applications,
Third Edition, Amos Gilat (John Wiley & Sons, Inc., 2008) Reprinted with per-
mission of John Wiley & Sons, Inc.
P4.28, P4.30: Adapted from Introduction to Engineering Programming: Solving
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COVER
TITLE PAGE
COPYRIGHT
PREFACE
A WALKTHOUGH OF THE LEARNING AIDS
ACKNOWLEDGMENTS
CONTENTS
SPECIAL FEATURES
CHAPTER 1 INTRODUCTION
1.1 What is Programming
1.2 The Anatomy of a Computer
1.3 Machine Code and Programming Languages
1.4 Becoming Familiar with Your Programming Environment
1.5 Analyzing Your First Program
1.6 Errors
1.7 Problem Solving: Algorithm Design
CHAPTER SUMMARY
REVIEW EXERCISES
PROGRAMMING EXERCISES
ANSWERS TO SELF – CHECK QUESTIONS
CHAPTER 2 FUNDAMENTAL DATA TYPES
2.1 Variables
2.1.1 Variable Definitions
2.1.2 Number Types
2.1.3 Variable Names
2.1.4 The Assignment Statement
2.1.5 Constants
2.1.6 Comments
2.2 Arithmetic
2.2.1 Arithmetic Operators
2.2.2 Increment and Decrement
2.2.3 Integer Division and Remainder
2.2.4 Converting Floating-Point Numbers to Integers
2.2.5 Powers and Roots
2.3 Input and Output
2.3.1 Input
2.3.2 Formatted Output
2.4 Problem Solving: First Do It By Hand
2.5 Strings
2.5.1 The String Type
2.5.2 Concatenation
2.5.3 String Input
2.5.4 String Functions
CHAPTER SUMMARY
REVIEW EXERCISES
PROGRAMMING EXERCISES
ANSWERS TO SELF – CHECK QUESTIONS
CHAPTER 3 DECISIONS
3.1 The if Statement
3.2 Comparing Numbers and Strings
3.3 Multiple Alternatives
3.4 Nested Branches
3.5 Problem Solving: Flowcharts
3.6 Problem Solving: Test Cases
3.7 Boolean Variables and Operators
3.8 Application: Input Validation
CHAPTER SUMMARY
REVIEW EXERCISES
PROGRAMMING EXERCISES
ANSWERS TO SELF – CHECK QUESTIONS
CHAPTER 4 LOOPS
4.1 The while Loop
4.2 Problem Solving: Hand-Tracing
4.3 The for Loop
4.4 The do Loop
4.5 Processing Input
4.6 Problem Solving: Storyboards
4.7 Common Loop Algorithms
4.7.1 Sum and Average Value
4.7.2 Counting Matches
4.7.3 Finding the First Match
4.7.4 Maximum and Minimum
4.7.5 Comparing Adjacent Values
4.8 Nested Loops
4.9 Random Numbers and Simulations
4.9.1 Generating Random Numbers
4.9.2 Simulating Die Tosses
4.9.3 The Monte Carlo Method
CHAPTER SUMMARY
REVIEW EXERCISES
PROGRAMMING EXERCISES
ANSWERS TO SELF – CHECK QUESTIONS
CHAPTER 5 FUNCTIONS
5.1 Functions as Black Boxes
5.2 Implementing Functions
5.3 Parameter Passing
5.4 Return Values
5.5 Functions Without Return Values
5.6 Problem Solving: Reusable Functions
5.7 Problem Solving: Stepwise Refinement
5.8 Variable Scope and Global Variables
5.9 Reference Parameters
5.10 Recursive Functions (Optional)
CHAPTER SUMMARY
REVIEW EXERCISES
PROGRAMMING EXERCISES
ANSWERS TO SELF – CHECK QUESTIONS
CHAPTER 6 ARRAYS AND VECTORS
6.1 Arrays
6.1.1 Defining Arrays
6.1.2 Accessing Array Elements
6.1.3 Partially Filled Arrays
6.2 Common Array Algorithms
6.2.1 Filling
6.2.2 Copying
6.2.3 Sum and Average Value
6.2.4 Maximum and Minimum
6.2.5 Element Separators
6.2.6 linear Search
6.2.7 Removing an Element
6.2.8 Inserting an Element
6.2.9 Swapping Elements
6.2.10 Reading Input
6.3 Arrays And Functions
6.4 Problem Solving: Adapting Algorithms
6.5 Problem Solving: Discovering Algorithms by Manipulating Physical Objects
6.6 Two-Dimensional Arrays
6.6.1 Defining Two-Dimensional Arrays
6.6.2 Accessing Elements
6.6.3 Computing Row and Column Totals
6.6.4 Two-Dimensional Array Parameters
6.7 Vectors
6.7.1 Defining Vectors
6.7.2 Growing and Shrinking Vectors
6.7.3 Vectors and Functions
6.7.4 Vector Algorithms
CHAPTER SUMMARY
REVIEW EXERCISES
PROGRAMMING EXERCISES
ANSWERS TO SELF – CHECK QUESTIONS
CHAPTER 7 POINTERS
7.1 Defining and Using Pointers
7.1.1 Defining Pointers
7.1.2 Accessing Variables Through Pointers
7.1.3 Initializing Pointers
7.2 Arrays and Pointers
7.2.1 Arrays as Pointers
7.2.2 Pointer Arithmetic
7.2.3 Array Parameter Variables are Pointers
7.3 C and C++ Strings
7.3.1 The Char Type
7.3.2 C Strings
7.3.3 Character Arrays
7.3.4 Converting Between C and C++ Strings
7.3.5 C++ Strings and The [] Operator
7.4 Dynamic Memory Allocation
7.5 Arrays and Vectors of Pointers
7.6 Problem Solving: Draw a Picture
7.7 Structures and Pointers (Optional)
7.7.1 Structures
7.7.2 Pointers to Structures
7.7.3 Structures with Pointer Members
CHAPTER SUMMARY
REVIEW EXERCISES
PROGRAMMING EXERCISES
ANSWERS TO SELF – CHECK QUESTIONS
CHAPTER 8 STREAMS
8.1 Reading and Writing Text Files
8.1.1 Opening a Stream
8.1.2 Reading from a File
8.1.3 Writing to a File
8.1.4 A File Processing Example
8.2 Reading Text Input
8.2.1 Reading Words
8.2.2 Reading Characters
8.2.3 Reading lines
8.3 Writing Text Output
8.4 String Streams
8.5 Command Line Arguments
8.6 Random Access and Binary Files
8.6.1 Random Access
8.6.2 Binary Files
8.6.3 Processing Image Files
CHAPTER SUMMARY
REVIEW EXERCISES
PROGRAMMING EXERCISES
ANSWERS TO SELF – CHECK QUESTIONS
CHAPTER 9 CLASSES
9.1 Object-Oriented Programming
9.2 Specifying the Public Interface of a Class
9.3 Data Members
9.4 Member Functions
9.4.1 Implementing Member Functions
9.4.2 Implicit and Explicit Parameters
9.4.3 Calling a Member Function from a Member Function
9.5 Constructors
9.6 Problem Solving: Tracing Objects
9.7 Problem Solving: Discovering Classes
9.8 Separate Compilation
9.9 Pointers to Objects
9.9.1 Dynamically allocating objects
9.9.2 The -> Operator
9.9.3 The This Pointer
CHAPTER SUMMARY
REVIEW EXERCISES
PROGRAMMING EXERCISES
ANSWERS TO SELF – CHECK QUESTIONS
CHAPTER 10 INHERITANCE
10.1 Inheritance Hierarchies
10.2 Implementing Derived Classes
10.3 Overriding Member Functions
10.4 Virtual Functions and Polymorphism
10.4.1 The Slicing Problem
10.4.2 Pointers to Base and Derived Classes
10.4.3 Virtual Functions
10.4.4 Polymorphism
CHAPTER SUMMARY
REVIEW EXERCISES
PROGRAMMING EXERCISES
ANSWERS TO SELF – CHECK QUESTIONS
APPENDICES
APPENDIX A: C++ LANGUAGE CODING GUIDELINES
APPENDIX B: RESERVED WORD SUMMARY
APPENDIX C: OPERATOR SUMMARY
APPENDIX D: CHARACTER CODES
APPENDIX E: C++ LIBRARY SUMMARY
GLOSSARY
INDEX
CREDITS
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