- return 0;
-}
-
-/// FactorizeBinOp - Simplify "LHS Opcode RHS" by factorizing out a common term
-/// using the operation OpCodeToExtract. For example, when Opcode is Add and
-/// OpCodeToExtract is Mul then this tries to turn "(A*B)+(A*C)" into "A*(B+C)".
-/// Returns the simplified value, or null if no simplification was performed.
-static Value *FactorizeBinOp(unsigned Opcode, Value *LHS, Value *RHS,
- unsigned OpcToExtract, const Query &Q,
- unsigned MaxRecurse) {
- Instruction::BinaryOps OpcodeToExtract = (Instruction::BinaryOps)OpcToExtract;
- // Recursion is always used, so bail out at once if we already hit the limit.
- if (!MaxRecurse--)
- return 0;
-
- BinaryOperator *Op0 = dyn_cast<BinaryOperator>(LHS);
- BinaryOperator *Op1 = dyn_cast<BinaryOperator>(RHS);
-
- if (!Op0 || Op0->getOpcode() != OpcodeToExtract ||
- !Op1 || Op1->getOpcode() != OpcodeToExtract)
- return 0;
-
- // The expression has the form "(A op' B) op (C op' D)".
- Value *A = Op0->getOperand(0), *B = Op0->getOperand(1);
- Value *C = Op1->getOperand(0), *D = Op1->getOperand(1);
-
- // Use left distributivity, i.e. "X op' (Y op Z) = (X op' Y) op (X op' Z)".
- // Does the instruction have the form "(A op' B) op (A op' D)" or, in the
- // commutative case, "(A op' B) op (C op' A)"?
- if (A == C || (Instruction::isCommutative(OpcodeToExtract) && A == D)) {
- Value *DD = A == C ? D : C;
- // Form "A op' (B op DD)" if it simplifies completely.
- // Does "B op DD" simplify?
- if (Value *V = SimplifyBinOp(Opcode, B, DD, Q, MaxRecurse)) {
- // It does! Return "A op' V" if it simplifies or is already available.
- // If V equals B then "A op' V" is just the LHS. If V equals DD then
- // "A op' V" is just the RHS.
- if (V == B || V == DD) {
- ++NumFactor;
- return V == B ? LHS : RHS;
- }
- // Otherwise return "A op' V" if it simplifies.
- if (Value *W = SimplifyBinOp(OpcodeToExtract, A, V, Q, MaxRecurse)) {
- ++NumFactor;
- return W;
- }
- }
- }
-
- // Use right distributivity, i.e. "(X op Y) op' Z = (X op' Z) op (Y op' Z)".
- // Does the instruction have the form "(A op' B) op (C op' B)" or, in the
- // commutative case, "(A op' B) op (B op' D)"?
- if (B == D || (Instruction::isCommutative(OpcodeToExtract) && B == C)) {
- Value *CC = B == D ? C : D;
- // Form "(A op CC) op' B" if it simplifies completely..
- // Does "A op CC" simplify?
- if (Value *V = SimplifyBinOp(Opcode, A, CC, Q, MaxRecurse)) {
- // It does! Return "V op' B" if it simplifies or is already available.
- // If V equals A then "V op' B" is just the LHS. If V equals CC then
- // "V op' B" is just the RHS.
- if (V == A || V == CC) {
- ++NumFactor;
- return V == A ? LHS : RHS;
- }
- // Otherwise return "V op' B" if it simplifies.
- if (Value *W = SimplifyBinOp(OpcodeToExtract, V, B, Q, MaxRecurse)) {
- ++NumFactor;
- return W;
- }
- }
- }
-
- return 0;