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

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Message-ID: <f73f7ab80901281127t191c9c97nba99d859224e14bd@mail.gmail.com>
Date:	Wed, 28 Jan 2009 14:27:22 -0500
From:	Kyle Moffett <kyle@...fetthome.net>
To:	"H. Peter Anvin" <hpa@...or.com>
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

On Wed, Jan 28, 2009 at 12:29 PM, H. Peter Anvin <hpa@...or.com> wrote:
> 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.)

On a BE 32-bit machine, the "output register" technically ought to be
"64-bit" anyways, since it's constrained to be the same as the 64-bit
"input register".  That means that you ought to make sure to set
*both* output registers appropriately, one of them being 0 and the
other being the 32-bit number.  I think that's the only answer that
actually makes any sense from a holistic code-generation sense.  So it
seems we are in violent agreement :-D.

Cheers,
Kyle Moffett
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