Understanding bitcast in LLVM IR

user_9278434 0
Viewed 3063

I am trying to understand the LLVM IR generated from a C++ program

int add(int *x);
int func()
{
        int T;
        T=25;
        return add(&T);    
}

The generated IR is:

define i32 @_Z4funcv() local_unnamed_addr #0 {
entry:
  %T = alloca i32, align 4
  %0 = bitcast i32* %T to i8*
  call void @llvm.lifetime.start.p0i8(i64 4, i8* nonnull %0) #3
  store i32 25, i32* %T, align 4, !tbaa !2
  %call = call i32 @_Z3addPi(i32* nonnull %T)
  call void @llvm.lifetime.end.p0i8(i64 4, i8* nonnull %0) #3
  ret i32 %call
}

I do not understand this line %0 = bitcast i32* %T to i8*. What is the purpose of converting %T from i32 to i8?

llvm-ir llvm
1 Answers

Assuming you know about intrinsics

llvm.lifetime.start / llvm.lifetime.end

and its uses as memory uses marker for MemoryDependenceAnalysis.

About the choice of pointer(address of variable) as i8 was made to make it more generic as byte addressable memory region with first arguments as number of bytes same as we use in malloc.

so to generate the intrinsic call we need a memory byte address and the number of bytes that is sizeof(T). that is why we need to convert i32* to i8*.

by the way the signature if lifetime intrinsics used in your examples are

declare void @llvm.lifetime.start(i64 , i8* nocapture )

declare void @llvm.lifetime.end(i64 , i8* nocapture )

go through Lang ref for more info.

user_58717480
Related