zhen jiang dept. of computer science west chester university west chester, pa 19383
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Zhen Jiang
Dept. of Computer Science
West Chester University
West Chester, PA 19383
zjiang@wcupa.edu
CSC530 Data Structures - LOOP
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Price is right. Sample execution (click on this link to
try) Each button in the above sample
REPEAT …?
Loop
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While loop Format & Logic
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<initialization>;while (<test>) {
<body>;}
Do-while loop Format Logic
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How does this differ from the while loop?
The controlled <statement(s)> will always execute the first time, regardless of whether the <test> is true or false.
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For loop Format Logic
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for (<init>; <test>; <update>) { <body>;
}
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Summary
Body first, and thenevent change/update
Development process
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Controlling Number of Loop Iterations If the number of iterations is known
before the loop starts, the loop is called a count-controlled loop. Counter =0, counter++, counter
<number Counter = 1, counter++, counter
<=number Use for loop for an easy
development.
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Code:for (int i = 1; i <= 4; i++) { cout << i << " squared is " << (i * i) << endl;}
Output:1 squared is 12 squared is 43 squared is 94 squared is 16
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Code:cin >> n; // try 6!for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { cout << “*”; } cout <<endl;}
Output:************************************
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Code: cin >> n; // try 5!
for (int i = 1; i <= n; i++) { for (int j = 1; j <= 10; j++) { cout <<(i * j) << " "; } cout << endl; }
Output:1 2 3 4 5 6 7 8 9 102 4 6 8 10 12 14 16 18 203 6 9 12 15 18 21 24 27 304 8 12 16 20 24 28 32 36 405 10 15 20 25 30 35 40 45 50
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Code:cin >> n; // try 6!for (i = 1; i<=n; i++) cout << “*”;cout <<endl;for (i = 1; i <= n-2; i++) {
cout << “*”; for (int j = 1; j <= n-2; j++)
cout <<“ ”; cout <<“*” <<endl;}for (i = 1; i<=n; i++) cout <<“*”;cout << endl;
Output:******* ** ** ** *******
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Code:cin >> n; // try 6!for (int i = 1; i <= n; i++) { for (int j = 1; j <= i; j++) { cout << "*"; } cout <<endl;}
Output:*********************
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Code:cin >> n; // try 6!for (int i = 1; i <= n; i++) { for (int j = 1; j <= i; j++) { cout << i; } cout << endl;}
Output:122333444455555666666
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Code:cin >> n; // try 5!for (int i = 1; i <= n; i++) { for (int j = 1; j <= (n - i); j++) { cout << " "; } for (int k = 1; k <= i; k++) { cout << i; } cout << endl;}
Output: 1 22 333 44445555504/21/23
Otherwise (unknown or unclear), the loop is called a event-controlled loop. Use a while loop or a do-while loop for an
easy checkpoint development. Asking the user before each iteration if it is
time to end the loop is called the ask-before-iterating technique.
Appropriate status update (or event initializing) for a sequence of iterations
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Controlling Event of Loop Iterations
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Finds and prints a number's first factor other than 1:
cin >> n; // try 91int f = 2;while (n % f != 0) { f++;}cout << "First factor:“ << f <<endl;
Sample run:First factor:7
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Write a program that will repeatedly prompt the user to type a number until the user types a non-negative number, then square it.
Example log:
Type a non-negative integer: -5
Invalid number, try again: -1
Invalid number, try again: -235
Invalid number, try again: -87
Invalid number, try again: 11
11 squared is 121
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cout << "Type a non-negative integer: ";cin >> n;
while (n < 0) { cout << "Invalid number, try again: "; cin >> n;}
int square = n * n;cout << n << " squared is " << square<<endl;
Notice that the number variable had to be declared outside the while loop in order to remain in scope.
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Write a class named DigitSum that reads an integer from the user and prints the sum of the digits of that number. You may assume that the number is non-negative.
Example:
Enter a nonnegative number: 29107prints out 19 (i.e.,2+9+1+0+7 )
Hint: Use the % operator to extract the last digit of a number. If we do this repeatedly, when should we stop?04/21/23
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cin >> n; int sum = 0; while (n > 0) { sum += n % 10; // add last digit to sum n = n / 10; // remove last digit } cout << “sum = “ << sum << endl;
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Write a program named CountFactors that reads in an integer and displays its number of factors.
For example, if the user enters 60, CountFactors displays 12 because 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60 are all factors of 60.
cin >> n; int sum = 0, k = ?; while ( ) {
} cout << “sum = “ << sum <<endl;
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Exercise
population TV purchase 1+2+4+8+...1+2+3+4+...+99
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Solution
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