C Left Shift Operator

Question

I have the following program:

#include <stdio.h>
int main ()
{
    
    int x = 8;
    printf("%d %d %d ", x++, x << 2, x >> 1);
 
}

The way I feel like this is supposed to go is this way: The first number should be 8. Then it gets incremented to 9. 9<<2 is 36 so the second %d is 36. The last one is 8 >> 1 which is 4.

However, when I put this into the compiler I get '8 32 4' and not '8 36 4'.

Can someone explain why please?

Thank you!

Function arguments can be evaluated in any order. So the order of side effects is also not guaranteed. — wildplasser, Jul 24 '20 at 12:24
@P__J__ you are right, it's the wrong dupe. Better would be: https://stackoverflow.com/questions/4176328/undefined-behavior-and-sequence-points although that's labelled C++ rather than C (although in this case, it's the same). — Paul Hankin, Jul 24 '20 at 12:35

score 2 · Answer 1 · answered Jul 24 '20 at 12:24

This operation invokes an Undefined Behavior.

The gcc executes the way you observe, clang does it another way as the result of this program is undefined.

clang also issues a warning

https://godbolt.org/z/GzhPKn

The make program result defined:

int main ()
{    
    int x = 8;
    x++;
    printf("%d %d %d ", x, x << 2, x >> 1);
}

score 0 · Answer 2 · answered Jul 24 '20 at 12:28

Argument evaluation order in C and C++ is undefined. Whether the increment or one of the shifts happens first depends on implementation and compiler optimization. It is never advisable to write code like this.

int a() { printf("A"); return 0; }
int b() { printf("B"); return 0; }
int c() { printf("C"); return 0; }

If you do:

printf("%d%d%d", a(), b(), c());

It doesn't necessarily mean you'll get an ABC on your console. Ignore the 0s.

C Left Shift Operator

2 Answers2