1 //===- Reassociate.cpp - Reassociate binary expressions -------------------===//
3 // The LLVM Compiler Infrastructure
5 // This file was developed by the LLVM research group and is distributed under
6 // the University of Illinois Open Source License. See LICENSE.TXT for details.
8 //===----------------------------------------------------------------------===//
10 // This pass reassociates commutative expressions in an order that is designed
11 // to promote better constant propagation, GCSE, LICM, PRE...
13 // For example: 4 + (x + 5) -> x + (4 + 5)
15 // In the implementation of this algorithm, constants are assigned rank = 0,
16 // function arguments are rank = 1, and other values are assigned ranks
17 // corresponding to the reverse post order traversal of current function
18 // (starting at 2), which effectively gives values in deep loops higher rank
19 // than values not in loops.
21 //===----------------------------------------------------------------------===//
23 #define DEBUG_TYPE "reassociate"
24 #include "llvm/Transforms/Scalar.h"
25 #include "llvm/Constants.h"
26 #include "llvm/DerivedTypes.h"
27 #include "llvm/Function.h"
28 #include "llvm/Instructions.h"
29 #include "llvm/Pass.h"
30 #include "llvm/Assembly/Writer.h"
31 #include "llvm/Support/CFG.h"
32 #include "llvm/Support/Debug.h"
33 #include "llvm/ADT/PostOrderIterator.h"
34 #include "llvm/ADT/Statistic.h"
38 STATISTIC(NumLinear , "Number of insts linearized");
39 STATISTIC(NumChanged, "Number of insts reassociated");
40 STATISTIC(NumAnnihil, "Number of expr tree annihilated");
41 STATISTIC(NumFactor , "Number of multiplies factored");
47 ValueEntry(unsigned R, Value *O) : Rank(R), Op(O) {}
49 inline bool operator<(const ValueEntry &LHS, const ValueEntry &RHS) {
50 return LHS.Rank > RHS.Rank; // Sort so that highest rank goes to start.
54 /// PrintOps - Print out the expression identified in the Ops list.
56 static void PrintOps(Instruction *I, const std::vector<ValueEntry> &Ops) {
57 Module *M = I->getParent()->getParent()->getParent();
58 cerr << Instruction::getOpcodeName(I->getOpcode()) << " "
59 << *Ops[0].Op->getType();
60 for (unsigned i = 0, e = Ops.size(); i != e; ++i)
61 WriteAsOperand(*cerr.stream() << " ", Ops[i].Op, false, M)
62 << "," << Ops[i].Rank;
66 class Reassociate : public FunctionPass {
67 std::map<BasicBlock*, unsigned> RankMap;
68 std::map<Value*, unsigned> ValueRankMap;
71 bool runOnFunction(Function &F);
73 virtual void getAnalysisUsage(AnalysisUsage &AU) const {
77 void BuildRankMap(Function &F);
78 unsigned getRank(Value *V);
79 void ReassociateExpression(BinaryOperator *I);
80 void RewriteExprTree(BinaryOperator *I, std::vector<ValueEntry> &Ops,
82 Value *OptimizeExpression(BinaryOperator *I, std::vector<ValueEntry> &Ops);
83 void LinearizeExprTree(BinaryOperator *I, std::vector<ValueEntry> &Ops);
84 void LinearizeExpr(BinaryOperator *I);
85 Value *RemoveFactorFromExpression(Value *V, Value *Factor);
86 void ReassociateBB(BasicBlock *BB);
88 void RemoveDeadBinaryOp(Value *V);
91 RegisterPass<Reassociate> X("reassociate", "Reassociate expressions");
94 // Public interface to the Reassociate pass
95 FunctionPass *llvm::createReassociatePass() { return new Reassociate(); }
97 void Reassociate::RemoveDeadBinaryOp(Value *V) {
98 Instruction *Op = dyn_cast<Instruction>(V);
99 if (!Op || !isa<BinaryOperator>(Op) || !isa<CmpInst>(Op) || !Op->use_empty())
102 Value *LHS = Op->getOperand(0), *RHS = Op->getOperand(1);
103 RemoveDeadBinaryOp(LHS);
104 RemoveDeadBinaryOp(RHS);
108 static bool isUnmovableInstruction(Instruction *I) {
109 if (I->getOpcode() == Instruction::PHI ||
110 I->getOpcode() == Instruction::Alloca ||
111 I->getOpcode() == Instruction::Load ||
112 I->getOpcode() == Instruction::Malloc ||
113 I->getOpcode() == Instruction::Invoke ||
114 I->getOpcode() == Instruction::Call ||
115 I->getOpcode() == Instruction::UDiv ||
116 I->getOpcode() == Instruction::SDiv ||
117 I->getOpcode() == Instruction::FDiv ||
118 I->getOpcode() == Instruction::URem ||
119 I->getOpcode() == Instruction::SRem ||
120 I->getOpcode() == Instruction::FRem)
125 void Reassociate::BuildRankMap(Function &F) {
128 // Assign distinct ranks to function arguments
129 for (Function::arg_iterator I = F.arg_begin(), E = F.arg_end(); I != E; ++I)
130 ValueRankMap[I] = ++i;
132 ReversePostOrderTraversal<Function*> RPOT(&F);
133 for (ReversePostOrderTraversal<Function*>::rpo_iterator I = RPOT.begin(),
134 E = RPOT.end(); I != E; ++I) {
136 unsigned BBRank = RankMap[BB] = ++i << 16;
138 // Walk the basic block, adding precomputed ranks for any instructions that
139 // we cannot move. This ensures that the ranks for these instructions are
140 // all different in the block.
141 for (BasicBlock::iterator I = BB->begin(), E = BB->end(); I != E; ++I)
142 if (isUnmovableInstruction(I))
143 ValueRankMap[I] = ++BBRank;
147 unsigned Reassociate::getRank(Value *V) {
148 if (isa<Argument>(V)) return ValueRankMap[V]; // Function argument...
150 Instruction *I = dyn_cast<Instruction>(V);
151 if (I == 0) return 0; // Otherwise it's a global or constant, rank 0.
153 unsigned &CachedRank = ValueRankMap[I];
154 if (CachedRank) return CachedRank; // Rank already known?
156 // If this is an expression, return the 1+MAX(rank(LHS), rank(RHS)) so that
157 // we can reassociate expressions for code motion! Since we do not recurse
158 // for PHI nodes, we cannot have infinite recursion here, because there
159 // cannot be loops in the value graph that do not go through PHI nodes.
160 unsigned Rank = 0, MaxRank = RankMap[I->getParent()];
161 for (unsigned i = 0, e = I->getNumOperands();
162 i != e && Rank != MaxRank; ++i)
163 Rank = std::max(Rank, getRank(I->getOperand(i)));
165 // If this is a not or neg instruction, do not count it for rank. This
166 // assures us that X and ~X will have the same rank.
167 if (!I->getType()->isIntegral() ||
168 (!BinaryOperator::isNot(I) && !BinaryOperator::isNeg(I)))
171 //DOUT << "Calculated Rank[" << V->getName() << "] = "
174 return CachedRank = Rank;
177 /// isReassociableOp - Return true if V is an instruction of the specified
178 /// opcode and if it only has one use.
179 static BinaryOperator *isReassociableOp(Value *V, unsigned Opcode) {
180 if ((V->hasOneUse() || V->use_empty()) && isa<Instruction>(V) &&
181 cast<Instruction>(V)->getOpcode() == Opcode)
182 return cast<BinaryOperator>(V);
186 /// LowerNegateToMultiply - Replace 0-X with X*-1.
188 static Instruction *LowerNegateToMultiply(Instruction *Neg) {
190 if (Neg->getType()->isFloatingPoint())
191 Cst = ConstantFP::get(Neg->getType(), -1);
193 Cst = ConstantInt::getAllOnesValue(Neg->getType());
195 std::string NegName = Neg->getName(); Neg->setName("");
196 Instruction *Res = BinaryOperator::createMul(Neg->getOperand(1), Cst, NegName,
198 Neg->replaceAllUsesWith(Res);
199 Neg->eraseFromParent();
203 // Given an expression of the form '(A+B)+(D+C)', turn it into '(((A+B)+C)+D)'.
204 // Note that if D is also part of the expression tree that we recurse to
205 // linearize it as well. Besides that case, this does not recurse into A,B, or
207 void Reassociate::LinearizeExpr(BinaryOperator *I) {
208 BinaryOperator *LHS = cast<BinaryOperator>(I->getOperand(0));
209 BinaryOperator *RHS = cast<BinaryOperator>(I->getOperand(1));
210 assert(isReassociableOp(LHS, I->getOpcode()) &&
211 isReassociableOp(RHS, I->getOpcode()) &&
212 "Not an expression that needs linearization?");
214 DOUT << "Linear" << *LHS << *RHS << *I;
216 // Move the RHS instruction to live immediately before I, avoiding breaking
217 // dominator properties.
220 // Move operands around to do the linearization.
221 I->setOperand(1, RHS->getOperand(0));
222 RHS->setOperand(0, LHS);
223 I->setOperand(0, RHS);
227 DOUT << "Linearized: " << *I;
229 // If D is part of this expression tree, tail recurse.
230 if (isReassociableOp(I->getOperand(1), I->getOpcode()))
235 /// LinearizeExprTree - Given an associative binary expression tree, traverse
236 /// all of the uses putting it into canonical form. This forces a left-linear
237 /// form of the the expression (((a+b)+c)+d), and collects information about the
238 /// rank of the non-tree operands.
240 /// NOTE: These intentionally destroys the expression tree operands (turning
241 /// them into undef values) to reduce #uses of the values. This means that the
242 /// caller MUST use something like RewriteExprTree to put the values back in.
244 void Reassociate::LinearizeExprTree(BinaryOperator *I,
245 std::vector<ValueEntry> &Ops) {
246 Value *LHS = I->getOperand(0), *RHS = I->getOperand(1);
247 unsigned Opcode = I->getOpcode();
249 // First step, linearize the expression if it is in ((A+B)+(C+D)) form.
250 BinaryOperator *LHSBO = isReassociableOp(LHS, Opcode);
251 BinaryOperator *RHSBO = isReassociableOp(RHS, Opcode);
253 // If this is a multiply expression tree and it contains internal negations,
254 // transform them into multiplies by -1 so they can be reassociated.
255 if (I->getOpcode() == Instruction::Mul) {
256 if (!LHSBO && LHS->hasOneUse() && BinaryOperator::isNeg(LHS)) {
257 LHS = LowerNegateToMultiply(cast<Instruction>(LHS));
258 LHSBO = isReassociableOp(LHS, Opcode);
260 if (!RHSBO && RHS->hasOneUse() && BinaryOperator::isNeg(RHS)) {
261 RHS = LowerNegateToMultiply(cast<Instruction>(RHS));
262 RHSBO = isReassociableOp(RHS, Opcode);
268 // Neither the LHS or RHS as part of the tree, thus this is a leaf. As
269 // such, just remember these operands and their rank.
270 Ops.push_back(ValueEntry(getRank(LHS), LHS));
271 Ops.push_back(ValueEntry(getRank(RHS), RHS));
273 // Clear the leaves out.
274 I->setOperand(0, UndefValue::get(I->getType()));
275 I->setOperand(1, UndefValue::get(I->getType()));
278 // Turn X+(Y+Z) -> (Y+Z)+X
279 std::swap(LHSBO, RHSBO);
281 bool Success = !I->swapOperands();
282 assert(Success && "swapOperands failed");
286 // Turn (A+B)+(C+D) -> (((A+B)+C)+D). This guarantees the the RHS is not
287 // part of the expression tree.
289 LHS = LHSBO = cast<BinaryOperator>(I->getOperand(0));
290 RHS = I->getOperand(1);
294 // Okay, now we know that the LHS is a nested expression and that the RHS is
295 // not. Perform reassociation.
296 assert(!isReassociableOp(RHS, Opcode) && "LinearizeExpr failed!");
298 // Move LHS right before I to make sure that the tree expression dominates all
300 LHSBO->moveBefore(I);
302 // Linearize the expression tree on the LHS.
303 LinearizeExprTree(LHSBO, Ops);
305 // Remember the RHS operand and its rank.
306 Ops.push_back(ValueEntry(getRank(RHS), RHS));
308 // Clear the RHS leaf out.
309 I->setOperand(1, UndefValue::get(I->getType()));
312 // RewriteExprTree - Now that the operands for this expression tree are
313 // linearized and optimized, emit them in-order. This function is written to be
315 void Reassociate::RewriteExprTree(BinaryOperator *I,
316 std::vector<ValueEntry> &Ops,
318 if (i+2 == Ops.size()) {
319 if (I->getOperand(0) != Ops[i].Op ||
320 I->getOperand(1) != Ops[i+1].Op) {
321 Value *OldLHS = I->getOperand(0);
322 DOUT << "RA: " << *I;
323 I->setOperand(0, Ops[i].Op);
324 I->setOperand(1, Ops[i+1].Op);
325 DOUT << "TO: " << *I;
329 // If we reassociated a tree to fewer operands (e.g. (1+a+2) -> (a+3)
330 // delete the extra, now dead, nodes.
331 RemoveDeadBinaryOp(OldLHS);
335 assert(i+2 < Ops.size() && "Ops index out of range!");
337 if (I->getOperand(1) != Ops[i].Op) {
338 DOUT << "RA: " << *I;
339 I->setOperand(1, Ops[i].Op);
340 DOUT << "TO: " << *I;
345 BinaryOperator *LHS = cast<BinaryOperator>(I->getOperand(0));
346 assert(LHS->getOpcode() == I->getOpcode() &&
347 "Improper expression tree!");
349 // Compactify the tree instructions together with each other to guarantee
350 // that the expression tree is dominated by all of Ops.
352 RewriteExprTree(LHS, Ops, i+1);
357 // NegateValue - Insert instructions before the instruction pointed to by BI,
358 // that computes the negative version of the value specified. The negative
359 // version of the value is returned, and BI is left pointing at the instruction
360 // that should be processed next by the reassociation pass.
362 static Value *NegateValue(Value *V, Instruction *BI) {
363 // We are trying to expose opportunity for reassociation. One of the things
364 // that we want to do to achieve this is to push a negation as deep into an
365 // expression chain as possible, to expose the add instructions. In practice,
366 // this means that we turn this:
367 // X = -(A+12+C+D) into X = -A + -12 + -C + -D = -12 + -A + -C + -D
368 // so that later, a: Y = 12+X could get reassociated with the -12 to eliminate
369 // the constants. We assume that instcombine will clean up the mess later if
370 // we introduce tons of unnecessary negation instructions...
372 if (Instruction *I = dyn_cast<Instruction>(V))
373 if (I->getOpcode() == Instruction::Add && I->hasOneUse()) {
374 // Push the negates through the add.
375 I->setOperand(0, NegateValue(I->getOperand(0), BI));
376 I->setOperand(1, NegateValue(I->getOperand(1), BI));
378 // We must move the add instruction here, because the neg instructions do
379 // not dominate the old add instruction in general. By moving it, we are
380 // assured that the neg instructions we just inserted dominate the
381 // instruction we are about to insert after them.
384 I->setName(I->getName()+".neg");
388 // Insert a 'neg' instruction that subtracts the value from zero to get the
391 return BinaryOperator::createNeg(V, V->getName() + ".neg", BI);
394 /// BreakUpSubtract - If we have (X-Y), and if either X is an add, or if this is
395 /// only used by an add, transform this into (X+(0-Y)) to promote better
397 static Instruction *BreakUpSubtract(Instruction *Sub) {
398 // Don't bother to break this up unless either the LHS is an associable add or
399 // if this is only used by one.
400 if (!isReassociableOp(Sub->getOperand(0), Instruction::Add) &&
401 !isReassociableOp(Sub->getOperand(1), Instruction::Add) &&
402 !(Sub->hasOneUse() &&isReassociableOp(Sub->use_back(), Instruction::Add)))
405 // Convert a subtract into an add and a neg instruction... so that sub
406 // instructions can be commuted with other add instructions...
408 // Calculate the negative value of Operand 1 of the sub instruction...
409 // and set it as the RHS of the add instruction we just made...
411 std::string Name = Sub->getName();
413 Value *NegVal = NegateValue(Sub->getOperand(1), Sub);
415 BinaryOperator::createAdd(Sub->getOperand(0), NegVal, Name, Sub);
417 // Everyone now refers to the add instruction.
418 Sub->replaceAllUsesWith(New);
419 Sub->eraseFromParent();
421 DOUT << "Negated: " << *New;
425 /// ConvertShiftToMul - If this is a shift of a reassociable multiply or is used
426 /// by one, change this into a multiply by a constant to assist with further
428 static Instruction *ConvertShiftToMul(Instruction *Shl) {
429 // If an operand of this shift is a reassociable multiply, or if the shift
430 // is used by a reassociable multiply or add, turn into a multiply.
431 if (isReassociableOp(Shl->getOperand(0), Instruction::Mul) ||
433 (isReassociableOp(Shl->use_back(), Instruction::Mul) ||
434 isReassociableOp(Shl->use_back(), Instruction::Add)))) {
435 Constant *MulCst = ConstantInt::get(Shl->getType(), 1);
436 MulCst = ConstantExpr::getShl(MulCst, cast<Constant>(Shl->getOperand(1)));
438 std::string Name = Shl->getName(); Shl->setName("");
439 Instruction *Mul = BinaryOperator::createMul(Shl->getOperand(0), MulCst,
441 Shl->replaceAllUsesWith(Mul);
442 Shl->eraseFromParent();
448 // Scan backwards and forwards among values with the same rank as element i to
449 // see if X exists. If X does not exist, return i.
450 static unsigned FindInOperandList(std::vector<ValueEntry> &Ops, unsigned i,
452 unsigned XRank = Ops[i].Rank;
453 unsigned e = Ops.size();
454 for (unsigned j = i+1; j != e && Ops[j].Rank == XRank; ++j)
458 for (unsigned j = i-1; j != ~0U && Ops[j].Rank == XRank; --j)
464 /// EmitAddTreeOfValues - Emit a tree of add instructions, summing Ops together
465 /// and returning the result. Insert the tree before I.
466 static Value *EmitAddTreeOfValues(Instruction *I, std::vector<Value*> &Ops) {
467 if (Ops.size() == 1) return Ops.back();
469 Value *V1 = Ops.back();
471 Value *V2 = EmitAddTreeOfValues(I, Ops);
472 return BinaryOperator::createAdd(V2, V1, "tmp", I);
475 /// RemoveFactorFromExpression - If V is an expression tree that is a
476 /// multiplication sequence, and if this sequence contains a multiply by Factor,
477 /// remove Factor from the tree and return the new tree.
478 Value *Reassociate::RemoveFactorFromExpression(Value *V, Value *Factor) {
479 BinaryOperator *BO = isReassociableOp(V, Instruction::Mul);
482 std::vector<ValueEntry> Factors;
483 LinearizeExprTree(BO, Factors);
485 bool FoundFactor = false;
486 for (unsigned i = 0, e = Factors.size(); i != e; ++i)
487 if (Factors[i].Op == Factor) {
489 Factors.erase(Factors.begin()+i);
493 // Make sure to restore the operands to the expression tree.
494 RewriteExprTree(BO, Factors);
498 if (Factors.size() == 1) return Factors[0].Op;
500 RewriteExprTree(BO, Factors);
504 /// FindSingleUseMultiplyFactors - If V is a single-use multiply, recursively
505 /// add its operands as factors, otherwise add V to the list of factors.
506 static void FindSingleUseMultiplyFactors(Value *V,
507 std::vector<Value*> &Factors) {
509 if ((!V->hasOneUse() && !V->use_empty()) ||
510 !(BO = dyn_cast<BinaryOperator>(V)) ||
511 BO->getOpcode() != Instruction::Mul) {
512 Factors.push_back(V);
516 // Otherwise, add the LHS and RHS to the list of factors.
517 FindSingleUseMultiplyFactors(BO->getOperand(1), Factors);
518 FindSingleUseMultiplyFactors(BO->getOperand(0), Factors);
523 Value *Reassociate::OptimizeExpression(BinaryOperator *I,
524 std::vector<ValueEntry> &Ops) {
525 // Now that we have the linearized expression tree, try to optimize it.
526 // Start by folding any constants that we found.
527 bool IterateOptimization = false;
528 if (Ops.size() == 1) return Ops[0].Op;
530 unsigned Opcode = I->getOpcode();
532 if (Constant *V1 = dyn_cast<Constant>(Ops[Ops.size()-2].Op))
533 if (Constant *V2 = dyn_cast<Constant>(Ops.back().Op)) {
535 Ops.back().Op = ConstantExpr::get(Opcode, V1, V2);
536 return OptimizeExpression(I, Ops);
539 // Check for destructive annihilation due to a constant being used.
540 if (ConstantIntegral *CstVal = dyn_cast<ConstantIntegral>(Ops.back().Op))
543 case Instruction::And:
544 if (CstVal->isNullValue()) { // ... & 0 -> 0
547 } else if (CstVal->isAllOnesValue()) { // ... & -1 -> ...
551 case Instruction::Mul:
552 if (CstVal->isNullValue()) { // ... * 0 -> 0
555 } else if (cast<ConstantInt>(CstVal)->getZExtValue() == 1) {
556 Ops.pop_back(); // ... * 1 -> ...
559 case Instruction::Or:
560 if (CstVal->isAllOnesValue()) { // ... | -1 -> -1
565 case Instruction::Add:
566 case Instruction::Xor:
567 if (CstVal->isNullValue()) // ... [|^+] 0 -> ...
571 if (Ops.size() == 1) return Ops[0].Op;
573 // Handle destructive annihilation do to identities between elements in the
574 // argument list here.
577 case Instruction::And:
578 case Instruction::Or:
579 case Instruction::Xor:
580 // Scan the operand lists looking for X and ~X pairs, along with X,X pairs.
581 // If we find any, we can simplify the expression. X&~X == 0, X|~X == -1.
582 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
583 // First, check for X and ~X in the operand list.
584 assert(i < Ops.size());
585 if (BinaryOperator::isNot(Ops[i].Op)) { // Cannot occur for ^.
586 Value *X = BinaryOperator::getNotArgument(Ops[i].Op);
587 unsigned FoundX = FindInOperandList(Ops, i, X);
589 if (Opcode == Instruction::And) { // ...&X&~X = 0
591 return Constant::getNullValue(X->getType());
592 } else if (Opcode == Instruction::Or) { // ...|X|~X = -1
594 return ConstantIntegral::getAllOnesValue(X->getType());
599 // Next, check for duplicate pairs of values, which we assume are next to
600 // each other, due to our sorting criteria.
601 assert(i < Ops.size());
602 if (i+1 != Ops.size() && Ops[i+1].Op == Ops[i].Op) {
603 if (Opcode == Instruction::And || Opcode == Instruction::Or) {
604 // Drop duplicate values.
605 Ops.erase(Ops.begin()+i);
607 IterateOptimization = true;
610 assert(Opcode == Instruction::Xor);
613 return Constant::getNullValue(Ops[0].Op->getType());
616 Ops.erase(Ops.begin()+i, Ops.begin()+i+2);
618 IterateOptimization = true;
625 case Instruction::Add:
626 // Scan the operand lists looking for X and -X pairs. If we find any, we
627 // can simplify the expression. X+-X == 0.
628 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
629 assert(i < Ops.size());
630 // Check for X and -X in the operand list.
631 if (BinaryOperator::isNeg(Ops[i].Op)) {
632 Value *X = BinaryOperator::getNegArgument(Ops[i].Op);
633 unsigned FoundX = FindInOperandList(Ops, i, X);
635 // Remove X and -X from the operand list.
636 if (Ops.size() == 2) {
638 return Constant::getNullValue(X->getType());
640 Ops.erase(Ops.begin()+i);
644 --i; // Need to back up an extra one.
645 Ops.erase(Ops.begin()+FoundX);
646 IterateOptimization = true;
648 --i; // Revisit element.
649 e -= 2; // Removed two elements.
656 // Scan the operand list, checking to see if there are any common factors
657 // between operands. Consider something like A*A+A*B*C+D. We would like to
658 // reassociate this to A*(A+B*C)+D, which reduces the number of multiplies.
659 // To efficiently find this, we count the number of times a factor occurs
660 // for any ADD operands that are MULs.
661 std::map<Value*, unsigned> FactorOccurrences;
663 Value *MaxOccVal = 0;
664 if (!I->getType()->isFloatingPoint()) {
665 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
666 if (BinaryOperator *BOp = dyn_cast<BinaryOperator>(Ops[i].Op))
667 if (BOp->getOpcode() == Instruction::Mul && BOp->use_empty()) {
668 // Compute all of the factors of this added value.
669 std::vector<Value*> Factors;
670 FindSingleUseMultiplyFactors(BOp, Factors);
671 assert(Factors.size() > 1 && "Bad linearize!");
673 // Add one to FactorOccurrences for each unique factor in this op.
674 if (Factors.size() == 2) {
675 unsigned Occ = ++FactorOccurrences[Factors[0]];
676 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[0]; }
677 if (Factors[0] != Factors[1]) { // Don't double count A*A.
678 Occ = ++FactorOccurrences[Factors[1]];
679 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[1]; }
682 std::set<Value*> Duplicates;
683 for (unsigned i = 0, e = Factors.size(); i != e; ++i)
684 if (Duplicates.insert(Factors[i]).second) {
685 unsigned Occ = ++FactorOccurrences[Factors[i]];
686 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[i]; }
693 // If any factor occurred more than one time, we can pull it out.
695 DOUT << "\nFACTORING [" << MaxOcc << "]: " << *MaxOccVal << "\n";
697 // Create a new instruction that uses the MaxOccVal twice. If we don't do
698 // this, we could otherwise run into situations where removing a factor
699 // from an expression will drop a use of maxocc, and this can cause
700 // RemoveFactorFromExpression on successive values to behave differently.
701 Instruction *DummyInst = BinaryOperator::createAdd(MaxOccVal, MaxOccVal);
702 std::vector<Value*> NewMulOps;
703 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
704 if (Value *V = RemoveFactorFromExpression(Ops[i].Op, MaxOccVal)) {
705 NewMulOps.push_back(V);
706 Ops.erase(Ops.begin()+i);
711 // No need for extra uses anymore.
714 unsigned NumAddedValues = NewMulOps.size();
715 Value *V = EmitAddTreeOfValues(I, NewMulOps);
716 Value *V2 = BinaryOperator::createMul(V, MaxOccVal, "tmp", I);
718 // Now that we have inserted V and its sole use, optimize it. This allows
719 // us to handle cases that require multiple factoring steps, such as this:
720 // A*A*B + A*A*C --> A*(A*B+A*C) --> A*(A*(B+C))
721 if (NumAddedValues > 1)
722 ReassociateExpression(cast<BinaryOperator>(V));
729 // Add the new value to the list of things being added.
730 Ops.insert(Ops.begin(), ValueEntry(getRank(V2), V2));
732 // Rewrite the tree so that there is now a use of V.
733 RewriteExprTree(I, Ops);
734 return OptimizeExpression(I, Ops);
737 //case Instruction::Mul:
740 if (IterateOptimization)
741 return OptimizeExpression(I, Ops);
746 /// ReassociateBB - Inspect all of the instructions in this basic block,
747 /// reassociating them as we go.
748 void Reassociate::ReassociateBB(BasicBlock *BB) {
749 for (BasicBlock::iterator BBI = BB->begin(); BBI != BB->end(); ) {
750 Instruction *BI = BBI++;
751 if (BI->getOpcode() == Instruction::Shl &&
752 isa<ConstantInt>(BI->getOperand(1)))
753 if (Instruction *NI = ConvertShiftToMul(BI)) {
758 // Reject cases where it is pointless to do this.
759 if (!isa<BinaryOperator>(BI) || BI->getType()->isFloatingPoint() ||
760 isa<PackedType>(BI->getType()))
761 continue; // Floating point ops are not associative.
763 // If this is a subtract instruction which is not already in negate form,
764 // see if we can convert it to X+-Y.
765 if (BI->getOpcode() == Instruction::Sub) {
766 if (!BinaryOperator::isNeg(BI)) {
767 if (Instruction *NI = BreakUpSubtract(BI)) {
772 // Otherwise, this is a negation. See if the operand is a multiply tree
773 // and if this is not an inner node of a multiply tree.
774 if (isReassociableOp(BI->getOperand(1), Instruction::Mul) &&
776 !isReassociableOp(BI->use_back(), Instruction::Mul))) {
777 BI = LowerNegateToMultiply(BI);
783 // If this instruction is a commutative binary operator, process it.
784 if (!BI->isAssociative()) continue;
785 BinaryOperator *I = cast<BinaryOperator>(BI);
787 // If this is an interior node of a reassociable tree, ignore it until we
788 // get to the root of the tree, to avoid N^2 analysis.
789 if (I->hasOneUse() && isReassociableOp(I->use_back(), I->getOpcode()))
792 // If this is an add tree that is used by a sub instruction, ignore it
793 // until we process the subtract.
794 if (I->hasOneUse() && I->getOpcode() == Instruction::Add &&
795 cast<Instruction>(I->use_back())->getOpcode() == Instruction::Sub)
798 ReassociateExpression(I);
802 void Reassociate::ReassociateExpression(BinaryOperator *I) {
804 // First, walk the expression tree, linearizing the tree, collecting
805 std::vector<ValueEntry> Ops;
806 LinearizeExprTree(I, Ops);
808 DOUT << "RAIn:\t"; DEBUG(PrintOps(I, Ops)); DOUT << "\n";
810 // Now that we have linearized the tree to a list and have gathered all of
811 // the operands and their ranks, sort the operands by their rank. Use a
812 // stable_sort so that values with equal ranks will have their relative
813 // positions maintained (and so the compiler is deterministic). Note that
814 // this sorts so that the highest ranking values end up at the beginning of
816 std::stable_sort(Ops.begin(), Ops.end());
818 // OptimizeExpression - Now that we have the expression tree in a convenient
819 // sorted form, optimize it globally if possible.
820 if (Value *V = OptimizeExpression(I, Ops)) {
821 // This expression tree simplified to something that isn't a tree,
823 DOUT << "Reassoc to scalar: " << *V << "\n";
824 I->replaceAllUsesWith(V);
825 RemoveDeadBinaryOp(I);
829 // We want to sink immediates as deeply as possible except in the case where
830 // this is a multiply tree used only by an add, and the immediate is a -1.
831 // In this case we reassociate to put the negation on the outside so that we
832 // can fold the negation into the add: (-X)*Y + Z -> Z-X*Y
833 if (I->getOpcode() == Instruction::Mul && I->hasOneUse() &&
834 cast<Instruction>(I->use_back())->getOpcode() == Instruction::Add &&
835 isa<ConstantInt>(Ops.back().Op) &&
836 cast<ConstantInt>(Ops.back().Op)->isAllOnesValue()) {
837 Ops.insert(Ops.begin(), Ops.back());
841 DOUT << "RAOut:\t"; DEBUG(PrintOps(I, Ops)); DOUT << "\n";
843 if (Ops.size() == 1) {
844 // This expression tree simplified to something that isn't a tree,
846 I->replaceAllUsesWith(Ops[0].Op);
847 RemoveDeadBinaryOp(I);
849 // Now that we ordered and optimized the expressions, splat them back into
850 // the expression tree, removing any unneeded nodes.
851 RewriteExprTree(I, Ops);
856 bool Reassociate::runOnFunction(Function &F) {
857 // Recalculate the rank map for F
861 for (Function::iterator FI = F.begin(), FE = F.end(); FI != FE; ++FI)
864 // We are done with the rank map...
866 ValueRankMap.clear();