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-rw-r--r--llvm/lib/Transforms/Vectorize/SLPVectorizer.cpp15
1 files changed, 11 insertions, 4 deletions
diff --git a/llvm/lib/Transforms/Vectorize/SLPVectorizer.cpp b/llvm/lib/Transforms/Vectorize/SLPVectorizer.cpp
index 3e6f84ae7e0..9931c78fcbc 100644
--- a/llvm/lib/Transforms/Vectorize/SLPVectorizer.cpp
+++ b/llvm/lib/Transforms/Vectorize/SLPVectorizer.cpp
@@ -416,6 +416,9 @@ public:
private:
struct TreeEntry;
+ /// Checks if all users of \p I are the part of the vectorization tree.
+ bool areAllUsersVectorized(Instruction *I) const;
+
/// \returns the cost of the vectorizable entry.
int getEntryCost(TreeEntry *E);
@@ -1702,6 +1705,13 @@ bool BoUpSLP::canReuseExtract(ArrayRef<Value *> VL, Value *OpValue) const {
return true;
}
+bool BoUpSLP::areAllUsersVectorized(Instruction *I) const {
+ return I->hasOneUse() ||
+ std::all_of(I->user_begin(), I->user_end(), [this](User *U) {
+ return ScalarToTreeEntry.count(U) > 0;
+ });
+}
+
int BoUpSLP::getEntryCost(TreeEntry *E) {
ArrayRef<Value*> VL = E->Scalars;
@@ -1742,10 +1752,7 @@ int BoUpSLP::getEntryCost(TreeEntry *E) {
// If all users are going to be vectorized, instruction can be
// considered as dead.
// The same, if have only one user, it will be vectorized for sure.
- if (E->hasOneUse() ||
- std::all_of(E->user_begin(), E->user_end(), [this](User *U) {
- return ScalarToTreeEntry.count(U) > 0;
- }))
+ if (areAllUsersVectorized(E))
// Take credit for instruction that will become dead.
DeadCost +=
TTI->getVectorInstrCost(Instruction::ExtractElement, VecTy, i);
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