Why this program outputs 4?

Question

Using define preprocessor directive why following program outputs 4 instead of 3?

#include <stdio.h>
#define MAX(A,B)((A)>(B)) ? A : B

int max_t(int a, int  b)
{
  return ((a)>(b)) ? a : b; 
}

int main()
{
  int i=1;
  int j=2;
  int val = MAX(++i, ++j); // this outputs 4 but why?
  //int val = max_t(++i, ++j);
  printf("%d\n", val);
  return 0;
}

@eyllanesc *no*. Not that one. Here there is no undefined behaviour. — Antti Haapala -- Слава Україні, Aug 31 '18 at 04:35

score 7 · Accepted Answer · answered Aug 31 '18 at 04:26

7

Macros are just replacements by tokens so

MAX(++i, ++j);

expands to

((++i) > (++j)) ? ++i : ++j;

As you see the parameters are evaluated (and incremented) twice in this case and therefore the output is 4.

answered Aug 31 '18 at 04:26

Osiris

2,783
9
17

That is what I wanted to know. – Tushar Aug 31 '18 at 04:29

score 3 · Answer 2 · answered Aug 31 '18 at 04:36

 int main()
 {
   int i=1;
  int j=2;
  int val = ((++i)>(++j)) ? ++i : ++j;

   printf("%d\n", val);
  return 0;
 }

This is what happening after preprocessor stage (After Macro expansion) and you can check this by "cc -E file.c -o file.i" command in linux gcc compiler , this expanded code is at last of the file .

so here : val = (2 > 3 ? 3 : 4 ) , this 4 get stored into val , thats why val = 4.

Why this program outputs 4?

2 Answers2