llvm-6502/lib/Transforms/Scalar/Reassociate.cpp
Chris Lattner 7b4ad94282 Fix a problem that Dan Berlin noticed, where reassociation would not succeed
in building maximal expressions before simplifying them.  In particular, i
cases like this:

X-(A+B+X)

the code would consider A+B+X to be a maximal expression (not understanding
that the single use '-' would be turned into a + later), simplify it (a noop)
then later get simplified again.

Each of these simplify steps is where the cost of reassociation comes from,
so this patch should speed up the already fast pass a bit.

Thanks to Dan for noticing this!


git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@23214 91177308-0d34-0410-b5e6-96231b3b80d8
2005-09-02 07:07:58 +00:00

682 lines
25 KiB
C++

//===- Reassociate.cpp - Reassociate binary expressions -------------------===//
//
// The LLVM Compiler Infrastructure
//
// This file was developed by the LLVM research group and is distributed under
// the University of Illinois Open Source License. See LICENSE.TXT for details.
//
//===----------------------------------------------------------------------===//
//
// This pass reassociates commutative expressions in an order that is designed
// to promote better constant propagation, GCSE, LICM, PRE...
//
// For example: 4 + (x + 5) -> x + (4 + 5)
//
// In the implementation of this algorithm, constants are assigned rank = 0,
// function arguments are rank = 1, and other values are assigned ranks
// corresponding to the reverse post order traversal of current function
// (starting at 2), which effectively gives values in deep loops higher rank
// than values not in loops.
//
//===----------------------------------------------------------------------===//
#define DEBUG_TYPE "reassociate"
#include "llvm/Transforms/Scalar.h"
#include "llvm/Constants.h"
#include "llvm/Function.h"
#include "llvm/Instructions.h"
#include "llvm/Pass.h"
#include "llvm/Type.h"
#include "llvm/Assembly/Writer.h"
#include "llvm/Support/CFG.h"
#include "llvm/Support/Debug.h"
#include "llvm/ADT/PostOrderIterator.h"
#include "llvm/ADT/Statistic.h"
#include <algorithm>
using namespace llvm;
namespace {
Statistic<> NumLinear ("reassociate","Number of insts linearized");
Statistic<> NumChanged("reassociate","Number of insts reassociated");
Statistic<> NumSwapped("reassociate","Number of insts with operands swapped");
Statistic<> NumAnnihil("reassociate","Number of expr tree annihilated");
struct ValueEntry {
unsigned Rank;
Value *Op;
ValueEntry(unsigned R, Value *O) : Rank(R), Op(O) {}
};
inline bool operator<(const ValueEntry &LHS, const ValueEntry &RHS) {
return LHS.Rank > RHS.Rank; // Sort so that highest rank goes to start.
}
class Reassociate : public FunctionPass {
std::map<BasicBlock*, unsigned> RankMap;
std::map<Value*, unsigned> ValueRankMap;
bool MadeChange;
public:
bool runOnFunction(Function &F);
virtual void getAnalysisUsage(AnalysisUsage &AU) const {
AU.setPreservesCFG();
}
private:
void BuildRankMap(Function &F);
unsigned getRank(Value *V);
void RewriteExprTree(BinaryOperator *I, unsigned Idx,
std::vector<ValueEntry> &Ops);
void OptimizeExpression(unsigned Opcode, std::vector<ValueEntry> &Ops);
void LinearizeExprTree(BinaryOperator *I, std::vector<ValueEntry> &Ops);
void LinearizeExpr(BinaryOperator *I);
void ReassociateBB(BasicBlock *BB);
};
RegisterOpt<Reassociate> X("reassociate", "Reassociate expressions");
}
// Public interface to the Reassociate pass
FunctionPass *llvm::createReassociatePass() { return new Reassociate(); }
static bool isUnmovableInstruction(Instruction *I) {
if (I->getOpcode() == Instruction::PHI ||
I->getOpcode() == Instruction::Alloca ||
I->getOpcode() == Instruction::Load ||
I->getOpcode() == Instruction::Malloc ||
I->getOpcode() == Instruction::Invoke ||
I->getOpcode() == Instruction::Call ||
I->getOpcode() == Instruction::Div ||
I->getOpcode() == Instruction::Rem)
return true;
return false;
}
void Reassociate::BuildRankMap(Function &F) {
unsigned i = 2;
// Assign distinct ranks to function arguments
for (Function::arg_iterator I = F.arg_begin(), E = F.arg_end(); I != E; ++I)
ValueRankMap[I] = ++i;
ReversePostOrderTraversal<Function*> RPOT(&F);
for (ReversePostOrderTraversal<Function*>::rpo_iterator I = RPOT.begin(),
E = RPOT.end(); I != E; ++I) {
BasicBlock *BB = *I;
unsigned BBRank = RankMap[BB] = ++i << 16;
// Walk the basic block, adding precomputed ranks for any instructions that
// we cannot move. This ensures that the ranks for these instructions are
// all different in the block.
for (BasicBlock::iterator I = BB->begin(), E = BB->end(); I != E; ++I)
if (isUnmovableInstruction(I))
ValueRankMap[I] = ++BBRank;
}
}
unsigned Reassociate::getRank(Value *V) {
if (isa<Argument>(V)) return ValueRankMap[V]; // Function argument...
Instruction *I = dyn_cast<Instruction>(V);
if (I == 0) return 0; // Otherwise it's a global or constant, rank 0.
unsigned &CachedRank = ValueRankMap[I];
if (CachedRank) return CachedRank; // Rank already known?
// If this is an expression, return the 1+MAX(rank(LHS), rank(RHS)) so that
// we can reassociate expressions for code motion! Since we do not recurse
// for PHI nodes, we cannot have infinite recursion here, because there
// cannot be loops in the value graph that do not go through PHI nodes.
unsigned Rank = 0, MaxRank = RankMap[I->getParent()];
for (unsigned i = 0, e = I->getNumOperands();
i != e && Rank != MaxRank; ++i)
Rank = std::max(Rank, getRank(I->getOperand(i)));
// If this is a not or neg instruction, do not count it for rank. This
// assures us that X and ~X will have the same rank.
if (!I->getType()->isIntegral() ||
(!BinaryOperator::isNot(I) && !BinaryOperator::isNeg(I)))
++Rank;
//DEBUG(std::cerr << "Calculated Rank[" << V->getName() << "] = "
//<< Rank << "\n");
return CachedRank = Rank;
}
/// isReassociableOp - Return true if V is an instruction of the specified
/// opcode and if it only has one use.
static BinaryOperator *isReassociableOp(Value *V, unsigned Opcode) {
if (V->hasOneUse() && isa<Instruction>(V) &&
cast<Instruction>(V)->getOpcode() == Opcode)
return cast<BinaryOperator>(V);
return 0;
}
/// LowerNegateToMultiply - Replace 0-X with X*-1.
///
static Instruction *LowerNegateToMultiply(Instruction *Neg) {
Constant *Cst;
if (Neg->getType()->isFloatingPoint())
Cst = ConstantFP::get(Neg->getType(), -1);
else
Cst = ConstantInt::getAllOnesValue(Neg->getType());
std::string NegName = Neg->getName(); Neg->setName("");
Instruction *Res = BinaryOperator::createMul(Neg->getOperand(1), Cst, NegName,
Neg);
Neg->replaceAllUsesWith(Res);
Neg->eraseFromParent();
return Res;
}
// Given an expression of the form '(A+B)+(D+C)', turn it into '(((A+B)+C)+D)'.
// Note that if D is also part of the expression tree that we recurse to
// linearize it as well. Besides that case, this does not recurse into A,B, or
// C.
void Reassociate::LinearizeExpr(BinaryOperator *I) {
BinaryOperator *LHS = cast<BinaryOperator>(I->getOperand(0));
BinaryOperator *RHS = cast<BinaryOperator>(I->getOperand(1));
assert(isReassociableOp(LHS, I->getOpcode()) &&
isReassociableOp(RHS, I->getOpcode()) &&
"Not an expression that needs linearization?");
DEBUG(std::cerr << "Linear" << *LHS << *RHS << *I);
// Move the RHS instruction to live immediately before I, avoiding breaking
// dominator properties.
RHS->moveBefore(I);
// Move operands around to do the linearization.
I->setOperand(1, RHS->getOperand(0));
RHS->setOperand(0, LHS);
I->setOperand(0, RHS);
++NumLinear;
MadeChange = true;
DEBUG(std::cerr << "Linearized: " << *I);
// If D is part of this expression tree, tail recurse.
if (isReassociableOp(I->getOperand(1), I->getOpcode()))
LinearizeExpr(I);
}
/// LinearizeExprTree - Given an associative binary expression tree, traverse
/// all of the uses putting it into canonical form. This forces a left-linear
/// form of the the expression (((a+b)+c)+d), and collects information about the
/// rank of the non-tree operands.
///
/// This returns the rank of the RHS operand, which is known to be the highest
/// rank value in the expression tree.
///
void Reassociate::LinearizeExprTree(BinaryOperator *I,
std::vector<ValueEntry> &Ops) {
Value *LHS = I->getOperand(0), *RHS = I->getOperand(1);
unsigned Opcode = I->getOpcode();
// First step, linearize the expression if it is in ((A+B)+(C+D)) form.
BinaryOperator *LHSBO = isReassociableOp(LHS, Opcode);
BinaryOperator *RHSBO = isReassociableOp(RHS, Opcode);
// If this is a multiply expression tree and it contains internal negations,
// transform them into multiplies by -1 so they can be reassociated.
if (I->getOpcode() == Instruction::Mul) {
if (!LHSBO && LHS->hasOneUse() && BinaryOperator::isNeg(LHS)) {
LHS = LowerNegateToMultiply(cast<Instruction>(LHS));
LHSBO = isReassociableOp(LHS, Opcode);
}
if (!RHSBO && RHS->hasOneUse() && BinaryOperator::isNeg(RHS)) {
RHS = LowerNegateToMultiply(cast<Instruction>(RHS));
RHSBO = isReassociableOp(RHS, Opcode);
}
}
if (!LHSBO) {
if (!RHSBO) {
// Neither the LHS or RHS as part of the tree, thus this is a leaf. As
// such, just remember these operands and their rank.
Ops.push_back(ValueEntry(getRank(LHS), LHS));
Ops.push_back(ValueEntry(getRank(RHS), RHS));
return;
} else {
// Turn X+(Y+Z) -> (Y+Z)+X
std::swap(LHSBO, RHSBO);
std::swap(LHS, RHS);
bool Success = !I->swapOperands();
assert(Success && "swapOperands failed");
MadeChange = true;
}
} else if (RHSBO) {
// Turn (A+B)+(C+D) -> (((A+B)+C)+D). This guarantees the the RHS is not
// part of the expression tree.
LinearizeExpr(I);
LHS = LHSBO = cast<BinaryOperator>(I->getOperand(0));
RHS = I->getOperand(1);
RHSBO = 0;
}
// Okay, now we know that the LHS is a nested expression and that the RHS is
// not. Perform reassociation.
assert(!isReassociableOp(RHS, Opcode) && "LinearizeExpr failed!");
// Move LHS right before I to make sure that the tree expression dominates all
// values.
LHSBO->moveBefore(I);
// Linearize the expression tree on the LHS.
LinearizeExprTree(LHSBO, Ops);
// Remember the RHS operand and its rank.
Ops.push_back(ValueEntry(getRank(RHS), RHS));
}
// RewriteExprTree - Now that the operands for this expression tree are
// linearized and optimized, emit them in-order. This function is written to be
// tail recursive.
void Reassociate::RewriteExprTree(BinaryOperator *I, unsigned i,
std::vector<ValueEntry> &Ops) {
if (i+2 == Ops.size()) {
if (I->getOperand(0) != Ops[i].Op ||
I->getOperand(1) != Ops[i+1].Op) {
DEBUG(std::cerr << "RA: " << *I);
I->setOperand(0, Ops[i].Op);
I->setOperand(1, Ops[i+1].Op);
DEBUG(std::cerr << "TO: " << *I);
MadeChange = true;
++NumChanged;
}
return;
}
assert(i+2 < Ops.size() && "Ops index out of range!");
if (I->getOperand(1) != Ops[i].Op) {
DEBUG(std::cerr << "RA: " << *I);
I->setOperand(1, Ops[i].Op);
DEBUG(std::cerr << "TO: " << *I);
MadeChange = true;
++NumChanged;
}
RewriteExprTree(cast<BinaryOperator>(I->getOperand(0)), i+1, Ops);
}
// NegateValue - Insert instructions before the instruction pointed to by BI,
// that computes the negative version of the value specified. The negative
// version of the value is returned, and BI is left pointing at the instruction
// that should be processed next by the reassociation pass.
//
static Value *NegateValue(Value *V, Instruction *BI) {
// We are trying to expose opportunity for reassociation. One of the things
// that we want to do to achieve this is to push a negation as deep into an
// expression chain as possible, to expose the add instructions. In practice,
// this means that we turn this:
// X = -(A+12+C+D) into X = -A + -12 + -C + -D = -12 + -A + -C + -D
// so that later, a: Y = 12+X could get reassociated with the -12 to eliminate
// the constants. We assume that instcombine will clean up the mess later if
// we introduce tons of unnecessary negation instructions...
//
if (Instruction *I = dyn_cast<Instruction>(V))
if (I->getOpcode() == Instruction::Add && I->hasOneUse()) {
// Push the negates through the add.
I->setOperand(0, NegateValue(I->getOperand(0), BI));
I->setOperand(1, NegateValue(I->getOperand(1), BI));
// We must move the add instruction here, because the neg instructions do
// not dominate the old add instruction in general. By moving it, we are
// assured that the neg instructions we just inserted dominate the
// instruction we are about to insert after them.
//
I->moveBefore(BI);
I->setName(I->getName()+".neg");
return I;
}
// Insert a 'neg' instruction that subtracts the value from zero to get the
// negation.
//
return BinaryOperator::createNeg(V, V->getName() + ".neg", BI);
}
/// BreakUpSubtract - If we have (X-Y), and if either X is an add, or if this is
/// only used by an add, transform this into (X+(0-Y)) to promote better
/// reassociation.
static Instruction *BreakUpSubtract(Instruction *Sub) {
// Don't bother to break this up unless either the LHS is an associable add or
// if this is only used by one.
if (!isReassociableOp(Sub->getOperand(0), Instruction::Add) &&
!isReassociableOp(Sub->getOperand(1), Instruction::Add) &&
!(Sub->hasOneUse() &&isReassociableOp(Sub->use_back(), Instruction::Add)))
return 0;
// Convert a subtract into an add and a neg instruction... so that sub
// instructions can be commuted with other add instructions...
//
// Calculate the negative value of Operand 1 of the sub instruction...
// and set it as the RHS of the add instruction we just made...
//
std::string Name = Sub->getName();
Sub->setName("");
Value *NegVal = NegateValue(Sub->getOperand(1), Sub);
Instruction *New =
BinaryOperator::createAdd(Sub->getOperand(0), NegVal, Name, Sub);
// Everyone now refers to the add instruction.
Sub->replaceAllUsesWith(New);
Sub->eraseFromParent();
DEBUG(std::cerr << "Negated: " << *New);
return New;
}
/// ConvertShiftToMul - If this is a shift of a reassociable multiply or is used
/// by one, change this into a multiply by a constant to assist with further
/// reassociation.
static Instruction *ConvertShiftToMul(Instruction *Shl) {
if (!isReassociableOp(Shl->getOperand(0), Instruction::Mul) &&
!(Shl->hasOneUse() && isReassociableOp(Shl->use_back(),Instruction::Mul)))
return 0;
Constant *MulCst = ConstantInt::get(Shl->getType(), 1);
MulCst = ConstantExpr::getShl(MulCst, cast<Constant>(Shl->getOperand(1)));
std::string Name = Shl->getName(); Shl->setName("");
Instruction *Mul = BinaryOperator::createMul(Shl->getOperand(0), MulCst,
Name, Shl);
Shl->replaceAllUsesWith(Mul);
Shl->eraseFromParent();
return Mul;
}
// Scan backwards and forwards among values with the same rank as element i to
// see if X exists. If X does not exist, return i.
static unsigned FindInOperandList(std::vector<ValueEntry> &Ops, unsigned i,
Value *X) {
unsigned XRank = Ops[i].Rank;
unsigned e = Ops.size();
for (unsigned j = i+1; j != e && Ops[j].Rank == XRank; ++j)
if (Ops[j].Op == X)
return j;
// Scan backwards
for (unsigned j = i-1; j != ~0U && Ops[j].Rank == XRank; --j)
if (Ops[j].Op == X)
return j;
return i;
}
void Reassociate::OptimizeExpression(unsigned Opcode,
std::vector<ValueEntry> &Ops) {
// Now that we have the linearized expression tree, try to optimize it.
// Start by folding any constants that we found.
bool IterateOptimization = false;
if (Ops.size() == 1) return;
if (Constant *V1 = dyn_cast<Constant>(Ops[Ops.size()-2].Op))
if (Constant *V2 = dyn_cast<Constant>(Ops.back().Op)) {
Ops.pop_back();
Ops.back().Op = ConstantExpr::get(Opcode, V1, V2);
OptimizeExpression(Opcode, Ops);
return;
}
// Check for destructive annihilation due to a constant being used.
if (ConstantIntegral *CstVal = dyn_cast<ConstantIntegral>(Ops.back().Op))
switch (Opcode) {
default: break;
case Instruction::And:
if (CstVal->isNullValue()) { // ... & 0 -> 0
Ops[0].Op = CstVal;
Ops.erase(Ops.begin()+1, Ops.end());
++NumAnnihil;
return;
} else if (CstVal->isAllOnesValue()) { // ... & -1 -> ...
Ops.pop_back();
}
break;
case Instruction::Mul:
if (CstVal->isNullValue()) { // ... * 0 -> 0
Ops[0].Op = CstVal;
Ops.erase(Ops.begin()+1, Ops.end());
++NumAnnihil;
return;
} else if (cast<ConstantInt>(CstVal)->getRawValue() == 1) {
Ops.pop_back(); // ... * 1 -> ...
}
break;
case Instruction::Or:
if (CstVal->isAllOnesValue()) { // ... | -1 -> -1
Ops[0].Op = CstVal;
Ops.erase(Ops.begin()+1, Ops.end());
++NumAnnihil;
return;
}
// FALLTHROUGH!
case Instruction::Add:
case Instruction::Xor:
if (CstVal->isNullValue()) // ... [|^+] 0 -> ...
Ops.pop_back();
break;
}
if (Ops.size() == 1) return;
// Handle destructive annihilation do to identities between elements in the
// argument list here.
switch (Opcode) {
default: break;
case Instruction::And:
case Instruction::Or:
case Instruction::Xor:
// Scan the operand lists looking for X and ~X pairs, along with X,X pairs.
// If we find any, we can simplify the expression. X&~X == 0, X|~X == -1.
for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
// First, check for X and ~X in the operand list.
assert(i < Ops.size());
if (BinaryOperator::isNot(Ops[i].Op)) { // Cannot occur for ^.
Value *X = BinaryOperator::getNotArgument(Ops[i].Op);
unsigned FoundX = FindInOperandList(Ops, i, X);
if (FoundX != i) {
if (Opcode == Instruction::And) { // ...&X&~X = 0
Ops[0].Op = Constant::getNullValue(X->getType());
Ops.erase(Ops.begin()+1, Ops.end());
++NumAnnihil;
return;
} else if (Opcode == Instruction::Or) { // ...|X|~X = -1
Ops[0].Op = ConstantIntegral::getAllOnesValue(X->getType());
Ops.erase(Ops.begin()+1, Ops.end());
++NumAnnihil;
return;
}
}
}
// Next, check for duplicate pairs of values, which we assume are next to
// each other, due to our sorting criteria.
assert(i < Ops.size());
if (i+1 != Ops.size() && Ops[i+1].Op == Ops[i].Op) {
if (Opcode == Instruction::And || Opcode == Instruction::Or) {
// Drop duplicate values.
Ops.erase(Ops.begin()+i);
--i; --e;
IterateOptimization = true;
++NumAnnihil;
} else {
assert(Opcode == Instruction::Xor);
if (e == 2) {
Ops[0].Op = Constant::getNullValue(Ops[0].Op->getType());
Ops.erase(Ops.begin()+1, Ops.end());
++NumAnnihil;
return;
}
// ... X^X -> ...
Ops.erase(Ops.begin()+i, Ops.begin()+i+2);
i -= 1; e -= 2;
IterateOptimization = true;
++NumAnnihil;
}
}
}
break;
case Instruction::Add:
// Scan the operand lists looking for X and -X pairs. If we find any, we
// can simplify the expression. X+-X == 0
for (unsigned i = 0, e = Ops.size(); i != e; ++i) {
assert(i < Ops.size());
// Check for X and -X in the operand list.
if (BinaryOperator::isNeg(Ops[i].Op)) {
Value *X = BinaryOperator::getNegArgument(Ops[i].Op);
unsigned FoundX = FindInOperandList(Ops, i, X);
if (FoundX != i) {
// Remove X and -X from the operand list.
if (Ops.size() == 2) {
Ops[0].Op = Constant::getNullValue(X->getType());
Ops.pop_back();
++NumAnnihil;
return;
} else {
Ops.erase(Ops.begin()+i);
if (i < FoundX)
--FoundX;
else
--i; // Need to back up an extra one.
Ops.erase(Ops.begin()+FoundX);
IterateOptimization = true;
++NumAnnihil;
--i; // Revisit element.
e -= 2; // Removed two elements.
}
}
}
}
break;
//case Instruction::Mul:
}
if (IterateOptimization)
OptimizeExpression(Opcode, Ops);
}
/// PrintOps - Print out the expression identified in the Ops list.
///
static void PrintOps(unsigned Opcode, const std::vector<ValueEntry> &Ops,
BasicBlock *BB) {
Module *M = BB->getParent()->getParent();
std::cerr << Instruction::getOpcodeName(Opcode) << " "
<< *Ops[0].Op->getType();
for (unsigned i = 0, e = Ops.size(); i != e; ++i)
WriteAsOperand(std::cerr << " ", Ops[i].Op, false, true, M)
<< "," << Ops[i].Rank;
}
/// ReassociateBB - Inspect all of the instructions in this basic block,
/// reassociating them as we go.
void Reassociate::ReassociateBB(BasicBlock *BB) {
for (BasicBlock::iterator BI = BB->begin(); BI != BB->end(); ++BI) {
if (BI->getOpcode() == Instruction::Shl &&
isa<ConstantInt>(BI->getOperand(1)))
if (Instruction *NI = ConvertShiftToMul(BI)) {
MadeChange = true;
BI = NI;
}
// Reject cases where it is pointless to do this.
if (!isa<BinaryOperator>(BI) || BI->getType()->isFloatingPoint())
continue; // Floating point ops are not associative.
// If this is a subtract instruction which is not already in negate form,
// see if we can convert it to X+-Y.
if (BI->getOpcode() == Instruction::Sub) {
if (!BinaryOperator::isNeg(BI)) {
if (Instruction *NI = BreakUpSubtract(BI)) {
MadeChange = true;
BI = NI;
}
} else {
// Otherwise, this is a negation. See if the operand is a multiply tree
// and if this is not an inner node of a multiply tree.
if (isReassociableOp(BI->getOperand(1), Instruction::Mul) &&
(!BI->hasOneUse() ||
!isReassociableOp(BI->use_back(), Instruction::Mul))) {
BI = LowerNegateToMultiply(BI);
MadeChange = true;
}
}
}
// If this instruction is a commutative binary operator, process it.
if (!BI->isAssociative()) continue;
BinaryOperator *I = cast<BinaryOperator>(BI);
// If this is an interior node of a reassociable tree, ignore it until we
// get to the root of the tree, to avoid N^2 analysis.
if (I->hasOneUse() && isReassociableOp(I->use_back(), I->getOpcode()))
continue;
// If this is an add tree that is used by a sub instruction, ignore it
// until we process the subtract.
if (I->hasOneUse() && I->getOpcode() == Instruction::Add &&
cast<Instruction>(I->use_back())->getOpcode() == Instruction::Sub)
continue;
// First, walk the expression tree, linearizing the tree, collecting
std::vector<ValueEntry> Ops;
LinearizeExprTree(I, Ops);
DEBUG(std::cerr << "RAIn:\t"; PrintOps(I->getOpcode(), Ops, BB);
std::cerr << "\n");
// Now that we have linearized the tree to a list and have gathered all of
// the operands and their ranks, sort the operands by their rank. Use a
// stable_sort so that values with equal ranks will have their relative
// positions maintained (and so the compiler is deterministic). Note that
// this sorts so that the highest ranking values end up at the beginning of
// the vector.
std::stable_sort(Ops.begin(), Ops.end());
// OptimizeExpression - Now that we have the expression tree in a convenient
// sorted form, optimize it globally if possible.
OptimizeExpression(I->getOpcode(), Ops);
// We want to sink immediates as deeply as possible except in the case where
// this is a multiply tree used only by an add, and the immediate is a -1.
// In this case we reassociate to put the negation on the outside so that we
// can fold the negation into the add: (-X)*Y + Z -> Z-X*Y
if (I->getOpcode() == Instruction::Mul && I->hasOneUse() &&
cast<Instruction>(I->use_back())->getOpcode() == Instruction::Add &&
isa<ConstantInt>(Ops.back().Op) &&
cast<ConstantInt>(Ops.back().Op)->isAllOnesValue()) {
Ops.insert(Ops.begin(), Ops.back());
Ops.pop_back();
}
DEBUG(std::cerr << "RAOut:\t"; PrintOps(I->getOpcode(), Ops, BB);
std::cerr << "\n");
if (Ops.size() == 1) {
// This expression tree simplified to something that isn't a tree,
// eliminate it.
I->replaceAllUsesWith(Ops[0].Op);
} else {
// Now that we ordered and optimized the expressions, splat them back into
// the expression tree, removing any unneeded nodes.
RewriteExprTree(I, 0, Ops);
}
}
}
bool Reassociate::runOnFunction(Function &F) {
// Recalculate the rank map for F
BuildRankMap(F);
MadeChange = false;
for (Function::iterator FI = F.begin(), FE = F.end(); FI != FE; ++FI)
ReassociateBB(FI);
// We are done with the rank map...
RankMap.clear();
ValueRankMap.clear();
return MadeChange;
}