[llvm-commits] Fwd: [llvm] r48223 - in /llvm/trunk/lib: CodeGen/LowerSubregs.cpp CodeGen/SelectionDAG/ScheduleDAG.cpp Target/Target.td Target/X86/X86ISelDAGToDAG.cpp Target/X86/X86InstrInfo.h Target/X86/X86InstrInfo.td Target/X86/X86RegisterInfo.h Target/X86/X86RegisterInfo.td

Evan Cheng evan.cheng at apple.com
Fri Mar 14 10:44:31 PDT 2008

On Mar 14, 2008, at 10:32 AM, Chris Lattner wrote:

> On Mar 14, 2008, at 10:02 AM, Evan Cheng wrote:
>> Think about insert_subreg in the literal sense. It's modifying part
>> of a register. That is, the input and output super register must be
>> the same. Insert into zero would be modeled like this.
>> vr1024 = mov 0
>> vr1024 = insert_subreg vr1024, vr1025, x86_subreg32bit
>> Ideally we model insert into undef the same way:
>> vr1024 = insert_subreg undef, vr1025, x86_subreg32bit
> This doesn't work because it breaks the invariant that the first
> operand is a register class.

Huh? You are talking about insert_subreg or in general? Can't be the  
case in general.

>> However, it's unclear what issues we would run into if insert_subreg
>> is a two address instruction.
> Right, that is one effect.
>> That's why I suggest making into into a separate insert_subreg_undef:
>> vr1024 = insert_subreg_undef vr1025, x86_subreg32bit.
> Why not:
> vr1027 = implicit_def
> vr1024 = insert_subreg vr1027, vr1025, x86_subreg32bit
> We already have implicit_def, so this reduce the number of concepts
> the code generator has to know about.

This will prevent coalescing. insert_subreg should be a two address  

vr1027 = implicit_def
vr1027 = insert_subreg vr1027, vr1025, x86_subreg32bit

vr1027 and vr1025 live intervals conflict.

Let's make sure we agree or disagree on whether insert_subreg is a two  
address instruction. To me, it's natural and makes practical sense.  
After all, insert_subreg means read / modify / write, it should not  
create a *new* register.


> -Chris
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