Compiler optimization causing program to run slower

Question

I have the following piece of code that I wrote in C. Its fairly simple as it just right bit-shifts x for every loop of for.

int main() {
   int x = 1;
   for (int i = 0; i > -2; i++) {
      x >> 2;
   }
}

Now the strange thing that is happening is that when I just compile it without any optimizations or with first level optimization (-O), it runs just fine (I am timing the executable and its about 1.4s with -O and 5.4s without any optimizations.

Now when I add -O2 or -O3 switch for compilation and time the resulting executable, it doesn't stop (I have tested for up to 60s).

Any ideas on what might be causing this?

Answer 1

The optimized loop is producing an infinite loop which is a result of you depending on signed integer overflow. Signed integer overflow is undefined behavior in C and should not be depended on. Not only can it confuse developers it may also be optimized out by the compiler.

Assembly (no optimizations): gcc -std=c99 -S -O0 main.c

_main:
LFB2:
    pushq   %rbp
LCFI0:
    movq    %rsp, %rbp
LCFI1:
    movl    $1, -4(%rbp)
    movl    $0, -8(%rbp)
    jmp L2
L3:
    incl    -8(%rbp)
L2:
    cmpl    $-2, -8(%rbp)
    jg  L3
    movl    $0, %eax
    leave
    ret

Assembly (optimized level 3): gcc -std=c99 -S -O3 main.c

_main:
LFB2:
    pushq   %rbp
LCFI0:
    movq    %rsp, %rbp
LCFI1:
L2:
    jmp L2  #<- infinite loop

Answer 2

You will get the definitive answer by looking at the binary that's produced (using objdump or something).

But as others have noted, this is probably because you're relying on undefined behaviour. One possible explanation is that the compiler is free to assume that i will never be less than -2, and so will eliminate the conditional entirely, and convert this into an infinite loop.

Also, your code has no observable side effects, so the compiler is also free to optimise the entire program away to nothing, if it likes.