llvm dev - May 2008 - [LLVMdev] Troubling promotion of return value to Integer ...

Hi Evan,
> If you are using a non-C ABI, then you may not need to do the  
> promotion at all. However, if you do need to do the promotion, you  
> need to know the type of "int".
how does this work for parameters.  Suppose I have an i1 parameter.
The AMD64 ABI says it is passed in one of %rsi, %rdx, %rcx, %r8 and
%r9 registers.  In the case of _Bool (i1) it says that the upper 63
bits will be zero.  How does this work?  I thought i1 would be legalized
to i8 (which is legal, right?).  Due to being marked zext, the upper
7 bits of the i8 will be zero.  But where does it get extended to
an i64 with upper 63 bits zero?

Thanks,

Duncan.

On May 15, 2008, at 12:14 PM, Duncan Sands wrote:
> Hi Evan,
>
>> If you are using a non-C ABI, then you may not need to do the
>> promotion at all. However, if you do need to do the promotion, you
>> need to know the type of "int".
>
> how does this work for parameters.  Suppose I have an i1 parameter.
> The AMD64 ABI says it is passed in one of %rsi, %rdx, %rcx, %r8 and
> %r9 registers.  In the case of _Bool (i1) it says that the upper 63
> bits will be zero.  How does this work?  I thought i1 would be  
> legalized
> to i8 (which is legal, right?).  Due to being marked zext, the upper
> 7 bits of the i8 will be zero.  But where does it get extended to
> an i64 with upper 63 bits zero?
X86ISelLowering will promote the i8 parameters to i32.

Evan
>
>
> Thanks,
>
> Duncan.

> >> If you are using a non-C ABI, then you may not need to do the
> >> promotion at all. However, if you do need to do the promotion, you
> >> need to know the type of "int".
> >
> > how does this work for parameters.  Suppose I have an i1 parameter.
> > The AMD64 ABI says it is passed in one of %rsi, %rdx, %rcx, %r8 and
> > %r9 registers.  In the case of _Bool (i1) it says that the upper 63
> > bits will be zero.  How does this work?  I thought i1 would be  
> > legalized
> > to i8 (which is legal, right?).  Due to being marked zext, the upper
> > 7 bits of the i8 will be zero.  But where does it get extended to
> > an i64 with upper 63 bits zero?
> 
> X86ISelLowering will promote the i8 parameters to i32.
OK, so why doesn't it work the same for return values?

Ciao,

Duncan.