[SelectionDAG] rename/move isKnownToBeAPowerOfTwo() from TargetLowering (NFC)
There are at least 2 places (DAGCombiner, X86ISelLowering) where this could be used instead
of ad-hoc and watered down code that is trying to match a power-of-2 pattern.
Differential Revision: http://reviews.llvm.org/D20439
llvm-svn: 270073
diff --git a/llvm/lib/CodeGen/SelectionDAG/TargetLowering.cpp b/llvm/lib/CodeGen/SelectionDAG/TargetLowering.cpp
index 7010347..923d822 100644
--- a/llvm/lib/CodeGen/SelectionDAG/TargetLowering.cpp
+++ b/llvm/lib/CodeGen/SelectionDAG/TargetLowering.cpp
@@ -1201,37 +1201,6 @@
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 @@
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