InstSimplify: Fix a bug when INT_MIN is in a sdiv

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.

llvm-svn: 212307

1 files changed, 9 insertions, 3 deletions

diff --git a/llvm/lib/Analysis/InstructionSimplify.cpp b/llvm/lib/Analysis/InstructionSimplify.cpp
index 31263bb7de8..bd42af15da3 100644
--- a/llvm/lib/Analysis/InstructionSimplify.cpp
+++ b/llvm/lib/Analysis/InstructionSimplify.cpp
@@ -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);