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Refactored and updated SimplifyUsingDistributiveLaws() to

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* Find factorization opportunities using identity values.
 * Find factorization opportunities by treating shl(X, C) as mul (X, shl(C))
 * Keep NSW flag while simplifying instruction using factorization.

This fixes PR19263.

Differential Revision: http://reviews.llvm.org/D3799



git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@211261 91177308-0d34-0410-b5e6-96231b3b80d8
This commit is contained in:
Dinesh Dwivedi
2014-06-19 08:29:18 +00:00
parent c28beb254d
commit 8317509052
3 changed files with 210 additions and 105 deletions
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195
lib/Transforms/InstCombine/InstructionCombining.cpp
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@ -395,6 +395,127 @@ static bool RightDistributesOverLeft(Instruction::BinaryOps LOp,
return false;
}
/// This function returns identity value for given opcode, which can be used to
/// factor patterns like (X * 2) + X ==> (X * 2) + (X * 1) ==> X * (2 + 1).
static Value *getIdentityValue(Instruction::BinaryOps OpCode, Value *V) {
if (isa<Constant>(V))
return nullptr;
if (OpCode == Instruction::Mul)
return ConstantInt::get(V->getType(), 1);
// TODO: We can handle other cases e.g. Instruction::And, Instruction::Or etc.
return nullptr;
}
/// This function factors binary ops which can be combined using distributive
/// laws. This also factor SHL as MUL e.g. SHL(X, 2) ==> MUL(X, 4).
Instruction::BinaryOps getBinOpsForFactorization(BinaryOperator *Op,
Value *&LHS, Value *&RHS) {
if (!Op)
return Instruction::BinaryOpsEnd;
if (Op->getOpcode() == Instruction::Shl) {
if (Constant *CST = dyn_cast<Constant>(Op->getOperand(1))) {
// The multiplier is really 1 << CST.
RHS = ConstantExpr::getShl(ConstantInt::get(Op->getType(), 1), CST);
LHS = Op->getOperand(0);
return Instruction::Mul;
}
}
// TODO: We can add other conversions e.g. shr => div etc.
LHS = Op->getOperand(0);
RHS = Op->getOperand(1);
return Op->getOpcode();
}
/// This tries to simplify binary operations by factorizing out common terms
/// (e. g. "(A*B)+(A*C)" -> "A*(B+C)").
static Value *tryFactorization(InstCombiner::BuilderTy *Builder,
const DataLayout *DL, BinaryOperator &I,
Instruction::BinaryOps InnerOpcode, Value *A,
Value *B, Value *C, Value *D) {
// If any of A, B, C, D are null, we can not factor I, return early.
// Checking A and C should be enough.
if (!A || !C || !B || !D)
return nullptr;
Value *SimplifiedInst = nullptr;
Value *LHS = I.getOperand(0), *RHS = I.getOperand(1);
Instruction::BinaryOps TopLevelOpcode = I.getOpcode();
// Does "X op' Y" always equal "Y op' X"?
bool InnerCommutative = Instruction::isCommutative(InnerOpcode);
// Does "X op' (Y op Z)" always equal "(X op' Y) op (X op' Z)"?
if (LeftDistributesOverRight(InnerOpcode, TopLevelOpcode))
// Does the instruction have the form "(A op' B) op (A op' D)" or, in the
// commutative case, "(A op' B) op (C op' A)"?
if (A == C || (InnerCommutative && A == D)) {
if (A != C)
std::swap(C, D);
// Consider forming "A op' (B op D)".
// If "B op D" simplifies then it can be formed with no cost.
Value *V = SimplifyBinOp(TopLevelOpcode, B, D, DL);
// If "B op D" doesn't simplify then only go on if both of the existing
// operations "A op' B" and "C op' D" will be zapped as no longer used.
if (!V && LHS->hasOneUse() && RHS->hasOneUse())
V = Builder->CreateBinOp(TopLevelOpcode, B, D, RHS->getName());
if (V) {
SimplifiedInst = Builder->CreateBinOp(InnerOpcode, A, V);
}
}
// Does "(X op Y) op' Z" always equal "(X op' Z) op (Y op' Z)"?
if (!SimplifiedInst && RightDistributesOverLeft(TopLevelOpcode, InnerOpcode))
// Does the instruction have the form "(A op' B) op (C op' B)" or, in the
// commutative case, "(A op' B) op (B op' D)"?
if (B == D || (InnerCommutative && B == C)) {
if (B != D)
std::swap(C, D);
// Consider forming "(A op C) op' B".
// If "A op C" simplifies then it can be formed with no cost.
Value *V = SimplifyBinOp(TopLevelOpcode, A, C, DL);
// If "A op C" doesn't simplify then only go on if both of the existing
// operations "A op' B" and "C op' D" will be zapped as no longer used.
if (!V && LHS->hasOneUse() && RHS->hasOneUse())
V = Builder->CreateBinOp(TopLevelOpcode, A, C, LHS->getName());
if (V) {
SimplifiedInst = Builder->CreateBinOp(InnerOpcode, V, B);
}
}
if (SimplifiedInst) {
++NumFactor;
SimplifiedInst->takeName(&I);
// Check if we can add NSW flag to SimplifiedInst. If so, set NSW flag.
// TODO: Check for NUW.
if (BinaryOperator *BO = dyn_cast<BinaryOperator>(SimplifiedInst)) {
if (isa<OverflowingBinaryOperator>(SimplifiedInst)) {
bool HasNSW = false;
if (isa<OverflowingBinaryOperator>(&I))
HasNSW = I.hasNoSignedWrap();
if (BinaryOperator *Op0 = dyn_cast<BinaryOperator>(LHS))
if (isa<OverflowingBinaryOperator>(Op0))
HasNSW &= Op0->hasNoSignedWrap();
if (BinaryOperator *Op1 = dyn_cast<BinaryOperator>(RHS))
if (isa<OverflowingBinaryOperator>(Op1))
HasNSW &= Op1->hasNoSignedWrap();
BO->setHasNoSignedWrap(HasNSW);
}
}
}
return SimplifiedInst;
}
/// SimplifyUsingDistributiveLaws - This tries to simplify binary operations
/// which some other binary operation distributes over either by factorizing
/// out common terms (eg "(A*B)+(A*C)" -> "A*(B+C)") or expanding out if this
@ -404,65 +525,33 @@ Value *InstCombiner::SimplifyUsingDistributiveLaws(BinaryOperator &I) {
Value *LHS = I.getOperand(0), *RHS = I.getOperand(1);
BinaryOperator *Op0 = dyn_cast<BinaryOperator>(LHS);
BinaryOperator *Op1 = dyn_cast<BinaryOperator>(RHS);
Instruction::BinaryOps TopLevelOpcode = I.getOpcode(); // op
// Factorization.
if (Op0 && Op1 && Op0->getOpcode() == Op1->getOpcode()) {
// The instruction has the form "(A op' B) op (C op' D)". Try to factorize
// a common term.
Value *A = Op0->getOperand(0), *B = Op0->getOperand(1);
Value *C = Op1->getOperand(0), *D = Op1->getOperand(1);
Instruction::BinaryOps InnerOpcode = Op0->getOpcode(); // op'
Value *A = nullptr, *B = nullptr, *C = nullptr, *D = nullptr;
Instruction::BinaryOps LHSOpcode = getBinOpsForFactorization(Op0, A, B);
Instruction::BinaryOps RHSOpcode = getBinOpsForFactorization(Op1, C, D);
// Does "X op' Y" always equal "Y op' X"?
bool InnerCommutative = Instruction::isCommutative(InnerOpcode);
// Does "X op' (Y op Z)" always equal "(X op' Y) op (X op' Z)"?
if (LeftDistributesOverRight(InnerOpcode, TopLevelOpcode))
// Does the instruction have the form "(A op' B) op (A op' D)" or, in the
// commutative case, "(A op' B) op (C op' A)"?
if (A == C || (InnerCommutative && A == D)) {
if (A != C)
std::swap(C, D);
// Consider forming "A op' (B op D)".
// If "B op D" simplifies then it can be formed with no cost.
Value *V = SimplifyBinOp(TopLevelOpcode, B, D, DL);
// If "B op D" doesn't simplify then only go on if both of the existing
// operations "A op' B" and "C op' D" will be zapped as no longer used.
if (!V && Op0->hasOneUse() && Op1->hasOneUse())
V = Builder->CreateBinOp(TopLevelOpcode, B, D, Op1->getName());
if (V) {
++NumFactor;
V = Builder->CreateBinOp(InnerOpcode, A, V);
V->takeName(&I);
return V;
}
}
// Does "(X op Y) op' Z" always equal "(X op' Z) op (Y op' Z)"?
if (RightDistributesOverLeft(TopLevelOpcode, InnerOpcode))
// Does the instruction have the form "(A op' B) op (C op' B)" or, in the
// commutative case, "(A op' B) op (B op' D)"?
if (B == D || (InnerCommutative && B == C)) {
if (B != D)
std::swap(C, D);
// Consider forming "(A op C) op' B".
// If "A op C" simplifies then it can be formed with no cost.
Value *V = SimplifyBinOp(TopLevelOpcode, A, C, DL);
// If "A op C" doesn't simplify then only go on if both of the existing
// operations "A op' B" and "C op' D" will be zapped as no longer used.
if (!V && Op0->hasOneUse() && Op1->hasOneUse())
V = Builder->CreateBinOp(TopLevelOpcode, A, C, Op0->getName());
if (V) {
++NumFactor;
V = Builder->CreateBinOp(InnerOpcode, V, B);
V->takeName(&I);
return V;
}
}
// The instruction has the form "(A op' B) op (C op' D)". Try to factorize
// a common term.
if (LHSOpcode == RHSOpcode) {
if (Value *V = tryFactorization(Builder, DL, I, LHSOpcode, A, B, C, D))
return V;
}
// The instruction has the form "(A op' B) op (C)". Try to factorize common
// term.
if (Value *V = tryFactorization(Builder, DL, I, LHSOpcode, A, B, RHS,
getIdentityValue(LHSOpcode, RHS)))
return V;
// The instruction has the form "(B) op (C op' D)". Try to factorize common
// term.
if (Value *V = tryFactorization(Builder, DL, I, RHSOpcode, LHS,
getIdentityValue(RHSOpcode, LHS), C, D))
return V;
// Expansion.
Instruction::BinaryOps TopLevelOpcode = I.getOpcode();
if (Op0 && RightDistributesOverLeft(Op0->getOpcode(), TopLevelOpcode)) {
// The instruction has the form "(A op' B) op C". See if expanding it out
// to "(A op C) op' (B op C)" results in simplifications.