The 'constexpr implies const' rule for non-static member functions is gone in
C++1y, so stop adding the 'const' there. Provide a compatibility warning for
code relying on this in C++11, with a fix-it hint. Update our lazily-written
tests to add the const, except for those ones which were testing our
implementation of this rule.
llvm-svn: 179969
diff --git a/clang/test/SemaCXX/enum-unscoped-nonexistent.cpp b/clang/test/SemaCXX/enum-unscoped-nonexistent.cpp
index d49800c..e9da38f 100644
--- a/clang/test/SemaCXX/enum-unscoped-nonexistent.cpp
+++ b/clang/test/SemaCXX/enum-unscoped-nonexistent.cpp
@@ -5,8 +5,8 @@
};
template<typename T> struct S : Base {
enum E : int;
- constexpr int f();
- constexpr int g(); // expected-note {{declared here}}
+ constexpr int f() const;
+ constexpr int g() const; // expected-note {{declared here}}
void h();
};
template<> enum S<char>::E : int {}; // expected-note {{enum 'S<char>::E' was explicitly specialized here}}
@@ -16,13 +16,13 @@
// The unqualified-id here names a member of the non-dependent base class Base
// and not the injected enumerator name 'a' from the specialization.
-template<typename T> constexpr int S<T>::f() { return a; }
+template<typename T> constexpr int S<T>::f() const { return a; }
static_assert(S<char>().f() == 1, "");
static_assert(S<int>().f() == 1, "");
// The unqualified-id here names a member of the current instantiation, which
// bizarrely might not exist in some instantiations.
-template<typename T> constexpr int S<T>::g() { return b; } // expected-error {{enumerator 'b' does not exist in instantiation of 'S<char>'}}
+template<typename T> constexpr int S<T>::g() const { return b; } // expected-error {{enumerator 'b' does not exist in instantiation of 'S<char>'}}
static_assert(S<char>().g() == 1, ""); // expected-note {{here}} expected-error {{not an integral constant expression}} expected-note {{undefined}}
static_assert(S<short>().g() == 2, "");
static_assert(S<long>().g() == 8, "");