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

From: H. Peter Anvin
Date: Wed Jan 28 2009 - 16:01:27 EST


Kyle Moffett wrote:

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.


No.

This is wrong on two accounts.

First of all, THERE ARE NO "TWO OUTPUT REGISTERS". Period. There is only one. The question is: which of the two *input* registers does it correspond to?

Second of all, "making sense from a holistic code-generation sense" doesn't apply here. This is about mimicing a gcc construct, regardless of which amount of sense it makes. Therefore, the only thing that actually makes sense is to mimic gcc behavior, no matter how stupid it happens to be.

-hpa

--
To unsubscribe from this list: send the line "unsubscribe linux-kernel" in
the body of a message to majordomo@xxxxxxxxxxxxxxx
More majordomo info at http://vger.kernel.org/majordomo-info.html
Please read the FAQ at http://www.tux.org/lkml/