Re: [x86] Strange 64-bit put_user ?

[Jeff Garzik]

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




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