Re: [x86] Strange 64-bit put_user ?

["H. Peter Anvin" <hpa@kernel.org>]

Jeff Garzik wrote:
> In arch/x86/include/asm/uaccess.h, if !CONFIG_X86_32, we see
> 
>> #define __put_user_x8(x, ptr, __ret_pu) \
>>         ({ u64 __ret_pu; __put_user_x(8, x, ptr, __ret_pu);
>> (int)__ret_pu; })
> 
> which was preceded by
> 
>> #define __put_user_x(size, x, ptr, __ret_pu)                    \
>>         asm volatile("call __put_user_" #size : "=a" (__ret_pu) \
>>                      :"0" ((typeof(*(ptr)))(x)), "c" (ptr) : "ebx")
> 
> 
> My question, from an admitted inline asm newbie:
> 
> Why is 32-bit register 'ebx' being used for a 64-bit put_user?
> 
> And a dumb-question follow-up, probably easy, for any x86 expert:  why
> are registers 'bl' and 'bx' not used for 8-bit and 16-bit put_user,
> respectively?
> 

The answer is simply that gcc doesn't make a distinction between bl, bx,
ebx, and rbx -- it considers it a single regioster which can contain an
8-, 16-, 32- or 64-bit number.  In particular, gcc can't use ah, bh, ch,
and dh as independent registers at all.

	-hpa