On Tue, 21 May 2019, Uros Bizjak wrote:
> --cut here--
> #include "cpuid.h"
>
> int main ()
> {
> unsigned int eax, ebx, ecx, edx;
>
> if (!__get_cpuid (1, &eax, &ebx, &ecx, &edx))
> __builtin_abort ();
>
> printf ("%#x, %#x, %#x, %#x\n", eax, ebx, ecx, edx);
> return 0;
> }
> --cut here--
>
> results in:
>
> 2f: 31 f6 xor %esi,%esi
> 31: 89 f0 mov %esi,%eax
> 33: 0f a2 cpuid
> 35: 85 c0 test %eax,%eax
> 37: 0f 84 fc ff ff ff je 39 <main+0x39>
> 3d: 83 ec 0c sub $0xc,%esp
> 40: 89 f3 mov %esi,%ebx
> 42: 89 f1 mov %esi,%ecx
> 44: b8 01 00 00 00 mov $0x1,%eax
> 49: 0f a2 cpuid
Hmm, I wonder why this produces larger code which relies on a single
`xor' instruction where it could save two bytes by doing all the zeroing
inline:
xor %eax,%eax
cpuid
test %eax,%eax
je 39 <main+0x39>
sub $0xc,%esp
xor %ebx,%ebx
xor %ecx,%ecx
mov $0x1,%eax
cpuid
Is `xor' considered more computatively expensive than `mov' on the i386?
Maciej
