[x86] Strange 64-bit put_user ?

[Jeff Garzik <jeff@garzik.org>]

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?

Thanks,

	Jeff