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[SCEV] Simplify SolveLinEquationWithOverflow a bit.

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Cleanup in preparation for generalizing it.

llvm-svn: 291808
This commit is contained in:
Eli Friedman 2017-01-12 20:21:00 +00:00
parent 3a4f22c55f
commit e1ae4eef15
1 changed files with 8 additions and 7 deletions
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15
lib/Analysis/ScalarEvolution.cpp
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@ -7032,20 +7032,21 @@ static const SCEV *SolveLinEquationWithOverflow(const APInt &A, const APInt &B,
// 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.
// If D == 1, (N / D) == N == 2^BW, so we need one extra bit to represent
// (N / D) in general. The inverse itself always fits into BW bits, though,
// so we immediately truncate it.
APInt AD = A.lshr(Mult2).zext(BW + 1); // AD = A / D
APInt Mod(BW + 1, 0);
Mod.setBit(BW - Mult2); // Mod = N / D
APInt I = AD.multiplicativeInverse(Mod);
APInt I = AD.multiplicativeInverse(Mod).trunc(BW);
// 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);
// To simplify the computation, we factor out the divide by D:
// (I * B mod N) / D
APInt Result = (I * B).lshr(Mult2);
// The result is guaranteed to be less than 2^BW so we may truncate it to BW
// bits.
return SE.getConstant(Result.trunc(BW));
return SE.getConstant(Result);
}
/// Find the roots of the quadratic equation for the given quadratic chrec