Reassociate.cpp revision ae74f555522298bef3be8a173163bf778d59adf9
1//===- Reassociate.cpp - Reassociate binary expressions -------------------===// 2// 3// The LLVM Compiler Infrastructure 4// 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. 7// 8//===----------------------------------------------------------------------===// 9// 10// This pass reassociates commutative expressions in an order that is designed 11// to promote better constant propagation, GCSE, LICM, PRE... 12// 13// For example: 4 + (x + 5) -> x + (4 + 5) 14// 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. 20// 21//===----------------------------------------------------------------------===// 22 23#define DEBUG_TYPE "reassociate" 24#include "llvm/Transforms/Scalar.h" 25#include "llvm/Constants.h" 26#include "llvm/DerivedTypes.h" 27#include "llvm/Function.h" 28#include "llvm/Instructions.h" 29#include "llvm/Pass.h" 30#include "llvm/Assembly/Writer.h" 31#include "llvm/Support/CFG.h" 32#include "llvm/Support/Debug.h" 33#include "llvm/ADT/PostOrderIterator.h" 34#include "llvm/ADT/Statistic.h" 35#include <algorithm> 36#include <iostream> 37using namespace llvm; 38 39namespace { 40 Statistic<> NumLinear ("reassociate","Number of insts linearized"); 41 Statistic<> NumChanged("reassociate","Number of insts reassociated"); 42 Statistic<> NumSwapped("reassociate","Number of insts with operands swapped"); 43 Statistic<> NumAnnihil("reassociate","Number of expr tree annihilated"); 44 Statistic<> NumFactor ("reassociate","Number of multiplies factored"); 45 46 struct ValueEntry { 47 unsigned Rank; 48 Value *Op; 49 ValueEntry(unsigned R, Value *O) : Rank(R), Op(O) {} 50 }; 51 inline bool operator<(const ValueEntry &LHS, const ValueEntry &RHS) { 52 return LHS.Rank > RHS.Rank; // Sort so that highest rank goes to start. 53 } 54} 55 56/// PrintOps - Print out the expression identified in the Ops list. 57/// 58static void PrintOps(Instruction *I, const std::vector<ValueEntry> &Ops) { 59 Module *M = I->getParent()->getParent()->getParent(); 60 std::cerr << Instruction::getOpcodeName(I->getOpcode()) << " " 61 << *Ops[0].Op->getType(); 62 for (unsigned i = 0, e = Ops.size(); i != e; ++i) 63 WriteAsOperand(std::cerr << " ", Ops[i].Op, false, true, M) 64 << "," << Ops[i].Rank; 65} 66 67namespace { 68 class Reassociate : public FunctionPass { 69 std::map<BasicBlock*, unsigned> RankMap; 70 std::map<Value*, unsigned> ValueRankMap; 71 bool MadeChange; 72 public: 73 bool runOnFunction(Function &F); 74 75 virtual void getAnalysisUsage(AnalysisUsage &AU) const { 76 AU.setPreservesCFG(); 77 } 78 private: 79 void BuildRankMap(Function &F); 80 unsigned getRank(Value *V); 81 void ReassociateExpression(BinaryOperator *I); 82 void RewriteExprTree(BinaryOperator *I, std::vector<ValueEntry> &Ops, 83 unsigned Idx = 0); 84 Value *OptimizeExpression(BinaryOperator *I, std::vector<ValueEntry> &Ops); 85 void LinearizeExprTree(BinaryOperator *I, std::vector<ValueEntry> &Ops); 86 void LinearizeExpr(BinaryOperator *I); 87 Value *RemoveFactorFromExpression(Value *V, Value *Factor); 88 void ReassociateBB(BasicBlock *BB); 89 90 void RemoveDeadBinaryOp(Value *V); 91 }; 92 93 RegisterOpt<Reassociate> X("reassociate", "Reassociate expressions"); 94} 95 96// Public interface to the Reassociate pass 97FunctionPass *llvm::createReassociatePass() { return new Reassociate(); } 98 99void Reassociate::RemoveDeadBinaryOp(Value *V) { 100 BinaryOperator *BOp = dyn_cast<BinaryOperator>(V); 101 if (!BOp || !BOp->use_empty()) return; 102 103 Value *LHS = BOp->getOperand(0), *RHS = BOp->getOperand(1); 104 RemoveDeadBinaryOp(LHS); 105 RemoveDeadBinaryOp(RHS); 106} 107 108 109static 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::Div || 117 I->getOpcode() == Instruction::Rem) 118 return true; 119 return false; 120} 121 122void Reassociate::BuildRankMap(Function &F) { 123 unsigned i = 2; 124 125 // Assign distinct ranks to function arguments 126 for (Function::arg_iterator I = F.arg_begin(), E = F.arg_end(); I != E; ++I) 127 ValueRankMap[I] = ++i; 128 129 ReversePostOrderTraversal<Function*> RPOT(&F); 130 for (ReversePostOrderTraversal<Function*>::rpo_iterator I = RPOT.begin(), 131 E = RPOT.end(); I != E; ++I) { 132 BasicBlock *BB = *I; 133 unsigned BBRank = RankMap[BB] = ++i << 16; 134 135 // Walk the basic block, adding precomputed ranks for any instructions that 136 // we cannot move. This ensures that the ranks for these instructions are 137 // all different in the block. 138 for (BasicBlock::iterator I = BB->begin(), E = BB->end(); I != E; ++I) 139 if (isUnmovableInstruction(I)) 140 ValueRankMap[I] = ++BBRank; 141 } 142} 143 144unsigned Reassociate::getRank(Value *V) { 145 if (isa<Argument>(V)) return ValueRankMap[V]; // Function argument... 146 147 Instruction *I = dyn_cast<Instruction>(V); 148 if (I == 0) return 0; // Otherwise it's a global or constant, rank 0. 149 150 unsigned &CachedRank = ValueRankMap[I]; 151 if (CachedRank) return CachedRank; // Rank already known? 152 153 // If this is an expression, return the 1+MAX(rank(LHS), rank(RHS)) so that 154 // we can reassociate expressions for code motion! Since we do not recurse 155 // for PHI nodes, we cannot have infinite recursion here, because there 156 // cannot be loops in the value graph that do not go through PHI nodes. 157 unsigned Rank = 0, MaxRank = RankMap[I->getParent()]; 158 for (unsigned i = 0, e = I->getNumOperands(); 159 i != e && Rank != MaxRank; ++i) 160 Rank = std::max(Rank, getRank(I->getOperand(i))); 161 162 // If this is a not or neg instruction, do not count it for rank. This 163 // assures us that X and ~X will have the same rank. 164 if (!I->getType()->isIntegral() || 165 (!BinaryOperator::isNot(I) && !BinaryOperator::isNeg(I))) 166 ++Rank; 167 168 //DEBUG(std::cerr << "Calculated Rank[" << V->getName() << "] = " 169 //<< Rank << "\n"); 170 171 return CachedRank = Rank; 172} 173 174/// isReassociableOp - Return true if V is an instruction of the specified 175/// opcode and if it only has one use. 176static BinaryOperator *isReassociableOp(Value *V, unsigned Opcode) { 177 if ((V->hasOneUse() || V->use_empty()) && isa<Instruction>(V) && 178 cast<Instruction>(V)->getOpcode() == Opcode) 179 return cast<BinaryOperator>(V); 180 return 0; 181} 182 183/// LowerNegateToMultiply - Replace 0-X with X*-1. 184/// 185static Instruction *LowerNegateToMultiply(Instruction *Neg) { 186 Constant *Cst; 187 if (Neg->getType()->isFloatingPoint()) 188 Cst = ConstantFP::get(Neg->getType(), -1); 189 else 190 Cst = ConstantInt::getAllOnesValue(Neg->getType()); 191 192 std::string NegName = Neg->getName(); Neg->setName(""); 193 Instruction *Res = BinaryOperator::createMul(Neg->getOperand(1), Cst, NegName, 194 Neg); 195 Neg->replaceAllUsesWith(Res); 196 Neg->eraseFromParent(); 197 return Res; 198} 199 200// 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 203// C. 204void 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?"); 210 211 DEBUG(std::cerr << "Linear" << *LHS << *RHS << *I); 212 213 // Move the RHS instruction to live immediately before I, avoiding breaking 214 // dominator properties. 215 RHS->moveBefore(I); 216 217 // Move operands around to do the linearization. 218 I->setOperand(1, RHS->getOperand(0)); 219 RHS->setOperand(0, LHS); 220 I->setOperand(0, RHS); 221 222 ++NumLinear; 223 MadeChange = true; 224 DEBUG(std::cerr << "Linearized: " << *I); 225 226 // If D is part of this expression tree, tail recurse. 227 if (isReassociableOp(I->getOperand(1), I->getOpcode())) 228 LinearizeExpr(I); 229} 230 231 232/// 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. 236/// 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. 240/// 241void Reassociate::LinearizeExprTree(BinaryOperator *I, 242 std::vector<ValueEntry> &Ops) { 243 Value *LHS = I->getOperand(0), *RHS = I->getOperand(1); 244 unsigned Opcode = I->getOpcode(); 245 246 // 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); 249 250 // 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); 256 } 257 if (!RHSBO && RHS->hasOneUse() && BinaryOperator::isNeg(RHS)) { 258 RHS = LowerNegateToMultiply(cast<Instruction>(RHS)); 259 RHSBO = isReassociableOp(RHS, Opcode); 260 } 261 } 262 263 if (!LHSBO) { 264 if (!RHSBO) { 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)); 269 270 // Clear the leaves out. 271 I->setOperand(0, UndefValue::get(I->getType())); 272 I->setOperand(1, UndefValue::get(I->getType())); 273 return; 274 } else { 275 // Turn X+(Y+Z) -> (Y+Z)+X 276 std::swap(LHSBO, RHSBO); 277 std::swap(LHS, RHS); 278 bool Success = !I->swapOperands(); 279 assert(Success && "swapOperands failed"); 280 MadeChange = true; 281 } 282 } else if (RHSBO) { 283 // Turn (A+B)+(C+D) -> (((A+B)+C)+D). This guarantees the the RHS is not 284 // part of the expression tree. 285 LinearizeExpr(I); 286 LHS = LHSBO = cast<BinaryOperator>(I->getOperand(0)); 287 RHS = I->getOperand(1); 288 RHSBO = 0; 289 } 290 291 // 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!"); 294 295 // Move LHS right before I to make sure that the tree expression dominates all 296 // values. 297 LHSBO->moveBefore(I); 298 299 // Linearize the expression tree on the LHS. 300 LinearizeExprTree(LHSBO, Ops); 301 302 // Remember the RHS operand and its rank. 303 Ops.push_back(ValueEntry(getRank(RHS), RHS)); 304 305 // Clear the RHS leaf out. 306 I->setOperand(1, UndefValue::get(I->getType())); 307} 308 309// RewriteExprTree - Now that the operands for this expression tree are 310// linearized and optimized, emit them in-order. This function is written to be 311// tail recursive. 312void Reassociate::RewriteExprTree(BinaryOperator *I, 313 std::vector<ValueEntry> &Ops, 314 unsigned i) { 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 DEBUG(std::cerr << "RA: " << *I); 320 I->setOperand(0, Ops[i].Op); 321 I->setOperand(1, Ops[i+1].Op); 322 DEBUG(std::cerr << "TO: " << *I); 323 MadeChange = true; 324 ++NumChanged; 325 326 // 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); 329 } 330 return; 331 } 332 assert(i+2 < Ops.size() && "Ops index out of range!"); 333 334 if (I->getOperand(1) != Ops[i].Op) { 335 DEBUG(std::cerr << "RA: " << *I); 336 I->setOperand(1, Ops[i].Op); 337 DEBUG(std::cerr << "TO: " << *I); 338 MadeChange = true; 339 ++NumChanged; 340 } 341 342 BinaryOperator *LHS = cast<BinaryOperator>(I->getOperand(0)); 343 assert(LHS->getOpcode() == I->getOpcode() && 344 "Improper expression tree!"); 345 346 // Compactify the tree instructions together with each other to guarantee 347 // that the expression tree is dominated by all of Ops. 348 LHS->moveBefore(I); 349 RewriteExprTree(LHS, Ops, i+1); 350} 351 352 353 354// 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. 358// 359static 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... 368 // 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)); 374 375 // 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. 379 // 380 I->moveBefore(BI); 381 I->setName(I->getName()+".neg"); 382 return I; 383 } 384 385 // Insert a 'neg' instruction that subtracts the value from zero to get the 386 // negation. 387 // 388 return BinaryOperator::createNeg(V, V->getName() + ".neg", BI); 389} 390 391/// 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 393/// reassociation. 394static 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))) 400 return 0; 401 402 // Convert a subtract into an add and a neg instruction... so that sub 403 // instructions can be commuted with other add instructions... 404 // 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... 407 // 408 std::string Name = Sub->getName(); 409 Sub->setName(""); 410 Value *NegVal = NegateValue(Sub->getOperand(1), Sub); 411 Instruction *New = 412 BinaryOperator::createAdd(Sub->getOperand(0), NegVal, Name, Sub); 413 414 // Everyone now refers to the add instruction. 415 Sub->replaceAllUsesWith(New); 416 Sub->eraseFromParent(); 417 418 DEBUG(std::cerr << "Negated: " << *New); 419 return New; 420} 421 422/// 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 424/// reassociation. 425static 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) || 429 (Shl->hasOneUse() && 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))); 434 435 std::string Name = Shl->getName(); Shl->setName(""); 436 Instruction *Mul = BinaryOperator::createMul(Shl->getOperand(0), MulCst, 437 Name, Shl); 438 Shl->replaceAllUsesWith(Mul); 439 Shl->eraseFromParent(); 440 return Mul; 441 } 442 return 0; 443} 444 445// 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. 447static unsigned FindInOperandList(std::vector<ValueEntry> &Ops, unsigned i, 448 Value *X) { 449 unsigned XRank = Ops[i].Rank; 450 unsigned e = Ops.size(); 451 for (unsigned j = i+1; j != e && Ops[j].Rank == XRank; ++j) 452 if (Ops[j].Op == X) 453 return j; 454 // Scan backwards 455 for (unsigned j = i-1; j != ~0U && Ops[j].Rank == XRank; --j) 456 if (Ops[j].Op == X) 457 return j; 458 return i; 459} 460 461/// EmitAddTreeOfValues - Emit a tree of add instructions, summing Ops together 462/// and returning the result. Insert the tree before I. 463static Value *EmitAddTreeOfValues(Instruction *I, std::vector<Value*> &Ops) { 464 if (Ops.size() == 1) return Ops.back(); 465 466 Value *V1 = Ops.back(); 467 Ops.pop_back(); 468 Value *V2 = EmitAddTreeOfValues(I, Ops); 469 return BinaryOperator::createAdd(V2, V1, "tmp", I); 470} 471 472/// 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. 475Value *Reassociate::RemoveFactorFromExpression(Value *V, Value *Factor) { 476 BinaryOperator *BO = isReassociableOp(V, Instruction::Mul); 477 if (!BO) return 0; 478 479 std::vector<ValueEntry> Factors; 480 LinearizeExprTree(BO, Factors); 481 482 bool FoundFactor = false; 483 for (unsigned i = 0, e = Factors.size(); i != e; ++i) 484 if (Factors[i].Op == Factor) { 485 FoundFactor = true; 486 Factors.erase(Factors.begin()+i); 487 break; 488 } 489 if (!FoundFactor) { 490 // Make sure to restore the operands to the expression tree. 491 RewriteExprTree(BO, Factors); 492 return 0; 493 } 494 495 if (Factors.size() == 1) return Factors[0].Op; 496 497 RewriteExprTree(BO, Factors); 498 return BO; 499} 500 501/// FindSingleUseMultiplyFactors - If V is a single-use multiply, recursively 502/// add its operands as factors, otherwise add V to the list of factors. 503static void FindSingleUseMultiplyFactors(Value *V, 504 std::vector<Value*> &Factors) { 505 BinaryOperator *BO; 506 if ((!V->hasOneUse() && !V->use_empty()) || 507 !(BO = dyn_cast<BinaryOperator>(V)) || 508 BO->getOpcode() != Instruction::Mul) { 509 Factors.push_back(V); 510 return; 511 } 512 513 // Otherwise, add the LHS and RHS to the list of factors. 514 FindSingleUseMultiplyFactors(BO->getOperand(1), Factors); 515 FindSingleUseMultiplyFactors(BO->getOperand(0), Factors); 516} 517 518 519 520Value *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; 526 527 unsigned Opcode = I->getOpcode(); 528 529 if (Constant *V1 = dyn_cast<Constant>(Ops[Ops.size()-2].Op)) 530 if (Constant *V2 = dyn_cast<Constant>(Ops.back().Op)) { 531 Ops.pop_back(); 532 Ops.back().Op = ConstantExpr::get(Opcode, V1, V2); 533 return OptimizeExpression(I, Ops); 534 } 535 536 // Check for destructive annihilation due to a constant being used. 537 if (ConstantIntegral *CstVal = dyn_cast<ConstantIntegral>(Ops.back().Op)) 538 switch (Opcode) { 539 default: break; 540 case Instruction::And: 541 if (CstVal->isNullValue()) { // ... & 0 -> 0 542 ++NumAnnihil; 543 return CstVal; 544 } else if (CstVal->isAllOnesValue()) { // ... & -1 -> ... 545 Ops.pop_back(); 546 } 547 break; 548 case Instruction::Mul: 549 if (CstVal->isNullValue()) { // ... * 0 -> 0 550 ++NumAnnihil; 551 return CstVal; 552 } else if (cast<ConstantInt>(CstVal)->getRawValue() == 1) { 553 Ops.pop_back(); // ... * 1 -> ... 554 } 555 break; 556 case Instruction::Or: 557 if (CstVal->isAllOnesValue()) { // ... | -1 -> -1 558 ++NumAnnihil; 559 return CstVal; 560 } 561 // FALLTHROUGH! 562 case Instruction::Add: 563 case Instruction::Xor: 564 if (CstVal->isNullValue()) // ... [|^+] 0 -> ... 565 Ops.pop_back(); 566 break; 567 } 568 if (Ops.size() == 1) return Ops[0].Op; 569 570 // Handle destructive annihilation do to identities between elements in the 571 // argument list here. 572 switch (Opcode) { 573 default: break; 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); 585 if (FoundX != i) { 586 if (Opcode == Instruction::And) { // ...&X&~X = 0 587 ++NumAnnihil; 588 return Constant::getNullValue(X->getType()); 589 } else if (Opcode == Instruction::Or) { // ...|X|~X = -1 590 ++NumAnnihil; 591 return ConstantIntegral::getAllOnesValue(X->getType()); 592 } 593 } 594 } 595 596 // 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); 603 --i; --e; 604 IterateOptimization = true; 605 ++NumAnnihil; 606 } else { 607 assert(Opcode == Instruction::Xor); 608 if (e == 2) { 609 ++NumAnnihil; 610 return Constant::getNullValue(Ops[0].Op->getType()); 611 } 612 // ... X^X -> ... 613 Ops.erase(Ops.begin()+i, Ops.begin()+i+2); 614 i -= 1; e -= 2; 615 IterateOptimization = true; 616 ++NumAnnihil; 617 } 618 } 619 } 620 break; 621 622 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); 631 if (FoundX != i) { 632 // Remove X and -X from the operand list. 633 if (Ops.size() == 2) { 634 ++NumAnnihil; 635 return Constant::getNullValue(X->getType()); 636 } else { 637 Ops.erase(Ops.begin()+i); 638 if (i < FoundX) 639 --FoundX; 640 else 641 --i; // Need to back up an extra one. 642 Ops.erase(Ops.begin()+FoundX); 643 IterateOptimization = true; 644 ++NumAnnihil; 645 --i; // Revisit element. 646 e -= 2; // Removed two elements. 647 } 648 } 649 } 650 } 651 652 653 // 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; 659 unsigned MaxOcc = 0; 660 Value *MaxOccVal = 0; 661 if (!I->getType()->isFloatingPoint()) { 662 for (unsigned i = 0, e = Ops.size(); i != e; ++i) { 663 if (BinaryOperator *BOp = dyn_cast<BinaryOperator>(Ops[i].Op)) 664 if (BOp->getOpcode() == Instruction::Mul && BOp->use_empty()) { 665 // Compute all of the factors of this added value. 666 std::vector<Value*> Factors; 667 FindSingleUseMultiplyFactors(BOp, Factors); 668 assert(Factors.size() > 1 && "Bad linearize!"); 669 670 // Add one to FactorOccurrences for each unique factor in this op. 671 if (Factors.size() == 2) { 672 unsigned Occ = ++FactorOccurrences[Factors[0]]; 673 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[0]; } 674 if (Factors[0] != Factors[1]) { // Don't double count A*A. 675 Occ = ++FactorOccurrences[Factors[1]]; 676 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[1]; } 677 } 678 } else { 679 std::set<Value*> Duplicates; 680 for (unsigned i = 0, e = Factors.size(); i != e; ++i) 681 if (Duplicates.insert(Factors[i]).second) { 682 unsigned Occ = ++FactorOccurrences[Factors[i]]; 683 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[i]; } 684 } 685 } 686 } 687 } 688 } 689 690 // If any factor occurred more than one time, we can pull it out. 691 if (MaxOcc > 1) { 692 DEBUG(std::cerr << "\nFACTORING [" << MaxOcc << "]: " 693 << *MaxOccVal << "\n"); 694 695 // Create a new instruction that uses the MaxOccVal twice. If we don't do 696 // this, we could otherwise run into situations where removing a factor 697 // from an expression will drop a use of maxocc, and this can cause 698 // RemoveFactorFromExpression on successive values to behave differently. 699 Instruction *DummyInst = BinaryOperator::createAdd(MaxOccVal, MaxOccVal); 700 std::vector<Value*> NewMulOps; 701 for (unsigned i = 0, e = Ops.size(); i != e; ++i) { 702 if (Value *V = RemoveFactorFromExpression(Ops[i].Op, MaxOccVal)) { 703 NewMulOps.push_back(V); 704 Ops.erase(Ops.begin()+i); 705 --i; --e; 706 } 707 } 708 709 // No need for extra uses anymore. 710 delete DummyInst; 711 712 unsigned NumAddedValues = NewMulOps.size(); 713 Value *V = EmitAddTreeOfValues(I, NewMulOps); 714 Value *V2 = BinaryOperator::createMul(V, MaxOccVal, "tmp", I); 715 716 // Now that we have inserted V and its sole use, optimize it. This allows 717 // us to handle cases that require multiple factoring steps, such as this: 718 // A*A*B + A*A*C --> A*(A*B+A*C) --> A*(A*(B+C)) 719 if (NumAddedValues > 1) 720 ReassociateExpression(cast<BinaryOperator>(V)); 721 722 ++NumFactor; 723 724 if (Ops.size() == 0) 725 return V2; 726 727 // Add the new value to the list of things being added. 728 Ops.insert(Ops.begin(), ValueEntry(getRank(V2), V2)); 729 730 // Rewrite the tree so that there is now a use of V. 731 RewriteExprTree(I, Ops); 732 return OptimizeExpression(I, Ops); 733 } 734 break; 735 //case Instruction::Mul: 736 } 737 738 if (IterateOptimization) 739 return OptimizeExpression(I, Ops); 740 return 0; 741} 742 743 744/// ReassociateBB - Inspect all of the instructions in this basic block, 745/// reassociating them as we go. 746void Reassociate::ReassociateBB(BasicBlock *BB) { 747 for (BasicBlock::iterator BBI = BB->begin(); BBI != BB->end(); ) { 748 Instruction *BI = BBI++; 749 if (BI->getOpcode() == Instruction::Shl && 750 isa<ConstantInt>(BI->getOperand(1))) 751 if (Instruction *NI = ConvertShiftToMul(BI)) { 752 MadeChange = true; 753 BI = NI; 754 } 755 756 // Reject cases where it is pointless to do this. 757 if (!isa<BinaryOperator>(BI) || BI->getType()->isFloatingPoint() || 758 isa<PackedType>(BI->getType())) 759 continue; // Floating point ops are not associative. 760 761 // If this is a subtract instruction which is not already in negate form, 762 // see if we can convert it to X+-Y. 763 if (BI->getOpcode() == Instruction::Sub) { 764 if (!BinaryOperator::isNeg(BI)) { 765 if (Instruction *NI = BreakUpSubtract(BI)) { 766 MadeChange = true; 767 BI = NI; 768 } 769 } else { 770 // Otherwise, this is a negation. See if the operand is a multiply tree 771 // and if this is not an inner node of a multiply tree. 772 if (isReassociableOp(BI->getOperand(1), Instruction::Mul) && 773 (!BI->hasOneUse() || 774 !isReassociableOp(BI->use_back(), Instruction::Mul))) { 775 BI = LowerNegateToMultiply(BI); 776 MadeChange = true; 777 } 778 } 779 } 780 781 // If this instruction is a commutative binary operator, process it. 782 if (!BI->isAssociative()) continue; 783 BinaryOperator *I = cast<BinaryOperator>(BI); 784 785 // If this is an interior node of a reassociable tree, ignore it until we 786 // get to the root of the tree, to avoid N^2 analysis. 787 if (I->hasOneUse() && isReassociableOp(I->use_back(), I->getOpcode())) 788 continue; 789 790 // If this is an add tree that is used by a sub instruction, ignore it 791 // until we process the subtract. 792 if (I->hasOneUse() && I->getOpcode() == Instruction::Add && 793 cast<Instruction>(I->use_back())->getOpcode() == Instruction::Sub) 794 continue; 795 796 ReassociateExpression(I); 797 } 798} 799 800void Reassociate::ReassociateExpression(BinaryOperator *I) { 801 802 // First, walk the expression tree, linearizing the tree, collecting 803 std::vector<ValueEntry> Ops; 804 LinearizeExprTree(I, Ops); 805 806 DEBUG(std::cerr << "RAIn:\t"; PrintOps(I, Ops); 807 std::cerr << "\n"); 808 809 // Now that we have linearized the tree to a list and have gathered all of 810 // the operands and their ranks, sort the operands by their rank. Use a 811 // stable_sort so that values with equal ranks will have their relative 812 // positions maintained (and so the compiler is deterministic). Note that 813 // this sorts so that the highest ranking values end up at the beginning of 814 // the vector. 815 std::stable_sort(Ops.begin(), Ops.end()); 816 817 // OptimizeExpression - Now that we have the expression tree in a convenient 818 // sorted form, optimize it globally if possible. 819 if (Value *V = OptimizeExpression(I, Ops)) { 820 // This expression tree simplified to something that isn't a tree, 821 // eliminate it. 822 DEBUG(std::cerr << "Reassoc to scalar: " << *V << "\n"); 823 I->replaceAllUsesWith(V); 824 RemoveDeadBinaryOp(I); 825 return; 826 } 827 828 // We want to sink immediates as deeply as possible except in the case where 829 // this is a multiply tree used only by an add, and the immediate is a -1. 830 // In this case we reassociate to put the negation on the outside so that we 831 // can fold the negation into the add: (-X)*Y + Z -> Z-X*Y 832 if (I->getOpcode() == Instruction::Mul && I->hasOneUse() && 833 cast<Instruction>(I->use_back())->getOpcode() == Instruction::Add && 834 isa<ConstantInt>(Ops.back().Op) && 835 cast<ConstantInt>(Ops.back().Op)->isAllOnesValue()) { 836 Ops.insert(Ops.begin(), Ops.back()); 837 Ops.pop_back(); 838 } 839 840 DEBUG(std::cerr << "RAOut:\t"; PrintOps(I, Ops); 841 std::cerr << "\n"); 842 843 if (Ops.size() == 1) { 844 // This expression tree simplified to something that isn't a tree, 845 // eliminate it. 846 I->replaceAllUsesWith(Ops[0].Op); 847 RemoveDeadBinaryOp(I); 848 } else { 849 // Now that we ordered and optimized the expressions, splat them back into 850 // the expression tree, removing any unneeded nodes. 851 RewriteExprTree(I, Ops); 852 } 853} 854 855 856bool Reassociate::runOnFunction(Function &F) { 857 // Recalculate the rank map for F 858 BuildRankMap(F); 859 860 MadeChange = false; 861 for (Function::iterator FI = F.begin(), FE = F.end(); FI != FE; ++FI) 862 ReassociateBB(FI); 863 864 // We are done with the rank map... 865 RankMap.clear(); 866 ValueRankMap.clear(); 867 return MadeChange; 868} 869 870