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/Compiler.h"
33 #include "llvm/Support/Debug.h"
34 #include "llvm/ADT/PostOrderIterator.h"
35 #include "llvm/ADT/Statistic.h"
39 STATISTIC(NumLinear , "Number of insts linearized");
40 STATISTIC(NumChanged, "Number of insts reassociated");
41 STATISTIC(NumAnnihil, "Number of expr tree annihilated");
42 STATISTIC(NumFactor , "Number of multiplies factored");
45 struct VISIBILITY_HIDDEN ValueEntry {
48 ValueEntry(unsigned R, Value *O) : Rank(R), Op(O) {}
50 inline bool operator<(const ValueEntry &LHS, const ValueEntry &RHS) {
51 return LHS.Rank > RHS.Rank; // Sort so that highest rank goes to start.
55 /// PrintOps - Print out the expression identified in the Ops list.
57 static void PrintOps(Instruction *I, const std::vector<ValueEntry> &Ops) {
58 Module *M = I->getParent()->getParent()->getParent();
59 cerr << Instruction::getOpcodeName(I->getOpcode()) << " "
60 << *Ops[0].Op->getType();
61 for (unsigned i = 0, e = Ops.size(); i != e; ++i)
62 WriteAsOperand(*cerr.stream() << " ", Ops[i].Op, false, M)
63 << "," << Ops[i].Rank;
67 class VISIBILITY_HIDDEN Reassociate : public FunctionPass {
68 std::map<BasicBlock*, unsigned> RankMap;
69 std::map<Value*, unsigned> ValueRankMap;
72 bool runOnFunction(Function &F);
74 virtual void getAnalysisUsage(AnalysisUsage &AU) const {
78 void BuildRankMap(Function &F);
79 unsigned getRank(Value *V);
80 void ReassociateExpression(BinaryOperator *I);
81 void RewriteExprTree(BinaryOperator *I, std::vector<ValueEntry> &Ops,
83 Value *OptimizeExpression(BinaryOperator *I, std::vector<ValueEntry> &Ops);
84 void LinearizeExprTree(BinaryOperator *I, std::vector<ValueEntry> &Ops);
85 void LinearizeExpr(BinaryOperator *I);
86 Value *RemoveFactorFromExpression(Value *V, Value *Factor);
87 void ReassociateBB(BasicBlock *BB);
89 void RemoveDeadBinaryOp(Value *V);
92 RegisterPass<Reassociate> X("reassociate", "Reassociate expressions");
95 // Public interface to the Reassociate pass
96 FunctionPass *llvm::createReassociatePass() { return new Reassociate(); }
98 void Reassociate::RemoveDeadBinaryOp(Value *V) {
99 Instruction *Op = dyn_cast<Instruction>(V);
100 if (!Op || !isa<BinaryOperator>(Op) || !isa<CmpInst>(Op) || !Op->use_empty())
103 Value *LHS = Op->getOperand(0), *RHS = Op->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()->isInteger() ||
169 (!BinaryOperator::isNot(I) && !BinaryOperator::isNeg(I)))
172 //DOUT << "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) {
190 Constant *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 DOUT << "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 DOUT << "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 /// NOTE: These intentionally destroys the expression tree operands (turning
238 /// them into undef values) to reduce #uses of the values. This means that the
239 /// caller MUST use something like RewriteExprTree to put the values back in.
241 void Reassociate::LinearizeExprTree(BinaryOperator *I,
242 std::vector<ValueEntry> &Ops) {
243 Value *LHS = I->getOperand(0), *RHS = I->getOperand(1);
244 unsigned Opcode = I->getOpcode();
246 // First step, linearize the expression if it is in ((A+B)+(C+D)) form.
247 BinaryOperator *LHSBO = isReassociableOp(LHS, Opcode);
248 BinaryOperator *RHSBO = isReassociableOp(RHS, Opcode);
250 // If this is a multiply expression tree and it contains internal negations,
251 // transform them into multiplies by -1 so they can be reassociated.
252 if (I->getOpcode() == Instruction::Mul) {
253 if (!LHSBO && LHS->hasOneUse() && BinaryOperator::isNeg(LHS)) {
254 LHS = LowerNegateToMultiply(cast<Instruction>(LHS));
255 LHSBO = isReassociableOp(LHS, Opcode);
257 if (!RHSBO && RHS->hasOneUse() && BinaryOperator::isNeg(RHS)) {
258 RHS = LowerNegateToMultiply(cast<Instruction>(RHS));
259 RHSBO = isReassociableOp(RHS, Opcode);
265 // Neither the LHS or RHS as part of the tree, thus this is a leaf. As
266 // such, just remember these operands and their rank.
267 Ops.push_back(ValueEntry(getRank(LHS), LHS));
268 Ops.push_back(ValueEntry(getRank(RHS), RHS));
270 // Clear the leaves out.
271 I->setOperand(0, UndefValue::get(I->getType()));
272 I->setOperand(1, UndefValue::get(I->getType()));
275 // Turn X+(Y+Z) -> (Y+Z)+X
276 std::swap(LHSBO, RHSBO);
278 bool Success = !I->swapOperands();
279 assert(Success && "swapOperands failed");
283 // Turn (A+B)+(C+D) -> (((A+B)+C)+D). This guarantees the the RHS is not
284 // part of the expression tree.
286 LHS = LHSBO = cast<BinaryOperator>(I->getOperand(0));
287 RHS = I->getOperand(1);
291 // Okay, now we know that the LHS is a nested expression and that the RHS is
292 // not. Perform reassociation.
293 assert(!isReassociableOp(RHS, Opcode) && "LinearizeExpr failed!");
295 // Move LHS right before I to make sure that the tree expression dominates all
297 LHSBO->moveBefore(I);
299 // Linearize the expression tree on the LHS.
300 LinearizeExprTree(LHSBO, Ops);
302 // Remember the RHS operand and its rank.
303 Ops.push_back(ValueEntry(getRank(RHS), RHS));
305 // Clear the RHS leaf out.
306 I->setOperand(1, UndefValue::get(I->getType()));
309 // RewriteExprTree - Now that the operands for this expression tree are
310 // linearized and optimized, emit them in-order. This function is written to be
312 void Reassociate::RewriteExprTree(BinaryOperator *I,
313 std::vector<ValueEntry> &Ops,
315 if (i+2 == Ops.size()) {
316 if (I->getOperand(0) != Ops[i].Op ||
317 I->getOperand(1) != Ops[i+1].Op) {
318 Value *OldLHS = I->getOperand(0);
319 DOUT << "RA: " << *I;
320 I->setOperand(0, Ops[i].Op);
321 I->setOperand(1, Ops[i+1].Op);
322 DOUT << "TO: " << *I;
326 // If we reassociated a tree to fewer operands (e.g. (1+a+2) -> (a+3)
327 // delete the extra, now dead, nodes.
328 RemoveDeadBinaryOp(OldLHS);
332 assert(i+2 < Ops.size() && "Ops index out of range!");
334 if (I->getOperand(1) != Ops[i].Op) {
335 DOUT << "RA: " << *I;
336 I->setOperand(1, Ops[i].Op);
337 DOUT << "TO: " << *I;
342 BinaryOperator *LHS = cast<BinaryOperator>(I->getOperand(0));
343 assert(LHS->getOpcode() == I->getOpcode() &&
344 "Improper expression tree!");
346 // Compactify the tree instructions together with each other to guarantee
347 // that the expression tree is dominated by all of Ops.
349 RewriteExprTree(LHS, Ops, i+1);
354 // NegateValue - Insert instructions before the instruction pointed to by BI,
355 // that computes the negative version of the value specified. The negative
356 // version of the value is returned, and BI is left pointing at the instruction
357 // that should be processed next by the reassociation pass.
359 static Value *NegateValue(Value *V, Instruction *BI) {
360 // We are trying to expose opportunity for reassociation. One of the things
361 // that we want to do to achieve this is to push a negation as deep into an
362 // expression chain as possible, to expose the add instructions. In practice,
363 // this means that we turn this:
364 // X = -(A+12+C+D) into X = -A + -12 + -C + -D = -12 + -A + -C + -D
365 // so that later, a: Y = 12+X could get reassociated with the -12 to eliminate
366 // the constants. We assume that instcombine will clean up the mess later if
367 // we introduce tons of unnecessary negation instructions...
369 if (Instruction *I = dyn_cast<Instruction>(V))
370 if (I->getOpcode() == Instruction::Add && I->hasOneUse()) {
371 // Push the negates through the add.
372 I->setOperand(0, NegateValue(I->getOperand(0), BI));
373 I->setOperand(1, NegateValue(I->getOperand(1), BI));
375 // We must move the add instruction here, because the neg instructions do
376 // not dominate the old add instruction in general. By moving it, we are
377 // assured that the neg instructions we just inserted dominate the
378 // instruction we are about to insert after them.
381 I->setName(I->getName()+".neg");
385 // Insert a 'neg' instruction that subtracts the value from zero to get the
388 return BinaryOperator::createNeg(V, V->getName() + ".neg", BI);
391 /// BreakUpSubtract - If we have (X-Y), and if either X is an add, or if this is
392 /// only used by an add, transform this into (X+(0-Y)) to promote better
394 static Instruction *BreakUpSubtract(Instruction *Sub) {
395 // Don't bother to break this up unless either the LHS is an associable add or
396 // if this is only used by one.
397 if (!isReassociableOp(Sub->getOperand(0), Instruction::Add) &&
398 !isReassociableOp(Sub->getOperand(1), Instruction::Add) &&
399 !(Sub->hasOneUse() &&isReassociableOp(Sub->use_back(), Instruction::Add)))
402 // Convert a subtract into an add and a neg instruction... so that sub
403 // instructions can be commuted with other add instructions...
405 // Calculate the negative value of Operand 1 of the sub instruction...
406 // and set it as the RHS of the add instruction we just made...
408 std::string Name = Sub->getName();
410 Value *NegVal = NegateValue(Sub->getOperand(1), Sub);
412 BinaryOperator::createAdd(Sub->getOperand(0), NegVal, Name, Sub);
414 // Everyone now refers to the add instruction.
415 Sub->replaceAllUsesWith(New);
416 Sub->eraseFromParent();
418 DOUT << "Negated: " << *New;
422 /// ConvertShiftToMul - If this is a shift of a reassociable multiply or is used
423 /// by one, change this into a multiply by a constant to assist with further
425 static Instruction *ConvertShiftToMul(Instruction *Shl) {
426 // If an operand of this shift is a reassociable multiply, or if the shift
427 // is used by a reassociable multiply or add, turn into a multiply.
428 if (isReassociableOp(Shl->getOperand(0), Instruction::Mul) ||
430 (isReassociableOp(Shl->use_back(), Instruction::Mul) ||
431 isReassociableOp(Shl->use_back(), Instruction::Add)))) {
432 Constant *MulCst = ConstantInt::get(Shl->getType(), 1);
433 MulCst = ConstantExpr::getShl(MulCst, cast<Constant>(Shl->getOperand(1)));
435 std::string Name = Shl->getName(); Shl->setName("");
436 Instruction *Mul = BinaryOperator::createMul(Shl->getOperand(0), MulCst,
438 Shl->replaceAllUsesWith(Mul);
439 Shl->eraseFromParent();
445 // Scan backwards and forwards among values with the same rank as element i to
446 // see if X exists. If X does not exist, return i.
447 static unsigned FindInOperandList(std::vector<ValueEntry> &Ops, unsigned i,
449 unsigned XRank = Ops[i].Rank;
450 unsigned e = Ops.size();
451 for (unsigned j = i+1; j != e && Ops[j].Rank == XRank; ++j)
455 for (unsigned j = i-1; j != ~0U && Ops[j].Rank == XRank; --j)
461 /// EmitAddTreeOfValues - Emit a tree of add instructions, summing Ops together
462 /// and returning the result. Insert the tree before I.
463 static Value *EmitAddTreeOfValues(Instruction *I, std::vector<Value*> &Ops) {
464 if (Ops.size() == 1) return Ops.back();
466 Value *V1 = Ops.back();
468 Value *V2 = EmitAddTreeOfValues(I, Ops);
469 return BinaryOperator::createAdd(V2, V1, "tmp", I);
472 /// RemoveFactorFromExpression - If V is an expression tree that is a
473 /// multiplication sequence, and if this sequence contains a multiply by Factor,
474 /// remove Factor from the tree and return the new tree.
475 Value *Reassociate::RemoveFactorFromExpression(Value *V, Value *Factor) {
476 BinaryOperator *BO = isReassociableOp(V, Instruction::Mul);
479 std::vector<ValueEntry> Factors;
480 LinearizeExprTree(BO, Factors);
482 bool FoundFactor = false;
483 for (unsigned i = 0, e = Factors.size(); i != e; ++i)
484 if (Factors[i].Op == Factor) {
486 Factors.erase(Factors.begin()+i);
490 // Make sure to restore the operands to the expression tree.
491 RewriteExprTree(BO, Factors);
495 if (Factors.size() == 1) return Factors[0].Op;
497 RewriteExprTree(BO, Factors);
501 /// FindSingleUseMultiplyFactors - If V is a single-use multiply, recursively
502 /// add its operands as factors, otherwise add V to the list of factors.
503 static void FindSingleUseMultiplyFactors(Value *V,
504 std::vector<Value*> &Factors) {
506 if ((!V->hasOneUse() && !V->use_empty()) ||
507 !(BO = dyn_cast<BinaryOperator>(V)) ||
508 BO->getOpcode() != Instruction::Mul) {
509 Factors.push_back(V);
513 // Otherwise, add the LHS and RHS to the list of factors.
514 FindSingleUseMultiplyFactors(BO->getOperand(1), Factors);
515 FindSingleUseMultiplyFactors(BO->getOperand(0), Factors);
520 Value *Reassociate::OptimizeExpression(BinaryOperator *I,
521 std::vector<ValueEntry> &Ops) {
522 // Now that we have the linearized expression tree, try to optimize it.
523 // Start by folding any constants that we found.
524 bool IterateOptimization = false;
525 if (Ops.size() == 1) return Ops[0].Op;
527 unsigned Opcode = I->getOpcode();
529 if (Constant *V1 = dyn_cast<Constant>(Ops[Ops.size()-2].Op))
530 if (Constant *V2 = dyn_cast<Constant>(Ops.back().Op)) {
532 Ops.back().Op = ConstantExpr::get(Opcode, V1, V2);
533 return OptimizeExpression(I, Ops);
536 // Check for destructive annihilation due to a constant being used.
537 if (ConstantInt *CstVal = dyn_cast<ConstantInt>(Ops.back().Op))
540 case Instruction::And:
541 if (CstVal->isNullValue()) { // ... & 0 -> 0
544 } else if (CstVal->isAllOnesValue()) { // ... & -1 -> ...
548 case Instruction::Mul:
549 if (CstVal->isNullValue()) { // ... * 0 -> 0
552 } else if (cast<ConstantInt>(CstVal)->getZExtValue() == 1) {
553 Ops.pop_back(); // ... * 1 -> ...
556 case Instruction::Or:
557 if (CstVal->isAllOnesValue()) { // ... | -1 -> -1
562 case Instruction::Add:
563 case Instruction::Xor:
564 if (CstVal->isNullValue()) // ... [|^+] 0 -> ...
568 if (Ops.size() == 1) return Ops[0].Op;
570 // Handle destructive annihilation do to identities between elements in the
571 // argument list here.
574 case Instruction::And:
575 case Instruction::Or:
576 case Instruction::Xor:
577 // Scan the operand lists looking for X and ~X pairs, along with X,X pairs.
578 // If we find any, we can simplify the expression. X&~X == 0, X|~X == -1.
579 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
580 // First, check for X and ~X in the operand list.
581 assert(i < Ops.size());
582 if (BinaryOperator::isNot(Ops[i].Op)) { // Cannot occur for ^.
583 Value *X = BinaryOperator::getNotArgument(Ops[i].Op);
584 unsigned FoundX = FindInOperandList(Ops, i, X);
586 if (Opcode == Instruction::And) { // ...&X&~X = 0
588 return Constant::getNullValue(X->getType());
589 } else if (Opcode == Instruction::Or) { // ...|X|~X = -1
591 return ConstantInt::getAllOnesValue(X->getType());
596 // Next, check for duplicate pairs of values, which we assume are next to
597 // each other, due to our sorting criteria.
598 assert(i < Ops.size());
599 if (i+1 != Ops.size() && Ops[i+1].Op == Ops[i].Op) {
600 if (Opcode == Instruction::And || Opcode == Instruction::Or) {
601 // Drop duplicate values.
602 Ops.erase(Ops.begin()+i);
604 IterateOptimization = true;
607 assert(Opcode == Instruction::Xor);
610 return Constant::getNullValue(Ops[0].Op->getType());
613 Ops.erase(Ops.begin()+i, Ops.begin()+i+2);
615 IterateOptimization = true;
622 case Instruction::Add:
623 // Scan the operand lists looking for X and -X pairs. If we find any, we
624 // can simplify the expression. X+-X == 0.
625 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
626 assert(i < Ops.size());
627 // Check for X and -X in the operand list.
628 if (BinaryOperator::isNeg(Ops[i].Op)) {
629 Value *X = BinaryOperator::getNegArgument(Ops[i].Op);
630 unsigned FoundX = FindInOperandList(Ops, i, X);
632 // Remove X and -X from the operand list.
633 if (Ops.size() == 2) {
635 return Constant::getNullValue(X->getType());
637 Ops.erase(Ops.begin()+i);
641 --i; // Need to back up an extra one.
642 Ops.erase(Ops.begin()+FoundX);
643 IterateOptimization = true;
645 --i; // Revisit element.
646 e -= 2; // Removed two elements.
653 // Scan the operand list, checking to see if there are any common factors
654 // between operands. Consider something like A*A+A*B*C+D. We would like to
655 // reassociate this to A*(A+B*C)+D, which reduces the number of multiplies.
656 // To efficiently find this, we count the number of times a factor occurs
657 // for any ADD operands that are MULs.
658 std::map<Value*, unsigned> FactorOccurrences;
660 Value *MaxOccVal = 0;
661 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
662 if (BinaryOperator *BOp = dyn_cast<BinaryOperator>(Ops[i].Op)) {
663 if (BOp->getOpcode() == Instruction::Mul && BOp->use_empty()) {
664 // Compute all of the factors of this added value.
665 std::vector<Value*> Factors;
666 FindSingleUseMultiplyFactors(BOp, Factors);
667 assert(Factors.size() > 1 && "Bad linearize!");
669 // Add one to FactorOccurrences for each unique factor in this op.
670 if (Factors.size() == 2) {
671 unsigned Occ = ++FactorOccurrences[Factors[0]];
672 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[0]; }
673 if (Factors[0] != Factors[1]) { // Don't double count A*A.
674 Occ = ++FactorOccurrences[Factors[1]];
675 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[1]; }
678 std::set<Value*> Duplicates;
679 for (unsigned i = 0, e = Factors.size(); i != e; ++i) {
680 if (Duplicates.insert(Factors[i]).second) {
681 unsigned Occ = ++FactorOccurrences[Factors[i]];
682 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[i]; }
690 // If any factor occurred more than one time, we can pull it out.
692 DOUT << "\nFACTORING [" << MaxOcc << "]: " << *MaxOccVal << "\n";
694 // Create a new instruction that uses the MaxOccVal twice. If we don't do
695 // this, we could otherwise run into situations where removing a factor
696 // from an expression will drop a use of maxocc, and this can cause
697 // RemoveFactorFromExpression on successive values to behave differently.
698 Instruction *DummyInst = BinaryOperator::createAdd(MaxOccVal, MaxOccVal);
699 std::vector<Value*> NewMulOps;
700 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
701 if (Value *V = RemoveFactorFromExpression(Ops[i].Op, MaxOccVal)) {
702 NewMulOps.push_back(V);
703 Ops.erase(Ops.begin()+i);
708 // No need for extra uses anymore.
711 unsigned NumAddedValues = NewMulOps.size();
712 Value *V = EmitAddTreeOfValues(I, NewMulOps);
713 Value *V2 = BinaryOperator::createMul(V, MaxOccVal, "tmp", I);
715 // Now that we have inserted V and its sole use, optimize it. This allows
716 // us to handle cases that require multiple factoring steps, such as this:
717 // A*A*B + A*A*C --> A*(A*B+A*C) --> A*(A*(B+C))
718 if (NumAddedValues > 1)
719 ReassociateExpression(cast<BinaryOperator>(V));
726 // Add the new value to the list of things being added.
727 Ops.insert(Ops.begin(), ValueEntry(getRank(V2), V2));
729 // Rewrite the tree so that there is now a use of V.
730 RewriteExprTree(I, Ops);
731 return OptimizeExpression(I, Ops);
734 //case Instruction::Mul:
737 if (IterateOptimization)
738 return OptimizeExpression(I, Ops);
743 /// ReassociateBB - Inspect all of the instructions in this basic block,
744 /// reassociating them as we go.
745 void Reassociate::ReassociateBB(BasicBlock *BB) {
746 for (BasicBlock::iterator BBI = BB->begin(); BBI != BB->end(); ) {
747 Instruction *BI = BBI++;
748 if (BI->getOpcode() == Instruction::Shl &&
749 isa<ConstantInt>(BI->getOperand(1)))
750 if (Instruction *NI = ConvertShiftToMul(BI)) {
755 // Reject cases where it is pointless to do this.
756 if (!isa<BinaryOperator>(BI) || BI->getType()->isFloatingPoint() ||
757 isa<PackedType>(BI->getType()))
758 continue; // Floating point ops are not associative.
760 // If this is a subtract instruction which is not already in negate form,
761 // see if we can convert it to X+-Y.
762 if (BI->getOpcode() == Instruction::Sub) {
763 if (!BinaryOperator::isNeg(BI)) {
764 if (Instruction *NI = BreakUpSubtract(BI)) {
769 // Otherwise, this is a negation. See if the operand is a multiply tree
770 // and if this is not an inner node of a multiply tree.
771 if (isReassociableOp(BI->getOperand(1), Instruction::Mul) &&
773 !isReassociableOp(BI->use_back(), Instruction::Mul))) {
774 BI = LowerNegateToMultiply(BI);
780 // If this instruction is a commutative binary operator, process it.
781 if (!BI->isAssociative()) continue;
782 BinaryOperator *I = cast<BinaryOperator>(BI);
784 // If this is an interior node of a reassociable tree, ignore it until we
785 // get to the root of the tree, to avoid N^2 analysis.
786 if (I->hasOneUse() && isReassociableOp(I->use_back(), I->getOpcode()))
789 // If this is an add tree that is used by a sub instruction, ignore it
790 // until we process the subtract.
791 if (I->hasOneUse() && I->getOpcode() == Instruction::Add &&
792 cast<Instruction>(I->use_back())->getOpcode() == Instruction::Sub)
795 ReassociateExpression(I);
799 void Reassociate::ReassociateExpression(BinaryOperator *I) {
801 // First, walk the expression tree, linearizing the tree, collecting
802 std::vector<ValueEntry> Ops;
803 LinearizeExprTree(I, Ops);
805 DOUT << "RAIn:\t"; DEBUG(PrintOps(I, Ops)); DOUT << "\n";
807 // Now that we have linearized the tree to a list and have gathered all of
808 // the operands and their ranks, sort the operands by their rank. Use a
809 // stable_sort so that values with equal ranks will have their relative
810 // positions maintained (and so the compiler is deterministic). Note that
811 // this sorts so that the highest ranking values end up at the beginning of
813 std::stable_sort(Ops.begin(), Ops.end());
815 // OptimizeExpression - Now that we have the expression tree in a convenient
816 // sorted form, optimize it globally if possible.
817 if (Value *V = OptimizeExpression(I, Ops)) {
818 // This expression tree simplified to something that isn't a tree,
820 DOUT << "Reassoc to scalar: " << *V << "\n";
821 I->replaceAllUsesWith(V);
822 RemoveDeadBinaryOp(I);
826 // We want to sink immediates as deeply as possible except in the case where
827 // this is a multiply tree used only by an add, and the immediate is a -1.
828 // In this case we reassociate to put the negation on the outside so that we
829 // can fold the negation into the add: (-X)*Y + Z -> Z-X*Y
830 if (I->getOpcode() == Instruction::Mul && I->hasOneUse() &&
831 cast<Instruction>(I->use_back())->getOpcode() == Instruction::Add &&
832 isa<ConstantInt>(Ops.back().Op) &&
833 cast<ConstantInt>(Ops.back().Op)->isAllOnesValue()) {
834 Ops.insert(Ops.begin(), Ops.back());
838 DOUT << "RAOut:\t"; DEBUG(PrintOps(I, Ops)); DOUT << "\n";
840 if (Ops.size() == 1) {
841 // This expression tree simplified to something that isn't a tree,
843 I->replaceAllUsesWith(Ops[0].Op);
844 RemoveDeadBinaryOp(I);
846 // Now that we ordered and optimized the expressions, splat them back into
847 // the expression tree, removing any unneeded nodes.
848 RewriteExprTree(I, Ops);
853 bool Reassociate::runOnFunction(Function &F) {
854 // Recalculate the rank map for F
858 for (Function::iterator FI = F.begin(), FE = F.end(); FI != FE; ++FI)
861 // We are done with the rank map...
863 ValueRankMap.clear();