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/Function.h"
27 #include "llvm/Instructions.h"
28 #include "llvm/Pass.h"
29 #include "llvm/Type.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, unsigned Idx,
83 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 RegisterOpt<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::Div ||
117 I->getOpcode() == Instruction::Rem)
122 void Reassociate::BuildRankMap(Function &F) {
125 // Assign distinct ranks to function arguments
126 for (Function::arg_iterator I = F.arg_begin(), E = F.arg_end(); I != E; ++I)
127 ValueRankMap[I] = ++i;
129 ReversePostOrderTraversal<Function*> RPOT(&F);
130 for (ReversePostOrderTraversal<Function*>::rpo_iterator I = RPOT.begin(),
131 E = RPOT.end(); I != E; ++I) {
133 unsigned BBRank = RankMap[BB] = ++i << 16;
135 // Walk the basic block, adding precomputed ranks for any instructions that
136 // we cannot move. This ensures that the ranks for these instructions are
137 // all different in the block.
138 for (BasicBlock::iterator I = BB->begin(), E = BB->end(); I != E; ++I)
139 if (isUnmovableInstruction(I))
140 ValueRankMap[I] = ++BBRank;
144 unsigned Reassociate::getRank(Value *V) {
145 if (isa<Argument>(V)) return ValueRankMap[V]; // Function argument...
147 Instruction *I = dyn_cast<Instruction>(V);
148 if (I == 0) return 0; // Otherwise it's a global or constant, rank 0.
150 unsigned &CachedRank = ValueRankMap[I];
151 if (CachedRank) return CachedRank; // Rank already known?
153 // If this is an expression, return the 1+MAX(rank(LHS), rank(RHS)) so that
154 // we can reassociate expressions for code motion! Since we do not recurse
155 // for PHI nodes, we cannot have infinite recursion here, because there
156 // cannot be loops in the value graph that do not go through PHI nodes.
157 unsigned Rank = 0, MaxRank = RankMap[I->getParent()];
158 for (unsigned i = 0, e = I->getNumOperands();
159 i != e && Rank != MaxRank; ++i)
160 Rank = std::max(Rank, getRank(I->getOperand(i)));
162 // If this is a not or neg instruction, do not count it for rank. This
163 // assures us that X and ~X will have the same rank.
164 if (!I->getType()->isIntegral() ||
165 (!BinaryOperator::isNot(I) && !BinaryOperator::isNeg(I)))
168 //DEBUG(std::cerr << "Calculated Rank[" << V->getName() << "] = "
171 return CachedRank = Rank;
174 /// isReassociableOp - Return true if V is an instruction of the specified
175 /// opcode and if it only has one use.
176 static BinaryOperator *isReassociableOp(Value *V, unsigned Opcode) {
177 if (V->hasOneUse() && isa<Instruction>(V) &&
178 cast<Instruction>(V)->getOpcode() == Opcode)
179 return cast<BinaryOperator>(V);
183 /// LowerNegateToMultiply - Replace 0-X with X*-1.
185 static Instruction *LowerNegateToMultiply(Instruction *Neg) {
187 if (Neg->getType()->isFloatingPoint())
188 Cst = ConstantFP::get(Neg->getType(), -1);
190 Cst = ConstantInt::getAllOnesValue(Neg->getType());
192 std::string NegName = Neg->getName(); Neg->setName("");
193 Instruction *Res = BinaryOperator::createMul(Neg->getOperand(1), Cst, NegName,
195 Neg->replaceAllUsesWith(Res);
196 Neg->eraseFromParent();
200 // Given an expression of the form '(A+B)+(D+C)', turn it into '(((A+B)+C)+D)'.
201 // Note that if D is also part of the expression tree that we recurse to
202 // linearize it as well. Besides that case, this does not recurse into A,B, or
204 void Reassociate::LinearizeExpr(BinaryOperator *I) {
205 BinaryOperator *LHS = cast<BinaryOperator>(I->getOperand(0));
206 BinaryOperator *RHS = cast<BinaryOperator>(I->getOperand(1));
207 assert(isReassociableOp(LHS, I->getOpcode()) &&
208 isReassociableOp(RHS, I->getOpcode()) &&
209 "Not an expression that needs linearization?");
211 DEBUG(std::cerr << "Linear" << *LHS << *RHS << *I);
213 // Move the RHS instruction to live immediately before I, avoiding breaking
214 // dominator properties.
217 // Move operands around to do the linearization.
218 I->setOperand(1, RHS->getOperand(0));
219 RHS->setOperand(0, LHS);
220 I->setOperand(0, RHS);
224 DEBUG(std::cerr << "Linearized: " << *I);
226 // If D is part of this expression tree, tail recurse.
227 if (isReassociableOp(I->getOperand(1), I->getOpcode()))
232 /// LinearizeExprTree - Given an associative binary expression tree, traverse
233 /// all of the uses putting it into canonical form. This forces a left-linear
234 /// form of the the expression (((a+b)+c)+d), and collects information about the
235 /// rank of the non-tree operands.
237 void Reassociate::LinearizeExprTree(BinaryOperator *I,
238 std::vector<ValueEntry> &Ops) {
239 Value *LHS = I->getOperand(0), *RHS = I->getOperand(1);
240 unsigned Opcode = I->getOpcode();
242 // First step, linearize the expression if it is in ((A+B)+(C+D)) form.
243 BinaryOperator *LHSBO = isReassociableOp(LHS, Opcode);
244 BinaryOperator *RHSBO = isReassociableOp(RHS, Opcode);
246 // If this is a multiply expression tree and it contains internal negations,
247 // transform them into multiplies by -1 so they can be reassociated.
248 if (I->getOpcode() == Instruction::Mul) {
249 if (!LHSBO && LHS->hasOneUse() && BinaryOperator::isNeg(LHS)) {
250 LHS = LowerNegateToMultiply(cast<Instruction>(LHS));
251 LHSBO = isReassociableOp(LHS, Opcode);
253 if (!RHSBO && RHS->hasOneUse() && BinaryOperator::isNeg(RHS)) {
254 RHS = LowerNegateToMultiply(cast<Instruction>(RHS));
255 RHSBO = isReassociableOp(RHS, Opcode);
261 // Neither the LHS or RHS as part of the tree, thus this is a leaf. As
262 // such, just remember these operands and their rank.
263 Ops.push_back(ValueEntry(getRank(LHS), LHS));
264 Ops.push_back(ValueEntry(getRank(RHS), RHS));
267 // Turn X+(Y+Z) -> (Y+Z)+X
268 std::swap(LHSBO, RHSBO);
270 bool Success = !I->swapOperands();
271 assert(Success && "swapOperands failed");
275 // Turn (A+B)+(C+D) -> (((A+B)+C)+D). This guarantees the the RHS is not
276 // part of the expression tree.
278 LHS = LHSBO = cast<BinaryOperator>(I->getOperand(0));
279 RHS = I->getOperand(1);
283 // Okay, now we know that the LHS is a nested expression and that the RHS is
284 // not. Perform reassociation.
285 assert(!isReassociableOp(RHS, Opcode) && "LinearizeExpr failed!");
287 // Move LHS right before I to make sure that the tree expression dominates all
289 LHSBO->moveBefore(I);
291 // Linearize the expression tree on the LHS.
292 LinearizeExprTree(LHSBO, Ops);
294 // Remember the RHS operand and its rank.
295 Ops.push_back(ValueEntry(getRank(RHS), RHS));
298 // RewriteExprTree - Now that the operands for this expression tree are
299 // linearized and optimized, emit them in-order. This function is written to be
301 void Reassociate::RewriteExprTree(BinaryOperator *I, unsigned i,
302 std::vector<ValueEntry> &Ops) {
303 if (i+2 == Ops.size()) {
304 if (I->getOperand(0) != Ops[i].Op ||
305 I->getOperand(1) != Ops[i+1].Op) {
306 Value *OldLHS = I->getOperand(0);
307 DEBUG(std::cerr << "RA: " << *I);
308 I->setOperand(0, Ops[i].Op);
309 I->setOperand(1, Ops[i+1].Op);
310 DEBUG(std::cerr << "TO: " << *I);
314 // If we reassociated a tree to fewer operands (e.g. (1+a+2) -> (a+3)
315 // delete the extra, now dead, nodes.
316 RemoveDeadBinaryOp(OldLHS);
320 assert(i+2 < Ops.size() && "Ops index out of range!");
322 if (I->getOperand(1) != Ops[i].Op) {
323 DEBUG(std::cerr << "RA: " << *I);
324 I->setOperand(1, Ops[i].Op);
325 DEBUG(std::cerr << "TO: " << *I);
330 BinaryOperator *LHS = cast<BinaryOperator>(I->getOperand(0));
331 assert(LHS->getOpcode() == I->getOpcode() &&
332 "Improper expression tree!");
334 // Compactify the tree instructions together with each other to guarantee
335 // that the expression tree is dominated by all of Ops.
337 RewriteExprTree(LHS, i+1, Ops);
342 // NegateValue - Insert instructions before the instruction pointed to by BI,
343 // that computes the negative version of the value specified. The negative
344 // version of the value is returned, and BI is left pointing at the instruction
345 // that should be processed next by the reassociation pass.
347 static Value *NegateValue(Value *V, Instruction *BI) {
348 // We are trying to expose opportunity for reassociation. One of the things
349 // that we want to do to achieve this is to push a negation as deep into an
350 // expression chain as possible, to expose the add instructions. In practice,
351 // this means that we turn this:
352 // X = -(A+12+C+D) into X = -A + -12 + -C + -D = -12 + -A + -C + -D
353 // so that later, a: Y = 12+X could get reassociated with the -12 to eliminate
354 // the constants. We assume that instcombine will clean up the mess later if
355 // we introduce tons of unnecessary negation instructions...
357 if (Instruction *I = dyn_cast<Instruction>(V))
358 if (I->getOpcode() == Instruction::Add && I->hasOneUse()) {
359 // Push the negates through the add.
360 I->setOperand(0, NegateValue(I->getOperand(0), BI));
361 I->setOperand(1, NegateValue(I->getOperand(1), BI));
363 // We must move the add instruction here, because the neg instructions do
364 // not dominate the old add instruction in general. By moving it, we are
365 // assured that the neg instructions we just inserted dominate the
366 // instruction we are about to insert after them.
369 I->setName(I->getName()+".neg");
373 // Insert a 'neg' instruction that subtracts the value from zero to get the
376 return BinaryOperator::createNeg(V, V->getName() + ".neg", BI);
379 /// BreakUpSubtract - If we have (X-Y), and if either X is an add, or if this is
380 /// only used by an add, transform this into (X+(0-Y)) to promote better
382 static Instruction *BreakUpSubtract(Instruction *Sub) {
383 // Don't bother to break this up unless either the LHS is an associable add or
384 // if this is only used by one.
385 if (!isReassociableOp(Sub->getOperand(0), Instruction::Add) &&
386 !isReassociableOp(Sub->getOperand(1), Instruction::Add) &&
387 !(Sub->hasOneUse() &&isReassociableOp(Sub->use_back(), Instruction::Add)))
390 // Convert a subtract into an add and a neg instruction... so that sub
391 // instructions can be commuted with other add instructions...
393 // Calculate the negative value of Operand 1 of the sub instruction...
394 // and set it as the RHS of the add instruction we just made...
396 std::string Name = Sub->getName();
398 Value *NegVal = NegateValue(Sub->getOperand(1), Sub);
400 BinaryOperator::createAdd(Sub->getOperand(0), NegVal, Name, Sub);
402 // Everyone now refers to the add instruction.
403 Sub->replaceAllUsesWith(New);
404 Sub->eraseFromParent();
406 DEBUG(std::cerr << "Negated: " << *New);
410 /// ConvertShiftToMul - If this is a shift of a reassociable multiply or is used
411 /// by one, change this into a multiply by a constant to assist with further
413 static Instruction *ConvertShiftToMul(Instruction *Shl) {
414 // If an operand of this shift is a reassociable multiply, or if the shift
415 // is used by a reassociable multiply or add, turn into a multiply.
416 if (isReassociableOp(Shl->getOperand(0), Instruction::Mul) ||
418 (isReassociableOp(Shl->use_back(), Instruction::Mul) ||
419 isReassociableOp(Shl->use_back(), Instruction::Add)))) {
420 Constant *MulCst = ConstantInt::get(Shl->getType(), 1);
421 MulCst = ConstantExpr::getShl(MulCst, cast<Constant>(Shl->getOperand(1)));
423 std::string Name = Shl->getName(); Shl->setName("");
424 Instruction *Mul = BinaryOperator::createMul(Shl->getOperand(0), MulCst,
426 Shl->replaceAllUsesWith(Mul);
427 Shl->eraseFromParent();
433 // Scan backwards and forwards among values with the same rank as element i to
434 // see if X exists. If X does not exist, return i.
435 static unsigned FindInOperandList(std::vector<ValueEntry> &Ops, unsigned i,
437 unsigned XRank = Ops[i].Rank;
438 unsigned e = Ops.size();
439 for (unsigned j = i+1; j != e && Ops[j].Rank == XRank; ++j)
443 for (unsigned j = i-1; j != ~0U && Ops[j].Rank == XRank; --j)
449 /// EmitAddTreeOfValues - Emit a tree of add instructions, summing Ops together
450 /// and returning the result. Insert the tree before I.
451 static Value *EmitAddTreeOfValues(Instruction *I, std::vector<Value*> &Ops) {
452 if (Ops.size() == 1) return Ops.back();
454 Value *V1 = Ops.back();
456 Value *V2 = EmitAddTreeOfValues(I, Ops);
457 return BinaryOperator::createAdd(V2, V1, "tmp", I);
460 /// RemoveFactorFromExpression - If V is an expression tree that is a
461 /// multiplication sequence, and if this sequence contains a multiply by Factor,
462 /// remove Factor from the tree and return the new tree.
463 Value *Reassociate::RemoveFactorFromExpression(Value *V, Value *Factor) {
464 BinaryOperator *BO = isReassociableOp(V, Instruction::Mul);
467 std::vector<ValueEntry> Factors;
468 LinearizeExprTree(BO, Factors);
470 bool FoundFactor = false;
471 for (unsigned i = 0, e = Factors.size(); i != e; ++i)
472 if (Factors[i].Op == Factor) {
474 Factors.erase(Factors.begin()+i);
477 if (!FoundFactor) return 0;
479 if (Factors.size() == 1) return Factors[0].Op;
481 RewriteExprTree(BO, 0, Factors);
486 Value *Reassociate::OptimizeExpression(BinaryOperator *I,
487 std::vector<ValueEntry> &Ops) {
488 // Now that we have the linearized expression tree, try to optimize it.
489 // Start by folding any constants that we found.
490 bool IterateOptimization = false;
491 if (Ops.size() == 1) return Ops[0].Op;
493 unsigned Opcode = I->getOpcode();
495 if (Constant *V1 = dyn_cast<Constant>(Ops[Ops.size()-2].Op))
496 if (Constant *V2 = dyn_cast<Constant>(Ops.back().Op)) {
498 Ops.back().Op = ConstantExpr::get(Opcode, V1, V2);
499 return OptimizeExpression(I, Ops);
502 // Check for destructive annihilation due to a constant being used.
503 if (ConstantIntegral *CstVal = dyn_cast<ConstantIntegral>(Ops.back().Op))
506 case Instruction::And:
507 if (CstVal->isNullValue()) { // ... & 0 -> 0
510 } else if (CstVal->isAllOnesValue()) { // ... & -1 -> ...
514 case Instruction::Mul:
515 if (CstVal->isNullValue()) { // ... * 0 -> 0
518 } else if (cast<ConstantInt>(CstVal)->getRawValue() == 1) {
519 Ops.pop_back(); // ... * 1 -> ...
522 case Instruction::Or:
523 if (CstVal->isAllOnesValue()) { // ... | -1 -> -1
528 case Instruction::Add:
529 case Instruction::Xor:
530 if (CstVal->isNullValue()) // ... [|^+] 0 -> ...
534 if (Ops.size() == 1) return Ops[0].Op;
536 // Handle destructive annihilation do to identities between elements in the
537 // argument list here.
540 case Instruction::And:
541 case Instruction::Or:
542 case Instruction::Xor:
543 // Scan the operand lists looking for X and ~X pairs, along with X,X pairs.
544 // If we find any, we can simplify the expression. X&~X == 0, X|~X == -1.
545 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
546 // First, check for X and ~X in the operand list.
547 assert(i < Ops.size());
548 if (BinaryOperator::isNot(Ops[i].Op)) { // Cannot occur for ^.
549 Value *X = BinaryOperator::getNotArgument(Ops[i].Op);
550 unsigned FoundX = FindInOperandList(Ops, i, X);
552 if (Opcode == Instruction::And) { // ...&X&~X = 0
554 return Constant::getNullValue(X->getType());
555 } else if (Opcode == Instruction::Or) { // ...|X|~X = -1
557 return ConstantIntegral::getAllOnesValue(X->getType());
562 // Next, check for duplicate pairs of values, which we assume are next to
563 // each other, due to our sorting criteria.
564 assert(i < Ops.size());
565 if (i+1 != Ops.size() && Ops[i+1].Op == Ops[i].Op) {
566 if (Opcode == Instruction::And || Opcode == Instruction::Or) {
567 // Drop duplicate values.
568 Ops.erase(Ops.begin()+i);
570 IterateOptimization = true;
573 assert(Opcode == Instruction::Xor);
576 return Constant::getNullValue(Ops[0].Op->getType());
579 Ops.erase(Ops.begin()+i, Ops.begin()+i+2);
581 IterateOptimization = true;
588 case Instruction::Add:
589 // Scan the operand lists looking for X and -X pairs. If we find any, we
590 // can simplify the expression. X+-X == 0.
591 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
592 assert(i < Ops.size());
593 // Check for X and -X in the operand list.
594 if (BinaryOperator::isNeg(Ops[i].Op)) {
595 Value *X = BinaryOperator::getNegArgument(Ops[i].Op);
596 unsigned FoundX = FindInOperandList(Ops, i, X);
598 // Remove X and -X from the operand list.
599 if (Ops.size() == 2) {
601 return Constant::getNullValue(X->getType());
603 Ops.erase(Ops.begin()+i);
607 --i; // Need to back up an extra one.
608 Ops.erase(Ops.begin()+FoundX);
609 IterateOptimization = true;
611 --i; // Revisit element.
612 e -= 2; // Removed two elements.
619 // Scan the operand list, checking to see if there are any common factors
620 // between operands. Consider something like A*A+A*B*C+D. We would like to
621 // reassociate this to A*(A+B*C)+D, which reduces the number of multiplies.
622 // To efficiently find this, we count the number of times a factor occurs
623 // for any ADD operands that are MULs.
624 std::map<Value*, unsigned> FactorOccurrences;
626 Value *MaxOccVal = 0;
627 if (!I->getType()->isFloatingPoint()) {
628 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
629 if (BinaryOperator *BOp = dyn_cast<BinaryOperator>(Ops[i].Op))
630 if (BOp->getOpcode() == Instruction::Mul && BOp->hasOneUse()) {
631 // Compute all of the factors of this added value.
632 std::vector<ValueEntry> Factors;
633 LinearizeExprTree(BOp, Factors);
634 assert(Factors.size() > 1 && "Bad linearize!");
636 // Add one to FactorOccurrences for each unique factor in this op.
637 if (Factors.size() == 2) {
638 unsigned Occ = ++FactorOccurrences[Factors[0].Op];
639 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[0].Op; }
640 if (Factors[0].Op != Factors[1].Op) { // Don't double count A*A.
641 Occ = ++FactorOccurrences[Factors[1].Op];
642 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[1].Op; }
645 std::set<Value*> Duplicates;
646 for (unsigned i = 0, e = Factors.size(); i != e; ++i)
647 if (Duplicates.insert(Factors[i].Op).second) {
648 unsigned Occ = ++FactorOccurrences[Factors[i].Op];
649 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[i].Op; }
656 // If any factor occurred more than one time, we can pull it out.
658 DEBUG(std::cerr << "\nFACTORING [" << MaxOcc << "]: "
659 << *MaxOccVal << "\n");
661 // Create a new instruction that uses the MaxOccVal twice. If we don't do
662 // this, we could otherwise run into situations where removing a factor
663 // from an expression will drop a use of maxocc, and this can cause
664 // RemoveFactorFromExpression on successive values to behave differently.
665 Instruction *DummyInst = BinaryOperator::createAdd(MaxOccVal, MaxOccVal);
666 std::vector<Value*> NewMulOps;
667 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
668 if (Value *V = RemoveFactorFromExpression(Ops[i].Op, MaxOccVal)) {
669 NewMulOps.push_back(V);
670 Ops.erase(Ops.begin()+i);
675 // No need for extra uses anymore.
678 Value *V = EmitAddTreeOfValues(I, NewMulOps);
679 // FIXME: Must optimize V now, to handle this case:
680 // A*A*B + A*A*C -> A*(A*B+A*C) -> A*(A*(B+C))
681 V = BinaryOperator::createMul(V, MaxOccVal, "tmp", I);
688 // Add the new value to the list of things being added.
689 Ops.insert(Ops.begin(), ValueEntry(getRank(V), V));
691 // Rewrite the tree so that there is now a use of V.
692 RewriteExprTree(I, 0, Ops);
693 return OptimizeExpression(I, Ops);
696 //case Instruction::Mul:
699 if (IterateOptimization)
700 return OptimizeExpression(I, Ops);
705 /// ReassociateBB - Inspect all of the instructions in this basic block,
706 /// reassociating them as we go.
707 void Reassociate::ReassociateBB(BasicBlock *BB) {
708 for (BasicBlock::iterator BBI = BB->begin(); BBI != BB->end(); ) {
709 Instruction *BI = BBI++;
710 if (BI->getOpcode() == Instruction::Shl &&
711 isa<ConstantInt>(BI->getOperand(1)))
712 if (Instruction *NI = ConvertShiftToMul(BI)) {
717 // Reject cases where it is pointless to do this.
718 if (!isa<BinaryOperator>(BI) || BI->getType()->isFloatingPoint())
719 continue; // Floating point ops are not associative.
721 // If this is a subtract instruction which is not already in negate form,
722 // see if we can convert it to X+-Y.
723 if (BI->getOpcode() == Instruction::Sub) {
724 if (!BinaryOperator::isNeg(BI)) {
725 if (Instruction *NI = BreakUpSubtract(BI)) {
730 // Otherwise, this is a negation. See if the operand is a multiply tree
731 // and if this is not an inner node of a multiply tree.
732 if (isReassociableOp(BI->getOperand(1), Instruction::Mul) &&
734 !isReassociableOp(BI->use_back(), Instruction::Mul))) {
735 BI = LowerNegateToMultiply(BI);
741 // If this instruction is a commutative binary operator, process it.
742 if (!BI->isAssociative()) continue;
743 BinaryOperator *I = cast<BinaryOperator>(BI);
745 // If this is an interior node of a reassociable tree, ignore it until we
746 // get to the root of the tree, to avoid N^2 analysis.
747 if (I->hasOneUse() && isReassociableOp(I->use_back(), I->getOpcode()))
750 // If this is an add tree that is used by a sub instruction, ignore it
751 // until we process the subtract.
752 if (I->hasOneUse() && I->getOpcode() == Instruction::Add &&
753 cast<Instruction>(I->use_back())->getOpcode() == Instruction::Sub)
756 ReassociateExpression(I);
760 void Reassociate::ReassociateExpression(BinaryOperator *I) {
762 // First, walk the expression tree, linearizing the tree, collecting
763 std::vector<ValueEntry> Ops;
764 LinearizeExprTree(I, Ops);
766 DEBUG(std::cerr << "RAIn:\t"; PrintOps(I, Ops);
769 // Now that we have linearized the tree to a list and have gathered all of
770 // the operands and their ranks, sort the operands by their rank. Use a
771 // stable_sort so that values with equal ranks will have their relative
772 // positions maintained (and so the compiler is deterministic). Note that
773 // this sorts so that the highest ranking values end up at the beginning of
775 std::stable_sort(Ops.begin(), Ops.end());
777 // OptimizeExpression - Now that we have the expression tree in a convenient
778 // sorted form, optimize it globally if possible.
779 if (Value *V = OptimizeExpression(I, Ops)) {
780 // This expression tree simplified to something that isn't a tree,
782 DEBUG(std::cerr << "Reassoc to scalar: " << *V << "\n");
783 I->replaceAllUsesWith(V);
784 RemoveDeadBinaryOp(I);
788 // We want to sink immediates as deeply as possible except in the case where
789 // this is a multiply tree used only by an add, and the immediate is a -1.
790 // In this case we reassociate to put the negation on the outside so that we
791 // can fold the negation into the add: (-X)*Y + Z -> Z-X*Y
792 if (I->getOpcode() == Instruction::Mul && I->hasOneUse() &&
793 cast<Instruction>(I->use_back())->getOpcode() == Instruction::Add &&
794 isa<ConstantInt>(Ops.back().Op) &&
795 cast<ConstantInt>(Ops.back().Op)->isAllOnesValue()) {
796 Ops.insert(Ops.begin(), Ops.back());
800 DEBUG(std::cerr << "RAOut:\t"; PrintOps(I, Ops);
803 if (Ops.size() == 1) {
804 // This expression tree simplified to something that isn't a tree,
806 I->replaceAllUsesWith(Ops[0].Op);
807 RemoveDeadBinaryOp(I);
809 // Now that we ordered and optimized the expressions, splat them back into
810 // the expression tree, removing any unneeded nodes.
811 RewriteExprTree(I, 0, Ops);
816 bool Reassociate::runOnFunction(Function &F) {
817 // Recalculate the rank map for F
821 for (Function::iterator FI = F.begin(), FE = F.end(); FI != FE; ++FI)
824 // We are done with the rank map...
826 ValueRankMap.clear();