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C++ control loops
 

C++ control loops

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    C++ control loops C++ control loops Presentation Transcript

    • C++ Control Statements
    • ITERATION STATEMENTS Parts of a loop • Initialization expression – Control / Counter Variable has to be initialized before entering inside the loop. This expression is executed only once. • Test expression – The truth value of this expression decides whether the loop has to be further executed or not • Update expression – This changes the value of the control / counter variable. This is executed at the end of loop statements • Body of the loop – Set of loop statements are executed based on the condition.
    • ITERATION STATEMENTS - FOR• Syntax • Example This program prints 1 to 10 for (initialization expr ; test expr ; update expr) body of the loop; for (i = 1; i <= 10; i=i+1) { cout << i ; }
    • ITERATION STATEMENTS - FOR • This will print -2 • This will loop for i values 0, 1, 2, 3, 4 for (a= 10; a >= 0; a=a-3); cout << a; for (i = 0; i < 5; i=i+1) cout << i * i; Note the semicolon here. This means empty loop The loop has empty body
    • Misc Declaration of variables in the loop – Variables declared inside the loop are accessible inside the loop only. They cannot be accessed outside. This is called local scope. (This applies for the selection statements also) int c; for (c = 0; c <=10; c = c+1) { int j = c; cout << j << “ “ << c ; } cout << j << “ “<< c; Valid Invalid
    • ITERATION STATEMENTS - WHILE • The syntax is • In the while loop, the control variable should be initialized outside the loop and it should be updated inside the loop while (expression) Loop body int a = 0; while (a <= 10) { cout << a; a = a + 1; } This program will print 0 to 10
    • Nested Loops• A loop may contain another loop inside its body. It is called Nested Loop. It is used when one variable has to change values for each value of another variable. Output will be 1 table starts 1 x 1 = 1 1 x 2 = 2 …. 1 x 20 = 20 1 table ends 2 table starts 2 x 1 = 2 2 x 2 = 4 …. 2 x 20 = 40 2 table ends upto 10 table ends for (int i = 1 ; i <= 10; i=i+1) { cout << i << “ table starts ”; for (int j = 1 ; j <= 20; j=j+1) cout << i << “ x “ << j << “ = “ << i * j ; cout << i << “table ends” ; cout << endl; }
    • Nested Loops Same program with while loop int i = j = 1; while (i <= 10) { cout << i << “table starts” ; while (j <= 20) { cout << i << “ x “ << j << “ = “ << i * j ; j = j + 1; } cout << i << “table ends” ; cout << endl; i = i + 1; }
    • Nested Loop concept • Identify the varying factors • Design a loop for each varying factor • For example, for a pattern like ******* ***** *** * • The varying factors are the line numbers, the spaces in each line and the number of stars in each line. • In each line, the number of spaces varies from 1 to m • In each line, the number of stars varies from 1 to n • So the loop would look one given in the next slide Line No Total No of Spaces (m) Total No of stars (n) 1 0 7 2 1 5 3 2 3 4 3 1
    • int m = 0; int n = 7; for (int i = 1 ; i <= 4; i=i+1) { for (int j = 0 ; j < m; j=j+1) { cout << “ ” ; } m = m + 1; for (int k = 1 ; k <= n; k=k+1) { cout << “*” ; } n = n – 2; cout << endl ; } Loop starts for each line Loop starts for spaces Loop starts for stars For going to the next line
    • Nested Loop Example • For example, for a pattern like 1 121 12321 1234321 • The varying factors are the line numbers, the spaces in each line, the numbers increasing and the numbers decreasing. • In each line, the number of spaces varies from 1 to m • In each line, the number increases upto the line no • In each line, the number increases from the line no Line No Total No of Spaces (m) Number increasin g till Number decreasi ng from 1 3 1 - 2 2 2 1 3 1 3 2 4 0 4 3
    • int m = 3; for (int i = 1 ; i <= 4; i=i+1) { for (int j = 1 ; j <= m; j=j+1) { cout << “ ” ; } m = m - 1; for (int k = 1 ; k <= i; k=k+1) { cout << k ; } for (int z = i-1 ; z >= 1; z=z-1) { cout << z ; } cout << endl ; } Loop starts for each line Loop starts for spaces Loop starts for increasing numbers For going to the next line Loop starts for decreasing numbers
    • Question Paper pattern • Write a program – Simple loops • Print a series (odd numbers, even numbers, 1 2 4 8 … 1024, a …. z, A…. Z, Factorial of a number, Sum of a given series of numbers ) • Print a table (8 table) – Nested loops • Print a pattern • Print tables – Combination of if and loop
    • Find the mistakes Wrong Code for (int c = 0; c >= 10; c = c + 1); cout << c; Wrong Code int a; while (a <= 10) { cout << a; } Corrected Code for (int c = 0; c <= 10; c = c + 1) cout << c; Corrected Code int a = 1; while (a <= 10) { cout << a; a = a + 1; }
    • Find the Output int sum = 0; for (int c = 1; c < 10; c = c + 3) sum = sum + c; cout << sum << “ “ << c << endl; Ans: sum = 0 initially c = 1 sum = 1 in first iteration c = 4 sum = 5 in second iteration c = 7 sum = 12 in third iteration c = 10 Since c < 10, so next iteration will not go inside the loop Output will be :
    • Find the Output int k = 0, c; while ( k <= 10) { c = k + 3; k = k + 2; } cout << c << “ “ << k; Ans: k = 0 initially c = 3 k = 2 in first iteration c = 5 k = 4 in second iteration c = 7 k = 6 in third iteration c = 9 k = 8 in fourth iteration c = 11 k = 10 in fifth iteration c = 13 k = 12 in sixth iteration Since k <= 10, next iteration will not go inside the loop as k becomes 12 Output will be : 13 12
    • Fill in the blanks This is a program to find the sum of odd numbers till 100. int sum = _____; int i = 1; while ( _________ ) { sum = sum + i; ___________ } cout << ____________ ;