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Message-ID: <49F79FEF.80508@kernel.org>
Date: Tue, 28 Apr 2009 17:31:43 -0700
From: "H. Peter Anvin" <hpa@...nel.org>
To: Jeff Garzik <jeff@...zik.org>
CC: LKML <linux-kernel@...r.kernel.org>, x86@...nel.org
Subject: Re: [x86] Strange 64-bit put_user ?
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
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