================
@@ -911,6 +916,61 @@ static void setLiveRoot(ModuleSummaryIndex &Index,
StringRef Name) {
Summary->setLive(true);
}
+static bool hasNonVTableUsers(const User *U, CXXABI *ABI) {
+ LLVM_DEBUG(dbgs() << "Check if " << *U << "has vtable users\n");
+ if (isa<Instruction>(U)) {
+ // If the type info is used in dynamic_cast or exception handling,
+ // its user must be the instruction.
+ return true;
+ }
+
+ // The virtual table type is either a struct of arrays. For example:
+ // @vtable = constant { [3 x ptr] } { [3 x ptr] [ ptr null, ptr @rtti, ptr
@vf] }
+ //
+ // In this case, the user of @rtti is an anonymous ConstantArray.
+ // Therefore, if the user of the type information is anonymous,
+ // we need to perform a depth-first search (DFS) to locate its named users.
+ //
+ // And we also need to iterate its users if the current user is the type
+ // info global variable itself.
+ StringRef Name = U->getName();
+ if (Name.empty() || ABI->isTypeInfo(Name)) {
+ for (const User *It : U->users())
+ if (hasNonVTableUsers(It, ABI))
+ return true;
+ return false;
+ }
+
+ if (!ABI->isVTable(Name))
+ return true;
----------------
teresajohnson wrote:
But here we are not in the thin link, so why are we not able to obtain the
vtable GV?
https://github.com/llvm/llvm-project/pull/126336
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