diff options
Diffstat (limited to 'llvm/lib/CodeGen/SelectionDAG/TargetLowering.cpp')
| -rw-r--r-- | llvm/lib/CodeGen/SelectionDAG/TargetLowering.cpp | 33 |
1 files changed, 1 insertions, 32 deletions
diff --git a/llvm/lib/CodeGen/SelectionDAG/TargetLowering.cpp b/llvm/lib/CodeGen/SelectionDAG/TargetLowering.cpp index 7010347f12a..923d82264de 100644 --- a/llvm/lib/CodeGen/SelectionDAG/TargetLowering.cpp +++ b/llvm/lib/CodeGen/SelectionDAG/TargetLowering.cpp @@ -1201,37 +1201,6 @@ unsigned TargetLowering::ComputeNumSignBitsForTargetNode(SDValue Op, return 1; } -/// Test if the given value is known to have exactly one bit set. This differs -/// from computeKnownBits in that it doesn't need to determine which bit is set. -static bool valueHasExactlyOneBitSet(SDValue Val, const SelectionDAG &DAG) { - // A left-shift of a constant one will have exactly one bit set because - // shifting the bit off the end is undefined. - if (Val.getOpcode() == ISD::SHL) { - auto *C = dyn_cast<ConstantSDNode>(Val.getOperand(0)); - if (C && C->getAPIntValue() == 1) - return true; - } - - // Similarly, a logical right-shift of a constant sign-bit will have exactly - // one bit set. - if (Val.getOpcode() == ISD::SRL) { - auto *C = dyn_cast<ConstantSDNode>(Val.getOperand(0)); - if (C && C->getAPIntValue().isSignBit()) - return true; - } - - // More could be done here, though the above checks are enough - // to handle some common cases. - - // Fall back to computeKnownBits to catch other known cases. - EVT OpVT = Val.getValueType(); - unsigned BitWidth = OpVT.getScalarType().getSizeInBits(); - APInt KnownZero, KnownOne; - DAG.computeKnownBits(Val, KnownZero, KnownOne); - return (KnownZero.countPopulation() == BitWidth - 1) && - (KnownOne.countPopulation() == 1); -} - bool TargetLowering::isConstTrueVal(const SDNode *N) const { if (!N) return false; @@ -1334,7 +1303,7 @@ SDValue TargetLowering::simplifySetCCWithAnd(EVT VT, SDValue N0, SDValue N1, SelectionDAG &DAG = DCI.DAG; SDValue Zero = DAG.getConstant(0, DL, OpVT); - if (valueHasExactlyOneBitSet(Y, DAG)) { + if (DAG.isKnownToBeAPowerOfTwo(Y)) { // Simplify X & Y == Y to X & Y != 0 if Y has exactly one bit set. // Note that where Y is variable and is known to have at most one bit set // (for example, if it is Z & 1) we cannot do this; the expressions are not |
