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cgccc-preprocessorternary-operatorpreprocessor-directive

Why does this program output 8?

#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;
}
Solution

  • 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.