Question

I have the following C program:

#include <stdio.h>

int main()
{
double x=0;
double y=0/x;
if (y==1)
  printf("y=1\n");
else
  printf("y=%f\n",y);
if (y!=1)
  printf("y!=1\n");
else
  printf("y=%f\n",y);

return 0;
}

The output I get is

y=nan
y!=1

But when I change the line
double x=0;
to
int x=0;
the output becomes

Floating point exception

Can anyone explain why?

Answer 1

There’s a special bit-pattern in IEE754 which indicates NaN as the result of floating point division by zero errors.

However there’s no such representation when using integer arithmetic, so the system has to throw an exception instead of returning NaN.