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Teach LLVM to unravel the "swap idiom". This implements:
Regression/Transforms/InstCombine/xor.ll:test20 git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@11492 91177308-0d34-0410-b5e6-96231b3b80d8
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@ -1086,7 +1086,7 @@ Instruction *InstCombiner::visitXor(BinaryOperator &I) {
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ConstantIntegral::getAllOnesValue(I.getType()));
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if (Instruction *Op1I = dyn_cast<Instruction>(Op1))
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if (Op1I->getOpcode() == Instruction::Or)
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if (Op1I->getOpcode() == Instruction::Or) {
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if (Op1I->getOperand(0) == Op0) { // B^(B|A) == (A|B)^B
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cast<BinaryOperator>(Op1I)->swapOperands();
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I.swapOperands();
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@ -1094,7 +1094,13 @@ Instruction *InstCombiner::visitXor(BinaryOperator &I) {
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} else if (Op1I->getOperand(1) == Op0) { // B^(A|B) == (A|B)^B
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I.swapOperands();
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std::swap(Op0, Op1);
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}
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}
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} else if (Op1I->getOpcode() == Instruction::Xor) {
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if (Op0 == Op1I->getOperand(0)) // A^(A^B) == B
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return ReplaceInstUsesWith(I, Op1I->getOperand(1));
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else if (Op0 == Op1I->getOperand(1)) // A^(B^A) == B
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return ReplaceInstUsesWith(I, Op1I->getOperand(0));
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}
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if (Instruction *Op0I = dyn_cast<Instruction>(Op0))
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if (Op0I->getOpcode() == Instruction::Or && Op0I->hasOneUse()) {
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@ -1106,6 +1112,11 @@ Instruction *InstCombiner::visitXor(BinaryOperator &I) {
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return BinaryOperator::create(Instruction::And, Op0I->getOperand(0),
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NotB);
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}
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} else if (Op0I->getOpcode() == Instruction::Xor) {
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if (Op1 == Op0I->getOperand(0)) // (A^B)^A == B
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return ReplaceInstUsesWith(I, Op0I->getOperand(1));
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else if (Op1 == Op0I->getOperand(1)) // (B^A)^A == B
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return ReplaceInstUsesWith(I, Op0I->getOperand(0));
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}
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// (A & C1)^(B & C2) -> (A & C1)|(B & C2) iff C1^C2 == 0
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