[Sema] Don't skip function bodies with 'auto' without trailing return type

Summary:
Skipping them was clearly not intentional. It's impossible to
guarantee correctness if the bodies are skipped.
Also adds a test case for r327504, now that it does not produce
invalid errors that made the test fail.

Reviewers: aaron.ballman, sammccall, rsmith

Reviewed By: rsmith

Subscribers: rayglover-ibm, rwols, cfe-commits

Differential Revision: https://reviews.llvm.org/D44480

llvm-svn: 333538

1 files changed, 8 insertions, 2 deletions

diff --git a/clang/lib/Sema/SemaDecl.cpp b/clang/lib/Sema/SemaDecl.cpp
index e0be9efe966..cc44c034eae 100644
--- a/clang/lib/Sema/SemaDecl.cpp
+++ b/clang/lib/Sema/SemaDecl.cpp
@@ -12660,9 +12660,15 @@ bool Sema::canSkipFunctionBody(Decl *D) {
   // rest of the file.
   // We cannot skip the body of a function with an undeduced return type,
   // because any callers of that function need to know the type.
-  if (const FunctionDecl *FD = D->getAsFunction())
-    if (FD->isConstexpr() || FD->getReturnType()->isUndeducedType())
+  if (const FunctionDecl *FD = D->getAsFunction()) {
+    if (FD->isConstexpr())
       return false;
+    // We can't simply call Type::isUndeducedType here, because inside template
+    // auto can be deduced to a dependent type, which is not considered
+    // "undeduced".
+    if (FD->getReturnType()->getContainedDeducedType())
+      return false;
+  }
   return Consumer.shouldSkipFunctionBody(D);
 }