This PR was initially accepts-invalid, but I think it's actually valid
C++20 code. My reasoning is that in C++20 we no longer require the
declaration of operator== (#if-defed in the test), because C++20's
[temp.names]/2 says "A name is also considered to refer to a template
if it is an unqualified-id followed by a < and name lookup either finds
one or more functions or finds nothing." so when we're parsing
constexpr friend bool operator==<T>(T lhs, const Foo& rhs);
we treat "operator==" as a template name, because name lookup of
"operator==" found nothing and we have an operator-function-id, which is
an unqualified-id, and it's followed by a <. So the declaration isn't
needed to treat "operator==<T>" as a template-id.
PR c++/93807
* g++.dg/cpp2a/fn-template20.C: New test.
+2020-04-22 Marek Polacek <polacek@redhat.com>
+
+ PR c++/93807
+ * g++.dg/cpp2a/fn-template20.C: New test.
+
2020-04-22 Duan bo <duanbo3@huawei.com>
PR testsuite/94712
--- /dev/null
+// PR c++/93807
+// { dg-do compile { target c++11 } }
+
+// In C++17, we need the following declaration to treat operator== as
+// a template name. In C++20, this is handled by [temp.names]/2.
+#if __cplusplus <= 201703L
+template <typename T>
+class Foo;
+template <typename T>
+constexpr bool operator==(T lhs, const Foo<T>& rhs);
+#endif
+
+template <typename T>
+class Foo {
+public:
+ constexpr Foo(T k) : mK(k) {}
+
+ constexpr friend bool operator==<T>(T lhs, const Foo& rhs);
+private:
+ T mK;
+};
+
+template <typename T>
+constexpr bool
+operator==(T lhs, const Foo<T>& rhs)
+{
+ return lhs == rhs.mK;
+}
+
+int
+main ()
+{
+ return 1 == Foo<int>(1) ? 0 : 1;
+}