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

[Kyle Moffett]

On Wed, Jan 28, 2009 at 12:29 PM, H. Peter Anvin <hpa@xxxxxxxxx> 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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