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InstSimplify: Fix a bug when INT_MIN is in a sdiv

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When INT_MIN is the numerator in a sdiv, we would not properly handle
overflow when calculating the bounds of possible values; abs(INT_MIN) is
not a meaningful number.

Instead, check and handle INT_MIN by reasoning that the largest value is
INT_MIN/-2 and the smallest value is INT_MIN.

This fixes PR20199.

git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@212307 91177308-0d34-0410-b5e6-96231b3b80d8
This commit is contained in:
David Majnemer
2014-07-04 00:23:39 +00:00
parent 5c58257d39
commit 00428878bb
2 changed files with 20 additions and 3 deletions
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12
lib/Analysis/InstructionSimplify.cpp
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@ -1948,9 +1948,15 @@ static Value *SimplifyICmpInst(unsigned Predicate, Value *LHS, Value *RHS,
if (!CI2->isZero())
Upper = NegOne.udiv(CI2->getValue()) + 1;
} else if (match(LHS, m_SDiv(m_ConstantInt(CI2), m_Value()))) {
// 'sdiv CI2, x' produces [-|CI2|, |CI2|].
Upper = CI2->getValue().abs() + 1;
Lower = (-Upper) + 1;
if (CI2->isMinSignedValue()) {
// 'sdiv INT_MIN, x' produces [INT_MIN, INT_MIN / -2].
Lower = CI2->getValue();
Upper = Lower.lshr(1) + 1;
} else {
// 'sdiv CI2, x' produces [-|CI2|, |CI2|].
Upper = CI2->getValue().abs() + 1;
Lower = (-Upper) + 1;
}
} else if (match(LHS, m_SDiv(m_Value(), m_ConstantInt(CI2)))) {
// 'sdiv x, CI2' produces [INT_MIN / CI2, INT_MAX / CI2].
APInt IntMin = APInt::getSignedMinValue(Width);