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 Instruction *Res = BinaryOperator::createMul(Neg->getOperand(1), Cst, "",Neg);
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 Value *NegVal = NegateValue(Sub->getOperand(1), Sub);
409 BinaryOperator::createAdd(Sub->getOperand(0), NegVal, "", Sub);
412 // Everyone now refers to the add instruction.
413 Sub->replaceAllUsesWith(New);
414 Sub->eraseFromParent();
416 DOUT << "Negated: " << *New;
420 /// ConvertShiftToMul - If this is a shift of a reassociable multiply or is used
421 /// by one, change this into a multiply by a constant to assist with further
423 static Instruction *ConvertShiftToMul(Instruction *Shl) {
424 // If an operand of this shift is a reassociable multiply, or if the shift
425 // is used by a reassociable multiply or add, turn into a multiply.
426 if (isReassociableOp(Shl->getOperand(0), Instruction::Mul) ||
428 (isReassociableOp(Shl->use_back(), Instruction::Mul) ||
429 isReassociableOp(Shl->use_back(), Instruction::Add)))) {
430 Constant *MulCst = ConstantInt::get(Shl->getType(), 1);
431 MulCst = ConstantExpr::getShl(MulCst, cast<Constant>(Shl->getOperand(1)));
433 Instruction *Mul = BinaryOperator::createMul(Shl->getOperand(0), MulCst,
436 Shl->replaceAllUsesWith(Mul);
437 Shl->eraseFromParent();
443 // Scan backwards and forwards among values with the same rank as element i to
444 // see if X exists. If X does not exist, return i.
445 static unsigned FindInOperandList(std::vector<ValueEntry> &Ops, unsigned i,
447 unsigned XRank = Ops[i].Rank;
448 unsigned e = Ops.size();
449 for (unsigned j = i+1; j != e && Ops[j].Rank == XRank; ++j)
453 for (unsigned j = i-1; j != ~0U && Ops[j].Rank == XRank; --j)
459 /// EmitAddTreeOfValues - Emit a tree of add instructions, summing Ops together
460 /// and returning the result. Insert the tree before I.
461 static Value *EmitAddTreeOfValues(Instruction *I, std::vector<Value*> &Ops) {
462 if (Ops.size() == 1) return Ops.back();
464 Value *V1 = Ops.back();
466 Value *V2 = EmitAddTreeOfValues(I, Ops);
467 return BinaryOperator::createAdd(V2, V1, "tmp", I);
470 /// RemoveFactorFromExpression - If V is an expression tree that is a
471 /// multiplication sequence, and if this sequence contains a multiply by Factor,
472 /// remove Factor from the tree and return the new tree.
473 Value *Reassociate::RemoveFactorFromExpression(Value *V, Value *Factor) {
474 BinaryOperator *BO = isReassociableOp(V, Instruction::Mul);
477 std::vector<ValueEntry> Factors;
478 LinearizeExprTree(BO, Factors);
480 bool FoundFactor = false;
481 for (unsigned i = 0, e = Factors.size(); i != e; ++i)
482 if (Factors[i].Op == Factor) {
484 Factors.erase(Factors.begin()+i);
488 // Make sure to restore the operands to the expression tree.
489 RewriteExprTree(BO, Factors);
493 if (Factors.size() == 1) return Factors[0].Op;
495 RewriteExprTree(BO, Factors);
499 /// FindSingleUseMultiplyFactors - If V is a single-use multiply, recursively
500 /// add its operands as factors, otherwise add V to the list of factors.
501 static void FindSingleUseMultiplyFactors(Value *V,
502 std::vector<Value*> &Factors) {
504 if ((!V->hasOneUse() && !V->use_empty()) ||
505 !(BO = dyn_cast<BinaryOperator>(V)) ||
506 BO->getOpcode() != Instruction::Mul) {
507 Factors.push_back(V);
511 // Otherwise, add the LHS and RHS to the list of factors.
512 FindSingleUseMultiplyFactors(BO->getOperand(1), Factors);
513 FindSingleUseMultiplyFactors(BO->getOperand(0), Factors);
518 Value *Reassociate::OptimizeExpression(BinaryOperator *I,
519 std::vector<ValueEntry> &Ops) {
520 // Now that we have the linearized expression tree, try to optimize it.
521 // Start by folding any constants that we found.
522 bool IterateOptimization = false;
523 if (Ops.size() == 1) return Ops[0].Op;
525 unsigned Opcode = I->getOpcode();
527 if (Constant *V1 = dyn_cast<Constant>(Ops[Ops.size()-2].Op))
528 if (Constant *V2 = dyn_cast<Constant>(Ops.back().Op)) {
530 Ops.back().Op = ConstantExpr::get(Opcode, V1, V2);
531 return OptimizeExpression(I, Ops);
534 // Check for destructive annihilation due to a constant being used.
535 if (ConstantInt *CstVal = dyn_cast<ConstantInt>(Ops.back().Op))
538 case Instruction::And:
539 if (CstVal->isNullValue()) { // ... & 0 -> 0
542 } else if (CstVal->isAllOnesValue()) { // ... & -1 -> ...
546 case Instruction::Mul:
547 if (CstVal->isNullValue()) { // ... * 0 -> 0
550 } else if (cast<ConstantInt>(CstVal)->getZExtValue() == 1) {
551 Ops.pop_back(); // ... * 1 -> ...
554 case Instruction::Or:
555 if (CstVal->isAllOnesValue()) { // ... | -1 -> -1
560 case Instruction::Add:
561 case Instruction::Xor:
562 if (CstVal->isNullValue()) // ... [|^+] 0 -> ...
566 if (Ops.size() == 1) return Ops[0].Op;
568 // Handle destructive annihilation do to identities between elements in the
569 // argument list here.
572 case Instruction::And:
573 case Instruction::Or:
574 case Instruction::Xor:
575 // Scan the operand lists looking for X and ~X pairs, along with X,X pairs.
576 // If we find any, we can simplify the expression. X&~X == 0, X|~X == -1.
577 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
578 // First, check for X and ~X in the operand list.
579 assert(i < Ops.size());
580 if (BinaryOperator::isNot(Ops[i].Op)) { // Cannot occur for ^.
581 Value *X = BinaryOperator::getNotArgument(Ops[i].Op);
582 unsigned FoundX = FindInOperandList(Ops, i, X);
584 if (Opcode == Instruction::And) { // ...&X&~X = 0
586 return Constant::getNullValue(X->getType());
587 } else if (Opcode == Instruction::Or) { // ...|X|~X = -1
589 return ConstantInt::getAllOnesValue(X->getType());
594 // Next, check for duplicate pairs of values, which we assume are next to
595 // each other, due to our sorting criteria.
596 assert(i < Ops.size());
597 if (i+1 != Ops.size() && Ops[i+1].Op == Ops[i].Op) {
598 if (Opcode == Instruction::And || Opcode == Instruction::Or) {
599 // Drop duplicate values.
600 Ops.erase(Ops.begin()+i);
602 IterateOptimization = true;
605 assert(Opcode == Instruction::Xor);
608 return Constant::getNullValue(Ops[0].Op->getType());
611 Ops.erase(Ops.begin()+i, Ops.begin()+i+2);
613 IterateOptimization = true;
620 case Instruction::Add:
621 // Scan the operand lists looking for X and -X pairs. If we find any, we
622 // can simplify the expression. X+-X == 0.
623 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
624 assert(i < Ops.size());
625 // Check for X and -X in the operand list.
626 if (BinaryOperator::isNeg(Ops[i].Op)) {
627 Value *X = BinaryOperator::getNegArgument(Ops[i].Op);
628 unsigned FoundX = FindInOperandList(Ops, i, X);
630 // Remove X and -X from the operand list.
631 if (Ops.size() == 2) {
633 return Constant::getNullValue(X->getType());
635 Ops.erase(Ops.begin()+i);
639 --i; // Need to back up an extra one.
640 Ops.erase(Ops.begin()+FoundX);
641 IterateOptimization = true;
643 --i; // Revisit element.
644 e -= 2; // Removed two elements.
651 // Scan the operand list, checking to see if there are any common factors
652 // between operands. Consider something like A*A+A*B*C+D. We would like to
653 // reassociate this to A*(A+B*C)+D, which reduces the number of multiplies.
654 // To efficiently find this, we count the number of times a factor occurs
655 // for any ADD operands that are MULs.
656 std::map<Value*, unsigned> FactorOccurrences;
658 Value *MaxOccVal = 0;
659 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
660 if (BinaryOperator *BOp = dyn_cast<BinaryOperator>(Ops[i].Op)) {
661 if (BOp->getOpcode() == Instruction::Mul && BOp->use_empty()) {
662 // Compute all of the factors of this added value.
663 std::vector<Value*> Factors;
664 FindSingleUseMultiplyFactors(BOp, Factors);
665 assert(Factors.size() > 1 && "Bad linearize!");
667 // Add one to FactorOccurrences for each unique factor in this op.
668 if (Factors.size() == 2) {
669 unsigned Occ = ++FactorOccurrences[Factors[0]];
670 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[0]; }
671 if (Factors[0] != Factors[1]) { // Don't double count A*A.
672 Occ = ++FactorOccurrences[Factors[1]];
673 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[1]; }
676 std::set<Value*> Duplicates;
677 for (unsigned i = 0, e = Factors.size(); i != e; ++i) {
678 if (Duplicates.insert(Factors[i]).second) {
679 unsigned Occ = ++FactorOccurrences[Factors[i]];
680 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[i]; }
688 // If any factor occurred more than one time, we can pull it out.
690 DOUT << "\nFACTORING [" << MaxOcc << "]: " << *MaxOccVal << "\n";
692 // Create a new instruction that uses the MaxOccVal twice. If we don't do
693 // this, we could otherwise run into situations where removing a factor
694 // from an expression will drop a use of maxocc, and this can cause
695 // RemoveFactorFromExpression on successive values to behave differently.
696 Instruction *DummyInst = BinaryOperator::createAdd(MaxOccVal, MaxOccVal);
697 std::vector<Value*> NewMulOps;
698 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
699 if (Value *V = RemoveFactorFromExpression(Ops[i].Op, MaxOccVal)) {
700 NewMulOps.push_back(V);
701 Ops.erase(Ops.begin()+i);
706 // No need for extra uses anymore.
709 unsigned NumAddedValues = NewMulOps.size();
710 Value *V = EmitAddTreeOfValues(I, NewMulOps);
711 Value *V2 = BinaryOperator::createMul(V, MaxOccVal, "tmp", I);
713 // Now that we have inserted V and its sole use, optimize it. This allows
714 // us to handle cases that require multiple factoring steps, such as this:
715 // A*A*B + A*A*C --> A*(A*B+A*C) --> A*(A*(B+C))
716 if (NumAddedValues > 1)
717 ReassociateExpression(cast<BinaryOperator>(V));
724 // Add the new value to the list of things being added.
725 Ops.insert(Ops.begin(), ValueEntry(getRank(V2), V2));
727 // Rewrite the tree so that there is now a use of V.
728 RewriteExprTree(I, Ops);
729 return OptimizeExpression(I, Ops);
732 //case Instruction::Mul:
735 if (IterateOptimization)
736 return OptimizeExpression(I, Ops);
741 /// ReassociateBB - Inspect all of the instructions in this basic block,
742 /// reassociating them as we go.
743 void Reassociate::ReassociateBB(BasicBlock *BB) {
744 for (BasicBlock::iterator BBI = BB->begin(); BBI != BB->end(); ) {
745 Instruction *BI = BBI++;
746 if (BI->getOpcode() == Instruction::Shl &&
747 isa<ConstantInt>(BI->getOperand(1)))
748 if (Instruction *NI = ConvertShiftToMul(BI)) {
753 // Reject cases where it is pointless to do this.
754 if (!isa<BinaryOperator>(BI) || BI->getType()->isFloatingPoint() ||
755 isa<PackedType>(BI->getType()))
756 continue; // Floating point ops are not associative.
758 // If this is a subtract instruction which is not already in negate form,
759 // see if we can convert it to X+-Y.
760 if (BI->getOpcode() == Instruction::Sub) {
761 if (!BinaryOperator::isNeg(BI)) {
762 if (Instruction *NI = BreakUpSubtract(BI)) {
767 // Otherwise, this is a negation. See if the operand is a multiply tree
768 // and if this is not an inner node of a multiply tree.
769 if (isReassociableOp(BI->getOperand(1), Instruction::Mul) &&
771 !isReassociableOp(BI->use_back(), Instruction::Mul))) {
772 BI = LowerNegateToMultiply(BI);
778 // If this instruction is a commutative binary operator, process it.
779 if (!BI->isAssociative()) continue;
780 BinaryOperator *I = cast<BinaryOperator>(BI);
782 // If this is an interior node of a reassociable tree, ignore it until we
783 // get to the root of the tree, to avoid N^2 analysis.
784 if (I->hasOneUse() && isReassociableOp(I->use_back(), I->getOpcode()))
787 // If this is an add tree that is used by a sub instruction, ignore it
788 // until we process the subtract.
789 if (I->hasOneUse() && I->getOpcode() == Instruction::Add &&
790 cast<Instruction>(I->use_back())->getOpcode() == Instruction::Sub)
793 ReassociateExpression(I);
797 void Reassociate::ReassociateExpression(BinaryOperator *I) {
799 // First, walk the expression tree, linearizing the tree, collecting
800 std::vector<ValueEntry> Ops;
801 LinearizeExprTree(I, Ops);
803 DOUT << "RAIn:\t"; DEBUG(PrintOps(I, Ops)); DOUT << "\n";
805 // Now that we have linearized the tree to a list and have gathered all of
806 // the operands and their ranks, sort the operands by their rank. Use a
807 // stable_sort so that values with equal ranks will have their relative
808 // positions maintained (and so the compiler is deterministic). Note that
809 // this sorts so that the highest ranking values end up at the beginning of
811 std::stable_sort(Ops.begin(), Ops.end());
813 // OptimizeExpression - Now that we have the expression tree in a convenient
814 // sorted form, optimize it globally if possible.
815 if (Value *V = OptimizeExpression(I, Ops)) {
816 // This expression tree simplified to something that isn't a tree,
818 DOUT << "Reassoc to scalar: " << *V << "\n";
819 I->replaceAllUsesWith(V);
820 RemoveDeadBinaryOp(I);
824 // We want to sink immediates as deeply as possible except in the case where
825 // this is a multiply tree used only by an add, and the immediate is a -1.
826 // In this case we reassociate to put the negation on the outside so that we
827 // can fold the negation into the add: (-X)*Y + Z -> Z-X*Y
828 if (I->getOpcode() == Instruction::Mul && I->hasOneUse() &&
829 cast<Instruction>(I->use_back())->getOpcode() == Instruction::Add &&
830 isa<ConstantInt>(Ops.back().Op) &&
831 cast<ConstantInt>(Ops.back().Op)->isAllOnesValue()) {
832 Ops.insert(Ops.begin(), Ops.back());
836 DOUT << "RAOut:\t"; DEBUG(PrintOps(I, Ops)); DOUT << "\n";
838 if (Ops.size() == 1) {
839 // This expression tree simplified to something that isn't a tree,
841 I->replaceAllUsesWith(Ops[0].Op);
842 RemoveDeadBinaryOp(I);
844 // Now that we ordered and optimized the expressions, splat them back into
845 // the expression tree, removing any unneeded nodes.
846 RewriteExprTree(I, Ops);
851 bool Reassociate::runOnFunction(Function &F) {
852 // Recalculate the rank map for F
856 for (Function::iterator FI = F.begin(), FE = F.end(); FI != FE; ++FI)
859 // We are done with the rank map...
861 ValueRankMap.clear();