linux-kernel - Re: [LLVMdev] inline asm semantics: output constraint width smaller than input

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Message-ID: <498095F2.4060502@zytor.com>
Date:	Wed, 28 Jan 2009 09:29:22 -0800
From:	"H. Peter Anvin" <hpa@...or.com>
To:	Kyle Moffett <kyle@...fetthome.net>
CC:	Duncan Sands <baldrick@...e.fr>, llvmdev@...uiuc.edu,
	Ingo Molnar <mingo@...e.hu>,
	Török Edwin <edwintorok@...il.com>,
	Thomas Gleixner <tglx@...utronix.de>,
	Linux Kernel <linux-kernel@...r.kernel.org>
Subject: Re: [LLVMdev] inline asm semantics: output constraint width smaller
 	than input

Kyle Moffett wrote:
> 
> Even in the 64-bit-integer on 32-bit-CPU case, you still end up with
> the lower 32-bits in a standard integer GPR, and it's trivial to just
> ignore the "upper" register.  You also would not need to do any kind
> of bit-shift, so long as your inline assembly initializes both GPRs
> and puts the halves of the result where they belong.
> 

In this case, we're talking about what happens when the assembly takes a
64-bit input operand in the same register as a 32-bit output operand
(with a "0" constraint.)  Is the output operand the same register number
as the high register or the low register?  On an LE machine the answer
is trivial and obvious -- the low register; on a BE machine both
interpretations are possible (I actually suspect gcc will assign the
high register, just based on how gcc internals work in this case.)

	-hpa

-- 
H. Peter Anvin, Intel Open Source Technology Center
I work for Intel.  I don't speak on their behalf.

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