Factor FlattenCFG out from SimplifyCFG

Patch by: Mei Ye

git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@187764 91177308-0d34-0410-b5e6-96231b3b80d8
This commit is contained in:
Tom Stellard
2013-08-06 02:43:45 +00:00
parent 99c22aaeed
commit 01d7203ef8
21 changed files with 832 additions and 703 deletions
+101
View File
@@ -665,3 +665,104 @@ TerminatorInst *llvm::SplitBlockAndInsertIfThen(Instruction *Cmp,
ReplaceInstWithInst(HeadOldTerm, HeadNewTerm);
return CheckTerm;
}
/// GetIfCondition - Given a basic block (BB) with two predecessors,
/// check to see if the merge at this block is due
/// to an "if condition". If so, return the boolean condition that determines
/// which entry into BB will be taken. Also, return by references the block
/// that will be entered from if the condition is true, and the block that will
/// be entered if the condition is false.
///
/// This does no checking to see if the true/false blocks have large or unsavory
/// instructions in them.
Value *llvm::GetIfCondition(BasicBlock *BB, BasicBlock *&IfTrue,
BasicBlock *&IfFalse) {
PHINode *SomePHI = dyn_cast<PHINode>(BB->begin());
BasicBlock *Pred1 = NULL;
BasicBlock *Pred2 = NULL;
if (SomePHI) {
if (SomePHI->getNumIncomingValues() != 2)
return NULL;
Pred1 = SomePHI->getIncomingBlock(0);
Pred2 = SomePHI->getIncomingBlock(1);
} else {
pred_iterator PI = pred_begin(BB), PE = pred_end(BB);
if (PI == PE) // No predecessor
return NULL;
Pred1 = *PI++;
if (PI == PE) // Only one predecessor
return NULL;
Pred2 = *PI++;
if (PI != PE) // More than two predecessors
return NULL;
}
// We can only handle branches. Other control flow will be lowered to
// branches if possible anyway.
BranchInst *Pred1Br = dyn_cast<BranchInst>(Pred1->getTerminator());
BranchInst *Pred2Br = dyn_cast<BranchInst>(Pred2->getTerminator());
if (Pred1Br == 0 || Pred2Br == 0)
return 0;
// Eliminate code duplication by ensuring that Pred1Br is conditional if
// either are.
if (Pred2Br->isConditional()) {
// If both branches are conditional, we don't have an "if statement". In
// reality, we could transform this case, but since the condition will be
// required anyway, we stand no chance of eliminating it, so the xform is
// probably not profitable.
if (Pred1Br->isConditional())
return 0;
std::swap(Pred1, Pred2);
std::swap(Pred1Br, Pred2Br);
}
if (Pred1Br->isConditional()) {
// The only thing we have to watch out for here is to make sure that Pred2
// doesn't have incoming edges from other blocks. If it does, the condition
// doesn't dominate BB.
if (Pred2->getSinglePredecessor() == 0)
return 0;
// If we found a conditional branch predecessor, make sure that it branches
// to BB and Pred2Br. If it doesn't, this isn't an "if statement".
if (Pred1Br->getSuccessor(0) == BB &&
Pred1Br->getSuccessor(1) == Pred2) {
IfTrue = Pred1;
IfFalse = Pred2;
} else if (Pred1Br->getSuccessor(0) == Pred2 &&
Pred1Br->getSuccessor(1) == BB) {
IfTrue = Pred2;
IfFalse = Pred1;
} else {
// We know that one arm of the conditional goes to BB, so the other must
// go somewhere unrelated, and this must not be an "if statement".
return 0;
}
return Pred1Br->getCondition();
}
// Ok, if we got here, both predecessors end with an unconditional branch to
// BB. Don't panic! If both blocks only have a single (identical)
// predecessor, and THAT is a conditional branch, then we're all ok!
BasicBlock *CommonPred = Pred1->getSinglePredecessor();
if (CommonPred == 0 || CommonPred != Pred2->getSinglePredecessor())
return 0;
// Otherwise, if this is a conditional branch, then we can use it!
BranchInst *BI = dyn_cast<BranchInst>(CommonPred->getTerminator());
if (BI == 0) return 0;
assert(BI->isConditional() && "Two successors but not conditional?");
if (BI->getSuccessor(0) == Pred1) {
IfTrue = Pred1;
IfFalse = Pred2;
} else {
IfTrue = Pred2;
IfFalse = Pred1;
}
return BI->getCondition();
}