forked from OSchip/llvm-project
parent
6b15606051
commit
2ca72f4971
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@ -569,6 +569,37 @@ bool TargetLowering::SimplifyDemandedBits(SDValue Op,
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}
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return false; // Don't fall through, will infinitely loop.
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case ISD::AND:
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// If the RHS is a constant, check to see if the LHS would be zero without
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// using the bits from the RHS. Below, we use knowledge about the RHS to
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// simplify the LHS, here we're using information from the LHS to simplify
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// the RHS.
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if (ConstantSDNode *RHSC = dyn_cast<ConstantSDNode>(Op.getOperand(1))) {
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SDValue Op0 = Op.getOperand(0);
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APInt LHSZero, LHSOne;
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// Do not increment Depth here; that can cause an infinite loop.
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TLO.DAG.computeKnownBits(Op0, LHSZero, LHSOne, Depth);
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// If the LHS already has zeros where RHSC does, this and is dead.
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if ((LHSZero & NewMask) == (~RHSC->getAPIntValue() & NewMask))
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return TLO.CombineTo(Op, Op0);
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// If any of the set bits in the RHS are known zero on the LHS, shrink
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// the constant.
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if (TLO.ShrinkDemandedConstant(Op, ~LHSZero & NewMask))
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return true;
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// Bitwise-not (xor X, -1) is a special case: we don't usually shrink its
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// constant, but if this 'and' is only clearing bits that were just set by
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// the xor, then this 'and' can be eliminated by shrinking the mask of
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// the xor. For example, for a 32-bit X:
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// and (xor (srl X, 31), -1), 1 --> xor (srl X, 31), 1
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if (isBitwiseNot(Op0) && Op0.hasOneUse() &&
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LHSOne == ~RHSC->getAPIntValue()) {
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SDValue Xor = TLO.DAG.getNode(ISD::XOR, dl, Op.getValueType(),
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Op0.getOperand(0), Op.getOperand(1));
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return TLO.CombineTo(Op, Xor);
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}
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}
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if (SimplifyDemandedBits(Op.getOperand(1), NewMask, KnownZero,
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KnownOne, TLO, Depth+1))
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return true;
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@ -594,21 +625,6 @@ bool TargetLowering::SimplifyDemandedBits(SDValue Op,
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if (TLO.ShrinkDemandedOp(Op, BitWidth, NewMask, dl))
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return true;
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if (ConstantSDNode *RHSC = dyn_cast<ConstantSDNode>(Op.getOperand(1))) {
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SDValue Op0 = Op.getOperand(0);
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// Bitwise-not (xor X, -1) is a special case: we don't usually shrink its
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// constant, but if this 'and' is only clearing bits that were just set by
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// the xor, then this 'and' can be eliminated by shrinking the mask of
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// the xor. For example, for a 32-bit X:
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// and (xor (srl X, 31), -1), 1 --> xor (srl X, 31), 1
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if (isBitwiseNot(Op0) && Op0.hasOneUse() &&
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KnownOne2 == ~RHSC->getAPIntValue()) {
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SDValue Xor = TLO.DAG.getNode(ISD::XOR, dl, Op.getValueType(),
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Op0.getOperand(0), Op.getOperand(1));
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return TLO.CombineTo(Op, Xor);
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}
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}
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// Output known-1 bits are only known if set in both the LHS & RHS.
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KnownOne &= KnownOne2;
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// Output known-0 are known to be clear if zero in either the LHS | RHS.
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