01/05/100 1

Loops/Iteration

• Used to repeat an action • Must have a STOP condition• Three flavors - for, while, do/while

01/05/100 2

• 1 check the test • 2 if the test is true

– exec the body– when the body has finished

• go to step 1

• if the test is false– exit the loop

int n = ?; // try n as 6while (n >= 0) {

n -= 2;cout << n << endl;

}cout << “final n is “ << n << endl;

Anatomy of a while loop

test ?

loop body

stmt following loop

stmt before loop

The test is always a “keep going” condition. To determine the termination condition, negate the test.I.e. the loop will keep going as long as n >= 0 the loop will terminate when n becomes negative (n < 0)

01/05/100 3

while Loops

• The test is checked at the very beginning and then again each time after the after the entire loop body has been executed

• The test is NOT checked in the middle of the loop body• This is true for all the loops (for, while, and do/while), not just

the while loop

x = ?; // try x as 45

while (x < 50)

{

x++;

cout << x << endl;

x++;

cout << x << endl;

}

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Practice what’s the output?

int a = 20, b = 50;while (a < b) { if (a % 5 == 0)

{ cout << “s”; }

else if (a % 3 == 0) {

cout << “e”; }

else {

cout << “w”; } a += 4;}cout << “\n”;cout << “a’s final value is “ << a << endl;

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PracticeWhat’s the output?

int d = 90;

while (d < 80) {

d ++;

}

cout << “d is “ << d << endl;

int x = 90;

while (x < 100)

{

x -= 5;

}

cout << “final value for x is “ << x << endl;

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Summing (even) numbers with a while loop

Example of an indeterminate loop - the user’s input will determine how many times the loop executes

int sum = 0, evensum = 0, number;

cout << “ First number please “;

cin >> number;

while (number > 0)

{

sum += number;

if (number % 2 == 0)

{

evensum += number;

}

cout << "number please ";

cin >> number;

}

cout << "sum is " << sum << endl;

cout << “sum of even #’s is “ << evensum << endl;

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Error checking with a while loopif the number is out of range, enter the while loop

otherwise, skip the loopreenter loop when the user’s number continues to be out of range

const int cMax = 10, cMin = 5;

cout << "Pls enter a # between "

<< cMax << " and "

<< cMin << " please " << endl;

cin >> number;

while (number < cMin || number > cMax)

{

cout << "error in input \n";

cout << "Pls enter a # between "

<< cMax << " and "

<< cMin << " please " << endl;

cin >> number;

}

cout << “The user entered “

<< number << endl;

test ?

loop body

stmt following loop

stmt before loop

Block/Scope

• { } define a scope

void main()

{

int n = 9; // this n belongs to main

{

int n = 56; // this n belongs to this block nested in main

cout << n << endl; // use the inner most n

}

cout << n << endl; // don’t know about inner n, use n from main

}

this is not a good idea, having nested blocks that reuse the same variable name

Scope/Blocks

identifiers declared in a nested block cannot be referred to outside the block

but identifiers in an outer block are accessible in a nested block

void main() {

int a = 5;

{ int x = 9;

cout << “inside inner block\n”; cout << x << “ “ << a << endl; } // ILLEGAL, x is not visible in this block cout << x << “ “ << a << endl;}}

Where you really use blocks

• You won’t use free-standing nested or sequential blocks demo’ed in the last slides, but you do use blocks in constructs that have a pair of { }

void main() {

int sum = 0;while (sum < 80) {

int response;cout << “Please enter a number\n”;cin >> response;sum += response;

}cout << response << endl; // ILLEGAL , response is declared inside of the

while loop block and cannot be used outside of the while loop

cout << sum << endl; // sum can be used both in and out of the while loop

}