Insert an LLVM instruction using insertAfter() - c++

Question

I'm new to LLVM, and I'm doing some experiments on it such as inserting an instruction.
My main.c is shown below:
int foo(int e, int a) {
int b = a + 1;
int c = b * 2;
b = e << 1;
int d = b / 4;
return c * d;
}
I use the command below to generate the LLVM bytecode
clang-12 -O0 -Xclang -disable-O0-optnone -emit-llvm -c main.c -o main.bc
opt-12 -S -mem2reg main.bc -o main.ll
The bytecode is
; Function Attrs: noinline nounwind uwtable
define dso_local i32 #foo(i32 %0, i32 %1) #0 {
%3 = add nsw i32 %1, 1
%4 = mul nsw i32 %3, 2
%5 = shl i32 %0, 1
%6 = sdiv i32 %5, 4
%7 = mul nsw i32 %4, %6
ret i32 %7
}
And I use the code to insert an instruction after the first instruction:
bool runOnBasicBlock(BasicBlock &B) {
// get the first and second instruction
Instruction &Inst1st = *B.begin();
Instruction *NewInst = BinaryOperator::Create(
Instruction::Add, Inst1st.getOperand(0), Inst1st.getOperand(0));
NewInst->insertAfter(&Inst1st);
...
}
After I run this pass, the bytecode is changed to
; Function Attrs: noinline nounwind uwtable
define dso_local i32 #foo(i32 %0, i32 %1) #0 {
%3 = add nsw i32 %1, 1
%4 = add i32 %1, %1
%5 = mul nsw i32 %4, 2
%6 = shl i32 %0, 1
%7 = sdiv i32 %6, 4
%8 = mul nsw i32 %5, %7
ret i32 %8
}
It seems that the inserted instruction is equal to b = a + a;, so the instruction %4 = mul nsw i32 %3, 2 is changed to %5 = mul nsw i32 %4, 2. I cannot understand the reason. Any help?

Answer 1

As I know, NewInst->insertAfter(&Inst1st); makes from the block
int b = a + 1;
int c = b * 2;
the following block
int b = a + 1, a + a;
int c = b * 2;
therefore b drops off the previous value %3 and gets the new value %4 and further mul uses that new value of b.