C Number pattern 2 program with Explanation

Write "C"  program to print below Number Pattern 2:

To understand this example, you should have knowledge of the following topics:

· C if Else Statement

· C For loop

· C programming operators

C Number pattern 2 program using For loop:

#include <stdio.h>

int main()

{

    int i, j;

    for(i = 1; i <= 5; i++)

    {

        for(j = i; j <= 5; j++)

        {

            printf("%d",j);

        }

        printf("\n");

    }

    return 0;

}

Output:

12345

2345

345

45

5

Explanation:

1) First two integers "i" and "j" are declared of type int.

2) Then in the First for loop “i” value is initialized with value 1.

3) Now “i” value is checked with the condition i <= 5 that is (1 <= 5). So the condition is True.

4) Now the loop enters into the second for loop and checks the condition j <= 5. Where “j” value is “1”.

5) So the value of j=1 is less than “5” so the condition is true. So it's print the value of “J”. And the value of “j” increments after printing value. It prints the value of J up to “j = 5” when becomes 6 it comes out the loop and increment value i = 2 and the same process repeats.

6) Last where “i” value becomes “0” the first for loop condition becomes false and the program terminates.

See also: C number pattern1 program C number pattern3 program

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