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::Rem)
124 void Reassociate::BuildRankMap(Function &F) {
127 // Assign distinct ranks to function arguments
128 for (Function::arg_iterator I = F.arg_begin(), E = F.arg_end(); I != E; ++I)
129 ValueRankMap[I] = ++i;
131 ReversePostOrderTraversal<Function*> RPOT(&F);
132 for (ReversePostOrderTraversal<Function*>::rpo_iterator I = RPOT.begin(),
133 E = RPOT.end(); I != E; ++I) {
135 unsigned BBRank = RankMap[BB] = ++i << 16;
137 // Walk the basic block, adding precomputed ranks for any instructions that
138 // we cannot move. This ensures that the ranks for these instructions are
139 // all different in the block.
140 for (BasicBlock::iterator I = BB->begin(), E = BB->end(); I != E; ++I)
141 if (isUnmovableInstruction(I))
142 ValueRankMap[I] = ++BBRank;
146 unsigned Reassociate::getRank(Value *V) {
147 if (isa<Argument>(V)) return ValueRankMap[V]; // Function argument...
149 Instruction *I = dyn_cast<Instruction>(V);
150 if (I == 0) return 0; // Otherwise it's a global or constant, rank 0.
152 unsigned &CachedRank = ValueRankMap[I];
153 if (CachedRank) return CachedRank; // Rank already known?
155 // If this is an expression, return the 1+MAX(rank(LHS), rank(RHS)) so that
156 // we can reassociate expressions for code motion! Since we do not recurse
157 // for PHI nodes, we cannot have infinite recursion here, because there
158 // cannot be loops in the value graph that do not go through PHI nodes.
159 unsigned Rank = 0, MaxRank = RankMap[I->getParent()];
160 for (unsigned i = 0, e = I->getNumOperands();
161 i != e && Rank != MaxRank; ++i)
162 Rank = std::max(Rank, getRank(I->getOperand(i)));
164 // If this is a not or neg instruction, do not count it for rank. This
165 // assures us that X and ~X will have the same rank.
166 if (!I->getType()->isIntegral() ||
167 (!BinaryOperator::isNot(I) && !BinaryOperator::isNeg(I)))
170 //DEBUG(std::cerr << "Calculated Rank[" << V->getName() << "] = "
173 return CachedRank = Rank;
176 /// isReassociableOp - Return true if V is an instruction of the specified
177 /// opcode and if it only has one use.
178 static BinaryOperator *isReassociableOp(Value *V, unsigned Opcode) {
179 if ((V->hasOneUse() || V->use_empty()) && isa<Instruction>(V) &&
180 cast<Instruction>(V)->getOpcode() == Opcode)
181 return cast<BinaryOperator>(V);
185 /// LowerNegateToMultiply - Replace 0-X with X*-1.
187 static Instruction *LowerNegateToMultiply(Instruction *Neg) {
189 if (Neg->getType()->isFloatingPoint())
190 Cst = ConstantFP::get(Neg->getType(), -1);
192 Cst = ConstantInt::getAllOnesValue(Neg->getType());
194 std::string NegName = Neg->getName(); Neg->setName("");
195 Instruction *Res = BinaryOperator::createMul(Neg->getOperand(1), Cst, NegName,
197 Neg->replaceAllUsesWith(Res);
198 Neg->eraseFromParent();
202 // Given an expression of the form '(A+B)+(D+C)', turn it into '(((A+B)+C)+D)'.
203 // Note that if D is also part of the expression tree that we recurse to
204 // linearize it as well. Besides that case, this does not recurse into A,B, or
206 void Reassociate::LinearizeExpr(BinaryOperator *I) {
207 BinaryOperator *LHS = cast<BinaryOperator>(I->getOperand(0));
208 BinaryOperator *RHS = cast<BinaryOperator>(I->getOperand(1));
209 assert(isReassociableOp(LHS, I->getOpcode()) &&
210 isReassociableOp(RHS, I->getOpcode()) &&
211 "Not an expression that needs linearization?");
213 DEBUG(std::cerr << "Linear" << *LHS << *RHS << *I);
215 // Move the RHS instruction to live immediately before I, avoiding breaking
216 // dominator properties.
219 // Move operands around to do the linearization.
220 I->setOperand(1, RHS->getOperand(0));
221 RHS->setOperand(0, LHS);
222 I->setOperand(0, RHS);
226 DEBUG(std::cerr << "Linearized: " << *I);
228 // If D is part of this expression tree, tail recurse.
229 if (isReassociableOp(I->getOperand(1), I->getOpcode()))
234 /// LinearizeExprTree - Given an associative binary expression tree, traverse
235 /// all of the uses putting it into canonical form. This forces a left-linear
236 /// form of the the expression (((a+b)+c)+d), and collects information about the
237 /// rank of the non-tree operands.
239 /// NOTE: These intentionally destroys the expression tree operands (turning
240 /// them into undef values) to reduce #uses of the values. This means that the
241 /// caller MUST use something like RewriteExprTree to put the values back in.
243 void Reassociate::LinearizeExprTree(BinaryOperator *I,
244 std::vector<ValueEntry> &Ops) {
245 Value *LHS = I->getOperand(0), *RHS = I->getOperand(1);
246 unsigned Opcode = I->getOpcode();
248 // First step, linearize the expression if it is in ((A+B)+(C+D)) form.
249 BinaryOperator *LHSBO = isReassociableOp(LHS, Opcode);
250 BinaryOperator *RHSBO = isReassociableOp(RHS, Opcode);
252 // If this is a multiply expression tree and it contains internal negations,
253 // transform them into multiplies by -1 so they can be reassociated.
254 if (I->getOpcode() == Instruction::Mul) {
255 if (!LHSBO && LHS->hasOneUse() && BinaryOperator::isNeg(LHS)) {
256 LHS = LowerNegateToMultiply(cast<Instruction>(LHS));
257 LHSBO = isReassociableOp(LHS, Opcode);
259 if (!RHSBO && RHS->hasOneUse() && BinaryOperator::isNeg(RHS)) {
260 RHS = LowerNegateToMultiply(cast<Instruction>(RHS));
261 RHSBO = isReassociableOp(RHS, Opcode);
267 // Neither the LHS or RHS as part of the tree, thus this is a leaf. As
268 // such, just remember these operands and their rank.
269 Ops.push_back(ValueEntry(getRank(LHS), LHS));
270 Ops.push_back(ValueEntry(getRank(RHS), RHS));
272 // Clear the leaves out.
273 I->setOperand(0, UndefValue::get(I->getType()));
274 I->setOperand(1, UndefValue::get(I->getType()));
277 // Turn X+(Y+Z) -> (Y+Z)+X
278 std::swap(LHSBO, RHSBO);
280 bool Success = !I->swapOperands();
281 assert(Success && "swapOperands failed");
285 // Turn (A+B)+(C+D) -> (((A+B)+C)+D). This guarantees the the RHS is not
286 // part of the expression tree.
288 LHS = LHSBO = cast<BinaryOperator>(I->getOperand(0));
289 RHS = I->getOperand(1);
293 // Okay, now we know that the LHS is a nested expression and that the RHS is
294 // not. Perform reassociation.
295 assert(!isReassociableOp(RHS, Opcode) && "LinearizeExpr failed!");
297 // Move LHS right before I to make sure that the tree expression dominates all
299 LHSBO->moveBefore(I);
301 // Linearize the expression tree on the LHS.
302 LinearizeExprTree(LHSBO, Ops);
304 // Remember the RHS operand and its rank.
305 Ops.push_back(ValueEntry(getRank(RHS), RHS));
307 // Clear the RHS leaf out.
308 I->setOperand(1, UndefValue::get(I->getType()));
311 // RewriteExprTree - Now that the operands for this expression tree are
312 // linearized and optimized, emit them in-order. This function is written to be
314 void Reassociate::RewriteExprTree(BinaryOperator *I,
315 std::vector<ValueEntry> &Ops,
317 if (i+2 == Ops.size()) {
318 if (I->getOperand(0) != Ops[i].Op ||
319 I->getOperand(1) != Ops[i+1].Op) {
320 Value *OldLHS = I->getOperand(0);
321 DEBUG(std::cerr << "RA: " << *I);
322 I->setOperand(0, Ops[i].Op);
323 I->setOperand(1, Ops[i+1].Op);
324 DEBUG(std::cerr << "TO: " << *I);
328 // If we reassociated a tree to fewer operands (e.g. (1+a+2) -> (a+3)
329 // delete the extra, now dead, nodes.
330 RemoveDeadBinaryOp(OldLHS);
334 assert(i+2 < Ops.size() && "Ops index out of range!");
336 if (I->getOperand(1) != Ops[i].Op) {
337 DEBUG(std::cerr << "RA: " << *I);
338 I->setOperand(1, Ops[i].Op);
339 DEBUG(std::cerr << "TO: " << *I);
344 BinaryOperator *LHS = cast<BinaryOperator>(I->getOperand(0));
345 assert(LHS->getOpcode() == I->getOpcode() &&
346 "Improper expression tree!");
348 // Compactify the tree instructions together with each other to guarantee
349 // that the expression tree is dominated by all of Ops.
351 RewriteExprTree(LHS, Ops, i+1);
356 // NegateValue - Insert instructions before the instruction pointed to by BI,
357 // that computes the negative version of the value specified. The negative
358 // version of the value is returned, and BI is left pointing at the instruction
359 // that should be processed next by the reassociation pass.
361 static Value *NegateValue(Value *V, Instruction *BI) {
362 // We are trying to expose opportunity for reassociation. One of the things
363 // that we want to do to achieve this is to push a negation as deep into an
364 // expression chain as possible, to expose the add instructions. In practice,
365 // this means that we turn this:
366 // X = -(A+12+C+D) into X = -A + -12 + -C + -D = -12 + -A + -C + -D
367 // so that later, a: Y = 12+X could get reassociated with the -12 to eliminate
368 // the constants. We assume that instcombine will clean up the mess later if
369 // we introduce tons of unnecessary negation instructions...
371 if (Instruction *I = dyn_cast<Instruction>(V))
372 if (I->getOpcode() == Instruction::Add && I->hasOneUse()) {
373 // Push the negates through the add.
374 I->setOperand(0, NegateValue(I->getOperand(0), BI));
375 I->setOperand(1, NegateValue(I->getOperand(1), BI));
377 // We must move the add instruction here, because the neg instructions do
378 // not dominate the old add instruction in general. By moving it, we are
379 // assured that the neg instructions we just inserted dominate the
380 // instruction we are about to insert after them.
383 I->setName(I->getName()+".neg");
387 // Insert a 'neg' instruction that subtracts the value from zero to get the
390 return BinaryOperator::createNeg(V, V->getName() + ".neg", BI);
393 /// BreakUpSubtract - If we have (X-Y), and if either X is an add, or if this is
394 /// only used by an add, transform this into (X+(0-Y)) to promote better
396 static Instruction *BreakUpSubtract(Instruction *Sub) {
397 // Don't bother to break this up unless either the LHS is an associable add or
398 // if this is only used by one.
399 if (!isReassociableOp(Sub->getOperand(0), Instruction::Add) &&
400 !isReassociableOp(Sub->getOperand(1), Instruction::Add) &&
401 !(Sub->hasOneUse() &&isReassociableOp(Sub->use_back(), Instruction::Add)))
404 // Convert a subtract into an add and a neg instruction... so that sub
405 // instructions can be commuted with other add instructions...
407 // Calculate the negative value of Operand 1 of the sub instruction...
408 // and set it as the RHS of the add instruction we just made...
410 std::string Name = Sub->getName();
412 Value *NegVal = NegateValue(Sub->getOperand(1), Sub);
414 BinaryOperator::createAdd(Sub->getOperand(0), NegVal, Name, Sub);
416 // Everyone now refers to the add instruction.
417 Sub->replaceAllUsesWith(New);
418 Sub->eraseFromParent();
420 DEBUG(std::cerr << "Negated: " << *New);
424 /// ConvertShiftToMul - If this is a shift of a reassociable multiply or is used
425 /// by one, change this into a multiply by a constant to assist with further
427 static Instruction *ConvertShiftToMul(Instruction *Shl) {
428 // If an operand of this shift is a reassociable multiply, or if the shift
429 // is used by a reassociable multiply or add, turn into a multiply.
430 if (isReassociableOp(Shl->getOperand(0), Instruction::Mul) ||
432 (isReassociableOp(Shl->use_back(), Instruction::Mul) ||
433 isReassociableOp(Shl->use_back(), Instruction::Add)))) {
434 Constant *MulCst = ConstantInt::get(Shl->getType(), 1);
435 MulCst = ConstantExpr::getShl(MulCst, cast<Constant>(Shl->getOperand(1)));
437 std::string Name = Shl->getName(); Shl->setName("");
438 Instruction *Mul = BinaryOperator::createMul(Shl->getOperand(0), MulCst,
440 Shl->replaceAllUsesWith(Mul);
441 Shl->eraseFromParent();
447 // Scan backwards and forwards among values with the same rank as element i to
448 // see if X exists. If X does not exist, return i.
449 static unsigned FindInOperandList(std::vector<ValueEntry> &Ops, unsigned i,
451 unsigned XRank = Ops[i].Rank;
452 unsigned e = Ops.size();
453 for (unsigned j = i+1; j != e && Ops[j].Rank == XRank; ++j)
457 for (unsigned j = i-1; j != ~0U && Ops[j].Rank == XRank; --j)
463 /// EmitAddTreeOfValues - Emit a tree of add instructions, summing Ops together
464 /// and returning the result. Insert the tree before I.
465 static Value *EmitAddTreeOfValues(Instruction *I, std::vector<Value*> &Ops) {
466 if (Ops.size() == 1) return Ops.back();
468 Value *V1 = Ops.back();
470 Value *V2 = EmitAddTreeOfValues(I, Ops);
471 return BinaryOperator::createAdd(V2, V1, "tmp", I);
474 /// RemoveFactorFromExpression - If V is an expression tree that is a
475 /// multiplication sequence, and if this sequence contains a multiply by Factor,
476 /// remove Factor from the tree and return the new tree.
477 Value *Reassociate::RemoveFactorFromExpression(Value *V, Value *Factor) {
478 BinaryOperator *BO = isReassociableOp(V, Instruction::Mul);
481 std::vector<ValueEntry> Factors;
482 LinearizeExprTree(BO, Factors);
484 bool FoundFactor = false;
485 for (unsigned i = 0, e = Factors.size(); i != e; ++i)
486 if (Factors[i].Op == Factor) {
488 Factors.erase(Factors.begin()+i);
492 // Make sure to restore the operands to the expression tree.
493 RewriteExprTree(BO, Factors);
497 if (Factors.size() == 1) return Factors[0].Op;
499 RewriteExprTree(BO, Factors);
503 /// FindSingleUseMultiplyFactors - If V is a single-use multiply, recursively
504 /// add its operands as factors, otherwise add V to the list of factors.
505 static void FindSingleUseMultiplyFactors(Value *V,
506 std::vector<Value*> &Factors) {
508 if ((!V->hasOneUse() && !V->use_empty()) ||
509 !(BO = dyn_cast<BinaryOperator>(V)) ||
510 BO->getOpcode() != Instruction::Mul) {
511 Factors.push_back(V);
515 // Otherwise, add the LHS and RHS to the list of factors.
516 FindSingleUseMultiplyFactors(BO->getOperand(1), Factors);
517 FindSingleUseMultiplyFactors(BO->getOperand(0), Factors);
522 Value *Reassociate::OptimizeExpression(BinaryOperator *I,
523 std::vector<ValueEntry> &Ops) {
524 // Now that we have the linearized expression tree, try to optimize it.
525 // Start by folding any constants that we found.
526 bool IterateOptimization = false;
527 if (Ops.size() == 1) return Ops[0].Op;
529 unsigned Opcode = I->getOpcode();
531 if (Constant *V1 = dyn_cast<Constant>(Ops[Ops.size()-2].Op))
532 if (Constant *V2 = dyn_cast<Constant>(Ops.back().Op)) {
534 Ops.back().Op = ConstantExpr::get(Opcode, V1, V2);
535 return OptimizeExpression(I, Ops);
538 // Check for destructive annihilation due to a constant being used.
539 if (ConstantIntegral *CstVal = dyn_cast<ConstantIntegral>(Ops.back().Op))
542 case Instruction::And:
543 if (CstVal->isNullValue()) { // ... & 0 -> 0
546 } else if (CstVal->isAllOnesValue()) { // ... & -1 -> ...
550 case Instruction::Mul:
551 if (CstVal->isNullValue()) { // ... * 0 -> 0
554 } else if (cast<ConstantInt>(CstVal)->getZExtValue() == 1) {
555 Ops.pop_back(); // ... * 1 -> ...
558 case Instruction::Or:
559 if (CstVal->isAllOnesValue()) { // ... | -1 -> -1
564 case Instruction::Add:
565 case Instruction::Xor:
566 if (CstVal->isNullValue()) // ... [|^+] 0 -> ...
570 if (Ops.size() == 1) return Ops[0].Op;
572 // Handle destructive annihilation do to identities between elements in the
573 // argument list here.
576 case Instruction::And:
577 case Instruction::Or:
578 case Instruction::Xor:
579 // Scan the operand lists looking for X and ~X pairs, along with X,X pairs.
580 // If we find any, we can simplify the expression. X&~X == 0, X|~X == -1.
581 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
582 // First, check for X and ~X in the operand list.
583 assert(i < Ops.size());
584 if (BinaryOperator::isNot(Ops[i].Op)) { // Cannot occur for ^.
585 Value *X = BinaryOperator::getNotArgument(Ops[i].Op);
586 unsigned FoundX = FindInOperandList(Ops, i, X);
588 if (Opcode == Instruction::And) { // ...&X&~X = 0
590 return Constant::getNullValue(X->getType());
591 } else if (Opcode == Instruction::Or) { // ...|X|~X = -1
593 return ConstantIntegral::getAllOnesValue(X->getType());
598 // Next, check for duplicate pairs of values, which we assume are next to
599 // each other, due to our sorting criteria.
600 assert(i < Ops.size());
601 if (i+1 != Ops.size() && Ops[i+1].Op == Ops[i].Op) {
602 if (Opcode == Instruction::And || Opcode == Instruction::Or) {
603 // Drop duplicate values.
604 Ops.erase(Ops.begin()+i);
606 IterateOptimization = true;
609 assert(Opcode == Instruction::Xor);
612 return Constant::getNullValue(Ops[0].Op->getType());
615 Ops.erase(Ops.begin()+i, Ops.begin()+i+2);
617 IterateOptimization = true;
624 case Instruction::Add:
625 // Scan the operand lists looking for X and -X pairs. If we find any, we
626 // can simplify the expression. X+-X == 0.
627 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
628 assert(i < Ops.size());
629 // Check for X and -X in the operand list.
630 if (BinaryOperator::isNeg(Ops[i].Op)) {
631 Value *X = BinaryOperator::getNegArgument(Ops[i].Op);
632 unsigned FoundX = FindInOperandList(Ops, i, X);
634 // Remove X and -X from the operand list.
635 if (Ops.size() == 2) {
637 return Constant::getNullValue(X->getType());
639 Ops.erase(Ops.begin()+i);
643 --i; // Need to back up an extra one.
644 Ops.erase(Ops.begin()+FoundX);
645 IterateOptimization = true;
647 --i; // Revisit element.
648 e -= 2; // Removed two elements.
655 // Scan the operand list, checking to see if there are any common factors
656 // between operands. Consider something like A*A+A*B*C+D. We would like to
657 // reassociate this to A*(A+B*C)+D, which reduces the number of multiplies.
658 // To efficiently find this, we count the number of times a factor occurs
659 // for any ADD operands that are MULs.
660 std::map<Value*, unsigned> FactorOccurrences;
662 Value *MaxOccVal = 0;
663 if (!I->getType()->isFloatingPoint()) {
664 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
665 if (BinaryOperator *BOp = dyn_cast<BinaryOperator>(Ops[i].Op))
666 if (BOp->getOpcode() == Instruction::Mul && BOp->use_empty()) {
667 // Compute all of the factors of this added value.
668 std::vector<Value*> Factors;
669 FindSingleUseMultiplyFactors(BOp, Factors);
670 assert(Factors.size() > 1 && "Bad linearize!");
672 // Add one to FactorOccurrences for each unique factor in this op.
673 if (Factors.size() == 2) {
674 unsigned Occ = ++FactorOccurrences[Factors[0]];
675 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[0]; }
676 if (Factors[0] != Factors[1]) { // Don't double count A*A.
677 Occ = ++FactorOccurrences[Factors[1]];
678 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[1]; }
681 std::set<Value*> Duplicates;
682 for (unsigned i = 0, e = Factors.size(); i != e; ++i)
683 if (Duplicates.insert(Factors[i]).second) {
684 unsigned Occ = ++FactorOccurrences[Factors[i]];
685 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[i]; }
692 // If any factor occurred more than one time, we can pull it out.
694 DEBUG(std::cerr << "\nFACTORING [" << MaxOcc << "]: "
695 << *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 DEBUG(std::cerr << "RAIn:\t"; PrintOps(I, Ops);
811 // Now that we have linearized the tree to a list and have gathered all of
812 // the operands and their ranks, sort the operands by their rank. Use a
813 // stable_sort so that values with equal ranks will have their relative
814 // positions maintained (and so the compiler is deterministic). Note that
815 // this sorts so that the highest ranking values end up at the beginning of
817 std::stable_sort(Ops.begin(), Ops.end());
819 // OptimizeExpression - Now that we have the expression tree in a convenient
820 // sorted form, optimize it globally if possible.
821 if (Value *V = OptimizeExpression(I, Ops)) {
822 // This expression tree simplified to something that isn't a tree,
824 DEBUG(std::cerr << "Reassoc to scalar: " << *V << "\n");
825 I->replaceAllUsesWith(V);
826 RemoveDeadBinaryOp(I);
830 // We want to sink immediates as deeply as possible except in the case where
831 // this is a multiply tree used only by an add, and the immediate is a -1.
832 // In this case we reassociate to put the negation on the outside so that we
833 // can fold the negation into the add: (-X)*Y + Z -> Z-X*Y
834 if (I->getOpcode() == Instruction::Mul && I->hasOneUse() &&
835 cast<Instruction>(I->use_back())->getOpcode() == Instruction::Add &&
836 isa<ConstantInt>(Ops.back().Op) &&
837 cast<ConstantInt>(Ops.back().Op)->isAllOnesValue()) {
838 Ops.insert(Ops.begin(), Ops.back());
842 DEBUG(std::cerr << "RAOut:\t"; PrintOps(I, Ops);
845 if (Ops.size() == 1) {
846 // This expression tree simplified to something that isn't a tree,
848 I->replaceAllUsesWith(Ops[0].Op);
849 RemoveDeadBinaryOp(I);
851 // Now that we ordered and optimized the expressions, splat them back into
852 // the expression tree, removing any unneeded nodes.
853 RewriteExprTree(I, Ops);
858 bool Reassociate::runOnFunction(Function &F) {
859 // Recalculate the rank map for F
863 for (Function::iterator FI = F.begin(), FE = F.end(); FI != FE; ++FI)
866 // We are done with the rank map...
868 ValueRankMap.clear();