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/Function.h"
27 #include "llvm/Instructions.h"
28 #include "llvm/Pass.h"
29 #include "llvm/Type.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"
39 Statistic<> NumLinear ("reassociate","Number of insts linearized");
40 Statistic<> NumChanged("reassociate","Number of insts reassociated");
41 Statistic<> NumSwapped("reassociate","Number of insts with operands swapped");
42 Statistic<> NumAnnihil("reassociate","Number of expr tree annihilated");
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.
53 class Reassociate : public FunctionPass {
54 std::map<BasicBlock*, unsigned> RankMap;
55 std::map<Value*, unsigned> ValueRankMap;
58 bool runOnFunction(Function &F);
60 virtual void getAnalysisUsage(AnalysisUsage &AU) const {
64 void BuildRankMap(Function &F);
65 unsigned getRank(Value *V);
66 void RewriteExprTree(BinaryOperator *I, unsigned Idx,
67 std::vector<ValueEntry> &Ops);
68 void OptimizeExpression(unsigned Opcode, std::vector<ValueEntry> &Ops);
69 void LinearizeExprTree(BinaryOperator *I, std::vector<ValueEntry> &Ops);
70 void LinearizeExpr(BinaryOperator *I);
71 void ReassociateBB(BasicBlock *BB);
74 RegisterOpt<Reassociate> X("reassociate", "Reassociate expressions");
77 // Public interface to the Reassociate pass
78 FunctionPass *llvm::createReassociatePass() { return new Reassociate(); }
81 static bool isUnmovableInstruction(Instruction *I) {
82 if (I->getOpcode() == Instruction::PHI ||
83 I->getOpcode() == Instruction::Alloca ||
84 I->getOpcode() == Instruction::Load ||
85 I->getOpcode() == Instruction::Malloc ||
86 I->getOpcode() == Instruction::Invoke ||
87 I->getOpcode() == Instruction::Call ||
88 I->getOpcode() == Instruction::Div ||
89 I->getOpcode() == Instruction::Rem)
94 void Reassociate::BuildRankMap(Function &F) {
97 // Assign distinct ranks to function arguments
98 for (Function::arg_iterator I = F.arg_begin(), E = F.arg_end(); I != E; ++I)
99 ValueRankMap[I] = ++i;
101 ReversePostOrderTraversal<Function*> RPOT(&F);
102 for (ReversePostOrderTraversal<Function*>::rpo_iterator I = RPOT.begin(),
103 E = RPOT.end(); I != E; ++I) {
105 unsigned BBRank = RankMap[BB] = ++i << 16;
107 // Walk the basic block, adding precomputed ranks for any instructions that
108 // we cannot move. This ensures that the ranks for these instructions are
109 // all different in the block.
110 for (BasicBlock::iterator I = BB->begin(), E = BB->end(); I != E; ++I)
111 if (isUnmovableInstruction(I))
112 ValueRankMap[I] = ++BBRank;
116 unsigned Reassociate::getRank(Value *V) {
117 if (isa<Argument>(V)) return ValueRankMap[V]; // Function argument...
119 Instruction *I = dyn_cast<Instruction>(V);
120 if (I == 0) return 0; // Otherwise it's a global or constant, rank 0.
122 unsigned &CachedRank = ValueRankMap[I];
123 if (CachedRank) return CachedRank; // Rank already known?
125 // If this is an expression, return the 1+MAX(rank(LHS), rank(RHS)) so that
126 // we can reassociate expressions for code motion! Since we do not recurse
127 // for PHI nodes, we cannot have infinite recursion here, because there
128 // cannot be loops in the value graph that do not go through PHI nodes.
129 unsigned Rank = 0, MaxRank = RankMap[I->getParent()];
130 for (unsigned i = 0, e = I->getNumOperands();
131 i != e && Rank != MaxRank; ++i)
132 Rank = std::max(Rank, getRank(I->getOperand(i)));
134 // If this is a not or neg instruction, do not count it for rank. This
135 // assures us that X and ~X will have the same rank.
136 if (!I->getType()->isIntegral() ||
137 (!BinaryOperator::isNot(I) && !BinaryOperator::isNeg(I)))
140 //DEBUG(std::cerr << "Calculated Rank[" << V->getName() << "] = "
143 return CachedRank = Rank;
146 /// isReassociableOp - Return true if V is an instruction of the specified
147 /// opcode and if it only has one use.
148 static BinaryOperator *isReassociableOp(Value *V, unsigned Opcode) {
149 if (V->hasOneUse() && isa<Instruction>(V) &&
150 cast<Instruction>(V)->getOpcode() == Opcode)
151 return cast<BinaryOperator>(V);
155 /// LowerNegateToMultiply - Replace 0-X with X*-1.
157 static Instruction *LowerNegateToMultiply(Instruction *Neg) {
159 if (Neg->getType()->isFloatingPoint())
160 Cst = ConstantFP::get(Neg->getType(), -1);
162 Cst = ConstantInt::getAllOnesValue(Neg->getType());
164 std::string NegName = Neg->getName(); Neg->setName("");
165 Instruction *Res = BinaryOperator::createMul(Neg->getOperand(1), Cst, NegName,
167 Neg->replaceAllUsesWith(Res);
168 Neg->eraseFromParent();
172 // Given an expression of the form '(A+B)+(D+C)', turn it into '(((A+B)+C)+D)'.
173 // Note that if D is also part of the expression tree that we recurse to
174 // linearize it as well. Besides that case, this does not recurse into A,B, or
176 void Reassociate::LinearizeExpr(BinaryOperator *I) {
177 BinaryOperator *LHS = cast<BinaryOperator>(I->getOperand(0));
178 BinaryOperator *RHS = cast<BinaryOperator>(I->getOperand(1));
179 assert(isReassociableOp(LHS, I->getOpcode()) &&
180 isReassociableOp(RHS, I->getOpcode()) &&
181 "Not an expression that needs linearization?");
183 DEBUG(std::cerr << "Linear" << *LHS << *RHS << *I);
185 // Move the RHS instruction to live immediately before I, avoiding breaking
186 // dominator properties.
189 // Move operands around to do the linearization.
190 I->setOperand(1, RHS->getOperand(0));
191 RHS->setOperand(0, LHS);
192 I->setOperand(0, RHS);
196 DEBUG(std::cerr << "Linearized: " << *I);
198 // If D is part of this expression tree, tail recurse.
199 if (isReassociableOp(I->getOperand(1), I->getOpcode()))
204 /// LinearizeExprTree - Given an associative binary expression tree, traverse
205 /// all of the uses putting it into canonical form. This forces a left-linear
206 /// form of the the expression (((a+b)+c)+d), and collects information about the
207 /// rank of the non-tree operands.
209 /// This returns the rank of the RHS operand, which is known to be the highest
210 /// rank value in the expression tree.
212 void Reassociate::LinearizeExprTree(BinaryOperator *I,
213 std::vector<ValueEntry> &Ops) {
214 Value *LHS = I->getOperand(0), *RHS = I->getOperand(1);
215 unsigned Opcode = I->getOpcode();
217 // First step, linearize the expression if it is in ((A+B)+(C+D)) form.
218 BinaryOperator *LHSBO = isReassociableOp(LHS, Opcode);
219 BinaryOperator *RHSBO = isReassociableOp(RHS, Opcode);
221 // If this is a multiply expression tree and it contains internal negations,
222 // transform them into multiplies by -1 so they can be reassociated.
223 if (I->getOpcode() == Instruction::Mul) {
224 if (!LHSBO && LHS->hasOneUse() && BinaryOperator::isNeg(LHS)) {
225 LHS = LowerNegateToMultiply(cast<Instruction>(LHS));
226 LHSBO = isReassociableOp(LHS, Opcode);
228 if (!RHSBO && RHS->hasOneUse() && BinaryOperator::isNeg(RHS)) {
229 RHS = LowerNegateToMultiply(cast<Instruction>(RHS));
230 RHSBO = isReassociableOp(RHS, Opcode);
236 // Neither the LHS or RHS as part of the tree, thus this is a leaf. As
237 // such, just remember these operands and their rank.
238 Ops.push_back(ValueEntry(getRank(LHS), LHS));
239 Ops.push_back(ValueEntry(getRank(RHS), RHS));
242 // Turn X+(Y+Z) -> (Y+Z)+X
243 std::swap(LHSBO, RHSBO);
245 bool Success = !I->swapOperands();
246 assert(Success && "swapOperands failed");
250 // Turn (A+B)+(C+D) -> (((A+B)+C)+D). This guarantees the the RHS is not
251 // part of the expression tree.
253 LHS = LHSBO = cast<BinaryOperator>(I->getOperand(0));
254 RHS = I->getOperand(1);
258 // Okay, now we know that the LHS is a nested expression and that the RHS is
259 // not. Perform reassociation.
260 assert(!isReassociableOp(RHS, Opcode) && "LinearizeExpr failed!");
262 // Move LHS right before I to make sure that the tree expression dominates all
264 LHSBO->moveBefore(I);
266 // Linearize the expression tree on the LHS.
267 LinearizeExprTree(LHSBO, Ops);
269 // Remember the RHS operand and its rank.
270 Ops.push_back(ValueEntry(getRank(RHS), RHS));
273 // RewriteExprTree - Now that the operands for this expression tree are
274 // linearized and optimized, emit them in-order. This function is written to be
276 void Reassociate::RewriteExprTree(BinaryOperator *I, unsigned i,
277 std::vector<ValueEntry> &Ops) {
278 if (i+2 == Ops.size()) {
279 if (I->getOperand(0) != Ops[i].Op ||
280 I->getOperand(1) != Ops[i+1].Op) {
281 DEBUG(std::cerr << "RA: " << *I);
282 I->setOperand(0, Ops[i].Op);
283 I->setOperand(1, Ops[i+1].Op);
284 DEBUG(std::cerr << "TO: " << *I);
290 assert(i+2 < Ops.size() && "Ops index out of range!");
292 if (I->getOperand(1) != Ops[i].Op) {
293 DEBUG(std::cerr << "RA: " << *I);
294 I->setOperand(1, Ops[i].Op);
295 DEBUG(std::cerr << "TO: " << *I);
299 RewriteExprTree(cast<BinaryOperator>(I->getOperand(0)), i+1, Ops);
304 // NegateValue - Insert instructions before the instruction pointed to by BI,
305 // that computes the negative version of the value specified. The negative
306 // version of the value is returned, and BI is left pointing at the instruction
307 // that should be processed next by the reassociation pass.
309 static Value *NegateValue(Value *V, Instruction *BI) {
310 // We are trying to expose opportunity for reassociation. One of the things
311 // that we want to do to achieve this is to push a negation as deep into an
312 // expression chain as possible, to expose the add instructions. In practice,
313 // this means that we turn this:
314 // X = -(A+12+C+D) into X = -A + -12 + -C + -D = -12 + -A + -C + -D
315 // so that later, a: Y = 12+X could get reassociated with the -12 to eliminate
316 // the constants. We assume that instcombine will clean up the mess later if
317 // we introduce tons of unnecessary negation instructions...
319 if (Instruction *I = dyn_cast<Instruction>(V))
320 if (I->getOpcode() == Instruction::Add && I->hasOneUse()) {
321 Value *RHS = NegateValue(I->getOperand(1), BI);
322 Value *LHS = NegateValue(I->getOperand(0), BI);
324 // We must actually insert a new add instruction here, because the neg
325 // instructions do not dominate the old add instruction in general. By
326 // adding it now, we are assured that the neg instructions we just
327 // inserted dominate the instruction we are about to insert after them.
329 return BinaryOperator::create(Instruction::Add, LHS, RHS,
330 I->getName()+".neg", BI);
333 // Insert a 'neg' instruction that subtracts the value from zero to get the
336 return BinaryOperator::createNeg(V, V->getName() + ".neg", BI);
339 /// BreakUpSubtract - If we have (X-Y), and if either X is an add, or if this is
340 /// only used by an add, transform this into (X+(0-Y)) to promote better
342 static Instruction *BreakUpSubtract(Instruction *Sub) {
343 // Don't bother to break this up unless either the LHS is an associable add or
344 // if this is only used by one.
345 if (!isReassociableOp(Sub->getOperand(0), Instruction::Add) &&
346 !isReassociableOp(Sub->getOperand(1), Instruction::Add) &&
347 !(Sub->hasOneUse() &&isReassociableOp(Sub->use_back(), Instruction::Add)))
350 // Convert a subtract into an add and a neg instruction... so that sub
351 // instructions can be commuted with other add instructions...
353 // Calculate the negative value of Operand 1 of the sub instruction...
354 // and set it as the RHS of the add instruction we just made...
356 std::string Name = Sub->getName();
358 Value *NegVal = NegateValue(Sub->getOperand(1), Sub);
360 BinaryOperator::createAdd(Sub->getOperand(0), NegVal, Name, Sub);
362 // Everyone now refers to the add instruction.
363 Sub->replaceAllUsesWith(New);
364 Sub->eraseFromParent();
366 DEBUG(std::cerr << "Negated: " << *New);
370 /// ConvertShiftToMul - If this is a shift of a reassociable multiply or is used
371 /// by one, change this into a multiply by a constant to assist with further
373 static Instruction *ConvertShiftToMul(Instruction *Shl) {
374 if (!isReassociableOp(Shl->getOperand(0), Instruction::Mul) &&
375 !(Shl->hasOneUse() && isReassociableOp(Shl->use_back(),Instruction::Mul)))
378 Constant *MulCst = ConstantInt::get(Shl->getType(), 1);
379 MulCst = ConstantExpr::getShl(MulCst, cast<Constant>(Shl->getOperand(1)));
381 std::string Name = Shl->getName(); Shl->setName("");
382 Instruction *Mul = BinaryOperator::createMul(Shl->getOperand(0), MulCst,
384 Shl->replaceAllUsesWith(Mul);
385 Shl->eraseFromParent();
389 // Scan backwards and forwards among values with the same rank as element i to
390 // see if X exists. If X does not exist, return i.
391 static unsigned FindInOperandList(std::vector<ValueEntry> &Ops, unsigned i,
393 unsigned XRank = Ops[i].Rank;
394 unsigned e = Ops.size();
395 for (unsigned j = i+1; j != e && Ops[j].Rank == XRank; ++j)
399 for (unsigned j = i-1; j != ~0U && Ops[j].Rank == XRank; --j)
405 void Reassociate::OptimizeExpression(unsigned Opcode,
406 std::vector<ValueEntry> &Ops) {
407 // Now that we have the linearized expression tree, try to optimize it.
408 // Start by folding any constants that we found.
409 bool IterateOptimization = false;
410 if (Ops.size() == 1) return;
412 if (Constant *V1 = dyn_cast<Constant>(Ops[Ops.size()-2].Op))
413 if (Constant *V2 = dyn_cast<Constant>(Ops.back().Op)) {
415 Ops.back().Op = ConstantExpr::get(Opcode, V1, V2);
416 OptimizeExpression(Opcode, Ops);
420 // Check for destructive annihilation due to a constant being used.
421 if (ConstantIntegral *CstVal = dyn_cast<ConstantIntegral>(Ops.back().Op))
424 case Instruction::And:
425 if (CstVal->isNullValue()) { // ... & 0 -> 0
427 Ops.erase(Ops.begin()+1, Ops.end());
430 } else if (CstVal->isAllOnesValue()) { // ... & -1 -> ...
434 case Instruction::Mul:
435 if (CstVal->isNullValue()) { // ... * 0 -> 0
437 Ops.erase(Ops.begin()+1, Ops.end());
440 } else if (cast<ConstantInt>(CstVal)->getRawValue() == 1) {
441 Ops.pop_back(); // ... * 1 -> ...
444 case Instruction::Or:
445 if (CstVal->isAllOnesValue()) { // ... | -1 -> -1
447 Ops.erase(Ops.begin()+1, Ops.end());
452 case Instruction::Add:
453 case Instruction::Xor:
454 if (CstVal->isNullValue()) // ... [|^+] 0 -> ...
459 // Handle destructive annihilation do to identities between elements in the
460 // argument list here.
463 case Instruction::And:
464 case Instruction::Or:
465 case Instruction::Xor:
466 // Scan the operand lists looking for X and ~X pairs, along with X,X pairs.
467 // If we find any, we can simplify the expression. X&~X == 0, X|~X == -1.
468 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
469 // First, check for X and ~X in the operand list.
470 if (BinaryOperator::isNot(Ops[i].Op)) { // Cannot occur for ^.
471 Value *X = BinaryOperator::getNotArgument(Ops[i].Op);
472 unsigned FoundX = FindInOperandList(Ops, i, X);
474 if (Opcode == Instruction::And) { // ...&X&~X = 0
475 Ops[0].Op = Constant::getNullValue(X->getType());
476 Ops.erase(Ops.begin()+1, Ops.end());
479 } else if (Opcode == Instruction::Or) { // ...|X|~X = -1
480 Ops[0].Op = ConstantIntegral::getAllOnesValue(X->getType());
481 Ops.erase(Ops.begin()+1, Ops.end());
488 // Next, check for duplicate pairs of values, which we assume are next to
489 // each other, due to our sorting criteria.
490 if (i+1 != Ops.size() && Ops[i+1].Op == Ops[i].Op) {
491 if (Opcode == Instruction::And || Opcode == Instruction::Or) {
492 // Drop duplicate values.
493 Ops.erase(Ops.begin()+i);
495 IterateOptimization = true;
498 assert(Opcode == Instruction::Xor);
500 Ops.erase(Ops.begin()+i, Ops.begin()+i+2);
502 IterateOptimization = true;
509 case Instruction::Add:
510 // Scan the operand lists looking for X and -X pairs. If we find any, we
511 // can simplify the expression. X+-X == 0
512 for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
513 // Check for X and -X in the operand list.
514 if (BinaryOperator::isNeg(Ops[i].Op)) {
515 Value *X = BinaryOperator::getNegArgument(Ops[i].Op);
516 unsigned FoundX = FindInOperandList(Ops, i, X);
518 // Remove X and -X from the operand list.
519 if (Ops.size() == 2) {
520 Ops[0].Op = Constant::getNullValue(X->getType());
521 Ops.erase(Ops.begin()+1);
525 Ops.erase(Ops.begin()+i);
526 if (i < FoundX) --FoundX;
527 Ops.erase(Ops.begin()+FoundX);
528 IterateOptimization = true;
535 //case Instruction::Mul:
538 if (IterateOptimization)
539 OptimizeExpression(Opcode, Ops);
542 /// PrintOps - Print out the expression identified in the Ops list.
544 static void PrintOps(unsigned Opcode, const std::vector<ValueEntry> &Ops,
546 Module *M = BB->getParent()->getParent();
547 std::cerr << Instruction::getOpcodeName(Opcode) << " "
548 << *Ops[0].Op->getType();
549 for (unsigned i = 0, e = Ops.size(); i != e; ++i)
550 WriteAsOperand(std::cerr << " ", Ops[i].Op, false, true, M)
551 << "," << Ops[i].Rank;
554 /// ReassociateBB - Inspect all of the instructions in this basic block,
555 /// reassociating them as we go.
556 void Reassociate::ReassociateBB(BasicBlock *BB) {
557 for (BasicBlock::iterator BI = BB->begin(); BI != BB->end(); ++BI) {
558 if (BI->getOpcode() == Instruction::Shl &&
559 isa<ConstantInt>(BI->getOperand(1)))
560 if (Instruction *NI = ConvertShiftToMul(BI)) {
565 // Reject cases where it is pointless to do this.
566 if (!isa<BinaryOperator>(BI) || BI->getType()->isFloatingPoint())
567 continue; // Floating point ops are not associative.
569 // If this is a subtract instruction which is not already in negate form,
570 // see if we can convert it to X+-Y.
571 if (BI->getOpcode() == Instruction::Sub) {
572 if (!BinaryOperator::isNeg(BI)) {
573 if (Instruction *NI = BreakUpSubtract(BI)) {
578 // Otherwise, this is a negation. See if the operand is a multiply tree
579 // and if this is not an inner node of a multiply tree.
580 if (isReassociableOp(BI->getOperand(1), Instruction::Mul) &&
582 !isReassociableOp(BI->use_back(), Instruction::Mul))) {
583 BI = LowerNegateToMultiply(BI);
589 // If this instruction is a commutative binary operator, process it.
590 if (!BI->isAssociative()) continue;
591 BinaryOperator *I = cast<BinaryOperator>(BI);
593 // If this is an interior node of a reassociable tree, ignore it until we
594 // get to the root of the tree, to avoid N^2 analysis.
595 if (I->hasOneUse() && isReassociableOp(I->use_back(), I->getOpcode()))
598 // First, walk the expression tree, linearizing the tree, collecting
599 std::vector<ValueEntry> Ops;
600 LinearizeExprTree(I, Ops);
602 DEBUG(std::cerr << "RAIn:\t"; PrintOps(I->getOpcode(), Ops, BB);
605 // Now that we have linearized the tree to a list and have gathered all of
606 // the operands and their ranks, sort the operands by their rank. Use a
607 // stable_sort so that values with equal ranks will have their relative
608 // positions maintained (and so the compiler is deterministic). Note that
609 // this sorts so that the highest ranking values end up at the beginning of
611 std::stable_sort(Ops.begin(), Ops.end());
613 // OptimizeExpression - Now that we have the expression tree in a convenient
614 // sorted form, optimize it globally if possible.
615 OptimizeExpression(I->getOpcode(), Ops);
617 // We want to sink immediates as deeply as possible except in the case where
618 // this is a multiply tree used only by an add, and the immediate is a -1.
619 // In this case we reassociate to put the negation on the outside so that we
620 // can fold the negation into the add: (-X)*Y + Z -> Z-X*Y
621 if (I->getOpcode() == Instruction::Mul && I->hasOneUse() &&
622 cast<Instruction>(I->use_back())->getOpcode() == Instruction::Add &&
623 isa<ConstantInt>(Ops.back().Op) &&
624 cast<ConstantInt>(Ops.back().Op)->isAllOnesValue()) {
625 Ops.insert(Ops.begin(), Ops.back());
629 DEBUG(std::cerr << "RAOut:\t"; PrintOps(I->getOpcode(), Ops, BB);
632 if (Ops.size() == 1) {
633 // This expression tree simplified to something that isn't a tree,
635 I->replaceAllUsesWith(Ops[0].Op);
637 // Now that we ordered and optimized the expressions, splat them back into
638 // the expression tree, removing any unneeded nodes.
639 RewriteExprTree(I, 0, Ops);
645 bool Reassociate::runOnFunction(Function &F) {
646 // Recalculate the rank map for F
650 for (Function::iterator FI = F.begin(), FE = F.end(); FI != FE; ++FI)
653 // We are done with the rank map...
655 ValueRankMap.clear();