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rearrange some code, no functionality change.

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git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@123136 91177308-0d34-0410-b5e6-96231b3b80d8
This commit is contained in:
Chris Lattner 2011-01-09 22:39:48 +00:00
parent fb00e2715b
commit 7975e3ebba
1 changed files with 45 additions and 41 deletions
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86
lib/Analysis/ScalarEvolution.cpp
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@ -4807,47 +4807,14 @@ ScalarEvolution::HowFarToZero(const SCEV *V, const Loop *L) {
if (!AddRec || AddRec->getLoop() != L)
return getCouldNotCompute();
if (AddRec->isAffine()) {
// If this is an affine expression, the execution count of this branch is
// the minimum unsigned root of the following equation:
//
// Start + Step*N = 0 (mod 2^BW)
//
// equivalent to:
//
// Step*N = -Start (mod 2^BW)
//
// where BW is the common bit width of Start and Step.
// Get the initial value for the loop.
const SCEV *Start = getSCEVAtScope(AddRec->getStart(),
L->getParentLoop());
const SCEV *Step = getSCEVAtScope(AddRec->getOperand(1),
L->getParentLoop());
if (const SCEVConstant *StepC = dyn_cast<SCEVConstant>(Step)) {
// For now we handle only constant steps.
// First, handle unitary steps.
if (StepC->getValue()->equalsInt(1)) // 1*N = -Start (mod 2^BW), so:
return getNegativeSCEV(Start); // N = -Start (as unsigned)
if (StepC->getValue()->isAllOnesValue()) // -1*N = -Start (mod 2^BW), so:
return Start; // N = Start (as unsigned)
// Then, try to solve the above equation provided that Start is constant.
if (const SCEVConstant *StartC = dyn_cast<SCEVConstant>(Start))
return SolveLinEquationWithOverflow(StepC->getValue()->getValue(),
-StartC->getValue()->getValue(),
*this);
}
} else if (AddRec->isQuadratic() && AddRec->getType()->isIntegerTy()) {
// If this is a quadratic (3-term) AddRec {L,+,M,+,N}, find the roots of
// the quadratic equation to solve it.
std::pair<const SCEV *,const SCEV *> Roots = SolveQuadraticEquation(AddRec,
*this);
// If this is a quadratic (3-term) AddRec {L,+,M,+,N}, find the roots of
// the quadratic equation to solve it.
if (AddRec->isQuadratic() && AddRec->getType()->isIntegerTy()) {
std::pair<const SCEV *,const SCEV *> Roots =
SolveQuadraticEquation(AddRec, *this);
const SCEVConstant *R1 = dyn_cast<SCEVConstant>(Roots.first);
const SCEVConstant *R2 = dyn_cast<SCEVConstant>(Roots.second);
if (R1) {
if (R1 && R2) {
#if 0
dbgs() << "HFTZ: " << *V << " - sol#1: " << *R1
<< " sol#2: " << *R2 << "\n";
@ -4855,10 +4822,10 @@ ScalarEvolution::HowFarToZero(const SCEV *V, const Loop *L) {
// Pick the smallest positive root value.
if (ConstantInt *CB =
dyn_cast<ConstantInt>(ConstantExpr::getICmp(ICmpInst::ICMP_ULT,
R1->getValue(), R2->getValue()))) {
R1->getValue(), R2->getValue()))) {
if (CB->getZExtValue() == false)
std::swap(R1, R2); // R1 is the minimum root now.
// We can only use this value if the chrec ends up with an exact zero
// value at this index. When solving for "X*X != 5", for example, we
// should not accept a root of 2.
@ -4867,8 +4834,45 @@ ScalarEvolution::HowFarToZero(const SCEV *V, const Loop *L) {
return R1; // We found a quadratic root!
}
}
return getCouldNotCompute();
}
// Otherwise we can only handle this if it is affine.
if (!AddRec->isAffine())
return getCouldNotCompute();
// If this is an affine expression, the execution count of this branch is
// the minimum unsigned root of the following equation:
//
// Start + Step*N = 0 (mod 2^BW)
//
// equivalent to:
//
// Step*N = -Start (mod 2^BW)
//
// where BW is the common bit width of Start and Step.
// Get the initial value for the loop.
const SCEV *Start = getSCEVAtScope(AddRec->getStart(), L->getParentLoop());
const SCEV *Step = getSCEVAtScope(AddRec->getOperand(1), L->getParentLoop());
// For now we handle only constant steps.
const SCEVConstant *StepC = dyn_cast<SCEVConstant>(Step);
if (StepC == 0)
return getCouldNotCompute();
// First, handle unitary steps.
if (StepC->getValue()->equalsInt(1)) // 1*N = -Start (mod 2^BW), so:
return getNegativeSCEV(Start); // N = -Start (as unsigned)
if (StepC->getValue()->isAllOnesValue()) // -1*N = -Start (mod 2^BW), so:
return Start; // N = Start (as unsigned)
// Then, try to solve the above equation provided that Start is constant.
if (const SCEVConstant *StartC = dyn_cast<SCEVConstant>(Start))
return SolveLinEquationWithOverflow(StepC->getValue()->getValue(),
-StartC->getValue()->getValue(),
*this);
return getCouldNotCompute();
}