return false;
}
+/// \brief Return true if we can prove that:
+/// (mul LHS, RHS) === (mul nsw LHS, RHS)
+bool InstCombiner::WillNotOverflowSignedMul(Value *LHS, Value *RHS,
+ Instruction *CxtI) {
+ if (IntegerType *IT = dyn_cast<IntegerType>(LHS->getType())) {
+
+ // Multiplying n * m significant bits yields a result of n + m significant
+ // bits. If the total number of significant bits does not exceed the
+ // result bit width (minus 1), there is no overflow.
+ // This means if we have enough leading sign bits in the operands
+ // we can guarantee that the result does not overflow.
+ // Ref: "Hacker's Delight" by Henry Warren
+ unsigned BitWidth = IT->getBitWidth();
+
+ // Note that underestimating the number of sign bits gives a more
+ // conservative answer.
+ unsigned SignBits = ComputeNumSignBits(LHS, 0, CxtI) +
+ ComputeNumSignBits(RHS, 0, CxtI);
+
+ // First handle the easy case: if we have enough sign bits there's
+ // definitely no overflow.
+ if (SignBits > BitWidth + 1)
+ return true;
+
+ // There are two ambiguous cases where there can be no overflow:
+ // SignBits == BitWidth + 1 and
+ // SignBits == BitWidth
+ // The second case is difficult to check, therefore we only handle the
+ // first case.
+ if (SignBits == BitWidth + 1) {
+ // It overflows only when both arguments are negative and the true
+ // product is exactly the minimum negative number.
+ // E.g. mul i16 with 17 sign bits: 0xff00 * 0xff80 = 0x8000
+ // For simplicity we just check if at least one side is not negative.
+ bool LHSNonNegative, LHSNegative;
+ bool RHSNonNegative, RHSNegative;
+ ComputeSignBit(LHS, LHSNonNegative, LHSNegative, DL, 0, AT, CxtI, DT);
+ ComputeSignBit(RHS, RHSNonNegative, RHSNegative, DL, 0, AT, CxtI, DT);
+ if (LHSNonNegative || RHSNonNegative)
+ return true;
+ }
+ }
+ return false;
+}
+
// Checks if any operand is negative and we can convert add to sub.
// This function checks for following negative patterns
// ADD(XOR(OR(Z, NOT(C)), C)), 1) == NEG(AND(Z, C))