Fix PR2088. Use modulo linear equation solver to compute loop iteration

count.


git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@53810 91177308-0d34-0410-b5e6-96231b3b80d8
This commit is contained in:
Wojciech Matyjewicz 2008-07-20 15:55:14 +00:00
parent b8c44d3aac
commit de0f2382e4
5 changed files with 101 additions and 23 deletions

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@ -78,6 +78,8 @@
#include "llvm/Support/MathExtras.h"
#include "llvm/Support/Streams.h"
#include "llvm/ADT/Statistic.h"
//TMP:
#include "llvm/Support/Debug.h"
#include <ostream>
#include <algorithm>
#include <cmath>
@ -2461,6 +2463,53 @@ SCEVHandle ScalarEvolutionsImpl::getSCEVAtScope(SCEV *V, const Loop *L) {
return UnknownValue;
}
/// SolveLinEquationWithOverflow - Finds the minimum unsigned root of the
/// following equation:
///
/// A * X = B (mod N)
///
/// where N = 2^BW and BW is the common bit width of A and B. The signedness of
/// A and B isn't important.
///
/// If the equation does not have a solution, SCEVCouldNotCompute is returned.
static SCEVHandle SolveLinEquationWithOverflow(const APInt &A, const APInt &B,
ScalarEvolution &SE) {
uint32_t BW = A.getBitWidth();
assert(BW == B.getBitWidth() && "Bit widths must be the same.");
assert(A != 0 && "A must be non-zero.");
// 1. D = gcd(A, N)
//
// The gcd of A and N may have only one prime factor: 2. The number of
// trailing zeros in A is its multiplicity
uint32_t Mult2 = A.countTrailingZeros();
// D = 2^Mult2
// 2. Check if B is divisible by D.
//
// B is divisible by D if and only if the multiplicity of prime factor 2 for B
// is not less than multiplicity of this prime factor for D.
if (B.countTrailingZeros() < Mult2)
return new SCEVCouldNotCompute();
// 3. Compute I: the multiplicative inverse of (A / D) in arithmetic
// modulo (N / D).
//
// (N / D) may need BW+1 bits in its representation. Hence, we'll use this
// bit width during computations.
APInt AD = A.lshr(Mult2).zext(BW + 1); // AD = A / D
APInt Mod(BW + 1, 0);
Mod.set(BW - Mult2); // Mod = N / D
APInt I = AD.multiplicativeInverse(Mod);
// 4. Compute the minimum unsigned root of the equation:
// I * (B / D) mod (N / D)
APInt Result = (I * B.lshr(Mult2).zext(BW + 1)).urem(Mod);
// The result is guaranteed to be less than 2^BW so we may truncate it to BW
// bits.
return SE.getConstant(Result.trunc(BW));
}
/// SolveQuadraticEquation - Find the roots of the quadratic equation for the
/// given quadratic chrec {L,+,M,+,N}. This returns either the two roots (which
@ -2533,36 +2582,36 @@ SCEVHandle ScalarEvolutionsImpl::HowFarToZero(SCEV *V, const Loop *L) {
return UnknownValue;
if (AddRec->isAffine()) {
// If this is an affine expression the execution count of this branch is
// equal to:
// If this is an affine expression, the execution count of this branch is
// the minimum unsigned root of the following equation:
//
// (0 - Start/Step) iff Start % Step == 0
// 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.
SCEVHandle Start = getSCEVAtScope(AddRec->getStart(), L->getParentLoop());
if (isa<SCEVCouldNotCompute>(Start)) return UnknownValue;
SCEVHandle Step = AddRec->getOperand(1);
Step = getSCEVAtScope(Step, L->getParentLoop());
SCEVHandle Step = getSCEVAtScope(AddRec->getOperand(1), L->getParentLoop());
// Figure out if Start % Step == 0.
// FIXME: We should add DivExpr and RemExpr operations to our AST.
if (SCEVConstant *StepC = dyn_cast<SCEVConstant>(Step)) {
if (StepC->getValue()->equalsInt(1)) // N % 1 == 0
return SE.getNegativeSCEV(Start); // 0 - Start/1 == -Start
if (StepC->getValue()->isAllOnesValue()) // N % -1 == 0
return Start; // 0 - Start/-1 == Start
// For now we handle only constant steps.
// Check to see if Start is divisible by SC with no remainder.
if (SCEVConstant *StartC = dyn_cast<SCEVConstant>(Start)) {
ConstantInt *StartCC = StartC->getValue();
Constant *StartNegC = ConstantExpr::getNeg(StartCC);
Constant *Rem = ConstantExpr::getURem(StartNegC, StepC->getValue());
if (Rem->isNullValue()) {
Constant *Result = ConstantExpr::getUDiv(StartNegC,StepC->getValue());
return SE.getUnknown(Result);
}
}
// First, handle unitary steps.
if (StepC->getValue()->equalsInt(1)) // 1*N = -Start (mod 2^BW), so:
return SE.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 (SCEVConstant *StartC = dyn_cast<SCEVConstant>(Start))
return SolveLinEquationWithOverflow(StepC->getValue()->getValue(),
-StartC->getValue()->getValue(),SE);
}
} else if (AddRec->isQuadratic() && AddRec->getType()->isInteger()) {
// If this is a quadratic (3-term) AddRec {L,+,M,+,N}, find the roots of

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@ -1466,7 +1466,7 @@ APInt APInt::multiplicativeInverse(const APInt& modulo) const {
// The next-to-last t is the multiplicative inverse. However, we are
// interested in a positive inverse. Calcuate a positive one from a negative
// one if necessary. A simple addition of the modulo suffices because
// abs(t[i]) is known to less than *this/2 (see the link above).
// abs(t[i]) is known to be less than *this/2 (see the link above).
return t[i].isNegative() ? t[i] + modulo : t[i];
}

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@ -1,7 +1,6 @@
; RUN: llvm-as < %s | opt -analyze -scalar-evolution \
; RUN: -scalar-evolution-max-iterations=0 | grep {61 iterations}
; PR2364
; XFAIL: *
define i32 @func_6() nounwind {
entry:

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@ -0,0 +1,15 @@
; RUN: llvm-as < %s | opt -analyze -scalar-evolution \
; RUN: -scalar-evolution-max-iterations=0 | grep Unpredictable
; PR2088
define void @fun() {
entry:
br label %loop
loop:
%i = phi i8 [ 0, %entry ], [ %i.next, %loop ]
%i.next = add i8 %i, 4
%cond = icmp ne i8 %i.next, 6
br i1 %cond, label %loop, label %exit
exit:
ret void
}

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@ -0,0 +1,15 @@
; RUN: llvm-as < %s | opt -analyze -scalar-evolution \
; RUN: -scalar-evolution-max-iterations=0 | grep {113 iterations}
; PR2088
define void @fun() {
entry:
br label %loop
loop:
%i = phi i8 [ 0, %entry ], [ %i.next, %loop ]
%i.next = add i8 %i, 18
%cond = icmp ne i8 %i.next, 4
br i1 %cond, label %loop, label %exit
exit:
ret void
}