Reassociate.cpp revision ac83b0301ea5ce0e1092fad8f294fe7f046832ff
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/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" 35#include <algorithm> 36using namespace llvm; 37 38namespace { 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"); 43 44 struct ValueEntry { 45 unsigned Rank; 46 Value *Op; 47 ValueEntry(unsigned R, Value *O) : Rank(R), Op(O) {} 48 }; 49 inline bool operator<(const ValueEntry &LHS, const ValueEntry &RHS) { 50 return LHS.Rank > RHS.Rank; // Sort so that highest rank goes to start. 51 } 52 53 class Reassociate : public FunctionPass { 54 std::map<BasicBlock*, unsigned> RankMap; 55 std::map<Value*, unsigned> ValueRankMap; 56 bool MadeChange; 57 public: 58 bool runOnFunction(Function &F); 59 60 virtual void getAnalysisUsage(AnalysisUsage &AU) const { 61 AU.setPreservesCFG(); 62 } 63 private: 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); 72 }; 73 74 RegisterOpt<Reassociate> X("reassociate", "Reassociate expressions"); 75} 76 77// Public interface to the Reassociate pass 78FunctionPass *llvm::createReassociatePass() { return new Reassociate(); } 79 80 81static 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) 90 return true; 91 return false; 92} 93 94void Reassociate::BuildRankMap(Function &F) { 95 unsigned i = 2; 96 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; 100 101 ReversePostOrderTraversal<Function*> RPOT(&F); 102 for (ReversePostOrderTraversal<Function*>::rpo_iterator I = RPOT.begin(), 103 E = RPOT.end(); I != E; ++I) { 104 BasicBlock *BB = *I; 105 unsigned BBRank = RankMap[BB] = ++i << 16; 106 107 // 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; 113 } 114} 115 116unsigned Reassociate::getRank(Value *V) { 117 if (isa<Argument>(V)) return ValueRankMap[V]; // Function argument... 118 119 Instruction *I = dyn_cast<Instruction>(V); 120 if (I == 0) return 0; // Otherwise it's a global or constant, rank 0. 121 122 unsigned &CachedRank = ValueRankMap[I]; 123 if (CachedRank) return CachedRank; // Rank already known? 124 125 // 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))); 133 134 // 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))) 138 ++Rank; 139 140 //DEBUG(std::cerr << "Calculated Rank[" << V->getName() << "] = " 141 //<< Rank << "\n"); 142 143 return CachedRank = Rank; 144} 145 146/// isReassociableOp - Return true if V is an instruction of the specified 147/// opcode and if it only has one use. 148static BinaryOperator *isReassociableOp(Value *V, unsigned Opcode) { 149 if (V->hasOneUse() && isa<Instruction>(V) && 150 cast<Instruction>(V)->getOpcode() == Opcode) 151 return cast<BinaryOperator>(V); 152 return 0; 153} 154 155/// LowerNegateToMultiply - Replace 0-X with X*-1. 156/// 157static Instruction *LowerNegateToMultiply(Instruction *Neg) { 158 Constant *Cst; 159 if (Neg->getType()->isFloatingPoint()) 160 Cst = ConstantFP::get(Neg->getType(), -1); 161 else 162 Cst = ConstantInt::getAllOnesValue(Neg->getType()); 163 164 std::string NegName = Neg->getName(); Neg->setName(""); 165 Instruction *Res = BinaryOperator::createMul(Neg->getOperand(1), Cst, NegName, 166 Neg); 167 Neg->replaceAllUsesWith(Res); 168 Neg->eraseFromParent(); 169 return Res; 170} 171 172// 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 175// C. 176void 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?"); 182 183 DEBUG(std::cerr << "Linear" << *LHS << *RHS << *I); 184 185 // Move the RHS instruction to live immediately before I, avoiding breaking 186 // dominator properties. 187 RHS->moveBefore(I); 188 189 // Move operands around to do the linearization. 190 I->setOperand(1, RHS->getOperand(0)); 191 RHS->setOperand(0, LHS); 192 I->setOperand(0, RHS); 193 194 ++NumLinear; 195 MadeChange = true; 196 DEBUG(std::cerr << "Linearized: " << *I); 197 198 // If D is part of this expression tree, tail recurse. 199 if (isReassociableOp(I->getOperand(1), I->getOpcode())) 200 LinearizeExpr(I); 201} 202 203 204/// 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. 208/// 209/// This returns the rank of the RHS operand, which is known to be the highest 210/// rank value in the expression tree. 211/// 212void Reassociate::LinearizeExprTree(BinaryOperator *I, 213 std::vector<ValueEntry> &Ops) { 214 Value *LHS = I->getOperand(0), *RHS = I->getOperand(1); 215 unsigned Opcode = I->getOpcode(); 216 217 // 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); 220 221 // 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); 227 } 228 if (!RHSBO && RHS->hasOneUse() && BinaryOperator::isNeg(RHS)) { 229 RHS = LowerNegateToMultiply(cast<Instruction>(RHS)); 230 RHSBO = isReassociableOp(RHS, Opcode); 231 } 232 } 233 234 if (!LHSBO) { 235 if (!RHSBO) { 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)); 240 return; 241 } else { 242 // Turn X+(Y+Z) -> (Y+Z)+X 243 std::swap(LHSBO, RHSBO); 244 std::swap(LHS, RHS); 245 bool Success = !I->swapOperands(); 246 assert(Success && "swapOperands failed"); 247 MadeChange = true; 248 } 249 } else if (RHSBO) { 250 // Turn (A+B)+(C+D) -> (((A+B)+C)+D). This guarantees the the RHS is not 251 // part of the expression tree. 252 LinearizeExpr(I); 253 LHS = LHSBO = cast<BinaryOperator>(I->getOperand(0)); 254 RHS = I->getOperand(1); 255 RHSBO = 0; 256 } 257 258 // 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!"); 261 262 // Move LHS right before I to make sure that the tree expression dominates all 263 // values. 264 LHSBO->moveBefore(I); 265 266 // Linearize the expression tree on the LHS. 267 LinearizeExprTree(LHSBO, Ops); 268 269 // Remember the RHS operand and its rank. 270 Ops.push_back(ValueEntry(getRank(RHS), RHS)); 271} 272 273// RewriteExprTree - Now that the operands for this expression tree are 274// linearized and optimized, emit them in-order. This function is written to be 275// tail recursive. 276void 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); 285 MadeChange = true; 286 ++NumChanged; 287 } 288 return; 289 } 290 assert(i+2 < Ops.size() && "Ops index out of range!"); 291 292 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); 296 MadeChange = true; 297 ++NumChanged; 298 } 299 RewriteExprTree(cast<BinaryOperator>(I->getOperand(0)), i+1, Ops); 300} 301 302 303 304// 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. 308// 309static 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... 318 // 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); 323 324 // 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. 328 // 329 return BinaryOperator::create(Instruction::Add, LHS, RHS, 330 I->getName()+".neg", BI); 331 } 332 333 // Insert a 'neg' instruction that subtracts the value from zero to get the 334 // negation. 335 // 336 return BinaryOperator::createNeg(V, V->getName() + ".neg", BI); 337} 338 339/// 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 341/// reassociation. 342static 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))) 348 return 0; 349 350 // Convert a subtract into an add and a neg instruction... so that sub 351 // instructions can be commuted with other add instructions... 352 // 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... 355 // 356 std::string Name = Sub->getName(); 357 Sub->setName(""); 358 Value *NegVal = NegateValue(Sub->getOperand(1), Sub); 359 Instruction *New = 360 BinaryOperator::createAdd(Sub->getOperand(0), NegVal, Name, Sub); 361 362 // Everyone now refers to the add instruction. 363 Sub->replaceAllUsesWith(New); 364 Sub->eraseFromParent(); 365 366 DEBUG(std::cerr << "Negated: " << *New); 367 return New; 368} 369 370/// 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 372/// reassociation. 373static Instruction *ConvertShiftToMul(Instruction *Shl) { 374 if (!isReassociableOp(Shl->getOperand(0), Instruction::Mul) && 375 !(Shl->hasOneUse() && isReassociableOp(Shl->use_back(),Instruction::Mul))) 376 return 0; 377 378 Constant *MulCst = ConstantInt::get(Shl->getType(), 1); 379 MulCst = ConstantExpr::getShl(MulCst, cast<Constant>(Shl->getOperand(1))); 380 381 std::string Name = Shl->getName(); Shl->setName(""); 382 Instruction *Mul = BinaryOperator::createMul(Shl->getOperand(0), MulCst, 383 Name, Shl); 384 Shl->replaceAllUsesWith(Mul); 385 Shl->eraseFromParent(); 386 return Mul; 387} 388 389// 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. 391static unsigned FindInOperandList(std::vector<ValueEntry> &Ops, unsigned i, 392 Value *X) { 393 unsigned XRank = Ops[i].Rank; 394 unsigned e = Ops.size(); 395 for (unsigned j = i+1; j != e && Ops[j].Rank == XRank; ++j) 396 if (Ops[j].Op == X) 397 return j; 398 // Scan backwards 399 for (unsigned j = i-1; j != ~0U && Ops[j].Rank == XRank; --j) 400 if (Ops[j].Op == X) 401 return j; 402 return i; 403} 404 405void 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; 411 412 if (Constant *V1 = dyn_cast<Constant>(Ops[Ops.size()-2].Op)) 413 if (Constant *V2 = dyn_cast<Constant>(Ops.back().Op)) { 414 Ops.pop_back(); 415 Ops.back().Op = ConstantExpr::get(Opcode, V1, V2); 416 OptimizeExpression(Opcode, Ops); 417 return; 418 } 419 420 // Check for destructive annihilation due to a constant being used. 421 if (ConstantIntegral *CstVal = dyn_cast<ConstantIntegral>(Ops.back().Op)) 422 switch (Opcode) { 423 default: break; 424 case Instruction::And: 425 if (CstVal->isNullValue()) { // ... & 0 -> 0 426 Ops[0].Op = CstVal; 427 Ops.erase(Ops.begin()+1, Ops.end()); 428 ++NumAnnihil; 429 return; 430 } else if (CstVal->isAllOnesValue()) { // ... & -1 -> ... 431 Ops.pop_back(); 432 } 433 break; 434 case Instruction::Mul: 435 if (CstVal->isNullValue()) { // ... * 0 -> 0 436 Ops[0].Op = CstVal; 437 Ops.erase(Ops.begin()+1, Ops.end()); 438 ++NumAnnihil; 439 return; 440 } else if (cast<ConstantInt>(CstVal)->getRawValue() == 1) { 441 Ops.pop_back(); // ... * 1 -> ... 442 } 443 break; 444 case Instruction::Or: 445 if (CstVal->isAllOnesValue()) { // ... | -1 -> -1 446 Ops[0].Op = CstVal; 447 Ops.erase(Ops.begin()+1, Ops.end()); 448 ++NumAnnihil; 449 return; 450 } 451 // FALLTHROUGH! 452 case Instruction::Add: 453 case Instruction::Xor: 454 if (CstVal->isNullValue()) // ... [|^+] 0 -> ... 455 Ops.pop_back(); 456 break; 457 } 458 459 // Handle destructive annihilation do to identities between elements in the 460 // argument list here. 461 switch (Opcode) { 462 default: break; 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); 473 if (FoundX != i) { 474 if (Opcode == Instruction::And) { // ...&X&~X = 0 475 Ops[0].Op = Constant::getNullValue(X->getType()); 476 Ops.erase(Ops.begin()+1, Ops.end()); 477 ++NumAnnihil; 478 return; 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()); 482 ++NumAnnihil; 483 return; 484 } 485 } 486 } 487 488 // 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); 494 --i; --e; 495 IterateOptimization = true; 496 ++NumAnnihil; 497 } else { 498 assert(Opcode == Instruction::Xor); 499 if (e == 2) { 500 Ops[0].Op = Constant::getNullValue(Ops[0].Op->getType()); 501 Ops.erase(Ops.begin()+1, Ops.end()); 502 ++NumAnnihil; 503 return; 504 } 505 // ... X^X -> ... 506 Ops.erase(Ops.begin()+i, Ops.begin()+i+2); 507 i -= 1; e -= 2; 508 IterateOptimization = true; 509 ++NumAnnihil; 510 } 511 } 512 } 513 break; 514 515 case Instruction::Add: 516 // Scan the operand lists looking for X and -X pairs. If we find any, we 517 // can simplify the expression. X+-X == 0 518 for (unsigned i = 0, e = Ops.size(); i != e; ++i) { 519 // Check for X and -X in the operand list. 520 if (BinaryOperator::isNeg(Ops[i].Op)) { 521 Value *X = BinaryOperator::getNegArgument(Ops[i].Op); 522 unsigned FoundX = FindInOperandList(Ops, i, X); 523 if (FoundX != i) { 524 // Remove X and -X from the operand list. 525 if (Ops.size() == 2) { 526 Ops[0].Op = Constant::getNullValue(X->getType()); 527 Ops.erase(Ops.begin()+1); 528 ++NumAnnihil; 529 return; 530 } else { 531 Ops.erase(Ops.begin()+i); 532 if (i < FoundX) --FoundX; 533 Ops.erase(Ops.begin()+FoundX); 534 IterateOptimization = true; 535 ++NumAnnihil; 536 } 537 } 538 } 539 } 540 break; 541 //case Instruction::Mul: 542 } 543 544 if (IterateOptimization) 545 OptimizeExpression(Opcode, Ops); 546} 547 548/// PrintOps - Print out the expression identified in the Ops list. 549/// 550static void PrintOps(unsigned Opcode, const std::vector<ValueEntry> &Ops, 551 BasicBlock *BB) { 552 Module *M = BB->getParent()->getParent(); 553 std::cerr << Instruction::getOpcodeName(Opcode) << " " 554 << *Ops[0].Op->getType(); 555 for (unsigned i = 0, e = Ops.size(); i != e; ++i) 556 WriteAsOperand(std::cerr << " ", Ops[i].Op, false, true, M) 557 << "," << Ops[i].Rank; 558} 559 560/// ReassociateBB - Inspect all of the instructions in this basic block, 561/// reassociating them as we go. 562void Reassociate::ReassociateBB(BasicBlock *BB) { 563 for (BasicBlock::iterator BI = BB->begin(); BI != BB->end(); ++BI) { 564 if (BI->getOpcode() == Instruction::Shl && 565 isa<ConstantInt>(BI->getOperand(1))) 566 if (Instruction *NI = ConvertShiftToMul(BI)) { 567 MadeChange = true; 568 BI = NI; 569 } 570 571 // Reject cases where it is pointless to do this. 572 if (!isa<BinaryOperator>(BI) || BI->getType()->isFloatingPoint()) 573 continue; // Floating point ops are not associative. 574 575 // If this is a subtract instruction which is not already in negate form, 576 // see if we can convert it to X+-Y. 577 if (BI->getOpcode() == Instruction::Sub) { 578 if (!BinaryOperator::isNeg(BI)) { 579 if (Instruction *NI = BreakUpSubtract(BI)) { 580 MadeChange = true; 581 BI = NI; 582 } 583 } else { 584 // Otherwise, this is a negation. See if the operand is a multiply tree 585 // and if this is not an inner node of a multiply tree. 586 if (isReassociableOp(BI->getOperand(1), Instruction::Mul) && 587 (!BI->hasOneUse() || 588 !isReassociableOp(BI->use_back(), Instruction::Mul))) { 589 BI = LowerNegateToMultiply(BI); 590 MadeChange = true; 591 } 592 } 593 } 594 595 // If this instruction is a commutative binary operator, process it. 596 if (!BI->isAssociative()) continue; 597 BinaryOperator *I = cast<BinaryOperator>(BI); 598 599 // If this is an interior node of a reassociable tree, ignore it until we 600 // get to the root of the tree, to avoid N^2 analysis. 601 if (I->hasOneUse() && isReassociableOp(I->use_back(), I->getOpcode())) 602 continue; 603 604 // First, walk the expression tree, linearizing the tree, collecting 605 std::vector<ValueEntry> Ops; 606 LinearizeExprTree(I, Ops); 607 608 DEBUG(std::cerr << "RAIn:\t"; PrintOps(I->getOpcode(), Ops, BB); 609 std::cerr << "\n"); 610 611 // Now that we have linearized the tree to a list and have gathered all of 612 // the operands and their ranks, sort the operands by their rank. Use a 613 // stable_sort so that values with equal ranks will have their relative 614 // positions maintained (and so the compiler is deterministic). Note that 615 // this sorts so that the highest ranking values end up at the beginning of 616 // the vector. 617 std::stable_sort(Ops.begin(), Ops.end()); 618 619 // OptimizeExpression - Now that we have the expression tree in a convenient 620 // sorted form, optimize it globally if possible. 621 OptimizeExpression(I->getOpcode(), Ops); 622 623 // We want to sink immediates as deeply as possible except in the case where 624 // this is a multiply tree used only by an add, and the immediate is a -1. 625 // In this case we reassociate to put the negation on the outside so that we 626 // can fold the negation into the add: (-X)*Y + Z -> Z-X*Y 627 if (I->getOpcode() == Instruction::Mul && I->hasOneUse() && 628 cast<Instruction>(I->use_back())->getOpcode() == Instruction::Add && 629 isa<ConstantInt>(Ops.back().Op) && 630 cast<ConstantInt>(Ops.back().Op)->isAllOnesValue()) { 631 Ops.insert(Ops.begin(), Ops.back()); 632 Ops.pop_back(); 633 } 634 635 DEBUG(std::cerr << "RAOut:\t"; PrintOps(I->getOpcode(), Ops, BB); 636 std::cerr << "\n"); 637 638 if (Ops.size() == 1) { 639 // This expression tree simplified to something that isn't a tree, 640 // eliminate it. 641 I->replaceAllUsesWith(Ops[0].Op); 642 } else { 643 // Now that we ordered and optimized the expressions, splat them back into 644 // the expression tree, removing any unneeded nodes. 645 RewriteExprTree(I, 0, Ops); 646 } 647 } 648} 649 650 651bool Reassociate::runOnFunction(Function &F) { 652 // Recalculate the rank map for F 653 BuildRankMap(F); 654 655 MadeChange = false; 656 for (Function::iterator FI = F.begin(), FE = F.end(); FI != FE; ++FI) 657 ReassociateBB(FI); 658 659 // We are done with the rank map... 660 RankMap.clear(); 661 ValueRankMap.clear(); 662 return MadeChange; 663} 664 665