((A == C && B == D) || (A == D && B == C)))
return BinaryOperator::CreateXor(A, B);
- if (Op0->hasOneUse() &&
- match(Op0, m_Xor(m_Value(A), m_Value(B)))) {
- if (A == Op1) { // (A^B)&A -> A&(A^B)
- I.swapOperands(); // Simplify below
- std::swap(Op0, Op1);
- } else if (B == Op1) { // (A^B)&B -> B&(B^A)
- cast<BinaryOperator>(Op0)->swapOperands();
- I.swapOperands(); // Simplify below
- std::swap(Op0, Op1);
+ // A&(A^B) => A & ~B
+ {
+ Value *tmpOp0 = Op0;
+ Value *tmpOp1 = Op1;
+ if (Op0->hasOneUse() &&
+ match(Op0, m_Xor(m_Value(A), m_Value(B)))) {
+ if (A == Op1 || B == Op1 ) {
+ tmpOp1 = Op0;
+ tmpOp0 = Op1;
+ // Simplify below
+ }
}
- }
- if (Op1->hasOneUse() &&
- match(Op1, m_Xor(m_Value(A), m_Value(B)))) {
- if (B == Op0) { // B&(A^B) -> B&(B^A)
- cast<BinaryOperator>(Op1)->swapOperands();
- std::swap(A, B);
+ if (tmpOp1->hasOneUse() &&
+ match(tmpOp1, m_Xor(m_Value(A), m_Value(B)))) {
+ if (B == tmpOp0) {
+ std::swap(A, B);
+ }
+ // Notice that the patten (A&(~B)) is actually (A&(-1^B)), so if
+ // A is originally -1 (or a vector of -1 and undefs), then we enter
+ // an endless loop. By checking that A is non-constant we ensure that
+ // we will never get to the loop.
+ if (A == tmpOp0 && !isa<Constant>(A)) // A&(A^B) -> A & ~B
+ return BinaryOperator::CreateAnd(A, Builder->CreateNot(B, "tmp"));
}
- // Notice that the patten (A&(~B)) is actually (A&(-1^B)), so if
- // A is originally -1 (or a vector of -1 and undefs), then we enter
- // an endless loop. By checking that A is non-constant we ensure that
- // we will never get to the loop.
- if (A == Op0 && !isa<Constant>(A)) // A&(A^B) -> A & ~B
- return BinaryOperator::CreateAnd(A, Builder->CreateNot(B, "tmp"));
}
// (A&((~A)|B)) -> A&B
; CHECK: @test4
; CHECK-NEXT: ret i1 false
}
+
+; Make sure we don't go into an infinite loop with this test
+define <4 x i32> @test5(<4 x i32> %A) {
+ %1 = xor <4 x i32> %A, <i32 1, i32 2, i32 3, i32 4>
+ %2 = and <4 x i32> <i32 1, i32 2, i32 3, i32 4>, %1
+ ret <4 x i32> %2
+}