[PATCH] D26151: RegCall - Handling long double arguments

[Oren Ben Simhon via llvm-commits]

oren_ben_simhon added inline comments.


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Comment at: lib/Target/X86/X86FloatingPoint.cpp:973
   assert(STReturns == 0 || (isMask_32(STReturns) && N <= 2));
+  StackTop = 0;
 
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ahatanak wrote:
> oren_ben_simhon wrote:
> > ahatanak wrote:
> > > I'm not sure what this comment means.
> > > 
> > > It seems like you are clearing StackTop because the first FP register can be used to pass an argument when the calling convention is regcall. Is that correct?
> > The problem occurs when FP0 was passed as argument and we try to return a parameter in FP0 as well.
> > In that case the StackTop points to 1 and we get to the following loop and we call pushreg(0) when FP0 is already assigned and StackTop is 1. 
> > 
> > So what actually happened before pushreg:    
> >            Stack[] = { 0, ...}
> >            RegMap[] = { 0, ...}
> > While after pushreg[0] (No stacktop reset):      
> >            Stack[] = { 0, 1..}
> >            RegMap[] = { 1...}
> > 
> > The issue causes debug assertion in line 169 to fail because then regmap[stack[0]] != 0.
> > 
> > I changed the way I handle the issue.
> > Instead of reseting the StackTop to zero, I do not push registers if they were already pushed into the stack.
> Wouldn't this push an extra register when a function taking a long double argument and returning complex long double were called (in which case N==2 because it's returned in st0 and st1)?
AFAIK, Long doubles are saved inside a single ST register (size of 80 bit). 
So in the case of one FP argument and one FP retruned value, I expect N to be 1. 
Meaning, The value will be returned in ST0.


Repository:
  rL LLVM

https://reviews.llvm.org/D26151