[LLVMdev] LLVM constant propagation optimization question

[Seb]

Hi all,

I'm writting following LLVM assembly:

; ModuleID = 'structaccess.ll'

%struct._anon0 = type <{ i32, i32, i32 }>

@s = common global %struct._anon0 zeroinitializer

define arm_aapcscc void @foo() nounwind {
L.entry:
  store i32 5, i32* getelementptr inbounds (%struct._anon0* @s, i32 0, i32
0)
  store i32 10, i32* getelementptr inbounds (%struct._anon0* @s, i32 0, i32
1)
  %0 = load i32* getelementptr inbounds (%struct._anon0* @s, i32 0, i32 0)
  %1 = load i32* getelementptr inbounds (%struct._anon0* @s, i32 0, i32 1)
  %2 = add i32 %0, %1
  store i32 %2, i32* getelementptr inbounds (%struct._anon0* @s, i32 0, i32
2)
  ret void
}

Using 'opt' utility as follows :

opt -O2 structaccess.ll -S -o structaccess-opt.ll

I've got following code for structaccess-opt.ll file:

; ModuleID = 'structaccess.ll'

%struct._anon0 = type <{ i32, i32, i32 }>

@s = common global %struct._anon0 zeroinitializer

define arm_aapcscc void @foo() nounwind {
L.entry:
  store i32 5, i32* getelementptr inbounds (%struct._anon0* @s, i32 0, i32
0)
  store i32 10, i32* getelementptr inbounds (%struct._anon0* @s, i32 0, i32
1)
  %0 = load i32* getelementptr inbounds (%struct._anon0* @s, i32 0, i32 0)
  %1 = add i32 %0, 10
  store i32 %1, i32* getelementptr inbounds (%struct._anon0* @s, i32 0, i32
2)
  ret void
}

I would have expected constant 5 to be propagated by 'opt' to its use and
thus LLVM assembly after opt to be :

; ModuleID = 'structaccess.ll'

%struct._anon0 = type <{ i32, i32, i32 }>

@s = common global %struct._anon0 zeroinitializer

define arm_aapcscc void @foo() nounwind {
L.entry:
  store i32 5, i32* getelementptr inbounds (%struct._anon0* @s, i32 0, i32
0)
  store i32 10, i32* getelementptr inbounds (%struct._anon0* @s, i32 0, i32
1)
  store i32 15, i32* getelementptr inbounds (%struct._anon0* @s, i32 0, i32
2)
  ret void
}

Can someone explain me why this is not the case ?

Note that C equivalent would be something like:

struct {
    int x, y, z;
} s;

void foo()
{
   s.x = 5 ;
   s.y = 10 ;
   s.z = s.x + s.y ;
}
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