+ if (const SCEVAddRecExpr *AR = dyn_cast<SCEVAddRecExpr>(S)) {
+ // An addrec. This is the interesting part.
+ SmallVector<const SCEV *, 8> Operands;
+ const Loop *L = AR->getLoop();
+ // The addrec conceptually uses its operands at loop entry.
+ Instruction *LUser = L->getHeader()->begin();
+ // Transform each operand.
+ for (SCEVNAryExpr::op_iterator I = AR->op_begin(), E = AR->op_end();
+ I != E; ++I) {
+ Operands.push_back(TransformSubExpr(*I, LUser, nullptr));
+ }
+ // Conservatively use AnyWrap until/unless we need FlagNW.
+ const SCEV *Result = SE.getAddRecExpr(Operands, L, SCEV::FlagAnyWrap);
+ switch (Kind) {
+ case NormalizeAutodetect:
+ // Normalize this SCEV by subtracting the expression for the final step.
+ // We only allow affine AddRecs to be normalized, otherwise we would not
+ // be able to correctly denormalize.
+ // e.g. {1,+,3,+,2} == {-2,+,1,+,2} + {3,+,2}
+ // Normalized form: {-2,+,1,+,2}
+ // Denormalized form: {1,+,3,+,2}
+ //
+ // However, denormalization would use the a different step expression than
+ // normalization (see getPostIncExpr), generating the wrong final
+ // expression: {-2,+,1,+,2} + {1,+,2} => {-1,+,3,+,2}
+ if (AR->isAffine() &&
+ IVUseShouldUsePostIncValue(User, OperandValToReplace, L, &DT)) {
+ const SCEV *TransformedStep =
+ TransformSubExpr(AR->getStepRecurrence(SE),
+ User, OperandValToReplace);
+ Result = SE.getMinusSCEV(Result, TransformedStep);
+ Loops.insert(L);
+ }
+#if 0
+ // This assert is conceptually correct, but ScalarEvolution currently
+ // sometimes fails to canonicalize two equal SCEVs to exactly the same
+ // form. It's possibly a pessimization when this happens, but it isn't a
+ // correctness problem, so disable this assert for now.
+ assert(S == TransformSubExpr(Result, User, OperandValToReplace) &&
+ "SCEV normalization is not invertible!");
+#endif
+ break;
+ case Normalize:
+ // We want to normalize step expression, because otherwise we might not be
+ // able to denormalize to the original expression.
+ //
+ // Here is an example what will happen if we don't normalize step:
+ // ORIGINAL ISE:
+ // {(100 /u {1,+,1}<%bb16>),+,(100 /u {1,+,1}<%bb16>)}<%bb25>
+ // NORMALIZED ISE:
+ // {((-1 * (100 /u {1,+,1}<%bb16>)) + (100 /u {0,+,1}<%bb16>)),+,
+ // (100 /u {0,+,1}<%bb16>)}<%bb25>
+ // DENORMALIZED BACK ISE:
+ // {((2 * (100 /u {1,+,1}<%bb16>)) + (-1 * (100 /u {2,+,1}<%bb16>))),+,
+ // (100 /u {1,+,1}<%bb16>)}<%bb25>
+ // Note that the initial value changes after normalization +
+ // denormalization, which isn't correct.
+ if (Loops.count(L)) {
+ const SCEV *TransformedStep =
+ TransformSubExpr(AR->getStepRecurrence(SE),
+ User, OperandValToReplace);
+ Result = SE.getMinusSCEV(Result, TransformedStep);
+ }
+#if 0
+ // See the comment on the assert above.
+ assert(S == TransformSubExpr(Result, User, OperandValToReplace) &&
+ "SCEV normalization is not invertible!");
+#endif
+ break;
+ case Denormalize:
+ // Here we want to normalize step expressions for the same reasons, as
+ // stated above.
+ if (Loops.count(L)) {
+ const SCEV *TransformedStep =
+ TransformSubExpr(AR->getStepRecurrence(SE),
+ User, OperandValToReplace);
+ Result = SE.getAddExpr(Result, TransformedStep);
+ }
+ break;
+ }
+ return Result;
+ }
+