1 //===- Reassociate.cpp - Reassociate binary expressions -------------------===//
3 // The LLVM Compiler Infrastructure
5 // This file is distributed under the University of Illinois Open Source
6 // 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/IntrinsicInst.h"
30 #include "llvm/LLVMContext.h"
31 #include "llvm/Pass.h"
32 #include "llvm/Assembly/Writer.h"
33 #include "llvm/Support/CFG.h"
34 #include "llvm/Support/Compiler.h"
35 #include "llvm/Support/Debug.h"
36 #include "llvm/Support/ValueHandle.h"
37 #include "llvm/Support/raw_ostream.h"
38 #include "llvm/ADT/PostOrderIterator.h"
39 #include "llvm/ADT/Statistic.h"
44 STATISTIC(NumLinear , "Number of insts linearized");
45 STATISTIC(NumChanged, "Number of insts reassociated");
46 STATISTIC(NumAnnihil, "Number of expr tree annihilated");
47 STATISTIC(NumFactor , "Number of multiplies factored");
50 struct VISIBILITY_HIDDEN ValueEntry {
53 ValueEntry(unsigned R, Value *O) : Rank(R), Op(O) {}
55 inline bool operator<(const ValueEntry &LHS, const ValueEntry &RHS) {
56 return LHS.Rank > RHS.Rank; // Sort so that highest rank goes to start.
61 /// PrintOps - Print out the expression identified in the Ops list.
63 static void PrintOps(Instruction *I, const std::vector<ValueEntry> &Ops) {
64 Module *M = I->getParent()->getParent()->getParent();
65 cerr << Instruction::getOpcodeName(I->getOpcode()) << " "
66 << *Ops[0].Op->getType();
67 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
68 WriteAsOperand(*cerr.stream() << " ", Ops[i].Op, false, M);
69 cerr << "," << Ops[i].Rank;
75 class VISIBILITY_HIDDEN Reassociate : public FunctionPass {
76 std::map<BasicBlock*, unsigned> RankMap;
77 std::map<AssertingVH<>, unsigned> ValueRankMap;
80 static char ID; // Pass identification, replacement for typeid
81 Reassociate() : FunctionPass(&ID) {}
83 bool runOnFunction(Function &F);
85 virtual void getAnalysisUsage(AnalysisUsage &AU) const {
89 void BuildRankMap(Function &F);
90 unsigned getRank(Value *V);
91 void ReassociateExpression(BinaryOperator *I);
92 void RewriteExprTree(BinaryOperator *I, std::vector<ValueEntry> &Ops,
94 Value *OptimizeExpression(BinaryOperator *I, std::vector<ValueEntry> &Ops);
95 void LinearizeExprTree(BinaryOperator *I, std::vector<ValueEntry> &Ops);
96 void LinearizeExpr(BinaryOperator *I);
97 Value *RemoveFactorFromExpression(Value *V, Value *Factor);
98 void ReassociateBB(BasicBlock *BB);
100 void RemoveDeadBinaryOp(Value *V);
104 char Reassociate::ID = 0;
105 static RegisterPass<Reassociate> X("reassociate", "Reassociate expressions");
107 // Public interface to the Reassociate pass
108 FunctionPass *llvm::createReassociatePass() { return new Reassociate(); }
110 void Reassociate::RemoveDeadBinaryOp(Value *V) {
111 Instruction *Op = dyn_cast<Instruction>(V);
112 if (!Op || !isa<BinaryOperator>(Op) || !isa<CmpInst>(Op) || !Op->use_empty())
115 Value *LHS = Op->getOperand(0), *RHS = Op->getOperand(1);
116 RemoveDeadBinaryOp(LHS);
117 RemoveDeadBinaryOp(RHS);
121 static bool isUnmovableInstruction(Instruction *I) {
122 if (I->getOpcode() == Instruction::PHI ||
123 I->getOpcode() == Instruction::Alloca ||
124 I->getOpcode() == Instruction::Load ||
125 I->getOpcode() == Instruction::Malloc ||
126 I->getOpcode() == Instruction::Invoke ||
127 (I->getOpcode() == Instruction::Call &&
128 !isa<DbgInfoIntrinsic>(I)) ||
129 I->getOpcode() == Instruction::UDiv ||
130 I->getOpcode() == Instruction::SDiv ||
131 I->getOpcode() == Instruction::FDiv ||
132 I->getOpcode() == Instruction::URem ||
133 I->getOpcode() == Instruction::SRem ||
134 I->getOpcode() == Instruction::FRem)
139 void Reassociate::BuildRankMap(Function &F) {
142 // Assign distinct ranks to function arguments
143 for (Function::arg_iterator I = F.arg_begin(), E = F.arg_end(); I != E; ++I)
144 ValueRankMap[&*I] = ++i;
146 ReversePostOrderTraversal<Function*> RPOT(&F);
147 for (ReversePostOrderTraversal<Function*>::rpo_iterator I = RPOT.begin(),
148 E = RPOT.end(); I != E; ++I) {
150 unsigned BBRank = RankMap[BB] = ++i << 16;
152 // Walk the basic block, adding precomputed ranks for any instructions that
153 // we cannot move. This ensures that the ranks for these instructions are
154 // all different in the block.
155 for (BasicBlock::iterator I = BB->begin(), E = BB->end(); I != E; ++I)
156 if (isUnmovableInstruction(I))
157 ValueRankMap[&*I] = ++BBRank;
161 unsigned Reassociate::getRank(Value *V) {
162 if (isa<Argument>(V)) return ValueRankMap[V]; // Function argument...
164 Instruction *I = dyn_cast<Instruction>(V);
165 if (I == 0) return 0; // Otherwise it's a global or constant, rank 0.
167 unsigned &CachedRank = ValueRankMap[I];
168 if (CachedRank) return CachedRank; // Rank already known?
170 // If this is an expression, return the 1+MAX(rank(LHS), rank(RHS)) so that
171 // we can reassociate expressions for code motion! Since we do not recurse
172 // for PHI nodes, we cannot have infinite recursion here, because there
173 // cannot be loops in the value graph that do not go through PHI nodes.
174 unsigned Rank = 0, MaxRank = RankMap[I->getParent()];
175 for (unsigned i = 0, e = I->getNumOperands();
176 i != e && Rank != MaxRank; ++i)
177 Rank = std::max(Rank, getRank(I->getOperand(i)));
179 // If this is a not or neg instruction, do not count it for rank. This
180 // assures us that X and ~X will have the same rank.
181 if (!I->getType()->isInteger() ||
182 (!BinaryOperator::isNot(I) && !BinaryOperator::isNeg(I)))
185 //DEBUG(errs() << "Calculated Rank[" << V->getName() << "] = "
188 return CachedRank = Rank;
191 /// isReassociableOp - Return true if V is an instruction of the specified
192 /// opcode and if it only has one use.
193 static BinaryOperator *isReassociableOp(Value *V, unsigned Opcode) {
194 if ((V->hasOneUse() || V->use_empty()) && isa<Instruction>(V) &&
195 cast<Instruction>(V)->getOpcode() == Opcode)
196 return cast<BinaryOperator>(V);
200 /// LowerNegateToMultiply - Replace 0-X with X*-1.
202 static Instruction *LowerNegateToMultiply(Instruction *Neg,
203 std::map<AssertingVH<>, unsigned> &ValueRankMap,
204 LLVMContext &Context) {
205 Constant *Cst = Constant::getAllOnesValue(Neg->getType());
207 Instruction *Res = BinaryOperator::CreateMul(Neg->getOperand(1), Cst, "",Neg);
208 ValueRankMap.erase(Neg);
210 Neg->replaceAllUsesWith(Res);
211 Neg->eraseFromParent();
215 // Given an expression of the form '(A+B)+(D+C)', turn it into '(((A+B)+C)+D)'.
216 // Note that if D is also part of the expression tree that we recurse to
217 // linearize it as well. Besides that case, this does not recurse into A,B, or
219 void Reassociate::LinearizeExpr(BinaryOperator *I) {
220 BinaryOperator *LHS = cast<BinaryOperator>(I->getOperand(0));
221 BinaryOperator *RHS = cast<BinaryOperator>(I->getOperand(1));
222 assert(isReassociableOp(LHS, I->getOpcode()) &&
223 isReassociableOp(RHS, I->getOpcode()) &&
224 "Not an expression that needs linearization?");
226 DEBUG(errs() << "Linear" << *LHS << '\n' << *RHS << '\n' << *I << '\n');
228 // Move the RHS instruction to live immediately before I, avoiding breaking
229 // dominator properties.
232 // Move operands around to do the linearization.
233 I->setOperand(1, RHS->getOperand(0));
234 RHS->setOperand(0, LHS);
235 I->setOperand(0, RHS);
239 DEBUG(errs() << "Linearized: " << *I << '\n');
241 // If D is part of this expression tree, tail recurse.
242 if (isReassociableOp(I->getOperand(1), I->getOpcode()))
247 /// LinearizeExprTree - Given an associative binary expression tree, traverse
248 /// all of the uses putting it into canonical form. This forces a left-linear
249 /// form of the the expression (((a+b)+c)+d), and collects information about the
250 /// rank of the non-tree operands.
252 /// NOTE: These intentionally destroys the expression tree operands (turning
253 /// them into undef values) to reduce #uses of the values. This means that the
254 /// caller MUST use something like RewriteExprTree to put the values back in.
256 void Reassociate::LinearizeExprTree(BinaryOperator *I,
257 std::vector<ValueEntry> &Ops) {
258 Value *LHS = I->getOperand(0), *RHS = I->getOperand(1);
259 unsigned Opcode = I->getOpcode();
260 LLVMContext &Context = I->getContext();
262 // First step, linearize the expression if it is in ((A+B)+(C+D)) form.
263 BinaryOperator *LHSBO = isReassociableOp(LHS, Opcode);
264 BinaryOperator *RHSBO = isReassociableOp(RHS, Opcode);
266 // If this is a multiply expression tree and it contains internal negations,
267 // transform them into multiplies by -1 so they can be reassociated.
268 if (I->getOpcode() == Instruction::Mul) {
269 if (!LHSBO && LHS->hasOneUse() && BinaryOperator::isNeg(LHS)) {
270 LHS = LowerNegateToMultiply(cast<Instruction>(LHS),
271 ValueRankMap, Context);
272 LHSBO = isReassociableOp(LHS, Opcode);
274 if (!RHSBO && RHS->hasOneUse() && BinaryOperator::isNeg(RHS)) {
275 RHS = LowerNegateToMultiply(cast<Instruction>(RHS),
276 ValueRankMap, Context);
277 RHSBO = isReassociableOp(RHS, Opcode);
283 // Neither the LHS or RHS as part of the tree, thus this is a leaf. As
284 // such, just remember these operands and their rank.
285 Ops.push_back(ValueEntry(getRank(LHS), LHS));
286 Ops.push_back(ValueEntry(getRank(RHS), RHS));
288 // Clear the leaves out.
289 I->setOperand(0, UndefValue::get(I->getType()));
290 I->setOperand(1, UndefValue::get(I->getType()));
293 // Turn X+(Y+Z) -> (Y+Z)+X
294 std::swap(LHSBO, RHSBO);
296 bool Success = !I->swapOperands();
297 assert(Success && "swapOperands failed");
302 // Turn (A+B)+(C+D) -> (((A+B)+C)+D). This guarantees the the RHS is not
303 // part of the expression tree.
305 LHS = LHSBO = cast<BinaryOperator>(I->getOperand(0));
306 RHS = I->getOperand(1);
310 // Okay, now we know that the LHS is a nested expression and that the RHS is
311 // not. Perform reassociation.
312 assert(!isReassociableOp(RHS, Opcode) && "LinearizeExpr failed!");
314 // Move LHS right before I to make sure that the tree expression dominates all
316 LHSBO->moveBefore(I);
318 // Linearize the expression tree on the LHS.
319 LinearizeExprTree(LHSBO, Ops);
321 // Remember the RHS operand and its rank.
322 Ops.push_back(ValueEntry(getRank(RHS), RHS));
324 // Clear the RHS leaf out.
325 I->setOperand(1, UndefValue::get(I->getType()));
328 // RewriteExprTree - Now that the operands for this expression tree are
329 // linearized and optimized, emit them in-order. This function is written to be
331 void Reassociate::RewriteExprTree(BinaryOperator *I,
332 std::vector<ValueEntry> &Ops,
334 if (i+2 == Ops.size()) {
335 if (I->getOperand(0) != Ops[i].Op ||
336 I->getOperand(1) != Ops[i+1].Op) {
337 Value *OldLHS = I->getOperand(0);
338 DEBUG(errs() << "RA: " << *I << '\n');
339 I->setOperand(0, Ops[i].Op);
340 I->setOperand(1, Ops[i+1].Op);
341 DEBUG(errs() << "TO: " << *I << '\n');
345 // If we reassociated a tree to fewer operands (e.g. (1+a+2) -> (a+3)
346 // delete the extra, now dead, nodes.
347 RemoveDeadBinaryOp(OldLHS);
351 assert(i+2 < Ops.size() && "Ops index out of range!");
353 if (I->getOperand(1) != Ops[i].Op) {
354 DEBUG(errs() << "RA: " << *I << '\n');
355 I->setOperand(1, Ops[i].Op);
356 DEBUG(errs() << "TO: " << *I << '\n');
361 BinaryOperator *LHS = cast<BinaryOperator>(I->getOperand(0));
362 assert(LHS->getOpcode() == I->getOpcode() &&
363 "Improper expression tree!");
365 // Compactify the tree instructions together with each other to guarantee
366 // that the expression tree is dominated by all of Ops.
368 RewriteExprTree(LHS, Ops, i+1);
373 // NegateValue - Insert instructions before the instruction pointed to by BI,
374 // that computes the negative version of the value specified. The negative
375 // version of the value is returned, and BI is left pointing at the instruction
376 // that should be processed next by the reassociation pass.
378 static Value *NegateValue(LLVMContext &Context, Value *V, Instruction *BI) {
379 // We are trying to expose opportunity for reassociation. One of the things
380 // that we want to do to achieve this is to push a negation as deep into an
381 // expression chain as possible, to expose the add instructions. In practice,
382 // this means that we turn this:
383 // X = -(A+12+C+D) into X = -A + -12 + -C + -D = -12 + -A + -C + -D
384 // so that later, a: Y = 12+X could get reassociated with the -12 to eliminate
385 // the constants. We assume that instcombine will clean up the mess later if
386 // we introduce tons of unnecessary negation instructions...
388 if (Instruction *I = dyn_cast<Instruction>(V))
389 if (I->getOpcode() == Instruction::Add && I->hasOneUse()) {
390 // Push the negates through the add.
391 I->setOperand(0, NegateValue(Context, I->getOperand(0), BI));
392 I->setOperand(1, NegateValue(Context, I->getOperand(1), BI));
394 // We must move the add instruction here, because the neg instructions do
395 // not dominate the old add instruction in general. By moving it, we are
396 // assured that the neg instructions we just inserted dominate the
397 // instruction we are about to insert after them.
400 I->setName(I->getName()+".neg");
404 // Insert a 'neg' instruction that subtracts the value from zero to get the
407 return BinaryOperator::CreateNeg(V, V->getName() + ".neg", BI);
410 /// ShouldBreakUpSubtract - Return true if we should break up this subtract of
411 /// X-Y into (X + -Y).
412 static bool ShouldBreakUpSubtract(LLVMContext &Context, Instruction *Sub) {
413 // If this is a negation, we can't split it up!
414 if (BinaryOperator::isNeg(Sub))
417 // Don't bother to break this up unless either the LHS is an associable add or
418 // subtract or if this is only used by one.
419 if (isReassociableOp(Sub->getOperand(0), Instruction::Add) ||
420 isReassociableOp(Sub->getOperand(0), Instruction::Sub))
422 if (isReassociableOp(Sub->getOperand(1), Instruction::Add) ||
423 isReassociableOp(Sub->getOperand(1), Instruction::Sub))
425 if (Sub->hasOneUse() &&
426 (isReassociableOp(Sub->use_back(), Instruction::Add) ||
427 isReassociableOp(Sub->use_back(), Instruction::Sub)))
433 /// BreakUpSubtract - If we have (X-Y), and if either X is an add, or if this is
434 /// only used by an add, transform this into (X+(0-Y)) to promote better
436 static Instruction *BreakUpSubtract(LLVMContext &Context, Instruction *Sub,
437 std::map<AssertingVH<>, unsigned> &ValueRankMap) {
438 // Convert a subtract into an add and a neg instruction... so that sub
439 // instructions can be commuted with other add instructions...
441 // Calculate the negative value of Operand 1 of the sub instruction...
442 // and set it as the RHS of the add instruction we just made...
444 Value *NegVal = NegateValue(Context, Sub->getOperand(1), Sub);
446 BinaryOperator::CreateAdd(Sub->getOperand(0), NegVal, "", Sub);
449 // Everyone now refers to the add instruction.
450 ValueRankMap.erase(Sub);
451 Sub->replaceAllUsesWith(New);
452 Sub->eraseFromParent();
454 DEBUG(errs() << "Negated: " << *New << '\n');
458 /// ConvertShiftToMul - If this is a shift of a reassociable multiply or is used
459 /// by one, change this into a multiply by a constant to assist with further
461 static Instruction *ConvertShiftToMul(Instruction *Shl,
462 std::map<AssertingVH<>, unsigned> &ValueRankMap,
463 LLVMContext &Context) {
464 // If an operand of this shift is a reassociable multiply, or if the shift
465 // is used by a reassociable multiply or add, turn into a multiply.
466 if (isReassociableOp(Shl->getOperand(0), Instruction::Mul) ||
468 (isReassociableOp(Shl->use_back(), Instruction::Mul) ||
469 isReassociableOp(Shl->use_back(), Instruction::Add)))) {
470 Constant *MulCst = ConstantInt::get(Shl->getType(), 1);
472 ConstantExpr::getShl(MulCst, cast<Constant>(Shl->getOperand(1)));
474 Instruction *Mul = BinaryOperator::CreateMul(Shl->getOperand(0), MulCst,
476 ValueRankMap.erase(Shl);
478 Shl->replaceAllUsesWith(Mul);
479 Shl->eraseFromParent();
485 // Scan backwards and forwards among values with the same rank as element i to
486 // see if X exists. If X does not exist, return i.
487 static unsigned FindInOperandList(std::vector<ValueEntry> &Ops, unsigned i,
489 unsigned XRank = Ops[i].Rank;
490 unsigned e = Ops.size();
491 for (unsigned j = i+1; j != e && Ops[j].Rank == XRank; ++j)
495 for (unsigned j = i-1; j != ~0U && Ops[j].Rank == XRank; --j)
501 /// EmitAddTreeOfValues - Emit a tree of add instructions, summing Ops together
502 /// and returning the result. Insert the tree before I.
503 static Value *EmitAddTreeOfValues(Instruction *I, std::vector<Value*> &Ops) {
504 if (Ops.size() == 1) return Ops.back();
506 Value *V1 = Ops.back();
508 Value *V2 = EmitAddTreeOfValues(I, Ops);
509 return BinaryOperator::CreateAdd(V2, V1, "tmp", I);
512 /// RemoveFactorFromExpression - If V is an expression tree that is a
513 /// multiplication sequence, and if this sequence contains a multiply by Factor,
514 /// remove Factor from the tree and return the new tree.
515 Value *Reassociate::RemoveFactorFromExpression(Value *V, Value *Factor) {
516 BinaryOperator *BO = isReassociableOp(V, Instruction::Mul);
519 std::vector<ValueEntry> Factors;
520 LinearizeExprTree(BO, Factors);
522 bool FoundFactor = false;
523 for (unsigned i = 0, e = Factors.size(); i != e; ++i)
524 if (Factors[i].Op == Factor) {
526 Factors.erase(Factors.begin()+i);
530 // Make sure to restore the operands to the expression tree.
531 RewriteExprTree(BO, Factors);
535 if (Factors.size() == 1) return Factors[0].Op;
537 RewriteExprTree(BO, Factors);
541 /// FindSingleUseMultiplyFactors - If V is a single-use multiply, recursively
542 /// add its operands as factors, otherwise add V to the list of factors.
543 static void FindSingleUseMultiplyFactors(Value *V,
544 std::vector<Value*> &Factors) {
546 if ((!V->hasOneUse() && !V->use_empty()) ||
547 !(BO = dyn_cast<BinaryOperator>(V)) ||
548 BO->getOpcode() != Instruction::Mul) {
549 Factors.push_back(V);
553 // Otherwise, add the LHS and RHS to the list of factors.
554 FindSingleUseMultiplyFactors(BO->getOperand(1), Factors);
555 FindSingleUseMultiplyFactors(BO->getOperand(0), Factors);
560 Value *Reassociate::OptimizeExpression(BinaryOperator *I,
561 std::vector<ValueEntry> &Ops) {
562 // Now that we have the linearized expression tree, try to optimize it.
563 // Start by folding any constants that we found.
564 bool IterateOptimization = false;
565 if (Ops.size() == 1) return Ops[0].Op;
567 unsigned Opcode = I->getOpcode();
569 if (Constant *V1 = dyn_cast<Constant>(Ops[Ops.size()-2].Op))
570 if (Constant *V2 = dyn_cast<Constant>(Ops.back().Op)) {
572 Ops.back().Op = ConstantExpr::get(Opcode, V1, V2);
573 return OptimizeExpression(I, Ops);
576 // Check for destructive annihilation due to a constant being used.
577 if (ConstantInt *CstVal = dyn_cast<ConstantInt>(Ops.back().Op))
580 case Instruction::And:
581 if (CstVal->isZero()) { // ... & 0 -> 0
584 } else if (CstVal->isAllOnesValue()) { // ... & -1 -> ...
588 case Instruction::Mul:
589 if (CstVal->isZero()) { // ... * 0 -> 0
592 } else if (cast<ConstantInt>(CstVal)->isOne()) {
593 Ops.pop_back(); // ... * 1 -> ...
596 case Instruction::Or:
597 if (CstVal->isAllOnesValue()) { // ... | -1 -> -1
602 case Instruction::Add:
603 case Instruction::Xor:
604 if (CstVal->isZero()) // ... [|^+] 0 -> ...
608 if (Ops.size() == 1) return Ops[0].Op;
610 // Handle destructive annihilation do to identities between elements in the
611 // argument list here.
614 case Instruction::And:
615 case Instruction::Or:
616 case Instruction::Xor:
617 // Scan the operand lists looking for X and ~X pairs, along with X,X pairs.
618 // If we find any, we can simplify the expression. X&~X == 0, X|~X == -1.
619 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
620 // First, check for X and ~X in the operand list.
621 assert(i < Ops.size());
622 if (BinaryOperator::isNot(Ops[i].Op)) { // Cannot occur for ^.
623 Value *X = BinaryOperator::getNotArgument(Ops[i].Op);
624 unsigned FoundX = FindInOperandList(Ops, i, X);
626 if (Opcode == Instruction::And) { // ...&X&~X = 0
628 return Constant::getNullValue(X->getType());
629 } else if (Opcode == Instruction::Or) { // ...|X|~X = -1
631 return Constant::getAllOnesValue(X->getType());
636 // Next, check for duplicate pairs of values, which we assume are next to
637 // each other, due to our sorting criteria.
638 assert(i < Ops.size());
639 if (i+1 != Ops.size() && Ops[i+1].Op == Ops[i].Op) {
640 if (Opcode == Instruction::And || Opcode == Instruction::Or) {
641 // Drop duplicate values.
642 Ops.erase(Ops.begin()+i);
644 IterateOptimization = true;
647 assert(Opcode == Instruction::Xor);
650 return Constant::getNullValue(Ops[0].Op->getType());
653 Ops.erase(Ops.begin()+i, Ops.begin()+i+2);
655 IterateOptimization = true;
662 case Instruction::Add:
663 // Scan the operand lists looking for X and -X pairs. If we find any, we
664 // can simplify the expression. X+-X == 0.
665 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
666 assert(i < Ops.size());
667 // Check for X and -X in the operand list.
668 if (BinaryOperator::isNeg(Ops[i].Op)) {
669 Value *X = BinaryOperator::getNegArgument(Ops[i].Op);
670 unsigned FoundX = FindInOperandList(Ops, i, X);
672 // Remove X and -X from the operand list.
673 if (Ops.size() == 2) {
675 return Constant::getNullValue(X->getType());
677 Ops.erase(Ops.begin()+i);
681 --i; // Need to back up an extra one.
682 Ops.erase(Ops.begin()+FoundX);
683 IterateOptimization = true;
685 --i; // Revisit element.
686 e -= 2; // Removed two elements.
693 // Scan the operand list, checking to see if there are any common factors
694 // between operands. Consider something like A*A+A*B*C+D. We would like to
695 // reassociate this to A*(A+B*C)+D, which reduces the number of multiplies.
696 // To efficiently find this, we count the number of times a factor occurs
697 // for any ADD operands that are MULs.
698 std::map<Value*, unsigned> FactorOccurrences;
700 Value *MaxOccVal = 0;
701 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
702 if (BinaryOperator *BOp = dyn_cast<BinaryOperator>(Ops[i].Op)) {
703 if (BOp->getOpcode() == Instruction::Mul && BOp->use_empty()) {
704 // Compute all of the factors of this added value.
705 std::vector<Value*> Factors;
706 FindSingleUseMultiplyFactors(BOp, Factors);
707 assert(Factors.size() > 1 && "Bad linearize!");
709 // Add one to FactorOccurrences for each unique factor in this op.
710 if (Factors.size() == 2) {
711 unsigned Occ = ++FactorOccurrences[Factors[0]];
712 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[0]; }
713 if (Factors[0] != Factors[1]) { // Don't double count A*A.
714 Occ = ++FactorOccurrences[Factors[1]];
715 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[1]; }
718 std::set<Value*> Duplicates;
719 for (unsigned i = 0, e = Factors.size(); i != e; ++i) {
720 if (Duplicates.insert(Factors[i]).second) {
721 unsigned Occ = ++FactorOccurrences[Factors[i]];
722 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[i]; }
730 // If any factor occurred more than one time, we can pull it out.
732 DEBUG(errs() << "\nFACTORING [" << MaxOcc << "]: " << *MaxOccVal << "\n");
734 // Create a new instruction that uses the MaxOccVal twice. If we don't do
735 // this, we could otherwise run into situations where removing a factor
736 // from an expression will drop a use of maxocc, and this can cause
737 // RemoveFactorFromExpression on successive values to behave differently.
738 Instruction *DummyInst = BinaryOperator::CreateAdd(MaxOccVal, MaxOccVal);
739 std::vector<Value*> NewMulOps;
740 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
741 if (Value *V = RemoveFactorFromExpression(Ops[i].Op, MaxOccVal)) {
742 NewMulOps.push_back(V);
743 Ops.erase(Ops.begin()+i);
748 // No need for extra uses anymore.
751 unsigned NumAddedValues = NewMulOps.size();
752 Value *V = EmitAddTreeOfValues(I, NewMulOps);
753 Value *V2 = BinaryOperator::CreateMul(V, MaxOccVal, "tmp", I);
755 // Now that we have inserted V and its sole use, optimize it. This allows
756 // us to handle cases that require multiple factoring steps, such as this:
757 // A*A*B + A*A*C --> A*(A*B+A*C) --> A*(A*(B+C))
758 if (NumAddedValues > 1)
759 ReassociateExpression(cast<BinaryOperator>(V));
766 // Add the new value to the list of things being added.
767 Ops.insert(Ops.begin(), ValueEntry(getRank(V2), V2));
769 // Rewrite the tree so that there is now a use of V.
770 RewriteExprTree(I, Ops);
771 return OptimizeExpression(I, Ops);
774 //case Instruction::Mul:
777 if (IterateOptimization)
778 return OptimizeExpression(I, Ops);
783 /// ReassociateBB - Inspect all of the instructions in this basic block,
784 /// reassociating them as we go.
785 void Reassociate::ReassociateBB(BasicBlock *BB) {
786 LLVMContext &Context = BB->getContext();
788 for (BasicBlock::iterator BBI = BB->begin(); BBI != BB->end(); ) {
789 Instruction *BI = BBI++;
790 if (BI->getOpcode() == Instruction::Shl &&
791 isa<ConstantInt>(BI->getOperand(1)))
792 if (Instruction *NI = ConvertShiftToMul(BI, ValueRankMap, Context)) {
797 // Reject cases where it is pointless to do this.
798 if (!isa<BinaryOperator>(BI) || BI->getType()->isFloatingPoint() ||
799 isa<VectorType>(BI->getType()))
800 continue; // Floating point ops are not associative.
802 // If this is a subtract instruction which is not already in negate form,
803 // see if we can convert it to X+-Y.
804 if (BI->getOpcode() == Instruction::Sub) {
805 if (ShouldBreakUpSubtract(Context, BI)) {
806 BI = BreakUpSubtract(Context, BI, ValueRankMap);
808 } else if (BinaryOperator::isNeg(BI)) {
809 // Otherwise, this is a negation. See if the operand is a multiply tree
810 // and if this is not an inner node of a multiply tree.
811 if (isReassociableOp(BI->getOperand(1), Instruction::Mul) &&
813 !isReassociableOp(BI->use_back(), Instruction::Mul))) {
814 BI = LowerNegateToMultiply(BI, ValueRankMap, Context);
820 // If this instruction is a commutative binary operator, process it.
821 if (!BI->isAssociative()) continue;
822 BinaryOperator *I = cast<BinaryOperator>(BI);
824 // If this is an interior node of a reassociable tree, ignore it until we
825 // get to the root of the tree, to avoid N^2 analysis.
826 if (I->hasOneUse() && isReassociableOp(I->use_back(), I->getOpcode()))
829 // If this is an add tree that is used by a sub instruction, ignore it
830 // until we process the subtract.
831 if (I->hasOneUse() && I->getOpcode() == Instruction::Add &&
832 cast<Instruction>(I->use_back())->getOpcode() == Instruction::Sub)
835 ReassociateExpression(I);
839 void Reassociate::ReassociateExpression(BinaryOperator *I) {
841 // First, walk the expression tree, linearizing the tree, collecting
842 std::vector<ValueEntry> Ops;
843 LinearizeExprTree(I, Ops);
845 DEBUG(errs() << "RAIn:\t"; PrintOps(I, Ops); errs() << "\n");
847 // Now that we have linearized the tree to a list and have gathered all of
848 // the operands and their ranks, sort the operands by their rank. Use a
849 // stable_sort so that values with equal ranks will have their relative
850 // positions maintained (and so the compiler is deterministic). Note that
851 // this sorts so that the highest ranking values end up at the beginning of
853 std::stable_sort(Ops.begin(), Ops.end());
855 // OptimizeExpression - Now that we have the expression tree in a convenient
856 // sorted form, optimize it globally if possible.
857 if (Value *V = OptimizeExpression(I, Ops)) {
858 // This expression tree simplified to something that isn't a tree,
860 DEBUG(errs() << "Reassoc to scalar: " << *V << "\n");
861 I->replaceAllUsesWith(V);
862 RemoveDeadBinaryOp(I);
866 // We want to sink immediates as deeply as possible except in the case where
867 // this is a multiply tree used only by an add, and the immediate is a -1.
868 // In this case we reassociate to put the negation on the outside so that we
869 // can fold the negation into the add: (-X)*Y + Z -> Z-X*Y
870 if (I->getOpcode() == Instruction::Mul && I->hasOneUse() &&
871 cast<Instruction>(I->use_back())->getOpcode() == Instruction::Add &&
872 isa<ConstantInt>(Ops.back().Op) &&
873 cast<ConstantInt>(Ops.back().Op)->isAllOnesValue()) {
874 Ops.insert(Ops.begin(), Ops.back());
878 DEBUG(errs() << "RAOut:\t"; PrintOps(I, Ops); errs() << "\n");
880 if (Ops.size() == 1) {
881 // This expression tree simplified to something that isn't a tree,
883 I->replaceAllUsesWith(Ops[0].Op);
884 RemoveDeadBinaryOp(I);
886 // Now that we ordered and optimized the expressions, splat them back into
887 // the expression tree, removing any unneeded nodes.
888 RewriteExprTree(I, Ops);
893 bool Reassociate::runOnFunction(Function &F) {
894 // Recalculate the rank map for F
898 for (Function::iterator FI = F.begin(), FE = F.end(); FI != FE; ++FI)
901 // We are done with the rank map...
903 ValueRankMap.clear();