1 Copyright (C) 2008 by Dennis A. Fairclough all rights reserved.

More Control Statementsbased on the original work of Dr. Roger deBryVersion 1.0

1Copyright (C) 2008 by Dennis A. Fairclough all rights reserved.

Topics

LoopsBreak and continue statementsSoftware development process

Algorithms


ObjectivesAt the end of this topic, students should be able to:

Correctly use a while statement in a C++ programCorrectly use break and continue statements in a C++ programCorrectly use a do statement in a C++ programCorrectly use a for statement in a C++ programDescribe the stages of software developmentCreate simple algorithms to solve computing problems andcreate UML activity diagrams to describe their algorithms


The while StatementThe while statement allows the program to execute

the same statement multiple times.

while ( count <= MAX ){ cout << count << ‘ ‘; count ++;}

iscount <= MAX

?

cout << count << ‘ ‘;

count++;

true

this is called the limit

Inside the loop, you must alwaysdo something to the variable beingtested, so that the program will eventually exit the loop

false


// Using a While Statement with a limit

// main class declarationint main ( ) { const int MAX = 5; int count = 1; while ( count <= MAX ) { cout << count; count++; } }

while the limit has not been reached …

change the number being compared to the limit


// Using a While Statement with a sentinel

int main ( ) { int value = 1; while ( value != 0 ) { cout << "Enter in a digit between 0 and 9, use 0 to quit:"; cin >> value; cout << "You typed " << value; } cout << "Goodbye!"; }

the sentinel in this case is 0

get another value to compare with the sentinel


Using while to validate input. . .const int MAX = 5;cout << “Enter an integer between 0 and 5;

cin >> input;while ( input < 0 || input > MAX ){ cout << “Invalid input. Try again”; cin >> input;}. . .


// Nested While

int main ( ) { const int COUNT = 3; const int TIMES = 2; int valueOne, valueTwo; valueOne = 0; while ( valueOne <= COUNT ) { cout << "In outer loop, valueOne = " << valueOne; valueTwo = 0; while ( valueTwo <= TIMES ) { cout << " In inner loop, valueTwo = " << valueTwo; valueTwo++; } valueOne++; } cout << "Goodbye!"; }


break and continue

break – breaks immediately out of a loop/switch.

continue – skip the rest of the loop and go back and go through the loop another time.


The do-while Statement

do{ count++; cout << count;} while ( count < LIMIT);

count = count +1;

cout << count;

iscount < LIMIT

?

true

false

in a do statement, the loop willalways be executed at least once!


Validating Input

A common programming problem is to do something,and then ask the user if they want to do it again.

If the user answers “yes” you do it again. If the user answers “no” you stop.

If the user answers neither, tell him to try the answer again.


Write the code …..


First – our up front boilerplate

#include <iostream>using namespace std;

int main ( ){


Second – declare the variables we will use

int number; // a user entered valuechar yesNo; // store use response to do it againbool okay = true, notDone = false; // some flags


Prompt userFor avalue

Get inputFrom the

user

Display theresult

number


// prompt, get user input, and display itcout << <\nEnter an integer value: “;cin >> number;cout << “\nYou typed the number “ << number;



Get inputFrom the

user

Display theresult

Prompt“Play Again?”

Get inputFrom theuser

InputValid

?

Display anerror

message

no

yesNo


// prompt to play again – make sure response is validcout << <\nDo you want to play again(y or n)?: “;cin >> yesNo;if(cin.rdbuf()->in_avail()!=0) cin.ignore(80, ‘\n’); // get rid of extra characters

// see if response is valid – print a message if its notokay = (yesNo == ‘y’ || yesNo == ‘n’);if (!okay) cout << “\nSorry, but that is not a valid response.”;


// prompt to play again – make sure response is validcout << <\nDo you want to play again(y or n)?: “;cin >> yesNo;if(cin.rdbuf()->in_avail()!=0) cin.ignore(80, ‘\n’); // get rid of extra characters

// see if response is valid – print a message if its notokay = (yesNo == ‘y’ || yesNo == ‘n’);if (!okay) cout << “\nSorry, but that is not a valid response.”;

Then here is the first loop – stay in loop until there isa valid response

do{

} while (!okay);



Get inputFrom the

user

Display theresult

Prompt“Play Again?”

Get inputFrom the

user

InputValid

?

Display anerror

message

no Playagain

?

yes noend


#include <iostream>using namespace std;

int main ( ){ int number; char yesNo; bool okay = true, notDone = true;

do // main loop { cout << "\nEnter an integer value: "; cin >> number; cout << "\nYou types the number " << number; do { cout << "\nDo you want to play again (y or n)? "; cin >> yesNo; if(cin.rdbuf()->in_avail()!=0) cin.ignore ( 80, '\n'); // get rid of extra characters okay = ( yesNo == 'y' || yesNo == 'n'); if ( !okay ) cout << "\nSorry, that is not a valid response.” } while (!okay);

if (yesNo == 'n') notDone = false; } while (notDone);}


Apply this code to the guessing gamein the following slide

Practice


int main ( ) { // We will generate numbers between 0 and 10 const int MAX = 10; int answer, guess; randomize ( ); answer = rand( ) % 11; cout << "I am thinking of a number between 0 and "; cout << MAX << " What is it? "; cin >> guess; if(cin.rdbuf()->in_avail()!=0) cin.ignore(80,’\n’); if ( guess == answer ) cout << "Congratulations. You guessed it.\n"; else { cout << "That is not correct, sorry. "; cout << "The number was " << answer << ‘\n’; } cout << "Thanks for playing.\n"; }


The for StatementThe for statement is best used when you knowexactly how many times you want to executethe loop.

for ( ){ cout << count;}

initialization evaluation increment

int count = 0;count < LIMIT;count++

Initialize

evaluatecondition

body of loop

increment


The Comma OperatorThe comma operator is used to evaluate a list of expressions, but onlyreturn the value in the last expression.

goodNum = (first = 2, second = first + 1);

this expression is evaluated firstthen this expression is evaluatedand the result is assigned to goodNum.


establish the requirements

create the design

implement the code

test the implementation

Development Stages


Algorithm Development

The steps that we come up with to solvea problem are called an algorithm.

An algorithm is like a recipe in a cookbook.Once you come up with an algorithm tosolve a particular problem, you can apply itto other similar problems.


ProblemYou just purchased a new computer that cost $1000.00. Youdid not have to make a down payment, and your payments are$50.00 a month. The interest rate on your purchase is 18% peryear. How many payments will you have to make to pay off theloan and what is the total interest that you will pay over the lifeof the loan.

Each month when you make a payment, your payment first paysthe interest for that month. The monthly interest rate is 1.5%. Once the interest is paid, the balance of you payment goes towards the balance of the loan.


Step One

Write down everything you know about the problem.

Loan amount = 1000.00

Monthly Payment = 50.00

Interest Rate = 18%


Step Two

Write down what you are looking for

Months to Pay

Total Interest


Interest CalculationEach month when you make a payment, your payment first paysthe interest for that month. The monthly interest rate is 1.5%, sothe first month the interest will be 1.5% x $1000.00, or $15.00. Once the interest is paid, the balance of you payment goes towards the balance of the loan. The first month you will have $35.00 leftafter paying the interest, so subtracting this from the loan balancewill make the new balance $965.00.

The next month, repeat the process, starting with the new balance, $965.00.


Pseudo codePseudo code is an English-like description of the programmingsteps taken to solve a problem.

Write the pseudo code required to calculate the new balance each month

interest = balanceDue x monthlyRate

paymentBalance = payment – interest

balanceDue = balanceDue - paymentBalance


Activity DiagramWe can create what is called an Activity Diagram to show these steps pictorially.


paymentBalance =payment - interest

balanceDue = balanceDue –

paymentBalance


Building the LoopWe know that we will do this same calculation each month,for some number of months. This suggests that we put thecode we just developed into a loop of some kind. Thepseudo code to describe this might be something like

while the loan is unpaid, do the following:


paymentBalance = payment – interest

balanceDue = balanceDue - paymentBalance


Adding the Loop to the Diagram




paymentBalance

is the loanpaid off?

yesno


InitializationTo complete the pseudo code and the diagram, we just need totake care of initializing some things:




paymentBalance

is the loanpaid off?

yesno

monthlyRate = .015balanceDue = 1000.

payment = 50.


Convert to Codedouble balanceDue = 1000.00;double monthlyRate = .015;double payment = 50.00;

double interest = balanceDue * monthlyRate;double paymentBalance = payment – interest;balanceDue = balanceDue – paymentBalance;

What kind of loop?

If there is a balance on the loan, we must go through theloop. If not, we can quit.


double balanceDue = 1000.00;double monthlyRate = .015;double payment = 50.00;

double interest = balanceDue * monthlyRate;double paymentBalance = payment – interest;balanceDue = balanceDue – paymentBalance;

do{

} while (balanceDue > 0);


Are We Done?

Well, what did the problem ask for?

* The number of months it takes to pay off the loan

* the total interest paid during the life of the loan

Nope …. we haven’t calculated these yet.

How would you fix the program to add these calculations?


double balanceDue = 1000.00;double monthlyRate = .015;double payment = 50.00double totalInterest = 0.0;int monthsToPay = 0;

double interest = balanceDue * monthlyRate;totalInterest += interest;monthsToPay++;double paymentBalance = payment – interest;balanceDue = balanceDue – paymentBalance;

do{



Are We Done?

Well …. almost.

We probably ought to print out the results.




do{


cout << Months to pay off loan = “ << monthsToPay << “\n”;cout << “Total interest = “;cout.setf (ios::fixed);cout.setf (ios::showpoint);cout.precision (2);cout << totalInterest << “\n”;


Wrap it up and put a bow on it!




do{

} while (balanceDue > 0);cout << Months to pay off loan = “ << monthsToPay << “\n”;cout << “Total interest = “;cout.setf (ios::fixed);cout.setf (ios::showpoint);cout.precision (2);cout << totalInterest << “\n”;

#include <iostream>using std::cin;using std::cout;using std::ios;

int main( ){

{


PracticeWrite a program that displays the multiplicationtables between 2 and 12. Display the output incolumns, so that it all appears on one screen. Yourtable should be nicely lined up like the following:

2 3 4 5 6 7 8 9 10 11 122 4 6 8 10 12 14 16 18 20 22 243 6 9 12 15 18 21 24 27 30 33 364 8 12 16 20 24 28 32 36 40 44 48

etc . . .


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1 Copyright (C) 2008 by Dennis A. Fairclough all rights reserved.

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