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"
40 Statistic<> NumLinear ("reassociate","Number of insts linearized");
41 Statistic<> NumChanged("reassociate","Number of insts reassociated");
42 Statistic<> NumSwapped("reassociate","Number of insts with operands swapped");
43 Statistic<> NumAnnihil("reassociate","Number of expr tree annihilated");
44 Statistic<> NumFactor ("reassociate","Number of multiplies factored");
49 ValueEntry(unsigned R, Value *O) : Rank(R), Op(O) {}
51 inline bool operator<(const ValueEntry &LHS, const ValueEntry &RHS) {
52 return LHS.Rank > RHS.Rank; // Sort so that highest rank goes to start.
56 /// PrintOps - Print out the expression identified in the Ops list.
58 static void PrintOps(Instruction *I, const std::vector<ValueEntry> &Ops) {
59 Module *M = I->getParent()->getParent()->getParent();
60 std::cerr << Instruction::getOpcodeName(I->getOpcode()) << " "
61 << *Ops[0].Op->getType();
62 for (unsigned i = 0, e = Ops.size(); i != e; ++i)
63 WriteAsOperand(std::cerr << " ", Ops[i].Op, false, true, M)
64 << "," << Ops[i].Rank;
68 class Reassociate : public FunctionPass {
69 std::map<BasicBlock*, unsigned> RankMap;
70 std::map<Value*, unsigned> ValueRankMap;
73 bool runOnFunction(Function &F);
75 virtual void getAnalysisUsage(AnalysisUsage &AU) const {
79 void BuildRankMap(Function &F);
80 unsigned getRank(Value *V);
81 void ReassociateExpression(BinaryOperator *I);
82 void RewriteExprTree(BinaryOperator *I, std::vector<ValueEntry> &Ops,
84 Value *OptimizeExpression(BinaryOperator *I, std::vector<ValueEntry> &Ops);
85 void LinearizeExprTree(BinaryOperator *I, std::vector<ValueEntry> &Ops);
86 void LinearizeExpr(BinaryOperator *I);
87 Value *RemoveFactorFromExpression(Value *V, Value *Factor);
88 void ReassociateBB(BasicBlock *BB);
90 void RemoveDeadBinaryOp(Value *V);
93 RegisterPass<Reassociate> X("reassociate", "Reassociate expressions");
96 // Public interface to the Reassociate pass
97 FunctionPass *llvm::createReassociatePass() { return new Reassociate(); }
99 void Reassociate::RemoveDeadBinaryOp(Value *V) {
100 BinaryOperator *BOp = dyn_cast<BinaryOperator>(V);
101 if (!BOp || !BOp->use_empty()) return;
103 Value *LHS = BOp->getOperand(0), *RHS = BOp->getOperand(1);
104 RemoveDeadBinaryOp(LHS);
105 RemoveDeadBinaryOp(RHS);
109 static bool isUnmovableInstruction(Instruction *I) {
110 if (I->getOpcode() == Instruction::PHI ||
111 I->getOpcode() == Instruction::Alloca ||
112 I->getOpcode() == Instruction::Load ||
113 I->getOpcode() == Instruction::Malloc ||
114 I->getOpcode() == Instruction::Invoke ||
115 I->getOpcode() == Instruction::Call ||
116 I->getOpcode() == Instruction::UDiv ||
117 I->getOpcode() == Instruction::SDiv ||
118 I->getOpcode() == Instruction::FDiv ||
119 I->getOpcode() == Instruction::URem ||
120 I->getOpcode() == Instruction::SRem ||
121 I->getOpcode() == Instruction::FRem)
126 void Reassociate::BuildRankMap(Function &F) {
129 // Assign distinct ranks to function arguments
130 for (Function::arg_iterator I = F.arg_begin(), E = F.arg_end(); I != E; ++I)
131 ValueRankMap[I] = ++i;
133 ReversePostOrderTraversal<Function*> RPOT(&F);
134 for (ReversePostOrderTraversal<Function*>::rpo_iterator I = RPOT.begin(),
135 E = RPOT.end(); I != E; ++I) {
137 unsigned BBRank = RankMap[BB] = ++i << 16;
139 // Walk the basic block, adding precomputed ranks for any instructions that
140 // we cannot move. This ensures that the ranks for these instructions are
141 // all different in the block.
142 for (BasicBlock::iterator I = BB->begin(), E = BB->end(); I != E; ++I)
143 if (isUnmovableInstruction(I))
144 ValueRankMap[I] = ++BBRank;
148 unsigned Reassociate::getRank(Value *V) {
149 if (isa<Argument>(V)) return ValueRankMap[V]; // Function argument...
151 Instruction *I = dyn_cast<Instruction>(V);
152 if (I == 0) return 0; // Otherwise it's a global or constant, rank 0.
154 unsigned &CachedRank = ValueRankMap[I];
155 if (CachedRank) return CachedRank; // Rank already known?
157 // If this is an expression, return the 1+MAX(rank(LHS), rank(RHS)) so that
158 // we can reassociate expressions for code motion! Since we do not recurse
159 // for PHI nodes, we cannot have infinite recursion here, because there
160 // cannot be loops in the value graph that do not go through PHI nodes.
161 unsigned Rank = 0, MaxRank = RankMap[I->getParent()];
162 for (unsigned i = 0, e = I->getNumOperands();
163 i != e && Rank != MaxRank; ++i)
164 Rank = std::max(Rank, getRank(I->getOperand(i)));
166 // If this is a not or neg instruction, do not count it for rank. This
167 // assures us that X and ~X will have the same rank.
168 if (!I->getType()->isIntegral() ||
169 (!BinaryOperator::isNot(I) && !BinaryOperator::isNeg(I)))
172 //DEBUG(std::cerr << "Calculated Rank[" << V->getName() << "] = "
175 return CachedRank = Rank;
178 /// isReassociableOp - Return true if V is an instruction of the specified
179 /// opcode and if it only has one use.
180 static BinaryOperator *isReassociableOp(Value *V, unsigned Opcode) {
181 if ((V->hasOneUse() || V->use_empty()) && isa<Instruction>(V) &&
182 cast<Instruction>(V)->getOpcode() == Opcode)
183 return cast<BinaryOperator>(V);
187 /// LowerNegateToMultiply - Replace 0-X with X*-1.
189 static Instruction *LowerNegateToMultiply(Instruction *Neg) {
191 if (Neg->getType()->isFloatingPoint())
192 Cst = ConstantFP::get(Neg->getType(), -1);
194 Cst = ConstantInt::getAllOnesValue(Neg->getType());
196 std::string NegName = Neg->getName(); Neg->setName("");
197 Instruction *Res = BinaryOperator::createMul(Neg->getOperand(1), Cst, NegName,
199 Neg->replaceAllUsesWith(Res);
200 Neg->eraseFromParent();
204 // Given an expression of the form '(A+B)+(D+C)', turn it into '(((A+B)+C)+D)'.
205 // Note that if D is also part of the expression tree that we recurse to
206 // linearize it as well. Besides that case, this does not recurse into A,B, or
208 void Reassociate::LinearizeExpr(BinaryOperator *I) {
209 BinaryOperator *LHS = cast<BinaryOperator>(I->getOperand(0));
210 BinaryOperator *RHS = cast<BinaryOperator>(I->getOperand(1));
211 assert(isReassociableOp(LHS, I->getOpcode()) &&
212 isReassociableOp(RHS, I->getOpcode()) &&
213 "Not an expression that needs linearization?");
215 DEBUG(std::cerr << "Linear" << *LHS << *RHS << *I);
217 // Move the RHS instruction to live immediately before I, avoiding breaking
218 // dominator properties.
221 // Move operands around to do the linearization.
222 I->setOperand(1, RHS->getOperand(0));
223 RHS->setOperand(0, LHS);
224 I->setOperand(0, RHS);
228 DEBUG(std::cerr << "Linearized: " << *I);
230 // If D is part of this expression tree, tail recurse.
231 if (isReassociableOp(I->getOperand(1), I->getOpcode()))
236 /// LinearizeExprTree - Given an associative binary expression tree, traverse
237 /// all of the uses putting it into canonical form. This forces a left-linear
238 /// form of the the expression (((a+b)+c)+d), and collects information about the
239 /// rank of the non-tree operands.
241 /// NOTE: These intentionally destroys the expression tree operands (turning
242 /// them into undef values) to reduce #uses of the values. This means that the
243 /// caller MUST use something like RewriteExprTree to put the values back in.
245 void Reassociate::LinearizeExprTree(BinaryOperator *I,
246 std::vector<ValueEntry> &Ops) {
247 Value *LHS = I->getOperand(0), *RHS = I->getOperand(1);
248 unsigned Opcode = I->getOpcode();
250 // First step, linearize the expression if it is in ((A+B)+(C+D)) form.
251 BinaryOperator *LHSBO = isReassociableOp(LHS, Opcode);
252 BinaryOperator *RHSBO = isReassociableOp(RHS, Opcode);
254 // If this is a multiply expression tree and it contains internal negations,
255 // transform them into multiplies by -1 so they can be reassociated.
256 if (I->getOpcode() == Instruction::Mul) {
257 if (!LHSBO && LHS->hasOneUse() && BinaryOperator::isNeg(LHS)) {
258 LHS = LowerNegateToMultiply(cast<Instruction>(LHS));
259 LHSBO = isReassociableOp(LHS, Opcode);
261 if (!RHSBO && RHS->hasOneUse() && BinaryOperator::isNeg(RHS)) {
262 RHS = LowerNegateToMultiply(cast<Instruction>(RHS));
263 RHSBO = isReassociableOp(RHS, Opcode);
269 // Neither the LHS or RHS as part of the tree, thus this is a leaf. As
270 // such, just remember these operands and their rank.
271 Ops.push_back(ValueEntry(getRank(LHS), LHS));
272 Ops.push_back(ValueEntry(getRank(RHS), RHS));
274 // Clear the leaves out.
275 I->setOperand(0, UndefValue::get(I->getType()));
276 I->setOperand(1, UndefValue::get(I->getType()));
279 // Turn X+(Y+Z) -> (Y+Z)+X
280 std::swap(LHSBO, RHSBO);
282 bool Success = !I->swapOperands();
283 assert(Success && "swapOperands failed");
287 // Turn (A+B)+(C+D) -> (((A+B)+C)+D). This guarantees the the RHS is not
288 // part of the expression tree.
290 LHS = LHSBO = cast<BinaryOperator>(I->getOperand(0));
291 RHS = I->getOperand(1);
295 // Okay, now we know that the LHS is a nested expression and that the RHS is
296 // not. Perform reassociation.
297 assert(!isReassociableOp(RHS, Opcode) && "LinearizeExpr failed!");
299 // Move LHS right before I to make sure that the tree expression dominates all
301 LHSBO->moveBefore(I);
303 // Linearize the expression tree on the LHS.
304 LinearizeExprTree(LHSBO, Ops);
306 // Remember the RHS operand and its rank.
307 Ops.push_back(ValueEntry(getRank(RHS), RHS));
309 // Clear the RHS leaf out.
310 I->setOperand(1, UndefValue::get(I->getType()));
313 // RewriteExprTree - Now that the operands for this expression tree are
314 // linearized and optimized, emit them in-order. This function is written to be
316 void Reassociate::RewriteExprTree(BinaryOperator *I,
317 std::vector<ValueEntry> &Ops,
319 if (i+2 == Ops.size()) {
320 if (I->getOperand(0) != Ops[i].Op ||
321 I->getOperand(1) != Ops[i+1].Op) {
322 Value *OldLHS = I->getOperand(0);
323 DEBUG(std::cerr << "RA: " << *I);
324 I->setOperand(0, Ops[i].Op);
325 I->setOperand(1, Ops[i+1].Op);
326 DEBUG(std::cerr << "TO: " << *I);
330 // If we reassociated a tree to fewer operands (e.g. (1+a+2) -> (a+3)
331 // delete the extra, now dead, nodes.
332 RemoveDeadBinaryOp(OldLHS);
336 assert(i+2 < Ops.size() && "Ops index out of range!");
338 if (I->getOperand(1) != Ops[i].Op) {
339 DEBUG(std::cerr << "RA: " << *I);
340 I->setOperand(1, Ops[i].Op);
341 DEBUG(std::cerr << "TO: " << *I);
346 BinaryOperator *LHS = cast<BinaryOperator>(I->getOperand(0));
347 assert(LHS->getOpcode() == I->getOpcode() &&
348 "Improper expression tree!");
350 // Compactify the tree instructions together with each other to guarantee
351 // that the expression tree is dominated by all of Ops.
353 RewriteExprTree(LHS, Ops, i+1);
358 // NegateValue - Insert instructions before the instruction pointed to by BI,
359 // that computes the negative version of the value specified. The negative
360 // version of the value is returned, and BI is left pointing at the instruction
361 // that should be processed next by the reassociation pass.
363 static Value *NegateValue(Value *V, Instruction *BI) {
364 // We are trying to expose opportunity for reassociation. One of the things
365 // that we want to do to achieve this is to push a negation as deep into an
366 // expression chain as possible, to expose the add instructions. In practice,
367 // this means that we turn this:
368 // X = -(A+12+C+D) into X = -A + -12 + -C + -D = -12 + -A + -C + -D
369 // so that later, a: Y = 12+X could get reassociated with the -12 to eliminate
370 // the constants. We assume that instcombine will clean up the mess later if
371 // we introduce tons of unnecessary negation instructions...
373 if (Instruction *I = dyn_cast<Instruction>(V))
374 if (I->getOpcode() == Instruction::Add && I->hasOneUse()) {
375 // Push the negates through the add.
376 I->setOperand(0, NegateValue(I->getOperand(0), BI));
377 I->setOperand(1, NegateValue(I->getOperand(1), BI));
379 // We must move the add instruction here, because the neg instructions do
380 // not dominate the old add instruction in general. By moving it, we are
381 // assured that the neg instructions we just inserted dominate the
382 // instruction we are about to insert after them.
385 I->setName(I->getName()+".neg");
389 // Insert a 'neg' instruction that subtracts the value from zero to get the
392 return BinaryOperator::createNeg(V, V->getName() + ".neg", BI);
395 /// BreakUpSubtract - If we have (X-Y), and if either X is an add, or if this is
396 /// only used by an add, transform this into (X+(0-Y)) to promote better
398 static Instruction *BreakUpSubtract(Instruction *Sub) {
399 // Don't bother to break this up unless either the LHS is an associable add or
400 // if this is only used by one.
401 if (!isReassociableOp(Sub->getOperand(0), Instruction::Add) &&
402 !isReassociableOp(Sub->getOperand(1), Instruction::Add) &&
403 !(Sub->hasOneUse() &&isReassociableOp(Sub->use_back(), Instruction::Add)))
406 // Convert a subtract into an add and a neg instruction... so that sub
407 // instructions can be commuted with other add instructions...
409 // Calculate the negative value of Operand 1 of the sub instruction...
410 // and set it as the RHS of the add instruction we just made...
412 std::string Name = Sub->getName();
414 Value *NegVal = NegateValue(Sub->getOperand(1), Sub);
416 BinaryOperator::createAdd(Sub->getOperand(0), NegVal, Name, Sub);
418 // Everyone now refers to the add instruction.
419 Sub->replaceAllUsesWith(New);
420 Sub->eraseFromParent();
422 DEBUG(std::cerr << "Negated: " << *New);
426 /// ConvertShiftToMul - If this is a shift of a reassociable multiply or is used
427 /// by one, change this into a multiply by a constant to assist with further
429 static Instruction *ConvertShiftToMul(Instruction *Shl) {
430 // If an operand of this shift is a reassociable multiply, or if the shift
431 // is used by a reassociable multiply or add, turn into a multiply.
432 if (isReassociableOp(Shl->getOperand(0), Instruction::Mul) ||
434 (isReassociableOp(Shl->use_back(), Instruction::Mul) ||
435 isReassociableOp(Shl->use_back(), Instruction::Add)))) {
436 Constant *MulCst = ConstantInt::get(Shl->getType(), 1);
437 MulCst = ConstantExpr::getShl(MulCst, cast<Constant>(Shl->getOperand(1)));
439 std::string Name = Shl->getName(); Shl->setName("");
440 Instruction *Mul = BinaryOperator::createMul(Shl->getOperand(0), MulCst,
442 Shl->replaceAllUsesWith(Mul);
443 Shl->eraseFromParent();
449 // Scan backwards and forwards among values with the same rank as element i to
450 // see if X exists. If X does not exist, return i.
451 static unsigned FindInOperandList(std::vector<ValueEntry> &Ops, unsigned i,
453 unsigned XRank = Ops[i].Rank;
454 unsigned e = Ops.size();
455 for (unsigned j = i+1; j != e && Ops[j].Rank == XRank; ++j)
459 for (unsigned j = i-1; j != ~0U && Ops[j].Rank == XRank; --j)
465 /// EmitAddTreeOfValues - Emit a tree of add instructions, summing Ops together
466 /// and returning the result. Insert the tree before I.
467 static Value *EmitAddTreeOfValues(Instruction *I, std::vector<Value*> &Ops) {
468 if (Ops.size() == 1) return Ops.back();
470 Value *V1 = Ops.back();
472 Value *V2 = EmitAddTreeOfValues(I, Ops);
473 return BinaryOperator::createAdd(V2, V1, "tmp", I);
476 /// RemoveFactorFromExpression - If V is an expression tree that is a
477 /// multiplication sequence, and if this sequence contains a multiply by Factor,
478 /// remove Factor from the tree and return the new tree.
479 Value *Reassociate::RemoveFactorFromExpression(Value *V, Value *Factor) {
480 BinaryOperator *BO = isReassociableOp(V, Instruction::Mul);
483 std::vector<ValueEntry> Factors;
484 LinearizeExprTree(BO, Factors);
486 bool FoundFactor = false;
487 for (unsigned i = 0, e = Factors.size(); i != e; ++i)
488 if (Factors[i].Op == Factor) {
490 Factors.erase(Factors.begin()+i);
494 // Make sure to restore the operands to the expression tree.
495 RewriteExprTree(BO, Factors);
499 if (Factors.size() == 1) return Factors[0].Op;
501 RewriteExprTree(BO, Factors);
505 /// FindSingleUseMultiplyFactors - If V is a single-use multiply, recursively
506 /// add its operands as factors, otherwise add V to the list of factors.
507 static void FindSingleUseMultiplyFactors(Value *V,
508 std::vector<Value*> &Factors) {
510 if ((!V->hasOneUse() && !V->use_empty()) ||
511 !(BO = dyn_cast<BinaryOperator>(V)) ||
512 BO->getOpcode() != Instruction::Mul) {
513 Factors.push_back(V);
517 // Otherwise, add the LHS and RHS to the list of factors.
518 FindSingleUseMultiplyFactors(BO->getOperand(1), Factors);
519 FindSingleUseMultiplyFactors(BO->getOperand(0), Factors);
524 Value *Reassociate::OptimizeExpression(BinaryOperator *I,
525 std::vector<ValueEntry> &Ops) {
526 // Now that we have the linearized expression tree, try to optimize it.
527 // Start by folding any constants that we found.
528 bool IterateOptimization = false;
529 if (Ops.size() == 1) return Ops[0].Op;
531 unsigned Opcode = I->getOpcode();
533 if (Constant *V1 = dyn_cast<Constant>(Ops[Ops.size()-2].Op))
534 if (Constant *V2 = dyn_cast<Constant>(Ops.back().Op)) {
536 Ops.back().Op = ConstantExpr::get(Opcode, V1, V2);
537 return OptimizeExpression(I, Ops);
540 // Check for destructive annihilation due to a constant being used.
541 if (ConstantIntegral *CstVal = dyn_cast<ConstantIntegral>(Ops.back().Op))
544 case Instruction::And:
545 if (CstVal->isNullValue()) { // ... & 0 -> 0
548 } else if (CstVal->isAllOnesValue()) { // ... & -1 -> ...
552 case Instruction::Mul:
553 if (CstVal->isNullValue()) { // ... * 0 -> 0
556 } else if (cast<ConstantInt>(CstVal)->getZExtValue() == 1) {
557 Ops.pop_back(); // ... * 1 -> ...
560 case Instruction::Or:
561 if (CstVal->isAllOnesValue()) { // ... | -1 -> -1
566 case Instruction::Add:
567 case Instruction::Xor:
568 if (CstVal->isNullValue()) // ... [|^+] 0 -> ...
572 if (Ops.size() == 1) return Ops[0].Op;
574 // Handle destructive annihilation do to identities between elements in the
575 // argument list here.
578 case Instruction::And:
579 case Instruction::Or:
580 case Instruction::Xor:
581 // Scan the operand lists looking for X and ~X pairs, along with X,X pairs.
582 // If we find any, we can simplify the expression. X&~X == 0, X|~X == -1.
583 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
584 // First, check for X and ~X in the operand list.
585 assert(i < Ops.size());
586 if (BinaryOperator::isNot(Ops[i].Op)) { // Cannot occur for ^.
587 Value *X = BinaryOperator::getNotArgument(Ops[i].Op);
588 unsigned FoundX = FindInOperandList(Ops, i, X);
590 if (Opcode == Instruction::And) { // ...&X&~X = 0
592 return Constant::getNullValue(X->getType());
593 } else if (Opcode == Instruction::Or) { // ...|X|~X = -1
595 return ConstantIntegral::getAllOnesValue(X->getType());
600 // Next, check for duplicate pairs of values, which we assume are next to
601 // each other, due to our sorting criteria.
602 assert(i < Ops.size());
603 if (i+1 != Ops.size() && Ops[i+1].Op == Ops[i].Op) {
604 if (Opcode == Instruction::And || Opcode == Instruction::Or) {
605 // Drop duplicate values.
606 Ops.erase(Ops.begin()+i);
608 IterateOptimization = true;
611 assert(Opcode == Instruction::Xor);
614 return Constant::getNullValue(Ops[0].Op->getType());
617 Ops.erase(Ops.begin()+i, Ops.begin()+i+2);
619 IterateOptimization = true;
626 case Instruction::Add:
627 // Scan the operand lists looking for X and -X pairs. If we find any, we
628 // can simplify the expression. X+-X == 0.
629 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
630 assert(i < Ops.size());
631 // Check for X and -X in the operand list.
632 if (BinaryOperator::isNeg(Ops[i].Op)) {
633 Value *X = BinaryOperator::getNegArgument(Ops[i].Op);
634 unsigned FoundX = FindInOperandList(Ops, i, X);
636 // Remove X and -X from the operand list.
637 if (Ops.size() == 2) {
639 return Constant::getNullValue(X->getType());
641 Ops.erase(Ops.begin()+i);
645 --i; // Need to back up an extra one.
646 Ops.erase(Ops.begin()+FoundX);
647 IterateOptimization = true;
649 --i; // Revisit element.
650 e -= 2; // Removed two elements.
657 // Scan the operand list, checking to see if there are any common factors
658 // between operands. Consider something like A*A+A*B*C+D. We would like to
659 // reassociate this to A*(A+B*C)+D, which reduces the number of multiplies.
660 // To efficiently find this, we count the number of times a factor occurs
661 // for any ADD operands that are MULs.
662 std::map<Value*, unsigned> FactorOccurrences;
664 Value *MaxOccVal = 0;
665 if (!I->getType()->isFloatingPoint()) {
666 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
667 if (BinaryOperator *BOp = dyn_cast<BinaryOperator>(Ops[i].Op))
668 if (BOp->getOpcode() == Instruction::Mul && BOp->use_empty()) {
669 // Compute all of the factors of this added value.
670 std::vector<Value*> Factors;
671 FindSingleUseMultiplyFactors(BOp, Factors);
672 assert(Factors.size() > 1 && "Bad linearize!");
674 // Add one to FactorOccurrences for each unique factor in this op.
675 if (Factors.size() == 2) {
676 unsigned Occ = ++FactorOccurrences[Factors[0]];
677 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[0]; }
678 if (Factors[0] != Factors[1]) { // Don't double count A*A.
679 Occ = ++FactorOccurrences[Factors[1]];
680 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[1]; }
683 std::set<Value*> Duplicates;
684 for (unsigned i = 0, e = Factors.size(); i != e; ++i)
685 if (Duplicates.insert(Factors[i]).second) {
686 unsigned Occ = ++FactorOccurrences[Factors[i]];
687 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[i]; }
694 // If any factor occurred more than one time, we can pull it out.
696 DEBUG(std::cerr << "\nFACTORING [" << MaxOcc << "]: "
697 << *MaxOccVal << "\n");
699 // Create a new instruction that uses the MaxOccVal twice. If we don't do
700 // this, we could otherwise run into situations where removing a factor
701 // from an expression will drop a use of maxocc, and this can cause
702 // RemoveFactorFromExpression on successive values to behave differently.
703 Instruction *DummyInst = BinaryOperator::createAdd(MaxOccVal, MaxOccVal);
704 std::vector<Value*> NewMulOps;
705 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
706 if (Value *V = RemoveFactorFromExpression(Ops[i].Op, MaxOccVal)) {
707 NewMulOps.push_back(V);
708 Ops.erase(Ops.begin()+i);
713 // No need for extra uses anymore.
716 unsigned NumAddedValues = NewMulOps.size();
717 Value *V = EmitAddTreeOfValues(I, NewMulOps);
718 Value *V2 = BinaryOperator::createMul(V, MaxOccVal, "tmp", I);
720 // Now that we have inserted V and its sole use, optimize it. This allows
721 // us to handle cases that require multiple factoring steps, such as this:
722 // A*A*B + A*A*C --> A*(A*B+A*C) --> A*(A*(B+C))
723 if (NumAddedValues > 1)
724 ReassociateExpression(cast<BinaryOperator>(V));
731 // Add the new value to the list of things being added.
732 Ops.insert(Ops.begin(), ValueEntry(getRank(V2), V2));
734 // Rewrite the tree so that there is now a use of V.
735 RewriteExprTree(I, Ops);
736 return OptimizeExpression(I, Ops);
739 //case Instruction::Mul:
742 if (IterateOptimization)
743 return OptimizeExpression(I, Ops);
748 /// ReassociateBB - Inspect all of the instructions in this basic block,
749 /// reassociating them as we go.
750 void Reassociate::ReassociateBB(BasicBlock *BB) {
751 for (BasicBlock::iterator BBI = BB->begin(); BBI != BB->end(); ) {
752 Instruction *BI = BBI++;
753 if (BI->getOpcode() == Instruction::Shl &&
754 isa<ConstantInt>(BI->getOperand(1)))
755 if (Instruction *NI = ConvertShiftToMul(BI)) {
760 // Reject cases where it is pointless to do this.
761 if (!isa<BinaryOperator>(BI) || BI->getType()->isFloatingPoint() ||
762 isa<PackedType>(BI->getType()))
763 continue; // Floating point ops are not associative.
765 // If this is a subtract instruction which is not already in negate form,
766 // see if we can convert it to X+-Y.
767 if (BI->getOpcode() == Instruction::Sub) {
768 if (!BinaryOperator::isNeg(BI)) {
769 if (Instruction *NI = BreakUpSubtract(BI)) {
774 // Otherwise, this is a negation. See if the operand is a multiply tree
775 // and if this is not an inner node of a multiply tree.
776 if (isReassociableOp(BI->getOperand(1), Instruction::Mul) &&
778 !isReassociableOp(BI->use_back(), Instruction::Mul))) {
779 BI = LowerNegateToMultiply(BI);
785 // If this instruction is a commutative binary operator, process it.
786 if (!BI->isAssociative()) continue;
787 BinaryOperator *I = cast<BinaryOperator>(BI);
789 // If this is an interior node of a reassociable tree, ignore it until we
790 // get to the root of the tree, to avoid N^2 analysis.
791 if (I->hasOneUse() && isReassociableOp(I->use_back(), I->getOpcode()))
794 // If this is an add tree that is used by a sub instruction, ignore it
795 // until we process the subtract.
796 if (I->hasOneUse() && I->getOpcode() == Instruction::Add &&
797 cast<Instruction>(I->use_back())->getOpcode() == Instruction::Sub)
800 ReassociateExpression(I);
804 void Reassociate::ReassociateExpression(BinaryOperator *I) {
806 // First, walk the expression tree, linearizing the tree, collecting
807 std::vector<ValueEntry> Ops;
808 LinearizeExprTree(I, Ops);
810 DEBUG(std::cerr << "RAIn:\t"; PrintOps(I, Ops);
813 // Now that we have linearized the tree to a list and have gathered all of
814 // the operands and their ranks, sort the operands by their rank. Use a
815 // stable_sort so that values with equal ranks will have their relative
816 // positions maintained (and so the compiler is deterministic). Note that
817 // this sorts so that the highest ranking values end up at the beginning of
819 std::stable_sort(Ops.begin(), Ops.end());
821 // OptimizeExpression - Now that we have the expression tree in a convenient
822 // sorted form, optimize it globally if possible.
823 if (Value *V = OptimizeExpression(I, Ops)) {
824 // This expression tree simplified to something that isn't a tree,
826 DEBUG(std::cerr << "Reassoc to scalar: " << *V << "\n");
827 I->replaceAllUsesWith(V);
828 RemoveDeadBinaryOp(I);
832 // We want to sink immediates as deeply as possible except in the case where
833 // this is a multiply tree used only by an add, and the immediate is a -1.
834 // In this case we reassociate to put the negation on the outside so that we
835 // can fold the negation into the add: (-X)*Y + Z -> Z-X*Y
836 if (I->getOpcode() == Instruction::Mul && I->hasOneUse() &&
837 cast<Instruction>(I->use_back())->getOpcode() == Instruction::Add &&
838 isa<ConstantInt>(Ops.back().Op) &&
839 cast<ConstantInt>(Ops.back().Op)->isAllOnesValue()) {
840 Ops.insert(Ops.begin(), Ops.back());
844 DEBUG(std::cerr << "RAOut:\t"; PrintOps(I, Ops);
847 if (Ops.size() == 1) {
848 // This expression tree simplified to something that isn't a tree,
850 I->replaceAllUsesWith(Ops[0].Op);
851 RemoveDeadBinaryOp(I);
853 // Now that we ordered and optimized the expressions, splat them back into
854 // the expression tree, removing any unneeded nodes.
855 RewriteExprTree(I, Ops);
860 bool Reassociate::runOnFunction(Function &F) {
861 // Recalculate the rank map for F
865 for (Function::iterator FI = F.begin(), FE = F.end(); FI != FE; ++FI)
868 // We are done with the rank map...
870 ValueRankMap.clear();