[VPlan] Improve recursivelyDeleteDeadRecipes (NFC)

[artagnon]

Combine the worklist and the seen containers into a single container that acts as both the worklist and the seen-tracker.

[llvmbot]

@llvm/pr-subscribers-llvm-transforms

@llvm/pr-subscribers-vectorizers

Author: Ramkumar Ramachandra (artagnon)

Changes

Combine the worklist and the seen containers into a single container that acts as both the worklist and the seen-tracker.

---

Full diff: https://github.com/llvm/llvm-project/pull/162815.diff

1 Files Affected:

• (modified) llvm/lib/Transforms/Vectorize/VPlanTransforms.cpp (+6-11)

diff --git a/llvm/lib/Transforms/Vectorize/VPlanTransforms.cpp b/llvm/lib/Transforms/Vectorize/VPlanTransforms.cpp
index c8a2d84a535d3..b73a526f142b5 100644
--- a/llvm/lib/Transforms/Vectorize/VPlanTransforms.cpp
+++ b/llvm/lib/Transforms/Vectorize/VPlanTransforms.cpp
@@ -924,20 +924,15 @@ static void removeRedundantExpandSCEVRecipes(VPlan &Plan) {
 }
 
 static void recursivelyDeleteDeadRecipes(VPValue *V) {
-  SmallVector<VPValue *> WorkList;
-  SmallPtrSet<VPValue *, 8> Seen;
-  WorkList.push_back(V);
+  SetVector<VPValue *> WorkList;
+  WorkList.insert(V);
 
-  while (!WorkList.empty()) {
-    VPValue *Cur = WorkList.pop_back_val();
-    if (!Seen.insert(Cur).second)
-      continue;
+  for (unsigned I = 0; I < WorkList.size(); ++I) {
+    VPValue *Cur = WorkList[I];
     VPRecipeBase *R = Cur->getDefiningRecipe();
-    if (!R)
-      continue;
-    if (!isDeadRecipe(*R))
+    if (!R || !isDeadRecipe(*R))
       continue;
-    WorkList.append(R->op_begin(), R->op_end());
+    WorkList.insert_range(R->operands());
     R->eraseFromParent();
   }
 }

[fhahn]

Hmm, not sure if this is a strict improvement, it has slightly fewer lines, but SetVector probably is a bit less efficient than separate set + vector, as SetVector will continously grow?

[artagnon]

Hmm, not sure if this is a strict improvement, it has slightly fewer lines, but SetVector probably is a bit less efficient than separate set + vector, as SetVector will continously grow?

Hm, wouldn't it be exactly the size of the set, as we previously had to insert every element into the set to check for duplicates?