Willi Mentzel - 14 days ago 6

C Question

I saw the chosen answer to this post.

I was suprised that

`(x & 255) == (x % 256)`

`%`

`&`

`x % n`

`n = 2^a (a = [1, ...])`

Since this is a special case in which I as a human can decide because I know with which values the program will deal with and the compiler does not.

Sure, I could just compile and look at the dissassembly. But this would only answer my question for one compiler/architecture. I would like to know if this is in principle faster.

Answer

If your integral type is unsigned, the compiler will optimize it, and the result will be the same. If it's signed, something is different...

This program:

```
int mod_signed(int i) {
return i % 256;
}
int and_signed(int i) {
return i & 255;
}
unsigned mod_unsigned(unsigned int i) {
return i % 256;
}
unsigned and_unsigned(unsigned int i) {
return i & 255;
}
```

will be compiled (by GCC 6.2 with -O3; Clang 3.9 produces very similar code) into:

```
mod_signed(int):
mov edx, edi
sar edx, 31
shr edx, 24
lea eax, [rdi+rdx]
movzx eax, al
sub eax, edx
ret
and_signed(int):
movzx eax, dil
ret
mod_unsigned(unsigned int):
movzx eax, dil
ret
and_unsigned(unsigned int):
movzx eax, dil
ret
```

The result assembly of `mod_signed`

is different because

If both operands to a multiplication, division, or modulus expression have the same sign, the result is positive. Otherwise, the result is negative. The result of a modulus operation's sign is implementation-defined.

and AFAICT, most of implementation decided that the result of a modulus expression is always the same as the sign of the first operand. See this documentation.

Hence, `mod_signed`

is optimized to (from nwellnhof's comment):

```
int d = i < 0 ? 255 : 0;
return ((i + d) & 255) - d;
```

Logically, we can prove that `i % 256 == i & 255`

for all unsigned integers, hence, we can trust the compiler to do its job.