replace divide by shift

Question

I am taking a look at following two small functions:

int foo(int val) {
  return val / 2;
}

int bar(int val) {
  return val >> 1;
}

I expected compiler to generate the same code for both of them. However, here is the generated assembly at O3

foo(int): # @foo(int)
 mov eax, edi
 shr eax, 31
 lea eax, [rax + rdi]
 sar eax
 ret
bar(int): # @bar(int)
 sar edi
 mov eax, edi
 ret

Here is the godbolt link: https://godbolt.org/g/XrtdEY

I was wondering why there is difference between these two assemblies.

Answer 1

It's different assembly because division by 2 of a signed integer number must give a well-defined result (by the standard), even when faced with a negative input. The assembly has to account for that, and cannot just do a shift which may produce something "unexpected".

When you write an explicit shift, you acknowledge the looser contract the standard imposes on that operator. So the compiler has more leeway.