Why does this program output 8?

Question

#include <stdio.h>
#define abs(x) x > 0 ? x : -x

int main(void) {
  printf("%d\n", abs(abs(3 - 5)));
  return 0;
}

Why does the program above output 8 and not 2 while the program below outputs 2?

#include <stdio.h>

int abs(int x) {
  return x > 0 ? x : -x;
}

int main(void) {
  printf("%d\n", abs(abs(3 - 5)));
  return 0;
}

Change `#define abs(x) x > 0 ? x : -x` to `#define abs(x) ((x) > 0 ? (x) : -(x))` — ouah, Apr 14 '17 at 11:34
The program [outputs negative 8](http://ideone.com/H434lj), not 8. — Sergey Kalinichenko, Apr 14 '17 at 11:39
I have recently noticed that I posted the code I changed, if expression was `#define abs(x) (x > 0 ? x : -x)` then it would [output positive eight](http://ideone.com/7WY5Ie), @dasblinkenlight, thanks by the way. — Mehmet Eren Aldemir, Apr 14 '17 at 11:57

score 2 · Accepted Answer · answered Apr 14 '17 at 11:35

Short answer is "because a macro is not a function".

Long answer is that macro parameters are expanded into the text of the program, so C compiler sees this long expression:

3 - 5 > 0 ? 3 - 5 : -3 - 5 > 0 ? 3 - 5 > 0 ? 3 - 5 : -3 - 5 : -3 - 5 > 0 ? 3 - 5 : -3 - 5

In the expansion, negative sign applies to 3, not to (3-5), yielding negative 8.

Although you can work around this issue by placing parentheses around x in the macro definition, defining an inline function would be a better choice.

Why does this program output 8?

1 Answers1