+static void ComputeMaskedBitsAddSub(bool Add, Value *Op0, Value *Op1, bool NSW,
+ const APInt &Mask,
+ APInt &KnownZero, APInt &KnownOne,
+ APInt &KnownZero2, APInt &KnownOne2,
+ const TargetData *TD, unsigned Depth) {
+ if (!Add) {
+ if (ConstantInt *CLHS = dyn_cast<ConstantInt>(Op0)) {
+ // We know that the top bits of C-X are clear if X contains less bits
+ // than C (i.e. no wrap-around can happen). For example, 20-X is
+ // positive if we can prove that X is >= 0 and < 16.
+ if (!CLHS->getValue().isNegative()) {
+ unsigned BitWidth = Mask.getBitWidth();
+ unsigned NLZ = (CLHS->getValue()+1).countLeadingZeros();
+ // NLZ can't be BitWidth with no sign bit
+ APInt MaskV = APInt::getHighBitsSet(BitWidth, NLZ+1);
+ llvm::ComputeMaskedBits(Op1, MaskV, KnownZero2, KnownOne2, TD, Depth+1);
+
+ // If all of the MaskV bits are known to be zero, then we know the
+ // output top bits are zero, because we now know that the output is
+ // from [0-C].
+ if ((KnownZero2 & MaskV) == MaskV) {
+ unsigned NLZ2 = CLHS->getValue().countLeadingZeros();
+ // Top bits known zero.
+ KnownZero = APInt::getHighBitsSet(BitWidth, NLZ2) & Mask;
+ }
+ }
+ }
+ }
+
+ unsigned BitWidth = Mask.getBitWidth();
+
+ // If one of the operands has trailing zeros, then the bits that the
+ // other operand has in those bit positions will be preserved in the
+ // result. For an add, this works with either operand. For a subtract,
+ // this only works if the known zeros are in the right operand.
+ APInt LHSKnownZero(BitWidth, 0), LHSKnownOne(BitWidth, 0);
+ APInt Mask2 = APInt::getLowBitsSet(BitWidth,
+ BitWidth - Mask.countLeadingZeros());
+ llvm::ComputeMaskedBits(Op0, Mask2, LHSKnownZero, LHSKnownOne, TD, Depth+1);
+ assert((LHSKnownZero & LHSKnownOne) == 0 &&
+ "Bits known to be one AND zero?");
+ unsigned LHSKnownZeroOut = LHSKnownZero.countTrailingOnes();
+
+ llvm::ComputeMaskedBits(Op1, Mask2, KnownZero2, KnownOne2, TD, Depth+1);
+ assert((KnownZero2 & KnownOne2) == 0 && "Bits known to be one AND zero?");
+ unsigned RHSKnownZeroOut = KnownZero2.countTrailingOnes();
+
+ // Determine which operand has more trailing zeros, and use that
+ // many bits from the other operand.
+ if (LHSKnownZeroOut > RHSKnownZeroOut) {
+ if (Add) {
+ APInt Mask = APInt::getLowBitsSet(BitWidth, LHSKnownZeroOut);
+ KnownZero |= KnownZero2 & Mask;
+ KnownOne |= KnownOne2 & Mask;
+ } else {
+ // If the known zeros are in the left operand for a subtract,
+ // fall back to the minimum known zeros in both operands.
+ KnownZero |= APInt::getLowBitsSet(BitWidth,
+ std::min(LHSKnownZeroOut,
+ RHSKnownZeroOut));
+ }
+ } else if (RHSKnownZeroOut >= LHSKnownZeroOut) {
+ APInt Mask = APInt::getLowBitsSet(BitWidth, RHSKnownZeroOut);
+ KnownZero |= LHSKnownZero & Mask;
+ KnownOne |= LHSKnownOne & Mask;
+ }
+
+ // Are we still trying to solve for the sign bit?
+ if (Mask.isNegative() && !KnownZero.isNegative() && !KnownOne.isNegative()) {
+ if (NSW) {
+ if (Add) {
+ // Adding two positive numbers can't wrap into negative
+ if (LHSKnownZero.isNegative() && KnownZero2.isNegative())
+ KnownZero |= APInt::getSignBit(BitWidth);
+ // and adding two negative numbers can't wrap into positive.
+ else if (LHSKnownOne.isNegative() && KnownOne2.isNegative())
+ KnownOne |= APInt::getSignBit(BitWidth);
+ } else {
+ // Subtracting a negative number from a positive one can't wrap
+ if (LHSKnownZero.isNegative() && KnownOne2.isNegative())
+ KnownZero |= APInt::getSignBit(BitWidth);
+ // neither can subtracting a positive number from a negative one.
+ else if (LHSKnownOne.isNegative() && KnownZero2.isNegative())
+ KnownOne |= APInt::getSignBit(BitWidth);
+ }
+ }
+ }
+}
+