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

From: Kyle Moffett
Date: Tue Jan 27 2009 - 20:46:15 EST

On Tue, Jan 27, 2009 at 4:25 PM, H. Peter Anvin <hpa@xxxxxxxxx> wrote:
> However, things get a bit ugly in the case of different widths that affect
> individually scheduled registers, like 32- and 64-bit types on a 32-bit
> machine. Consider the case above where "bar" is a 64-bit type and "baz" is
> a 32-bit type, then you functionally have, at least on x86:
> uint64_t tmp = bar;
> asm("foo" : "+r" (tmp));
> baz = (uint32_t)tmp;
> One could possibly argue that the latter case should be
> "baz = (uint32_t)(tmp >> 32);" on a bigendian machine... since this is a gcc
> syntax it probably should be "whatever gcc does" in that case, as opposed to
> what might make sense.
> (I'm afraid I don't have a bigendian box readily available at the moment, so
> I can't test it out to see what gcc does. I have a powerpc machine, but
> it's at home and turned off.)

Actually, PPC64 boxes basically don't care... the usable GPRs are all
either 32-bit (for PPC32) or 64-bit (for PPC64), the <=32-bit
instructions are identical across both, they just
truncate/sign-extend/etc based on the lower 32-bits of the register.
Also, you would only do a right-shift if you were going all the way
out to memory as 64-bit and all the way back in as 32-bit... within a
single register it's kept coherent.

Structs are basically irrelevant for inline ASM as you can't pass a
struct to one... you can only pass the *address* of a struct, which is
always pointer-sized.

I think that really the only sane solution (which is hopefully what
GCC does) for integer types is to use a register the same size as the
larger of the two integers. Then you copy the value to/from the
smaller register (or just mask it on PPC64-alike architectures) before
or after the inline ASM.

Kyle Moffett
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