Precedence of || and && operators in C

Question

#include <stdio.h>
int main()
{
   int i=-3,j=2,k=0,m;
   m = ++i || ++j && ++k;
   printf("%d %d %d %d", i, j, k, m);
}

Output :

-2 2 0 1

Why should the expression m=++i||++j&&++k; not parsed as m=++i||(++j&&++k) as the precedence of && is higher than || ??

Quite a coincidence that your program uses the exact same variables and values as the duplicate question. :) — Barmar, May 26 '15 at 07:53
Ahh, they're identical except that the first `&&` has been changed to `||`. — Barmar, May 26 '15 at 08:06
Operator precedence != order of evaluation && left-to-right associativity != left-to-right evaluation. — Lundin, May 26 '15 at 08:34

Mohit Jain · Answer 1 · 2015-05-26T07:53:22.070

2

|| and && are short circuit operators. If the final result is evaluated from the left operand, right operand is not evaluated.

++i ||  /* Evaluate ++i which is -2, so the result of expression is 1 */
  ++j && ++k;  /* No need to evaluated this */

edited May 26 '15 at 07:53

answered May 26 '15 at 07:51

Mohit Jain

This is not guaranteed and is specific to compiler implementation. – Atmocreations May 26 '15 at 07:52
5

No, this is guaranteed by the specs – Mohit Jain May 26 '15 at 07:53
@Atmocreations logical operators in C-like languages are always shortcircuited by standard – phuclv May 26 '15 at 07:55
2

@Atmocreations 6.5.13/4 `"...the && operator guarantees left-to-right evaluation; if the second operand is evaluated, there is a sequence point between the evaluations of the first and second operands. If the first operand compares equal to 0, the second operand is not evaluated."` The standard can't really be more clear than that. – Lundin May 26 '15 at 08:36

1 Answers1