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()->isInteger() ||
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) {
189 Constant *Cst = ConstantInt::getAllOnesValue(Neg->getType());
191 std::string NegName = Neg->getName(); Neg->setName("");
192 Instruction *Res = BinaryOperator::createMul(Neg->getOperand(1), Cst, NegName,
194 Neg->replaceAllUsesWith(Res);
195 Neg->eraseFromParent();
199 // Given an expression of the form '(A+B)+(D+C)', turn it into '(((A+B)+C)+D)'.
200 // Note that if D is also part of the expression tree that we recurse to
201 // linearize it as well. Besides that case, this does not recurse into A,B, or
203 void Reassociate::LinearizeExpr(BinaryOperator *I) {
204 BinaryOperator *LHS = cast<BinaryOperator>(I->getOperand(0));
205 BinaryOperator *RHS = cast<BinaryOperator>(I->getOperand(1));
206 assert(isReassociableOp(LHS, I->getOpcode()) &&
207 isReassociableOp(RHS, I->getOpcode()) &&
208 "Not an expression that needs linearization?");
210 DOUT << "Linear" << *LHS << *RHS << *I;
212 // Move the RHS instruction to live immediately before I, avoiding breaking
213 // dominator properties.
216 // Move operands around to do the linearization.
217 I->setOperand(1, RHS->getOperand(0));
218 RHS->setOperand(0, LHS);
219 I->setOperand(0, RHS);
223 DOUT << "Linearized: " << *I;
225 // If D is part of this expression tree, tail recurse.
226 if (isReassociableOp(I->getOperand(1), I->getOpcode()))
231 /// LinearizeExprTree - Given an associative binary expression tree, traverse
232 /// all of the uses putting it into canonical form. This forces a left-linear
233 /// form of the the expression (((a+b)+c)+d), and collects information about the
234 /// rank of the non-tree operands.
236 /// NOTE: These intentionally destroys the expression tree operands (turning
237 /// them into undef values) to reduce #uses of the values. This means that the
238 /// caller MUST use something like RewriteExprTree to put the values back in.
240 void Reassociate::LinearizeExprTree(BinaryOperator *I,
241 std::vector<ValueEntry> &Ops) {
242 Value *LHS = I->getOperand(0), *RHS = I->getOperand(1);
243 unsigned Opcode = I->getOpcode();
245 // First step, linearize the expression if it is in ((A+B)+(C+D)) form.
246 BinaryOperator *LHSBO = isReassociableOp(LHS, Opcode);
247 BinaryOperator *RHSBO = isReassociableOp(RHS, Opcode);
249 // If this is a multiply expression tree and it contains internal negations,
250 // transform them into multiplies by -1 so they can be reassociated.
251 if (I->getOpcode() == Instruction::Mul) {
252 if (!LHSBO && LHS->hasOneUse() && BinaryOperator::isNeg(LHS)) {
253 LHS = LowerNegateToMultiply(cast<Instruction>(LHS));
254 LHSBO = isReassociableOp(LHS, Opcode);
256 if (!RHSBO && RHS->hasOneUse() && BinaryOperator::isNeg(RHS)) {
257 RHS = LowerNegateToMultiply(cast<Instruction>(RHS));
258 RHSBO = isReassociableOp(RHS, Opcode);
264 // Neither the LHS or RHS as part of the tree, thus this is a leaf. As
265 // such, just remember these operands and their rank.
266 Ops.push_back(ValueEntry(getRank(LHS), LHS));
267 Ops.push_back(ValueEntry(getRank(RHS), RHS));
269 // Clear the leaves out.
270 I->setOperand(0, UndefValue::get(I->getType()));
271 I->setOperand(1, UndefValue::get(I->getType()));
274 // Turn X+(Y+Z) -> (Y+Z)+X
275 std::swap(LHSBO, RHSBO);
277 bool Success = !I->swapOperands();
278 assert(Success && "swapOperands failed");
282 // Turn (A+B)+(C+D) -> (((A+B)+C)+D). This guarantees the the RHS is not
283 // part of the expression tree.
285 LHS = LHSBO = cast<BinaryOperator>(I->getOperand(0));
286 RHS = I->getOperand(1);
290 // Okay, now we know that the LHS is a nested expression and that the RHS is
291 // not. Perform reassociation.
292 assert(!isReassociableOp(RHS, Opcode) && "LinearizeExpr failed!");
294 // Move LHS right before I to make sure that the tree expression dominates all
296 LHSBO->moveBefore(I);
298 // Linearize the expression tree on the LHS.
299 LinearizeExprTree(LHSBO, Ops);
301 // Remember the RHS operand and its rank.
302 Ops.push_back(ValueEntry(getRank(RHS), RHS));
304 // Clear the RHS leaf out.
305 I->setOperand(1, UndefValue::get(I->getType()));
308 // RewriteExprTree - Now that the operands for this expression tree are
309 // linearized and optimized, emit them in-order. This function is written to be
311 void Reassociate::RewriteExprTree(BinaryOperator *I,
312 std::vector<ValueEntry> &Ops,
314 if (i+2 == Ops.size()) {
315 if (I->getOperand(0) != Ops[i].Op ||
316 I->getOperand(1) != Ops[i+1].Op) {
317 Value *OldLHS = I->getOperand(0);
318 DOUT << "RA: " << *I;
319 I->setOperand(0, Ops[i].Op);
320 I->setOperand(1, Ops[i+1].Op);
321 DOUT << "TO: " << *I;
325 // If we reassociated a tree to fewer operands (e.g. (1+a+2) -> (a+3)
326 // delete the extra, now dead, nodes.
327 RemoveDeadBinaryOp(OldLHS);
331 assert(i+2 < Ops.size() && "Ops index out of range!");
333 if (I->getOperand(1) != Ops[i].Op) {
334 DOUT << "RA: " << *I;
335 I->setOperand(1, Ops[i].Op);
336 DOUT << "TO: " << *I;
341 BinaryOperator *LHS = cast<BinaryOperator>(I->getOperand(0));
342 assert(LHS->getOpcode() == I->getOpcode() &&
343 "Improper expression tree!");
345 // Compactify the tree instructions together with each other to guarantee
346 // that the expression tree is dominated by all of Ops.
348 RewriteExprTree(LHS, Ops, i+1);
353 // NegateValue - Insert instructions before the instruction pointed to by BI,
354 // that computes the negative version of the value specified. The negative
355 // version of the value is returned, and BI is left pointing at the instruction
356 // that should be processed next by the reassociation pass.
358 static Value *NegateValue(Value *V, Instruction *BI) {
359 // We are trying to expose opportunity for reassociation. One of the things
360 // that we want to do to achieve this is to push a negation as deep into an
361 // expression chain as possible, to expose the add instructions. In practice,
362 // this means that we turn this:
363 // X = -(A+12+C+D) into X = -A + -12 + -C + -D = -12 + -A + -C + -D
364 // so that later, a: Y = 12+X could get reassociated with the -12 to eliminate
365 // the constants. We assume that instcombine will clean up the mess later if
366 // we introduce tons of unnecessary negation instructions...
368 if (Instruction *I = dyn_cast<Instruction>(V))
369 if (I->getOpcode() == Instruction::Add && I->hasOneUse()) {
370 // Push the negates through the add.
371 I->setOperand(0, NegateValue(I->getOperand(0), BI));
372 I->setOperand(1, NegateValue(I->getOperand(1), BI));
374 // We must move the add instruction here, because the neg instructions do
375 // not dominate the old add instruction in general. By moving it, we are
376 // assured that the neg instructions we just inserted dominate the
377 // instruction we are about to insert after them.
380 I->setName(I->getName()+".neg");
384 // Insert a 'neg' instruction that subtracts the value from zero to get the
387 return BinaryOperator::createNeg(V, V->getName() + ".neg", BI);
390 /// BreakUpSubtract - If we have (X-Y), and if either X is an add, or if this is
391 /// only used by an add, transform this into (X+(0-Y)) to promote better
393 static Instruction *BreakUpSubtract(Instruction *Sub) {
394 // Don't bother to break this up unless either the LHS is an associable add or
395 // if this is only used by one.
396 if (!isReassociableOp(Sub->getOperand(0), Instruction::Add) &&
397 !isReassociableOp(Sub->getOperand(1), Instruction::Add) &&
398 !(Sub->hasOneUse() &&isReassociableOp(Sub->use_back(), Instruction::Add)))
401 // Convert a subtract into an add and a neg instruction... so that sub
402 // instructions can be commuted with other add instructions...
404 // Calculate the negative value of Operand 1 of the sub instruction...
405 // and set it as the RHS of the add instruction we just made...
407 std::string Name = Sub->getName();
409 Value *NegVal = NegateValue(Sub->getOperand(1), Sub);
411 BinaryOperator::createAdd(Sub->getOperand(0), NegVal, Name, Sub);
413 // Everyone now refers to the add instruction.
414 Sub->replaceAllUsesWith(New);
415 Sub->eraseFromParent();
417 DOUT << "Negated: " << *New;
421 /// ConvertShiftToMul - If this is a shift of a reassociable multiply or is used
422 /// by one, change this into a multiply by a constant to assist with further
424 static Instruction *ConvertShiftToMul(Instruction *Shl) {
425 // If an operand of this shift is a reassociable multiply, or if the shift
426 // is used by a reassociable multiply or add, turn into a multiply.
427 if (isReassociableOp(Shl->getOperand(0), Instruction::Mul) ||
429 (isReassociableOp(Shl->use_back(), Instruction::Mul) ||
430 isReassociableOp(Shl->use_back(), Instruction::Add)))) {
431 Constant *MulCst = ConstantInt::get(Shl->getType(), 1);
432 MulCst = ConstantExpr::getShl(MulCst, cast<Constant>(Shl->getOperand(1)));
434 std::string Name = Shl->getName(); Shl->setName("");
435 Instruction *Mul = BinaryOperator::createMul(Shl->getOperand(0), MulCst,
437 Shl->replaceAllUsesWith(Mul);
438 Shl->eraseFromParent();
444 // Scan backwards and forwards among values with the same rank as element i to
445 // see if X exists. If X does not exist, return i.
446 static unsigned FindInOperandList(std::vector<ValueEntry> &Ops, unsigned i,
448 unsigned XRank = Ops[i].Rank;
449 unsigned e = Ops.size();
450 for (unsigned j = i+1; j != e && Ops[j].Rank == XRank; ++j)
454 for (unsigned j = i-1; j != ~0U && Ops[j].Rank == XRank; --j)
460 /// EmitAddTreeOfValues - Emit a tree of add instructions, summing Ops together
461 /// and returning the result. Insert the tree before I.
462 static Value *EmitAddTreeOfValues(Instruction *I, std::vector<Value*> &Ops) {
463 if (Ops.size() == 1) return Ops.back();
465 Value *V1 = Ops.back();
467 Value *V2 = EmitAddTreeOfValues(I, Ops);
468 return BinaryOperator::createAdd(V2, V1, "tmp", I);
471 /// RemoveFactorFromExpression - If V is an expression tree that is a
472 /// multiplication sequence, and if this sequence contains a multiply by Factor,
473 /// remove Factor from the tree and return the new tree.
474 Value *Reassociate::RemoveFactorFromExpression(Value *V, Value *Factor) {
475 BinaryOperator *BO = isReassociableOp(V, Instruction::Mul);
478 std::vector<ValueEntry> Factors;
479 LinearizeExprTree(BO, Factors);
481 bool FoundFactor = false;
482 for (unsigned i = 0, e = Factors.size(); i != e; ++i)
483 if (Factors[i].Op == Factor) {
485 Factors.erase(Factors.begin()+i);
489 // Make sure to restore the operands to the expression tree.
490 RewriteExprTree(BO, Factors);
494 if (Factors.size() == 1) return Factors[0].Op;
496 RewriteExprTree(BO, Factors);
500 /// FindSingleUseMultiplyFactors - If V is a single-use multiply, recursively
501 /// add its operands as factors, otherwise add V to the list of factors.
502 static void FindSingleUseMultiplyFactors(Value *V,
503 std::vector<Value*> &Factors) {
505 if ((!V->hasOneUse() && !V->use_empty()) ||
506 !(BO = dyn_cast<BinaryOperator>(V)) ||
507 BO->getOpcode() != Instruction::Mul) {
508 Factors.push_back(V);
512 // Otherwise, add the LHS and RHS to the list of factors.
513 FindSingleUseMultiplyFactors(BO->getOperand(1), Factors);
514 FindSingleUseMultiplyFactors(BO->getOperand(0), Factors);
519 Value *Reassociate::OptimizeExpression(BinaryOperator *I,
520 std::vector<ValueEntry> &Ops) {
521 // Now that we have the linearized expression tree, try to optimize it.
522 // Start by folding any constants that we found.
523 bool IterateOptimization = false;
524 if (Ops.size() == 1) return Ops[0].Op;
526 unsigned Opcode = I->getOpcode();
528 if (Constant *V1 = dyn_cast<Constant>(Ops[Ops.size()-2].Op))
529 if (Constant *V2 = dyn_cast<Constant>(Ops.back().Op)) {
531 Ops.back().Op = ConstantExpr::get(Opcode, V1, V2);
532 return OptimizeExpression(I, Ops);
535 // Check for destructive annihilation due to a constant being used.
536 if (ConstantInt *CstVal = dyn_cast<ConstantInt>(Ops.back().Op))
539 case Instruction::And:
540 if (CstVal->isNullValue()) { // ... & 0 -> 0
543 } else if (CstVal->isAllOnesValue()) { // ... & -1 -> ...
547 case Instruction::Mul:
548 if (CstVal->isNullValue()) { // ... * 0 -> 0
551 } else if (cast<ConstantInt>(CstVal)->getZExtValue() == 1) {
552 Ops.pop_back(); // ... * 1 -> ...
555 case Instruction::Or:
556 if (CstVal->isAllOnesValue()) { // ... | -1 -> -1
561 case Instruction::Add:
562 case Instruction::Xor:
563 if (CstVal->isNullValue()) // ... [|^+] 0 -> ...
567 if (Ops.size() == 1) return Ops[0].Op;
569 // Handle destructive annihilation do to identities between elements in the
570 // argument list here.
573 case Instruction::And:
574 case Instruction::Or:
575 case Instruction::Xor:
576 // Scan the operand lists looking for X and ~X pairs, along with X,X pairs.
577 // If we find any, we can simplify the expression. X&~X == 0, X|~X == -1.
578 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
579 // First, check for X and ~X in the operand list.
580 assert(i < Ops.size());
581 if (BinaryOperator::isNot(Ops[i].Op)) { // Cannot occur for ^.
582 Value *X = BinaryOperator::getNotArgument(Ops[i].Op);
583 unsigned FoundX = FindInOperandList(Ops, i, X);
585 if (Opcode == Instruction::And) { // ...&X&~X = 0
587 return Constant::getNullValue(X->getType());
588 } else if (Opcode == Instruction::Or) { // ...|X|~X = -1
590 return ConstantInt::getAllOnesValue(X->getType());
595 // Next, check for duplicate pairs of values, which we assume are next to
596 // each other, due to our sorting criteria.
597 assert(i < Ops.size());
598 if (i+1 != Ops.size() && Ops[i+1].Op == Ops[i].Op) {
599 if (Opcode == Instruction::And || Opcode == Instruction::Or) {
600 // Drop duplicate values.
601 Ops.erase(Ops.begin()+i);
603 IterateOptimization = true;
606 assert(Opcode == Instruction::Xor);
609 return Constant::getNullValue(Ops[0].Op->getType());
612 Ops.erase(Ops.begin()+i, Ops.begin()+i+2);
614 IterateOptimization = true;
621 case Instruction::Add:
622 // Scan the operand lists looking for X and -X pairs. If we find any, we
623 // can simplify the expression. X+-X == 0.
624 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
625 assert(i < Ops.size());
626 // Check for X and -X in the operand list.
627 if (BinaryOperator::isNeg(Ops[i].Op)) {
628 Value *X = BinaryOperator::getNegArgument(Ops[i].Op);
629 unsigned FoundX = FindInOperandList(Ops, i, X);
631 // Remove X and -X from the operand list.
632 if (Ops.size() == 2) {
634 return Constant::getNullValue(X->getType());
636 Ops.erase(Ops.begin()+i);
640 --i; // Need to back up an extra one.
641 Ops.erase(Ops.begin()+FoundX);
642 IterateOptimization = true;
644 --i; // Revisit element.
645 e -= 2; // Removed two elements.
652 // Scan the operand list, checking to see if there are any common factors
653 // between operands. Consider something like A*A+A*B*C+D. We would like to
654 // reassociate this to A*(A+B*C)+D, which reduces the number of multiplies.
655 // To efficiently find this, we count the number of times a factor occurs
656 // for any ADD operands that are MULs.
657 std::map<Value*, unsigned> FactorOccurrences;
659 Value *MaxOccVal = 0;
660 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
661 if (BinaryOperator *BOp = dyn_cast<BinaryOperator>(Ops[i].Op)) {
662 if (BOp->getOpcode() == Instruction::Mul && BOp->use_empty()) {
663 // Compute all of the factors of this added value.
664 std::vector<Value*> Factors;
665 FindSingleUseMultiplyFactors(BOp, Factors);
666 assert(Factors.size() > 1 && "Bad linearize!");
668 // Add one to FactorOccurrences for each unique factor in this op.
669 if (Factors.size() == 2) {
670 unsigned Occ = ++FactorOccurrences[Factors[0]];
671 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[0]; }
672 if (Factors[0] != Factors[1]) { // Don't double count A*A.
673 Occ = ++FactorOccurrences[Factors[1]];
674 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[1]; }
677 std::set<Value*> Duplicates;
678 for (unsigned i = 0, e = Factors.size(); i != e; ++i) {
679 if (Duplicates.insert(Factors[i]).second) {
680 unsigned Occ = ++FactorOccurrences[Factors[i]];
681 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[i]; }
689 // If any factor occurred more than one time, we can pull it out.
691 DOUT << "\nFACTORING [" << MaxOcc << "]: " << *MaxOccVal << "\n";
693 // Create a new instruction that uses the MaxOccVal twice. If we don't do
694 // this, we could otherwise run into situations where removing a factor
695 // from an expression will drop a use of maxocc, and this can cause
696 // RemoveFactorFromExpression on successive values to behave differently.
697 Instruction *DummyInst = BinaryOperator::createAdd(MaxOccVal, MaxOccVal);
698 std::vector<Value*> NewMulOps;
699 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
700 if (Value *V = RemoveFactorFromExpression(Ops[i].Op, MaxOccVal)) {
701 NewMulOps.push_back(V);
702 Ops.erase(Ops.begin()+i);
707 // No need for extra uses anymore.
710 unsigned NumAddedValues = NewMulOps.size();
711 Value *V = EmitAddTreeOfValues(I, NewMulOps);
712 Value *V2 = BinaryOperator::createMul(V, MaxOccVal, "tmp", I);
714 // Now that we have inserted V and its sole use, optimize it. This allows
715 // us to handle cases that require multiple factoring steps, such as this:
716 // A*A*B + A*A*C --> A*(A*B+A*C) --> A*(A*(B+C))
717 if (NumAddedValues > 1)
718 ReassociateExpression(cast<BinaryOperator>(V));
725 // Add the new value to the list of things being added.
726 Ops.insert(Ops.begin(), ValueEntry(getRank(V2), V2));
728 // Rewrite the tree so that there is now a use of V.
729 RewriteExprTree(I, Ops);
730 return OptimizeExpression(I, Ops);
733 //case Instruction::Mul:
736 if (IterateOptimization)
737 return OptimizeExpression(I, Ops);
742 /// ReassociateBB - Inspect all of the instructions in this basic block,
743 /// reassociating them as we go.
744 void Reassociate::ReassociateBB(BasicBlock *BB) {
745 for (BasicBlock::iterator BBI = BB->begin(); BBI != BB->end(); ) {
746 Instruction *BI = BBI++;
747 if (BI->getOpcode() == Instruction::Shl &&
748 isa<ConstantInt>(BI->getOperand(1)))
749 if (Instruction *NI = ConvertShiftToMul(BI)) {
754 // Reject cases where it is pointless to do this.
755 if (!isa<BinaryOperator>(BI) || BI->getType()->isFloatingPoint() ||
756 isa<PackedType>(BI->getType()))
757 continue; // Floating point ops are not associative.
759 // If this is a subtract instruction which is not already in negate form,
760 // see if we can convert it to X+-Y.
761 if (BI->getOpcode() == Instruction::Sub) {
762 if (!BinaryOperator::isNeg(BI)) {
763 if (Instruction *NI = BreakUpSubtract(BI)) {
768 // Otherwise, this is a negation. See if the operand is a multiply tree
769 // and if this is not an inner node of a multiply tree.
770 if (isReassociableOp(BI->getOperand(1), Instruction::Mul) &&
772 !isReassociableOp(BI->use_back(), Instruction::Mul))) {
773 BI = LowerNegateToMultiply(BI);
779 // If this instruction is a commutative binary operator, process it.
780 if (!BI->isAssociative()) continue;
781 BinaryOperator *I = cast<BinaryOperator>(BI);
783 // If this is an interior node of a reassociable tree, ignore it until we
784 // get to the root of the tree, to avoid N^2 analysis.
785 if (I->hasOneUse() && isReassociableOp(I->use_back(), I->getOpcode()))
788 // If this is an add tree that is used by a sub instruction, ignore it
789 // until we process the subtract.
790 if (I->hasOneUse() && I->getOpcode() == Instruction::Add &&
791 cast<Instruction>(I->use_back())->getOpcode() == Instruction::Sub)
794 ReassociateExpression(I);
798 void Reassociate::ReassociateExpression(BinaryOperator *I) {
800 // First, walk the expression tree, linearizing the tree, collecting
801 std::vector<ValueEntry> Ops;
802 LinearizeExprTree(I, Ops);
804 DOUT << "RAIn:\t"; DEBUG(PrintOps(I, Ops)); DOUT << "\n";
806 // Now that we have linearized the tree to a list and have gathered all of
807 // the operands and their ranks, sort the operands by their rank. Use a
808 // stable_sort so that values with equal ranks will have their relative
809 // positions maintained (and so the compiler is deterministic). Note that
810 // this sorts so that the highest ranking values end up at the beginning of
812 std::stable_sort(Ops.begin(), Ops.end());
814 // OptimizeExpression - Now that we have the expression tree in a convenient
815 // sorted form, optimize it globally if possible.
816 if (Value *V = OptimizeExpression(I, Ops)) {
817 // This expression tree simplified to something that isn't a tree,
819 DOUT << "Reassoc to scalar: " << *V << "\n";
820 I->replaceAllUsesWith(V);
821 RemoveDeadBinaryOp(I);
825 // We want to sink immediates as deeply as possible except in the case where
826 // this is a multiply tree used only by an add, and the immediate is a -1.
827 // In this case we reassociate to put the negation on the outside so that we
828 // can fold the negation into the add: (-X)*Y + Z -> Z-X*Y
829 if (I->getOpcode() == Instruction::Mul && I->hasOneUse() &&
830 cast<Instruction>(I->use_back())->getOpcode() == Instruction::Add &&
831 isa<ConstantInt>(Ops.back().Op) &&
832 cast<ConstantInt>(Ops.back().Op)->isAllOnesValue()) {
833 Ops.insert(Ops.begin(), Ops.back());
837 DOUT << "RAOut:\t"; DEBUG(PrintOps(I, Ops)); DOUT << "\n";
839 if (Ops.size() == 1) {
840 // This expression tree simplified to something that isn't a tree,
842 I->replaceAllUsesWith(Ops[0].Op);
843 RemoveDeadBinaryOp(I);
845 // Now that we ordered and optimized the expressions, splat them back into
846 // the expression tree, removing any unneeded nodes.
847 RewriteExprTree(I, Ops);
852 bool Reassociate::runOnFunction(Function &F) {
853 // Recalculate the rank map for F
857 for (Function::iterator FI = F.begin(), FE = F.end(); FI != FE; ++FI)
860 // We are done with the rank map...
862 ValueRankMap.clear();