Reassociate.cpp revision 42a75517250017a52afb03a0ade03cbd49559fe5
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> 36using namespace llvm; 37 38STATISTIC(NumLinear , "Number of insts linearized"); 39STATISTIC(NumChanged, "Number of insts reassociated"); 40STATISTIC(NumAnnihil, "Number of expr tree annihilated"); 41STATISTIC(NumFactor , "Number of multiplies factored"); 42 43namespace { 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 54/// PrintOps - Print out the expression identified in the Ops list. 55/// 56static void PrintOps(Instruction *I, const std::vector<ValueEntry> &Ops) { 57 Module *M = I->getParent()->getParent()->getParent(); 58 cerr << Instruction::getOpcodeName(I->getOpcode()) << " " 59 << *Ops[0].Op->getType(); 60 for (unsigned i = 0, e = Ops.size(); i != e; ++i) 61 WriteAsOperand(*cerr.stream() << " ", Ops[i].Op, false, M) 62 << "," << Ops[i].Rank; 63} 64 65namespace { 66 class Reassociate : public FunctionPass { 67 std::map<BasicBlock*, unsigned> RankMap; 68 std::map<Value*, unsigned> ValueRankMap; 69 bool MadeChange; 70 public: 71 bool runOnFunction(Function &F); 72 73 virtual void getAnalysisUsage(AnalysisUsage &AU) const { 74 AU.setPreservesCFG(); 75 } 76 private: 77 void BuildRankMap(Function &F); 78 unsigned getRank(Value *V); 79 void ReassociateExpression(BinaryOperator *I); 80 void RewriteExprTree(BinaryOperator *I, std::vector<ValueEntry> &Ops, 81 unsigned Idx = 0); 82 Value *OptimizeExpression(BinaryOperator *I, std::vector<ValueEntry> &Ops); 83 void LinearizeExprTree(BinaryOperator *I, std::vector<ValueEntry> &Ops); 84 void LinearizeExpr(BinaryOperator *I); 85 Value *RemoveFactorFromExpression(Value *V, Value *Factor); 86 void ReassociateBB(BasicBlock *BB); 87 88 void RemoveDeadBinaryOp(Value *V); 89 }; 90 91 RegisterPass<Reassociate> X("reassociate", "Reassociate expressions"); 92} 93 94// Public interface to the Reassociate pass 95FunctionPass *llvm::createReassociatePass() { return new Reassociate(); } 96 97void Reassociate::RemoveDeadBinaryOp(Value *V) { 98 Instruction *Op = dyn_cast<Instruction>(V); 99 if (!Op || !isa<BinaryOperator>(Op) || !isa<CmpInst>(Op) || !Op->use_empty()) 100 return; 101 102 Value *LHS = Op->getOperand(0), *RHS = Op->getOperand(1); 103 RemoveDeadBinaryOp(LHS); 104 RemoveDeadBinaryOp(RHS); 105} 106 107 108static bool isUnmovableInstruction(Instruction *I) { 109 if (I->getOpcode() == Instruction::PHI || 110 I->getOpcode() == Instruction::Alloca || 111 I->getOpcode() == Instruction::Load || 112 I->getOpcode() == Instruction::Malloc || 113 I->getOpcode() == Instruction::Invoke || 114 I->getOpcode() == Instruction::Call || 115 I->getOpcode() == Instruction::UDiv || 116 I->getOpcode() == Instruction::SDiv || 117 I->getOpcode() == Instruction::FDiv || 118 I->getOpcode() == Instruction::URem || 119 I->getOpcode() == Instruction::SRem || 120 I->getOpcode() == Instruction::FRem) 121 return true; 122 return false; 123} 124 125void Reassociate::BuildRankMap(Function &F) { 126 unsigned i = 2; 127 128 // Assign distinct ranks to function arguments 129 for (Function::arg_iterator I = F.arg_begin(), E = F.arg_end(); I != E; ++I) 130 ValueRankMap[I] = ++i; 131 132 ReversePostOrderTraversal<Function*> RPOT(&F); 133 for (ReversePostOrderTraversal<Function*>::rpo_iterator I = RPOT.begin(), 134 E = RPOT.end(); I != E; ++I) { 135 BasicBlock *BB = *I; 136 unsigned BBRank = RankMap[BB] = ++i << 16; 137 138 // Walk the basic block, adding precomputed ranks for any instructions that 139 // we cannot move. This ensures that the ranks for these instructions are 140 // all different in the block. 141 for (BasicBlock::iterator I = BB->begin(), E = BB->end(); I != E; ++I) 142 if (isUnmovableInstruction(I)) 143 ValueRankMap[I] = ++BBRank; 144 } 145} 146 147unsigned Reassociate::getRank(Value *V) { 148 if (isa<Argument>(V)) return ValueRankMap[V]; // Function argument... 149 150 Instruction *I = dyn_cast<Instruction>(V); 151 if (I == 0) return 0; // Otherwise it's a global or constant, rank 0. 152 153 unsigned &CachedRank = ValueRankMap[I]; 154 if (CachedRank) return CachedRank; // Rank already known? 155 156 // If this is an expression, return the 1+MAX(rank(LHS), rank(RHS)) so that 157 // we can reassociate expressions for code motion! Since we do not recurse 158 // for PHI nodes, we cannot have infinite recursion here, because there 159 // cannot be loops in the value graph that do not go through PHI nodes. 160 unsigned Rank = 0, MaxRank = RankMap[I->getParent()]; 161 for (unsigned i = 0, e = I->getNumOperands(); 162 i != e && Rank != MaxRank; ++i) 163 Rank = std::max(Rank, getRank(I->getOperand(i))); 164 165 // If this is a not or neg instruction, do not count it for rank. This 166 // assures us that X and ~X will have the same rank. 167 if (!I->getType()->isInteger() || 168 (!BinaryOperator::isNot(I) && !BinaryOperator::isNeg(I))) 169 ++Rank; 170 171 //DOUT << "Calculated Rank[" << V->getName() << "] = " 172 // << Rank << "\n"; 173 174 return CachedRank = Rank; 175} 176 177/// isReassociableOp - Return true if V is an instruction of the specified 178/// opcode and if it only has one use. 179static BinaryOperator *isReassociableOp(Value *V, unsigned Opcode) { 180 if ((V->hasOneUse() || V->use_empty()) && isa<Instruction>(V) && 181 cast<Instruction>(V)->getOpcode() == Opcode) 182 return cast<BinaryOperator>(V); 183 return 0; 184} 185 186/// LowerNegateToMultiply - Replace 0-X with X*-1. 187/// 188static Instruction *LowerNegateToMultiply(Instruction *Neg) { 189 Constant *Cst; 190 if (Neg->getType()->isFloatingPoint()) 191 Cst = ConstantFP::get(Neg->getType(), -1); 192 else 193 Cst = ConstantInt::getAllOnesValue(Neg->getType()); 194 195 std::string NegName = Neg->getName(); Neg->setName(""); 196 Instruction *Res = BinaryOperator::createMul(Neg->getOperand(1), Cst, NegName, 197 Neg); 198 Neg->replaceAllUsesWith(Res); 199 Neg->eraseFromParent(); 200 return Res; 201} 202 203// Given an expression of the form '(A+B)+(D+C)', turn it into '(((A+B)+C)+D)'. 204// Note that if D is also part of the expression tree that we recurse to 205// linearize it as well. Besides that case, this does not recurse into A,B, or 206// C. 207void Reassociate::LinearizeExpr(BinaryOperator *I) { 208 BinaryOperator *LHS = cast<BinaryOperator>(I->getOperand(0)); 209 BinaryOperator *RHS = cast<BinaryOperator>(I->getOperand(1)); 210 assert(isReassociableOp(LHS, I->getOpcode()) && 211 isReassociableOp(RHS, I->getOpcode()) && 212 "Not an expression that needs linearization?"); 213 214 DOUT << "Linear" << *LHS << *RHS << *I; 215 216 // Move the RHS instruction to live immediately before I, avoiding breaking 217 // dominator properties. 218 RHS->moveBefore(I); 219 220 // Move operands around to do the linearization. 221 I->setOperand(1, RHS->getOperand(0)); 222 RHS->setOperand(0, LHS); 223 I->setOperand(0, RHS); 224 225 ++NumLinear; 226 MadeChange = true; 227 DOUT << "Linearized: " << *I; 228 229 // If D is part of this expression tree, tail recurse. 230 if (isReassociableOp(I->getOperand(1), I->getOpcode())) 231 LinearizeExpr(I); 232} 233 234 235/// LinearizeExprTree - Given an associative binary expression tree, traverse 236/// all of the uses putting it into canonical form. This forces a left-linear 237/// form of the the expression (((a+b)+c)+d), and collects information about the 238/// rank of the non-tree operands. 239/// 240/// NOTE: These intentionally destroys the expression tree operands (turning 241/// them into undef values) to reduce #uses of the values. This means that the 242/// caller MUST use something like RewriteExprTree to put the values back in. 243/// 244void Reassociate::LinearizeExprTree(BinaryOperator *I, 245 std::vector<ValueEntry> &Ops) { 246 Value *LHS = I->getOperand(0), *RHS = I->getOperand(1); 247 unsigned Opcode = I->getOpcode(); 248 249 // First step, linearize the expression if it is in ((A+B)+(C+D)) form. 250 BinaryOperator *LHSBO = isReassociableOp(LHS, Opcode); 251 BinaryOperator *RHSBO = isReassociableOp(RHS, Opcode); 252 253 // If this is a multiply expression tree and it contains internal negations, 254 // transform them into multiplies by -1 so they can be reassociated. 255 if (I->getOpcode() == Instruction::Mul) { 256 if (!LHSBO && LHS->hasOneUse() && BinaryOperator::isNeg(LHS)) { 257 LHS = LowerNegateToMultiply(cast<Instruction>(LHS)); 258 LHSBO = isReassociableOp(LHS, Opcode); 259 } 260 if (!RHSBO && RHS->hasOneUse() && BinaryOperator::isNeg(RHS)) { 261 RHS = LowerNegateToMultiply(cast<Instruction>(RHS)); 262 RHSBO = isReassociableOp(RHS, Opcode); 263 } 264 } 265 266 if (!LHSBO) { 267 if (!RHSBO) { 268 // Neither the LHS or RHS as part of the tree, thus this is a leaf. As 269 // such, just remember these operands and their rank. 270 Ops.push_back(ValueEntry(getRank(LHS), LHS)); 271 Ops.push_back(ValueEntry(getRank(RHS), RHS)); 272 273 // Clear the leaves out. 274 I->setOperand(0, UndefValue::get(I->getType())); 275 I->setOperand(1, UndefValue::get(I->getType())); 276 return; 277 } else { 278 // Turn X+(Y+Z) -> (Y+Z)+X 279 std::swap(LHSBO, RHSBO); 280 std::swap(LHS, RHS); 281 bool Success = !I->swapOperands(); 282 assert(Success && "swapOperands failed"); 283 MadeChange = true; 284 } 285 } else if (RHSBO) { 286 // Turn (A+B)+(C+D) -> (((A+B)+C)+D). This guarantees the the RHS is not 287 // part of the expression tree. 288 LinearizeExpr(I); 289 LHS = LHSBO = cast<BinaryOperator>(I->getOperand(0)); 290 RHS = I->getOperand(1); 291 RHSBO = 0; 292 } 293 294 // Okay, now we know that the LHS is a nested expression and that the RHS is 295 // not. Perform reassociation. 296 assert(!isReassociableOp(RHS, Opcode) && "LinearizeExpr failed!"); 297 298 // Move LHS right before I to make sure that the tree expression dominates all 299 // values. 300 LHSBO->moveBefore(I); 301 302 // Linearize the expression tree on the LHS. 303 LinearizeExprTree(LHSBO, Ops); 304 305 // Remember the RHS operand and its rank. 306 Ops.push_back(ValueEntry(getRank(RHS), RHS)); 307 308 // Clear the RHS leaf out. 309 I->setOperand(1, UndefValue::get(I->getType())); 310} 311 312// RewriteExprTree - Now that the operands for this expression tree are 313// linearized and optimized, emit them in-order. This function is written to be 314// tail recursive. 315void Reassociate::RewriteExprTree(BinaryOperator *I, 316 std::vector<ValueEntry> &Ops, 317 unsigned i) { 318 if (i+2 == Ops.size()) { 319 if (I->getOperand(0) != Ops[i].Op || 320 I->getOperand(1) != Ops[i+1].Op) { 321 Value *OldLHS = I->getOperand(0); 322 DOUT << "RA: " << *I; 323 I->setOperand(0, Ops[i].Op); 324 I->setOperand(1, Ops[i+1].Op); 325 DOUT << "TO: " << *I; 326 MadeChange = true; 327 ++NumChanged; 328 329 // If we reassociated a tree to fewer operands (e.g. (1+a+2) -> (a+3) 330 // delete the extra, now dead, nodes. 331 RemoveDeadBinaryOp(OldLHS); 332 } 333 return; 334 } 335 assert(i+2 < Ops.size() && "Ops index out of range!"); 336 337 if (I->getOperand(1) != Ops[i].Op) { 338 DOUT << "RA: " << *I; 339 I->setOperand(1, Ops[i].Op); 340 DOUT << "TO: " << *I; 341 MadeChange = true; 342 ++NumChanged; 343 } 344 345 BinaryOperator *LHS = cast<BinaryOperator>(I->getOperand(0)); 346 assert(LHS->getOpcode() == I->getOpcode() && 347 "Improper expression tree!"); 348 349 // Compactify the tree instructions together with each other to guarantee 350 // that the expression tree is dominated by all of Ops. 351 LHS->moveBefore(I); 352 RewriteExprTree(LHS, Ops, i+1); 353} 354 355 356 357// NegateValue - Insert instructions before the instruction pointed to by BI, 358// that computes the negative version of the value specified. The negative 359// version of the value is returned, and BI is left pointing at the instruction 360// that should be processed next by the reassociation pass. 361// 362static Value *NegateValue(Value *V, Instruction *BI) { 363 // We are trying to expose opportunity for reassociation. One of the things 364 // that we want to do to achieve this is to push a negation as deep into an 365 // expression chain as possible, to expose the add instructions. In practice, 366 // this means that we turn this: 367 // X = -(A+12+C+D) into X = -A + -12 + -C + -D = -12 + -A + -C + -D 368 // so that later, a: Y = 12+X could get reassociated with the -12 to eliminate 369 // the constants. We assume that instcombine will clean up the mess later if 370 // we introduce tons of unnecessary negation instructions... 371 // 372 if (Instruction *I = dyn_cast<Instruction>(V)) 373 if (I->getOpcode() == Instruction::Add && I->hasOneUse()) { 374 // Push the negates through the add. 375 I->setOperand(0, NegateValue(I->getOperand(0), BI)); 376 I->setOperand(1, NegateValue(I->getOperand(1), BI)); 377 378 // We must move the add instruction here, because the neg instructions do 379 // not dominate the old add instruction in general. By moving it, we are 380 // assured that the neg instructions we just inserted dominate the 381 // instruction we are about to insert after them. 382 // 383 I->moveBefore(BI); 384 I->setName(I->getName()+".neg"); 385 return I; 386 } 387 388 // Insert a 'neg' instruction that subtracts the value from zero to get the 389 // negation. 390 // 391 return BinaryOperator::createNeg(V, V->getName() + ".neg", BI); 392} 393 394/// BreakUpSubtract - If we have (X-Y), and if either X is an add, or if this is 395/// only used by an add, transform this into (X+(0-Y)) to promote better 396/// reassociation. 397static Instruction *BreakUpSubtract(Instruction *Sub) { 398 // Don't bother to break this up unless either the LHS is an associable add or 399 // if this is only used by one. 400 if (!isReassociableOp(Sub->getOperand(0), Instruction::Add) && 401 !isReassociableOp(Sub->getOperand(1), Instruction::Add) && 402 !(Sub->hasOneUse() &&isReassociableOp(Sub->use_back(), Instruction::Add))) 403 return 0; 404 405 // Convert a subtract into an add and a neg instruction... so that sub 406 // instructions can be commuted with other add instructions... 407 // 408 // Calculate the negative value of Operand 1 of the sub instruction... 409 // and set it as the RHS of the add instruction we just made... 410 // 411 std::string Name = Sub->getName(); 412 Sub->setName(""); 413 Value *NegVal = NegateValue(Sub->getOperand(1), Sub); 414 Instruction *New = 415 BinaryOperator::createAdd(Sub->getOperand(0), NegVal, Name, Sub); 416 417 // Everyone now refers to the add instruction. 418 Sub->replaceAllUsesWith(New); 419 Sub->eraseFromParent(); 420 421 DOUT << "Negated: " << *New; 422 return New; 423} 424 425/// ConvertShiftToMul - If this is a shift of a reassociable multiply or is used 426/// by one, change this into a multiply by a constant to assist with further 427/// reassociation. 428static Instruction *ConvertShiftToMul(Instruction *Shl) { 429 // If an operand of this shift is a reassociable multiply, or if the shift 430 // is used by a reassociable multiply or add, turn into a multiply. 431 if (isReassociableOp(Shl->getOperand(0), Instruction::Mul) || 432 (Shl->hasOneUse() && 433 (isReassociableOp(Shl->use_back(), Instruction::Mul) || 434 isReassociableOp(Shl->use_back(), Instruction::Add)))) { 435 Constant *MulCst = ConstantInt::get(Shl->getType(), 1); 436 MulCst = ConstantExpr::getShl(MulCst, cast<Constant>(Shl->getOperand(1))); 437 438 std::string Name = Shl->getName(); Shl->setName(""); 439 Instruction *Mul = BinaryOperator::createMul(Shl->getOperand(0), MulCst, 440 Name, Shl); 441 Shl->replaceAllUsesWith(Mul); 442 Shl->eraseFromParent(); 443 return Mul; 444 } 445 return 0; 446} 447 448// Scan backwards and forwards among values with the same rank as element i to 449// see if X exists. If X does not exist, return i. 450static unsigned FindInOperandList(std::vector<ValueEntry> &Ops, unsigned i, 451 Value *X) { 452 unsigned XRank = Ops[i].Rank; 453 unsigned e = Ops.size(); 454 for (unsigned j = i+1; j != e && Ops[j].Rank == XRank; ++j) 455 if (Ops[j].Op == X) 456 return j; 457 // Scan backwards 458 for (unsigned j = i-1; j != ~0U && Ops[j].Rank == XRank; --j) 459 if (Ops[j].Op == X) 460 return j; 461 return i; 462} 463 464/// EmitAddTreeOfValues - Emit a tree of add instructions, summing Ops together 465/// and returning the result. Insert the tree before I. 466static Value *EmitAddTreeOfValues(Instruction *I, std::vector<Value*> &Ops) { 467 if (Ops.size() == 1) return Ops.back(); 468 469 Value *V1 = Ops.back(); 470 Ops.pop_back(); 471 Value *V2 = EmitAddTreeOfValues(I, Ops); 472 return BinaryOperator::createAdd(V2, V1, "tmp", I); 473} 474 475/// RemoveFactorFromExpression - If V is an expression tree that is a 476/// multiplication sequence, and if this sequence contains a multiply by Factor, 477/// remove Factor from the tree and return the new tree. 478Value *Reassociate::RemoveFactorFromExpression(Value *V, Value *Factor) { 479 BinaryOperator *BO = isReassociableOp(V, Instruction::Mul); 480 if (!BO) return 0; 481 482 std::vector<ValueEntry> Factors; 483 LinearizeExprTree(BO, Factors); 484 485 bool FoundFactor = false; 486 for (unsigned i = 0, e = Factors.size(); i != e; ++i) 487 if (Factors[i].Op == Factor) { 488 FoundFactor = true; 489 Factors.erase(Factors.begin()+i); 490 break; 491 } 492 if (!FoundFactor) { 493 // Make sure to restore the operands to the expression tree. 494 RewriteExprTree(BO, Factors); 495 return 0; 496 } 497 498 if (Factors.size() == 1) return Factors[0].Op; 499 500 RewriteExprTree(BO, Factors); 501 return BO; 502} 503 504/// FindSingleUseMultiplyFactors - If V is a single-use multiply, recursively 505/// add its operands as factors, otherwise add V to the list of factors. 506static void FindSingleUseMultiplyFactors(Value *V, 507 std::vector<Value*> &Factors) { 508 BinaryOperator *BO; 509 if ((!V->hasOneUse() && !V->use_empty()) || 510 !(BO = dyn_cast<BinaryOperator>(V)) || 511 BO->getOpcode() != Instruction::Mul) { 512 Factors.push_back(V); 513 return; 514 } 515 516 // Otherwise, add the LHS and RHS to the list of factors. 517 FindSingleUseMultiplyFactors(BO->getOperand(1), Factors); 518 FindSingleUseMultiplyFactors(BO->getOperand(0), Factors); 519} 520 521 522 523Value *Reassociate::OptimizeExpression(BinaryOperator *I, 524 std::vector<ValueEntry> &Ops) { 525 // Now that we have the linearized expression tree, try to optimize it. 526 // Start by folding any constants that we found. 527 bool IterateOptimization = false; 528 if (Ops.size() == 1) return Ops[0].Op; 529 530 unsigned Opcode = I->getOpcode(); 531 532 if (Constant *V1 = dyn_cast<Constant>(Ops[Ops.size()-2].Op)) 533 if (Constant *V2 = dyn_cast<Constant>(Ops.back().Op)) { 534 Ops.pop_back(); 535 Ops.back().Op = ConstantExpr::get(Opcode, V1, V2); 536 return OptimizeExpression(I, Ops); 537 } 538 539 // Check for destructive annihilation due to a constant being used. 540 if (ConstantInt *CstVal = dyn_cast<ConstantInt>(Ops.back().Op)) 541 switch (Opcode) { 542 default: break; 543 case Instruction::And: 544 if (CstVal->isNullValue()) { // ... & 0 -> 0 545 ++NumAnnihil; 546 return CstVal; 547 } else if (CstVal->isAllOnesValue()) { // ... & -1 -> ... 548 Ops.pop_back(); 549 } 550 break; 551 case Instruction::Mul: 552 if (CstVal->isNullValue()) { // ... * 0 -> 0 553 ++NumAnnihil; 554 return CstVal; 555 } else if (cast<ConstantInt>(CstVal)->getZExtValue() == 1) { 556 Ops.pop_back(); // ... * 1 -> ... 557 } 558 break; 559 case Instruction::Or: 560 if (CstVal->isAllOnesValue()) { // ... | -1 -> -1 561 ++NumAnnihil; 562 return CstVal; 563 } 564 // FALLTHROUGH! 565 case Instruction::Add: 566 case Instruction::Xor: 567 if (CstVal->isNullValue()) // ... [|^+] 0 -> ... 568 Ops.pop_back(); 569 break; 570 } 571 if (Ops.size() == 1) return Ops[0].Op; 572 573 // Handle destructive annihilation do to identities between elements in the 574 // argument list here. 575 switch (Opcode) { 576 default: break; 577 case Instruction::And: 578 case Instruction::Or: 579 case Instruction::Xor: 580 // Scan the operand lists looking for X and ~X pairs, along with X,X pairs. 581 // If we find any, we can simplify the expression. X&~X == 0, X|~X == -1. 582 for (unsigned i = 0, e = Ops.size(); i != e; ++i) { 583 // First, check for X and ~X in the operand list. 584 assert(i < Ops.size()); 585 if (BinaryOperator::isNot(Ops[i].Op)) { // Cannot occur for ^. 586 Value *X = BinaryOperator::getNotArgument(Ops[i].Op); 587 unsigned FoundX = FindInOperandList(Ops, i, X); 588 if (FoundX != i) { 589 if (Opcode == Instruction::And) { // ...&X&~X = 0 590 ++NumAnnihil; 591 return Constant::getNullValue(X->getType()); 592 } else if (Opcode == Instruction::Or) { // ...|X|~X = -1 593 ++NumAnnihil; 594 return ConstantInt::getAllOnesValue(X->getType()); 595 } 596 } 597 } 598 599 // Next, check for duplicate pairs of values, which we assume are next to 600 // each other, due to our sorting criteria. 601 assert(i < Ops.size()); 602 if (i+1 != Ops.size() && Ops[i+1].Op == Ops[i].Op) { 603 if (Opcode == Instruction::And || Opcode == Instruction::Or) { 604 // Drop duplicate values. 605 Ops.erase(Ops.begin()+i); 606 --i; --e; 607 IterateOptimization = true; 608 ++NumAnnihil; 609 } else { 610 assert(Opcode == Instruction::Xor); 611 if (e == 2) { 612 ++NumAnnihil; 613 return Constant::getNullValue(Ops[0].Op->getType()); 614 } 615 // ... X^X -> ... 616 Ops.erase(Ops.begin()+i, Ops.begin()+i+2); 617 i -= 1; e -= 2; 618 IterateOptimization = true; 619 ++NumAnnihil; 620 } 621 } 622 } 623 break; 624 625 case Instruction::Add: 626 // Scan the operand lists looking for X and -X pairs. If we find any, we 627 // can simplify the expression. X+-X == 0. 628 for (unsigned i = 0, e = Ops.size(); i != e; ++i) { 629 assert(i < Ops.size()); 630 // Check for X and -X in the operand list. 631 if (BinaryOperator::isNeg(Ops[i].Op)) { 632 Value *X = BinaryOperator::getNegArgument(Ops[i].Op); 633 unsigned FoundX = FindInOperandList(Ops, i, X); 634 if (FoundX != i) { 635 // Remove X and -X from the operand list. 636 if (Ops.size() == 2) { 637 ++NumAnnihil; 638 return Constant::getNullValue(X->getType()); 639 } else { 640 Ops.erase(Ops.begin()+i); 641 if (i < FoundX) 642 --FoundX; 643 else 644 --i; // Need to back up an extra one. 645 Ops.erase(Ops.begin()+FoundX); 646 IterateOptimization = true; 647 ++NumAnnihil; 648 --i; // Revisit element. 649 e -= 2; // Removed two elements. 650 } 651 } 652 } 653 } 654 655 656 // Scan the operand list, checking to see if there are any common factors 657 // between operands. Consider something like A*A+A*B*C+D. We would like to 658 // reassociate this to A*(A+B*C)+D, which reduces the number of multiplies. 659 // To efficiently find this, we count the number of times a factor occurs 660 // for any ADD operands that are MULs. 661 std::map<Value*, unsigned> FactorOccurrences; 662 unsigned MaxOcc = 0; 663 Value *MaxOccVal = 0; 664 if (!I->getType()->isFloatingPoint()) { 665 for (unsigned i = 0, e = Ops.size(); i != e; ++i) { 666 if (BinaryOperator *BOp = dyn_cast<BinaryOperator>(Ops[i].Op)) 667 if (BOp->getOpcode() == Instruction::Mul && BOp->use_empty()) { 668 // Compute all of the factors of this added value. 669 std::vector<Value*> Factors; 670 FindSingleUseMultiplyFactors(BOp, Factors); 671 assert(Factors.size() > 1 && "Bad linearize!"); 672 673 // Add one to FactorOccurrences for each unique factor in this op. 674 if (Factors.size() == 2) { 675 unsigned Occ = ++FactorOccurrences[Factors[0]]; 676 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[0]; } 677 if (Factors[0] != Factors[1]) { // Don't double count A*A. 678 Occ = ++FactorOccurrences[Factors[1]]; 679 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[1]; } 680 } 681 } else { 682 std::set<Value*> Duplicates; 683 for (unsigned i = 0, e = Factors.size(); i != e; ++i) 684 if (Duplicates.insert(Factors[i]).second) { 685 unsigned Occ = ++FactorOccurrences[Factors[i]]; 686 if (Occ > MaxOcc) { MaxOcc = Occ; MaxOccVal = Factors[i]; } 687 } 688 } 689 } 690 } 691 } 692 693 // If any factor occurred more than one time, we can pull it out. 694 if (MaxOcc > 1) { 695 DOUT << "\nFACTORING [" << MaxOcc << "]: " << *MaxOccVal << "\n"; 696 697 // Create a new instruction that uses the MaxOccVal twice. If we don't do 698 // this, we could otherwise run into situations where removing a factor 699 // from an expression will drop a use of maxocc, and this can cause 700 // RemoveFactorFromExpression on successive values to behave differently. 701 Instruction *DummyInst = BinaryOperator::createAdd(MaxOccVal, MaxOccVal); 702 std::vector<Value*> NewMulOps; 703 for (unsigned i = 0, e = Ops.size(); i != e; ++i) { 704 if (Value *V = RemoveFactorFromExpression(Ops[i].Op, MaxOccVal)) { 705 NewMulOps.push_back(V); 706 Ops.erase(Ops.begin()+i); 707 --i; --e; 708 } 709 } 710 711 // No need for extra uses anymore. 712 delete DummyInst; 713 714 unsigned NumAddedValues = NewMulOps.size(); 715 Value *V = EmitAddTreeOfValues(I, NewMulOps); 716 Value *V2 = BinaryOperator::createMul(V, MaxOccVal, "tmp", I); 717 718 // Now that we have inserted V and its sole use, optimize it. This allows 719 // us to handle cases that require multiple factoring steps, such as this: 720 // A*A*B + A*A*C --> A*(A*B+A*C) --> A*(A*(B+C)) 721 if (NumAddedValues > 1) 722 ReassociateExpression(cast<BinaryOperator>(V)); 723 724 ++NumFactor; 725 726 if (Ops.size() == 0) 727 return V2; 728 729 // Add the new value to the list of things being added. 730 Ops.insert(Ops.begin(), ValueEntry(getRank(V2), V2)); 731 732 // Rewrite the tree so that there is now a use of V. 733 RewriteExprTree(I, Ops); 734 return OptimizeExpression(I, Ops); 735 } 736 break; 737 //case Instruction::Mul: 738 } 739 740 if (IterateOptimization) 741 return OptimizeExpression(I, Ops); 742 return 0; 743} 744 745 746/// ReassociateBB - Inspect all of the instructions in this basic block, 747/// reassociating them as we go. 748void Reassociate::ReassociateBB(BasicBlock *BB) { 749 for (BasicBlock::iterator BBI = BB->begin(); BBI != BB->end(); ) { 750 Instruction *BI = BBI++; 751 if (BI->getOpcode() == Instruction::Shl && 752 isa<ConstantInt>(BI->getOperand(1))) 753 if (Instruction *NI = ConvertShiftToMul(BI)) { 754 MadeChange = true; 755 BI = NI; 756 } 757 758 // Reject cases where it is pointless to do this. 759 if (!isa<BinaryOperator>(BI) || BI->getType()->isFloatingPoint() || 760 isa<PackedType>(BI->getType())) 761 continue; // Floating point ops are not associative. 762 763 // If this is a subtract instruction which is not already in negate form, 764 // see if we can convert it to X+-Y. 765 if (BI->getOpcode() == Instruction::Sub) { 766 if (!BinaryOperator::isNeg(BI)) { 767 if (Instruction *NI = BreakUpSubtract(BI)) { 768 MadeChange = true; 769 BI = NI; 770 } 771 } else { 772 // Otherwise, this is a negation. See if the operand is a multiply tree 773 // and if this is not an inner node of a multiply tree. 774 if (isReassociableOp(BI->getOperand(1), Instruction::Mul) && 775 (!BI->hasOneUse() || 776 !isReassociableOp(BI->use_back(), Instruction::Mul))) { 777 BI = LowerNegateToMultiply(BI); 778 MadeChange = true; 779 } 780 } 781 } 782 783 // If this instruction is a commutative binary operator, process it. 784 if (!BI->isAssociative()) continue; 785 BinaryOperator *I = cast<BinaryOperator>(BI); 786 787 // If this is an interior node of a reassociable tree, ignore it until we 788 // get to the root of the tree, to avoid N^2 analysis. 789 if (I->hasOneUse() && isReassociableOp(I->use_back(), I->getOpcode())) 790 continue; 791 792 // If this is an add tree that is used by a sub instruction, ignore it 793 // until we process the subtract. 794 if (I->hasOneUse() && I->getOpcode() == Instruction::Add && 795 cast<Instruction>(I->use_back())->getOpcode() == Instruction::Sub) 796 continue; 797 798 ReassociateExpression(I); 799 } 800} 801 802void Reassociate::ReassociateExpression(BinaryOperator *I) { 803 804 // First, walk the expression tree, linearizing the tree, collecting 805 std::vector<ValueEntry> Ops; 806 LinearizeExprTree(I, Ops); 807 808 DOUT << "RAIn:\t"; DEBUG(PrintOps(I, Ops)); DOUT << "\n"; 809 810 // Now that we have linearized the tree to a list and have gathered all of 811 // the operands and their ranks, sort the operands by their rank. Use a 812 // stable_sort so that values with equal ranks will have their relative 813 // positions maintained (and so the compiler is deterministic). Note that 814 // this sorts so that the highest ranking values end up at the beginning of 815 // the vector. 816 std::stable_sort(Ops.begin(), Ops.end()); 817 818 // OptimizeExpression - Now that we have the expression tree in a convenient 819 // sorted form, optimize it globally if possible. 820 if (Value *V = OptimizeExpression(I, Ops)) { 821 // This expression tree simplified to something that isn't a tree, 822 // eliminate it. 823 DOUT << "Reassoc to scalar: " << *V << "\n"; 824 I->replaceAllUsesWith(V); 825 RemoveDeadBinaryOp(I); 826 return; 827 } 828 829 // We want to sink immediates as deeply as possible except in the case where 830 // this is a multiply tree used only by an add, and the immediate is a -1. 831 // In this case we reassociate to put the negation on the outside so that we 832 // can fold the negation into the add: (-X)*Y + Z -> Z-X*Y 833 if (I->getOpcode() == Instruction::Mul && I->hasOneUse() && 834 cast<Instruction>(I->use_back())->getOpcode() == Instruction::Add && 835 isa<ConstantInt>(Ops.back().Op) && 836 cast<ConstantInt>(Ops.back().Op)->isAllOnesValue()) { 837 Ops.insert(Ops.begin(), Ops.back()); 838 Ops.pop_back(); 839 } 840 841 DOUT << "RAOut:\t"; DEBUG(PrintOps(I, Ops)); DOUT << "\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