Pre and post increment confusion

Question

I tried running this program :

#include<stdio.h>
int main(){
        int a=5;
        printf("%d %d", ++a,a++);
        return 0;
}

with gcc in arch-chroot on a armv7 device. I expect to get output 7 5 but i'm getting 7 6. Can anyone explain what's going on? running the program

You are assuming a left to right evaluation order of the arguments. — xenoid, Apr 14 '19 at 16:30
Also, if the arguments were evaluated from left to right, wouldn't the result be `6 6` — Jonathon K, Apr 14 '19 at 16:35
yes. It would be... I think it's neither following right to left order or left to right order. What's the matter? right to left `7 5` . left to right `6 6` — Abinash Dash, Apr 14 '19 at 16:37
Read assembler source if you're really curious. Many x86 compilers would produce 7 5. But it's another architecture. And it's UB anyway. — Matt, Apr 14 '19 at 16:47

gsamaras · Accepted Answer · 2019-04-14T16:37:58.777

Your code is invoking Undefined Behavior (UB)!

Use the warning flgas -Wall -Wextra during compilation, and the compiler will tell you the story:

prog.c: In function 'main':
prog.c:4:30: warning: operation on 'a' may be undefined [-Wsequence-point]
    4 |         printf("%d %d", ++a,a++);
      |                             ~^~
7 5

In that online demo, I got a different output, a characteristic of UB.

Read more in printf and ++ operator.

Petr Skocik · Answer 2 · 2019-04-18T13:15:35.583

6.5p2

If a side effect on a scalar object is unsequenced relative to either a different side effect on the same scalar object or a value computation using the value of the same scalar object, the behavior is undefined. If there are multiple allowable orderings of the subexpressions of an expression, the behavior is undefined if such an unsequenced side effect occurs in any of the orderings.84)

++a and a++ are unsequenced. Your program is ill-formed.

Pre and post increment confusion

2 Answers2