Question

int main(void)
{
  int x;
  float y;

  x=10;
  y=4.0;

  printf("%d\n",x/y);

  return 0;
}

I compiled this code using a gcc compiler and when run I get 0 as output.
Why is this code giving output as 0 instead of 2?

Answer 1

The result of x/y is a float and is transferred to printf() as such.
However, you told printf() using %d to assume the input is an int.
There is no type-checking or automatic type-conversion in this case. printf() will just execute what you asked. This, by sheer accident, results in a 0 being printed.

You should have specified %d in the format string and cast the result of the division to int.

printf("%d\n", (int)(x/y) );

Or you could have used %f, but then you would have gotten 2.5 as output. (You might want to take a look at the floor() function too.)