Messages in this thread | | | Date | Tue, 27 Jan 2009 20:45:55 -0500 | Subject | Re: [LLVMdev] inline asm semantics: output constraint width smaller than input | From | Kyle Moffett <> |
| |
On Tue, Jan 27, 2009 at 4:25 PM, H. Peter Anvin <hpa@zytor.com> 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.
Cheers, Kyle Moffett
| |