Why a floating point exception

Question

Hi,

Just trying out some C on some of the project euler questions.

My question is why am I getting a floating point exception at run time on the following code?

#include <stdio.h>
main()
{
int sum;
int counter;

sum = 0;
counter = 0;

for (counter = 0; counter <= 1000; counter++)
{
    if (1000 % counter == 0) 
    {
        sum = sum + counter;
    }
}

printf("Hello World");
printf("The answer should be %d", sum);
}

Thanks,

Mike

Answer 1

You start with counter = 0 and then do a "mod" with counter. This is a division by zero.

Hope this helps!

Answer 2

What happens when counter is 0? You try to compute 1000 % 0. You can't compute anything modulo 0, so you get an error.

Your if statement should read:

if (counter % 1000 == 0)

Answer 3

1000 % 0 is a division by zero

Answer 4

You are dividing 1000 by zero on the first iteration (calculating a reminder of something divided by 0). Why it crashed with a floating-point exception... I don't know. Maybe the compiler translated it into a floating-point operation. Maybe just a quirk of the implementation.