[llvm] [FuncAttrs] Relax norecurse attribute inference (PR #139943)

[David Sherwood via llvm-commits]

================
@@ -2322,8 +2343,39 @@ PreservedAnalyses PostOrderFunctionAttrsPass::run(LazyCallGraph::SCC &C,
     Functions.push_back(&N.getFunction());
   }
 
-  auto ChangedFunctions =
-      deriveAttrsInPostOrder(Functions, AARGetter, ArgAttrsOnly);
+  bool NoFunctionsAddressIsTaken = false;
+  // Check if any function in the whole program has its address taken or has
+  // potentially external linkage.
+  // We use this information when inferring norecurse attribute: If there is
+  // no function whose address is taken and all functions have internal
+  // linkage, there is no path for a callback to any user function.
+  if (IsLTOPostLink || ForceLTOFuncAttrs) {
+    bool AnyFunctionsAddressIsTaken = false;
+    // Get the parent Module of the Function
+    Module &M = *C.begin()->getFunction().getParent();
+    for (Function &F : M) {
+      // We only care about functions defined in user program whose addresses
+      // escape, making them potential callback targets.
+      if (F.isDeclaration())
+        continue;
+
+      // If the function is already marked as norecurse, this should not block
+      // norecurse inference even though it may have external linkage.
+      // For ex: main() in C++.
+      if (F.doesNotRecurse())
+        continue;
----------------
david-arm wrote:

Would the call graph actually help in all cases though? For the example you gave above, i.e.

```
/*external linkage*/void foo() {
  a(0);
}

static void a(int v) {
  if (v)
    bar(); /*external function*/
  return;
}

extern void bar();
```

the call graph will still say that foo is potentially recursive I think. Unless we have a call graph that takes control flow into account based on function arguments? I agree with you, that if we cannot use `norecurse` to infer `norecurse` then it sounds like the whole RPO pass is unsound.

Is it possible that the definition of `norecurse` in the LangRef refers to the call graph, which would simply show that foo() calls a(), which calls bar(), which could call foo()? In this definition adding `norecurse` to `foo` would be incorrect. 

https://github.com/llvm/llvm-project/pull/139943