CODEWITHSUNDEEP
c
Zack
Zack

Reputation: 11

What is the run time error in this program?

When Running this program it crushes at some point and i can't find why any help? I've been literally looking at it for the past 30 min and i can't quite find it 

#include <stdio.h>

void main(void)
{
    int loop_counter = -7;
    int input = 9;
    char c1 = '9';
    char c2 = 43;

    while(input == 9)
    {
        printf("%d", loop_counter+1);
        printf("%d", loop_counter+2);

        printf("%d", loop_counter);
        printf("%d", loop_counter+1);
        printf("%d", loop_counter+2);

        int b = 4*loop_counter; 
        int a = 4/loop_counter;
        double c = loop_counter / 9;
        printf("%d", loop_counter);
        printf("%d", loop_counter+1);
        printf("%d", loop_counter+2);       

        if (loop_counter > 10)
        {
            input = 10;
        }

        loop_counter++;
    }

    printf("loop exit\n\n");    
    getchar();
}

Upvotes: 0

Views: 24

Answers (1)

dbush
dbush

Reputation: 223699

You have loop_counter starting at -7 which then increases on each iteration. When loop_counter is equal to 0, you then do this:

int a = 4/loop_counter;

Which is division by zero, and causes a floating point exception.

Either add a check for 0 at this point, or remove the line altogether since the value of a is never being used.

Upvotes: 1

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