[llvm] [LSV] Merge contiguous chains across scalar types (PR #154069)

[Drew Kersnar via llvm-commits]

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@@ -480,49 +582,120 @@ bool Vectorizer::runOnPseudoBB(BasicBlock::iterator Begin,
   });
 
   bool Changed = false;
+  SmallVector<Chain> ContiguousSubChains;
+
   for (const auto &[EqClassKey, EqClass] :
-       collectEquivalenceClasses(Begin, End))
-    Changed |= runOnEquivalenceClass(EqClassKey, EqClass);
+       collectEquivalenceClasses(Begin, End)) {
 
-  return Changed;
-}
+    LLVM_DEBUG({
+      dbgs() << "LSV: Running on equivalence class of size " << EqClass.size()
+             << " keyed on " << EqClassKey << ":\n";
+      for (Instruction *I : EqClass)
+        dbgs() << "  " << *I << "\n";
+    });
 
-bool Vectorizer::runOnEquivalenceClass(const EqClassKey &EqClassKey,
-                                       ArrayRef<Instruction *> EqClass) {
-  bool Changed = false;
+    for (Chain &C : gatherChains(EqClass)) {
 
-  LLVM_DEBUG({
-    dbgs() << "LSV: Running on equivalence class of size " << EqClass.size()
-           << " keyed on " << EqClassKey << ":\n";
-    for (Instruction *I : EqClass)
-      dbgs() << "  " << *I << "\n";
-  });
+      // Split up the chain into increasingly smaller chains, until we can
+      // finally vectorize the chains.
+      //
+      // (Don't be scared by the depth of the loop nest here.  These operations
+      // are all at worst O(n lg n) in the number of instructions, and splitting
+      // chains doesn't change the number of instrs.  So the whole loop nest is
+      // O(n lg n).)
+      for (auto &C : splitChainByMayAliasInstrs(C)) {
+        for (auto &C : splitChainByContiguity(C)) {
+          ContiguousSubChains.emplace_back(C);
+        }
+      }
+    }
+  }
 
-  std::vector<Chain> Chains = gatherChains(EqClass);
-  LLVM_DEBUG(dbgs() << "LSV: Got " << Chains.size()
-                    << " nontrivial chains.\n";);
-  for (Chain &C : Chains)
-    Changed |= runOnChain(C);
-  return Changed;
-}
+  // Merge chains in reverse order, so that the first chain is the largest.
+  for (int I = ContiguousSubChains.size() - 1; I > 0; I--) {
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dakersnar wrote:

nit: it would nice if this whole for loop was inside a helper function, it would make reading the overall algorithm easier.

https://github.com/llvm/llvm-project/pull/154069