Teach LLVM to unravel the "swap idiom". This implements:

Regression/Transforms/InstCombine/xor.ll:test20

llvm-svn: 11492
This commit is contained in:
Chris Lattner 2004-02-16 03:54:20 +00:00
parent 98c26a7842
commit b36d908f7b
1 changed files with 13 additions and 2 deletions

View File

@ -1086,7 +1086,7 @@ Instruction *InstCombiner::visitXor(BinaryOperator &I) {
ConstantIntegral::getAllOnesValue(I.getType()));
if (Instruction *Op1I = dyn_cast<Instruction>(Op1))
if (Op1I->getOpcode() == Instruction::Or)
if (Op1I->getOpcode() == Instruction::Or) {
if (Op1I->getOperand(0) == Op0) { // B^(B|A) == (A|B)^B
cast<BinaryOperator>(Op1I)->swapOperands();
I.swapOperands();
@ -1094,7 +1094,13 @@ Instruction *InstCombiner::visitXor(BinaryOperator &I) {
} else if (Op1I->getOperand(1) == Op0) { // B^(A|B) == (A|B)^B
I.swapOperands();
std::swap(Op0, Op1);
}
}
} else if (Op1I->getOpcode() == Instruction::Xor) {
if (Op0 == Op1I->getOperand(0)) // A^(A^B) == B
return ReplaceInstUsesWith(I, Op1I->getOperand(1));
else if (Op0 == Op1I->getOperand(1)) // A^(B^A) == B
return ReplaceInstUsesWith(I, Op1I->getOperand(0));
}
if (Instruction *Op0I = dyn_cast<Instruction>(Op0))
if (Op0I->getOpcode() == Instruction::Or && Op0I->hasOneUse()) {
@ -1106,6 +1112,11 @@ Instruction *InstCombiner::visitXor(BinaryOperator &I) {
return BinaryOperator::create(Instruction::And, Op0I->getOperand(0),
NotB);
}
} else if (Op0I->getOpcode() == Instruction::Xor) {
if (Op1 == Op0I->getOperand(0)) // (A^B)^A == B
return ReplaceInstUsesWith(I, Op0I->getOperand(1));
else if (Op1 == Op0I->getOperand(1)) // (B^A)^A == B
return ReplaceInstUsesWith(I, Op0I->getOperand(0));
}
// (A & C1)^(B & C2) -> (A & C1)|(B & C2) iff C1^C2 == 0