PR18746: If a constexpr function has a dependent return type and no return

statements, don't diagnose; the return type might end up being 'void'.

Patch by Rahul Jain! Tiny tweaks by me.

llvm-svn: 206929

1 files changed, 4 insertions, 2 deletions

diff --git a/clang/lib/Sema/SemaDeclCXX.cpp b/clang/lib/Sema/SemaDeclCXX.cpp
index 96c63601148..197518f99ef 100644
--- a/clang/lib/Sema/SemaDeclCXX.cpp
+++ b/clang/lib/Sema/SemaDeclCXX.cpp
@@ -1175,10 +1175,12 @@ bool Sema::CheckConstexprFunctionBody(const FunctionDecl *Dcl, Stmt *Body) {
   } else {
     if (ReturnStmts.empty()) {
       // C++1y doesn't require constexpr functions to contain a 'return'
-      // statement. We still do, unless the return type is void, because
+      // statement. We still do, unless the return type might be void, because
       // otherwise if there's no return statement, the function cannot
       // be used in a core constant expression.
-      bool OK = getLangOpts().CPlusPlus1y && Dcl->getReturnType()->isVoidType();
+      bool OK = getLangOpts().CPlusPlus1y &&
+                (Dcl->getReturnType()->isVoidType() ||
+                 Dcl->getReturnType()->isDependentType());
       Diag(Dcl->getLocation(),
            OK ? diag::warn_cxx11_compat_constexpr_body_no_return
               : diag::err_constexpr_body_no_return);