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[InstCombine] Move portion of SimplifyDemandedUseBits that deals with instructions with multiple uses out to a separate method. NFCI

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git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@300082 91177308-0d34-0410-b5e6-96231b3b80d8
This commit is contained in:
Craig Topper 2017-04-12 18:05:21 +00:00
parent 73ba5e9415
commit c7bad98e0e
2 changed files with 103 additions and 76 deletions
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7
lib/Transforms/InstCombine/InstCombineInternal.h
View File

@ -542,6 +542,13 @@ private:
bool SimplifyDemandedBits(Instruction *I, unsigned Op,
const APInt &DemandedMask, APInt &KnownZero,
APInt &KnownOne, unsigned Depth = 0);
/// Helper routine of SimplifyDemandedUseBits. It computes KnownZero/KnownOne
/// bits. It also tries to handle simplifications that can be done based on
/// DemandedMask, but without modifying the Instruction.
Value *SimplifyMultipleUseDemandedBits(Instruction *I,
const APInt &DemandedMask,
APInt &KnownZero, APInt &KnownOne,
unsigned Depth, Instruction *CxtI);
/// Helper routine of SimplifyDemandedUseBits. It tries to simplify demanded
/// bit for "r1 = shr x, c1; r2 = shl r1, c2" instruction sequence.
Value *SimplifyShrShlDemandedBits(Instruction *Lsr, Instruction *Sftl,

172
lib/Transforms/InstCombine/InstCombineSimplifyDemanded.cpp
View File

@ -142,9 +142,6 @@ Value *InstCombiner::SimplifyDemandedUseBits(Value *V, APInt DemandedMask,
if (Depth == 6) // Limit search depth.
return nullptr;
APInt LHSKnownZero(BitWidth, 0), LHSKnownOne(BitWidth, 0);
APInt RHSKnownZero(BitWidth, 0), RHSKnownOne(BitWidth, 0);
Instruction *I = dyn_cast<Instruction>(V);
if (!I) {
computeKnownBits(V, KnownZero, KnownOne, Depth, CxtI);
@ -155,81 +152,13 @@ Value *InstCombiner::SimplifyDemandedUseBits(Value *V, APInt DemandedMask,
// we can't do any simplifications of the operands, because DemandedMask
// only reflects the bits demanded by *one* of the users.
if (Depth != 0 && !I->hasOneUse()) {
// Despite the fact that we can't simplify this instruction in all User's
// context, we can at least compute the knownzero/knownone bits, and we can
// do simplifications that apply to *just* the one user if we know that
// this instruction has a simpler value in that context.
if (I->getOpcode() == Instruction::And) {
// If either the LHS or the RHS are Zero, the result is zero.
computeKnownBits(I->getOperand(1), RHSKnownZero, RHSKnownOne, Depth + 1,
CxtI);
computeKnownBits(I->getOperand(0), LHSKnownZero, LHSKnownOne, Depth + 1,
CxtI);
// If all of the demanded bits are known 1 on one side, return the other.
// These bits cannot contribute to the result of the 'and' in this
// context.
if ((DemandedMask & ~LHSKnownZero & RHSKnownOne) ==
(DemandedMask & ~LHSKnownZero))
return I->getOperand(0);
if ((DemandedMask & ~RHSKnownZero & LHSKnownOne) ==
(DemandedMask & ~RHSKnownZero))
return I->getOperand(1);
// If all of the demanded bits in the inputs are known zeros, return zero.
if ((DemandedMask & (RHSKnownZero|LHSKnownZero)) == DemandedMask)
return Constant::getNullValue(VTy);
} else if (I->getOpcode() == Instruction::Or) {
// We can simplify (X|Y) -> X or Y in the user's context if we know that
// only bits from X or Y are demanded.
// If either the LHS or the RHS are One, the result is One.
computeKnownBits(I->getOperand(1), RHSKnownZero, RHSKnownOne, Depth + 1,
CxtI);
computeKnownBits(I->getOperand(0), LHSKnownZero, LHSKnownOne, Depth + 1,
CxtI);
// If all of the demanded bits are known zero on one side, return the
// other. These bits cannot contribute to the result of the 'or' in this
// context.
if ((DemandedMask & ~LHSKnownOne & RHSKnownZero) ==
(DemandedMask & ~LHSKnownOne))
return I->getOperand(0);
if ((DemandedMask & ~RHSKnownOne & LHSKnownZero) ==
(DemandedMask & ~RHSKnownOne))
return I->getOperand(1);
// If all of the potentially set bits on one side are known to be set on
// the other side, just use the 'other' side.
if ((DemandedMask & (~RHSKnownZero) & LHSKnownOne) ==
(DemandedMask & (~RHSKnownZero)))
return I->getOperand(0);
if ((DemandedMask & (~LHSKnownZero) & RHSKnownOne) ==
(DemandedMask & (~LHSKnownZero)))
return I->getOperand(1);
} else if (I->getOpcode() == Instruction::Xor) {
// We can simplify (X^Y) -> X or Y in the user's context if we know that
// only bits from X or Y are demanded.
computeKnownBits(I->getOperand(1), RHSKnownZero, RHSKnownOne, Depth + 1,
CxtI);
computeKnownBits(I->getOperand(0), LHSKnownZero, LHSKnownOne, Depth + 1,
CxtI);
// If all of the demanded bits are known zero on one side, return the
// other.
if ((DemandedMask & RHSKnownZero) == DemandedMask)
return I->getOperand(0);
if ((DemandedMask & LHSKnownZero) == DemandedMask)
return I->getOperand(1);
}
// Compute the KnownZero/KnownOne bits to simplify things downstream.
computeKnownBits(I, KnownZero, KnownOne, Depth, CxtI);
return nullptr;
return SimplifyMultipleUseDemandedBits(I, DemandedMask, KnownZero, KnownOne,
Depth, CxtI);
}
APInt LHSKnownZero(BitWidth, 0), LHSKnownOne(BitWidth, 0);
APInt RHSKnownZero(BitWidth, 0), RHSKnownOne(BitWidth, 0);
// If this is the root being simplified, allow it to have multiple uses,
// just set the DemandedMask to all bits so that we can try to simplify the
// operands. This allows visitTruncInst (for example) to simplify the
@ -818,6 +747,97 @@ Value *InstCombiner::SimplifyDemandedUseBits(Value *V, APInt DemandedMask,
return nullptr;
}
/// Helper routine of SimplifyDemandedUseBits. It computes KnownZero/KnownOne
/// bits. It also tries to handle simplifications that can be done based on
/// DemandedMask, but without modifying the Instruction.
Value *InstCombiner::SimplifyMultipleUseDemandedBits(Instruction *I,
const APInt &DemandedMask,
APInt &KnownZero,
APInt &KnownOne,
unsigned Depth,
Instruction *CxtI) {
unsigned BitWidth = DemandedMask.getBitWidth();
Type *ITy = I->getType();
APInt LHSKnownZero(BitWidth, 0), LHSKnownOne(BitWidth, 0);
APInt RHSKnownZero(BitWidth, 0), RHSKnownOne(BitWidth, 0);
// Despite the fact that we can't simplify this instruction in all User's
// context, we can at least compute the knownzero/knownone bits, and we can
// do simplifications that apply to *just* the one user if we know that
// this instruction has a simpler value in that context.
if (I->getOpcode() == Instruction::And) {
// If either the LHS or the RHS are Zero, the result is zero.
computeKnownBits(I->getOperand(1), RHSKnownZero, RHSKnownOne, Depth + 1,
CxtI);
computeKnownBits(I->getOperand(0), LHSKnownZero, LHSKnownOne, Depth + 1,
CxtI);
// If all of the demanded bits are known 1 on one side, return the other.
// These bits cannot contribute to the result of the 'and' in this
// context.
if ((DemandedMask & ~LHSKnownZero & RHSKnownOne) ==
(DemandedMask & ~LHSKnownZero))
return I->getOperand(0);
if ((DemandedMask & ~RHSKnownZero & LHSKnownOne) ==
(DemandedMask & ~RHSKnownZero))
return I->getOperand(1);
// If all of the demanded bits in the inputs are known zeros, return zero.
if ((DemandedMask & (RHSKnownZero|LHSKnownZero)) == DemandedMask)
return Constant::getNullValue(ITy);
} else if (I->getOpcode() == Instruction::Or) {
// We can simplify (X|Y) -> X or Y in the user's context if we know that
// only bits from X or Y are demanded.
// If either the LHS or the RHS are One, the result is One.
computeKnownBits(I->getOperand(1), RHSKnownZero, RHSKnownOne, Depth + 1,
CxtI);
computeKnownBits(I->getOperand(0), LHSKnownZero, LHSKnownOne, Depth + 1,
CxtI);
// If all of the demanded bits are known zero on one side, return the
// other. These bits cannot contribute to the result of the 'or' in this
// context.
if ((DemandedMask & ~LHSKnownOne & RHSKnownZero) ==
(DemandedMask & ~LHSKnownOne))
return I->getOperand(0);
if ((DemandedMask & ~RHSKnownOne & LHSKnownZero) ==
(DemandedMask & ~RHSKnownOne))
return I->getOperand(1);
// If all of the potentially set bits on one side are known to be set on
// the other side, just use the 'other' side.
if ((DemandedMask & (~RHSKnownZero) & LHSKnownOne) ==
(DemandedMask & (~RHSKnownZero)))
return I->getOperand(0);
if ((DemandedMask & (~LHSKnownZero) & RHSKnownOne) ==
(DemandedMask & (~LHSKnownZero)))
return I->getOperand(1);
} else if (I->getOpcode() == Instruction::Xor) {
// We can simplify (X^Y) -> X or Y in the user's context if we know that
// only bits from X or Y are demanded.
computeKnownBits(I->getOperand(1), RHSKnownZero, RHSKnownOne, Depth + 1,
CxtI);
computeKnownBits(I->getOperand(0), LHSKnownZero, LHSKnownOne, Depth + 1,
CxtI);
// If all of the demanded bits are known zero on one side, return the
// other.
if ((DemandedMask & RHSKnownZero) == DemandedMask)
return I->getOperand(0);
if ((DemandedMask & LHSKnownZero) == DemandedMask)
return I->getOperand(1);
}
// Compute the KnownZero/KnownOne bits to simplify things downstream.
computeKnownBits(I, KnownZero, KnownOne, Depth, CxtI);
return nullptr;
}
/// Helper routine of SimplifyDemandedUseBits. It tries to simplify
/// "E1 = (X lsr C1) << C2", where the C1 and C2 are constant, into
/// "E2 = X << (C2 - C1)" or "E2 = X >> (C1 - C2)", depending on the sign