Savitch i Mch 03

Chapter 3

Function Basics

0. Introduction

This chapter shows how to use standard functions provided by the development system

libraries, then shows the student how to write functions to take care of situations not

covered by library functions. During this last task, the chapter shows how to use local

variables and value parameters. Reference parameters are deferred to the next chapter.

Finally, scope rules are treated.

The idea of function is central to programming, whether Object Oriented Programming or

Procedure Oriented Programming. If the paradigm is Procedure Oriented Programming,

the algorithm at hand is decomposed into a hierarchical set of functions. If paradigm is

Object Oriented Programming, the function members of the classes still must be coded.

The material in this chapter will be familiar to anyone who has programmed in any other

language. The differences between functions in C++ and other languages will be treated

in Chapter 4.

1. Outline of topics in the chapter

3.1 Predefined Functions

Predefined Functions That Return a Value

Predefined void functions

A Random Number Generator

3.2 Programmer-Defined Functions

Defining Functions that Return a Value

Alternate Form for Function Declaration

Instructor’s Resource Manual for Savitch Absolute C++ 04/07/23 Chapter 3 Function Basics

Functions Calling Functions

Functions that Return a Boolean Value

Defining void Functions

Return Statements in void Functions

Preconditions and Postconditions

main is a function

Recursive Functions

3.3 Scope Rules

Local Variables

Procedural Abstraction

Global Constants and Global Variables

Blocks

Nested Scopes

Variables Declared in a for Loop

2. General remarks on the chapter

In this chapter, the student begins to learn the tools that will be used to build the members

of the classes, that is, the actions that the classes will take. This involves the tool of

divide and conquer, or stepwise refinement. This tool involves, in its simplest form,

determining subproblems of a nature that, once the subproblems are solved, all that

remains of the original problem is fitting the solutions to the subproblems together. The

top down design tool shows how to recognize subproblems and solve them in such a way

that they fit together automatically, making the last step trivial.

Procedural Abstraction and Information Hiding

David L. Parnas, circa 1972, introduced the notion of information hiding. In this chapter,

we see information hiding in the notion of procedural abstraction. Procedural abstraction

is probably the most important idea in this chapter. If the student does not internalize the


notion that how a task is done should be hidden, with only what is done being revealed,

that student, as a programmer, is crippled. Code written by such a programmer is likely to

be difficult to read and unmaintainable.

Predefined Functions.

C++ requires that every object used in a program be declared prior to any use of the

object. Hence, all functions must be declared prior to use. Predefined (library) functions

are declared in standard header files. Use of predefined (library) functions requires

inclusion of the header files that contain the declarations of these functions. Function

declaration is the C++ name for what ANSI C calls 'prototype'. We will use the term

function declaration, though there is much literature that refers to function declarations as

prototypes.

Potential Problems with Header Files

It is important that the student understand that the #include commands are carried out

by the preprocessor, a part of the C++ compiler that is not smart. The command

#include <filename>

expects the name of the file to contain exactly the characters between the < and the >,

including any spaces. In this include statement,

#include < iostream>

the preprocessor thinks the file name begins with a space followed by the characters of

the name iostream. The preprocessor does not find the header file (no header file has a

name that starts with a space.) This another of the “hard to find” errors that look right.

If you have used the .h header files in C or early C++, you will find that the older header

files have been replaced. The files that did have a .h extension have been replaced by files

with the name composed of the old name with the .h stripped off and prefixed by a ‘c’.

For example, the math.h file has been replaced in C++ by cmath, and stdlib.h has

been replaced by cstdlib. The files specific to C++ such as iostream.h and

iomanip.h are now iostream and iomanip. The C++ Standard specifies that


implementers should arrange for header files without the .h to put their names into

namespace std and the header files without the .h should put their names into the

global namespace. Some caveats: Not all implementations follow the standard in this

respect. Not all implementations have the new files. Most implementations retain the

older files (with the .h extension, names in the global namespace) to avoid breaking older

software. I strongly recommend using the newer C++ features.

Some Possible Errors:

A type of problem my students have is computing an arithmetic average using integer

division (which truncates). Student code typically does the integer division first, then

casts the integer quotient to double.:

double average;

int sum, count;

average = static_cast<double>(sum / count);

instead of this

double average;

int sum, count;

average = static_cast<double>(sum) / count;

It is hard for the student programmer to see this error, as it looks right. Encourage the

students to help debug each other's code. While there is the very real risk of the student

copying code from one another, in my opinion, the enhancement to learning from the

debugging effort is worth the risk.

Another problem the student may encounter is writing a prototype with different return

values but having the same signature such as:

int avg(int s1, int s2, int s3, int s4);

and

double avg(int s1, int s2, int s3, int s4);

The following error message from g++ is clearer than most compiler messages. The

conflict is between the return types:

g++ Prb.cc

In function `double avg(int, int, int, int)':

38: conflicting types for `double avg(int, int, int, int)'


10: previous declaration as `int avg(int, int, int, int)'

Compilation exited abnormally with code 1

Encourage the student to learn the messages your compilers give for the various kinds of

errors.

Increment Operators and Order of Evaluation

Side effects can be evil, particularly when the same variable is affected by a side effect

and the result depends on the order of evaluation. There are only three places in C++ (that

I know of) where the order of evaluation is guaranteed, regardless of side effects. The

arguments to the && operator, the || operator and the comma operator are guaranteed

by the C++ Standard to be evaluated left to right. (But be careful. Your compiler may not

comply with the standard in this regard.) The C++ Standard states that code written that

depends on the order of evaluation of terms in an expression, or of function arguments, is

incorrect. Most compilers do not catch this error. The code will work until you compile

your code with another compiler that evaluates the expressions in a different order. Then

you may find that the program gives different results than you expect. If you are

fortunate, you will find this during testing. If not, a client who is using your code on the

alternate platform will get bad answers.

Example:

//sideEffect.cpp

#include <iostream>

using namespace std;

int main()

{

int x = 2, y;

y = ++x + (++x)*(++x);

cout << " y " << y << endl;

return 0;

}


When compiled with either the MS VC++ 6.0 or BCC 5.5 command line compiler, the

output is 30. This is unreasonable. I do not know why, except that this is illegal code and

the compiler is permitted to do literally anything the compiler author wishes.

Predefined void functions

From the outset it is important that the students understand several

points about void functions. A void function does not return a value.

While a return; statement is allowed, a return; statement is not

necessary, and a return; statement cannot have an argument in

void functions. The call to a void function is an executable statement.

The result of a call to a void function may not be assigned to a

variable. (Technically, a void function's value at execution is not an r-

value.) Execution of a return; statement terminates the function, with

control being returned to the caller. Finally, any code after a return

statement is not executed. A call to a function with no arguments must

have the parentheses following the function name, in contrast to

Pascal, where the function name is sufficient.

For details, read the sections 7.2 and 7.4 of Ellis and Stroustrup, The

Annotated C++ Reference Manual, Addison Wesley, 1991, reprinted

with corrections 1994, ISBN 0-201-5149-1, and sections 11.3.1-11.3.3

in Stroustrup, The design and Evolution of C++ Addison Wesley, 1994

reprinted with corrections, May 1994, or Stroustrup, The C++

Programming Language, 3rd edition Addison Wesley Longman, 1997.

Other Issues

ISO/ANSI C++ Language Standard uses the term "argument" to mean the object that

appears in parentheses following the function name in a function call. The Standard uses

the term "parameter" to mean the placeholder that appears in the declaration (prototype)

and in the definition of a function. The text uses "formal parameter" where the Standard

uses parameter and "argument" as the Standard does. The author's experience, and mine,

is that students definitely do better when terminology is used consistently.


We recommend that regardless of the usage the instructor chooses, that on tests and

exams, instructors should be sure to make the usage clear by using "argument" AND

whatever term they normally use. For example, "formal parameter (argument)" or

"argument (formal parameter)" or something appropriate to your usage. It is our desire

that the text and this IRM support the instructor, not to impose any particular terminology

on the instructor.

While formal parameter names are not necessary in the declaration, these names should

be provided by any student writing at this level.

3. Solutions to, and remarks on, selected Programming Projects

1. Miles per gallon.

//1 liter = 0.264179 gallons

//Input: Number of liters consumed and Number of miles traveled.

//Output: Miles per gallon. Allow repeats

//Do MPG in a function, use global const gallons/liter

#include <iostream>


const double GAL_PER_LITER = 0.264179;

// Input: liters consumed, miles traveled

// Returns miles per gallon.

// requires global constant GAL_PER_LITER

double mpg(double miles, double liters);

int main()

{

double liters;

double miles;

char ans;

do

{

cout << "Enter the number of liters used: ";


cin >> liters;

cout <<"Enter the number of miles driven: ";

cin >> miles;

cout << "Your vehicle gave you "

<< mpg(miles, liters) << " miles per gallon\n";

cout << "Enter Y/y to do another calculation, anything else quits."

<< endl;

cin >> ans;

} while( ans == 'Y' || ans == 'y');

}

double mpg(double m, double l)

{

//convert l liters to gallons

double g = l * GAL_PER_LITER;

return m/g;

}

A typical run is:

Enter the number of liters used: 39

Enter the number of miles driven: 310

Your vehicle gave you 30.0884 miles per gallon

Enter Y/y to do another calculation, anything else quits. y

Enter the number of liters used: 25

Enter the number of miles driven: 210

Your vehicle gave you 31.7966 miles per gallon

Enter Y/y to do another calculation, anything else quits. n

2. Inflation Rate.

//Input: price of an item (hot dog or diamond) 1yr ago and today

//Output: estimate of inflation = (newPrice - oldPrice)/oldPrice

//Use a function to compute inflation rate. Output inflation as

//double in form of a percentage, as in 6.7 for 6.7%

//Repeat ad libitum

#include <iostream>



//Pre: Prices in dollar amounts

//Post: Output inflation as double in form of a

// percentage, as in 6.7 for 6.7%

double inflation(double oldCost, double newCost);

int main()

{

double oldPrice;

double newPrice;

char ans;

cout.setf(ios::showpoint);

cout.setf(ios::fixed);

cout.precision(1); //one decimal place is enough

do

{

cout << "Enter price from a year ago of the item in dollars.\n";

cin >> oldPrice;

cout << "Old Price is " << oldPrice << endl;

cout << "Enter current price of the item in dollars.\n";

cin >> newPrice;

cout << "New Price is " << newPrice << endl << endl;

cout << "Your prices indicate an inflation rate of "

<< inflation(oldPrice, newPrice) << " percent.\n";

cout << "Another calculate again? Y/y repeats, other quits.\n";

cin >> ans;

cout << "You answered " << ans << endl << endl;

} while(ans == 'y' || ans == 'Y');

return 0;

}

double inflation(double oldCost, double newCost)

{

return 100.0 * (newCost - oldCost) / oldCost;

}

/*


I tested this on the DOS command line with the following command, where

the file, data is listed at the bottom of this comment.

$ch2Prob2 < data

Output from a typical run is:

Enter price from a year ago of the item in dollars.

Old Price is 100.0

Enter current price of the item in dollars.

New Price is 107.8

Your prices indicate an inflation rate of 7.8 percent.

Another calculate again? Y/y repeats, other quits.

You answered y


Old Price is 16200.0


New Price is 18095.0



You answered n

The data is:

100.00

107.80

y

16200

18095

n

*/


3. (Enhanced 2.) Inflation Rate.

In addition the problem as stated in #2, this program should print out an estimates, based

on the calculated inflation rate, of the price of the item one and two years hence. Define a

second function that returns the estimate. Arguments should be the inflation rate, the

current cost of the item, and the inflation rate.

//Input: price of an item (hot dog or diamond) 1yr ago and today

//Output: estimate of inflation = (newPrice - oldPrice)/oldPrice

//Estimate of inflated price one and two years hence.

//Use a function to compute inflated price. Arguments are inflation rate

//and current cost of the item.

//Use a function to compute inflation rate. Output inflation rate

//as double in form of a percentage, as in 6.7 for 6.7%

//Repeat ad libitum

#include <iostream>


//Pre: Prices in dollar amounts

//Post: Output inflation as double in form of a

// percentage, as in 6.7 for 6.7%

double inflation(double oldCost, double newCost);

//Function to determine inflated cost

//Pre: Price is in a dollar amount, inflation rate is a percent as

//6.7 (per cent) for 0.067.

//Post: Output is inflated price

double inflatedPrice(double cost, double inflationRate, int years);

int main()

{

double oldPrice;

double newPrice;

char ans;




cout.precision(1); //one decimal place is enough

do

{

cout << "Enter price from a year ago of the item in dollars.\n";

cin >> oldPrice;

cout << "Old Price is " << oldPrice << endl;

cout << "Enter current price of the item in dollars.\n";

cin >> newPrice;

cout << "New Price is " << newPrice << endl << endl;

double inflationRate = inflation(oldPrice, newPrice) ;

cout << "Your prices indicate an inflation rate of "

<< inflationRate << " percent.\n";

cout << "Estimated price of item 1 year hence: "

<< inflatedPrice(newPrice, inflationRate, 1) << endl;

cout << "Estimated price of item 2 years hence: "

<< inflatedPrice(newPrice, inflationRate, 2) << endl;

cout << "Another calculate again? Y/y repeats, other quits.\n";

cin >> ans;

cout << "You answered " << ans << endl << endl;

} while(ans == 'y' || ans == 'Y');

return 0;

}

double inflatedPrice(double cost, double inflationRate, int years)

{

return cost * (1 + years * inflationRate / 100);

}

double inflation(double oldCost, double newCost)

{

return 100.0 * (newCost - oldCost) / oldCost;

}

I tested this on the DOS command line, where the file, data is listed at the bottom of this.

ch2Prob2 < data


Output from a typical run is:


Old Price is 100.0


New Price is 107.8


Estimated price of item 1 year hence: 116.2

Estimated price of item 2 years hence: 124.6


You answered y


Old Price is 16200.0


New Price is 18095.0


Estimated price of item 1 year hence: 20211.7

Estimated price of item 2 years hence: 22328.3


You answered n

The data is:

100.00

107.80

y

16200

18095

n

4. Gravitational Attraction

We do not provide a solution to this problem.


5. Clothes size calculation

The biggest problem students have in this, as in many ‘word problems’, is determining

the formulas from the problem statement.

Given height, weight, age, compute clothes sizes:

hatSize = weight (lbs.) / height (in.) 2.9

jacketSize (chest size, in.) =

height * weight / 288 +(1/8)(age-30)/10

Note carefully that the adjustment only occurs for complete 10 year interval after age 30,

i.e., if age < 40, there is no adjustment!

40 <= age < 49 gets 1/8 in. adjustment, etc.

waist (in.) = weight / 5.7 + (1/10) * (age - 28)/2

NB: adjustment only occurs for complete 2-year interval after 28

age = 29, no adjustment

30 <= age < 32, 1/10 inch adjustment.

Use a function for each calculation. Allow repetition at user option.

Now let's make some declarations:

int height; // inches

int weight; // lbs

int age; // years

double jacketSize; // inches at chest

double waist; // inches at waist

double hatSize;

Hat Size Calculation:

hat size = 2.9 * double(weight) / height;

The cast is clearer, but is not strictly necessary. The weight would be promoted

automatically when multiplication by 2.9 occurs. (This is why the 2.9 is first!)

Jacket Size Calculation:

jacket = double(height) * weight / 288

if (age > 40)

jacket = jacket + (age - 30)/10 * 0.125;


This depends on the behavior of the C/C++ language to obtain the results required. The

(age-30/10) arithmetic will be done as int, since there is nothing to require type

change. The int result will be then converted to double in the multiplication by the 0.125

(which is 1/8 as a decimal.)

Waist Size Calculation:

size = weight/5.7;

if (age>=30)

size = size + (age - 28)/2 * 0.1;

Again, the weight will be converted to double in the division by 5.7. The expression,

(age - 28)/2, will be computed as an int, then be promoted to double in the

multiplication by 0.1.

//problem: Clothes size calculation:

//given height (inches) weight (pounds) and age (years)

//compute jacket size, waist size, in inches, and hat size:

//returns hat size in standard hat size units

#include <iostream>


double hatSize (int height, int weight)

{

return 2.9 * double(weight) / height;

}

//returns jacketSize in inches at the chest

double jacketSize (int height, int weight, int age)

{

double jacket = double(height) * weight / 288;

if (age > 40)

jacket = jacket + (age - 30)/10 * 0.125;

return jacket;

}


// returns waist size in inches

double waistSize (int height, int weight, int age)

{

double size = weight/5.7;

if (age >= 30)

size = size + (age - 28)/2 * 0.1;

return size;

}

int main()

{

int height, weight, age;

double hat, jacket, waist;

char ans;

do

{

cout << "Give me your height in inches, weight in "

<< "pounds, and age in years" << endl

<< "and I will give you your hat size, jacket "

<< " size(inches at chest)" << endl

<< "and your waist size in inches." << endl;

cin >> height >> weight >> age;

hat = hatSize (height, weight);

jacket = jacketSize (height, weight, age);

waist = waistSize (height, weight, age);



cout.precision(2);

cout << "hat size = " << hat << endl;

cout << "jacket size = " << jacket << endl;

cout << "waist size = " << waist << endl;

cout << endl

<< "enter Y or y to repeat, “

<< “any other character ends." << endl;

cin >> ans;

} while ('Y' == ans || 'y' == ans);

return 0;

}


A typical run follows:

Give me your height in inches, weight in pounds, and age in years

and I will give you your hat size, jacket size (inches at chest)

and your waist size in inches.

69 185 50

hat size = 7.78

jacket size = 44.57

waist size = 33.56

enter Y or y to repeat, any other character ends.

y

Give me your height in inches, weight in pounds, and age in years

and I will give you your hat size, jacket size (inches at chest)

and your waist size in inches.

67 200 58

hat size = 8.66

jacket size = 46.78

waist size = 36.59

enter Y or y to repeat, any other character ends.

n

17:08:55:~/AW$

6. Average and standard deviation of scores

The problem as stated in the text requests a problem that is impossible to do within the

elements of C++ available in the chapter. See the errata on the text’s web site, URL:

http://www.aw.com/savitch.

Problem as stated in the text:

Write a function that computes the average (technically, the arithmetic mean) and

standard deviation of four scores, s1, s2, s3, and s4. The average is

a = = s1 + s2 + s3 + s4;

The standard deviation, stdDev1, is the square root of the average of the squares of the

differences of the s values and the average, a.

1 These differences are often called the deviations from the mean, and the standard deviation is then the square root of the mean of the squares of the deviations from the mean:


stdDev = sqrt((s1 – a)2 *(s2 – a)2 *(s3 – a)2 *(s4 – a)2

);

The function takes six parameters and calls two other functions to do the average and

standard deviation of the scores.

Discussion and Modified Problem:

Such a problem requires call-by-reference parameters for the return values. We modify

the problem by allowing the main function to call the average function and the standard

deviation functions directly. These take four arguments and return their one value

(average or standard deviation) as the function return value. Notice that the standard

deviation function calls the average function twice, once to get the average, and once

again to compute the average of the squared deviations of the data from the average.

Solution:

#include <iostream>

#include <cmath>


//returns the arithmetic mean of the four arguments.

double stdDev(double s1, double s2, double s3, double s4);

//returns the standard deviation of the four arguments.

double average(double s1, double s2, double s3, double s4);

int main()

{

double s1, s2, s3, s4;

double stdDeviation, avg;

char ans;

do

{

cout << "Enter four values, separated by white space and"

<< "terminated with <CR> \n"

<< "I will compute the average and standard "

<< "deviation.\n";

cin >> s1 >> s2 >> s3 >> s4;

avg = average(s1, s2, s3, s4);

stdDeviation = stdDev(s1, s2, s3, s4);


cout << "The Average is: " << avg << endl;

cout << "The Standard Deviation is: "

<< stdDeviation << endl;

cout << "Y/y continues, any other quits.\n";

cin >> ans;

} while(ans == 'y' || ans == 'Y');

return 0;

}

double stdDev(double s1, double s2, double s3, double s4)

{

double a = average(s1, s2, s3, s4);

return sqrt(average((s1-a) * (s1-a), (s2-a) * (s2-a),

(s3-a) * (s3-a), (s4-a) * (s4-a)));

}

7. Wind chill index

The wind chill index is measure of the increase in chilling effect of low temperature with

wind speed over the effect of the temperature alone. It is worth noting that wind chill

only affects animals and people.

The formula given in the text is both incorrect and out of date. See the errata on the text’s

web page. A corrected version is listed below.

Meteorologists have developed a replacement formula that is (presumably) more

accurate. We use the old (corrected) formula in one solution and the new formula in an

alternate solution.

In the following, we declare w, v, and t, and use these as follows:

double w; // wind chill factor

double v; // wind speed in meters/sec

double t ; // temperature in degrees Celsius, t <= 10;

The (incorrect) formula from the book is:

w = 33 - (10 * sqrt(v)– v + 10.5) * (33 – t) / 23.1;


The formula given in the text has been superceded by

w = 13.12 + 0.6215 * t – 11.37 * pow(v,0.16)

+ 0.3965 * t * pow(v,0.16);

In our solutions, we will use a both a corrected old formula from the literature, and the

new formula. The correct (old) formula (one that agrees with wind chill factor tables) is

w = 0.045*(5.27*sqrt(v) + 10.45 – 0.28*v)*(t –33) + 33;

This formula is available from many web sites, but this formula is taken from this URL

home.vicnet.net.au/~bbcbush/feature.htm

and from this URL

www.msc-smc.ec.gc.ca/windchill/Science_equations_e.cfm

The problem as presented in the text requires that we write a function that returns the

wind chill index. We are told to enforce the temperature restriction and to look up

weather reports and compare your result to the weather reports. (That is how I found that

the formula is wrong.)

#include <iostream>

#include <cmath>

#include <cstdlib>


//Pre: t in degrees Celsius. Required t <= 10

//v wind speed in meters/sec

//Post: returned value is the windChill index

double windChill(double v, double t);

int main()

{

double t, v;

char ans;

do

{

cout << "Enter wind speed and Celsius temperature <= 10 degrees\n";

cin >> v;

cin >> t;


cout << "Wind chill factor is "

<< windChill(v, t) << endl;

cout << "Y/y continues, other quits\n";

cin >> ans;

} while(ans == 'y' || ans == 'Y');

return 0;

}

double windChill(double v, double t)

{


if(t>10)

{

cout << "Quitting. windChill called with temperature > 10 "

<< "degrees\n";

exit(0);

}

w = 0.045 * (5.27 * sqrt(v) + 10.45 - 0.28 * v) * (t - 33) + 33;

return w;

}

The windChill function that uses the alternate formula follows. Note that I have

omitted the restriction on wind speed, as the references do not require it.

//This requires the user include cmath for the pow function.

double windChill(double v, double t)

{


w = 13.12 + 0.6215 * t – 11.37 * pow(v,0.16)

+ 0.3965 * t * pow(v,0.16);

return w;

}

Though I have tested this solution, I do not provide run results for this problem.

Savitch i Mch 03

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