Re: [x86] Strange 64-bit put_user ?

[Jeff Garzik <jeff@garzik.org>]

H. Peter Anvin wrote:
> 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.

So, it isn't really referring to 'ebx' register specifically, it is 
referring to the hardware register that contains 'ebx'.  Gotcha.

Thanks!

	Jeff