Hi,
is there any standard way to figure out the label of the basic block where a conditional branch merges?
If we have a branch statement without an else part this is straightforward:
br i1 %cmp, label %if.then, label %if.end, !dbg !24
We only need to check the operand names whether they contain "end". However things are getting
more complex when there is an else branch and in particular when we have nested branches.
Take for example the following C code:
if (x == NULL || y == NULL) {
do {
// ...
} while (0);
}
This yields the following IR:
br i1 %cmp, label %if.then, label %lor.lhs.false, !dbg !31
lor.lhs.false: ; preds = %entry
%1 = load %struct.s1*, %struct.s1** %s1.addr, align 8, !dbg !32
%cmp1 = icmp eq %struct.s1* %1, null, !dbg !33
br i1 %cmp1, label %if.then, label %if.end, !dbg !34
if.then: ; preds = %lor.lhs.false, %entry
br label %do.body, !dbg !35, !llvm.loop !37
do.body: ; preds = %if.then
br label %do.end, !dbg !39
do.end: ; preds = %do.body
br label %if.end, !dbg !41
if.end: ; preds = %do.end, %lor.lhs.false
call void @llvm.dbg.declare(metadata i32* %a, metadata !42, metadata !24), !dbg !43
The question now is how to obtain "if.end" given br i1 %cmp, label %if.then, label %lor.lhs.false, !dbg !31.
My current algorithm basically works for a lot of cases but fails here. What I am doing is to push the branch instruction to a stack and then always use the left most label of the next terminator instruction to move forward. Whenever I encounter a basic block that contains an "end" in its label I pop from the stack, whenever there is a conditional branch I push. If the stack is empty I have reached the correct basic block. For the example given it would however yield "do.end" which is not correct.
I am wondering whether there is any known graph algorithm I could use to solve the problem or even better something that is already implemented within LLVM?
--Sebastian