Combining invariant.group metadata bugfix

[reames]

Test case? Description of problem?

[Prazek]

I could use the same IR that I got the bug on, but it will highly depend on how GVN is combining the instructions. Unfortunatelly I don't know any better approach.

[llvm-commits]

If the GVN example is reasonable small, that would be fine. In general,
even complex scenarios can be reduced if care is taken to eliminate all
the redundant pieces.

This code is also exercised by simplify-cfg and instcombine. You might
have an easier time constructing a small case from one of those.

Philip

[Prazek]

Well, actually I have no idea why I made extra review. I forgot that http://reviews.llvm.org/D12992 is not yet upstream. I am closing this review. I will also add test case

lib/Transforms/Utils/Local.cpp

switch (Kind) {
         break;
       case LLVMContext::MD_invariant_group:
         // Preserve !invariant.group in K.
         break;
     }
   }
   // Set !invariant.group from J if J has it. If both instructions have it
   // then we will just pick it from J - even when they are different.
+  // Also make sure that K is load or store - f.e. combining bitcast with load
+  // could produce bitcast with invariant.group metadata, which is invalid.
   // FIXME: we should try to preserve both invariant.group md if they are
   // different, but right now instruction can only have one invariant.group.
   if (auto *JMD = J->getMetadata(LLVMContext::MD_invariant_group))
+    if (isa<LoadInst>(K) || isa<StoreInst>(K))
       K->setMetadata(LLVMContext::MD_invariant_group, JMD);
 
 }
 
 unsigned llvm::replaceDominatedUsesWith(Value *From, Value *To,
                                         DominatorTree &DT,
                                         const BasicBlockEdge &Root) {
   assert(From->getType() == To->getType());