Explicit specialization in non-namespace scope does not compile in GCC

Question

The following code compiles in Clang but does not in GCC:

template<typename T>
struct Widget
{
    template<typename U>
    void foo(U)
    {
    }

    template<>
    void foo(int*)
    {
    }
};

According to the C++ standard ([temp.expl.spec], paragraph 2):

An explicit specialization may be declared in any scope in which the corresponding primary template may be defined

Is this a bug in GCC and if so how can I find it in its bug tracker?

This is GCC's output:

prog.cc:13:14: error: explicit specialization in non-namespace scope 'struct Widget<T>'
     template<>
              ^

I'm using GCC HEAD 8.0.1, with -std=c++2a .

Answer 1

This should be a GCC bug. Full specialization should be allowed in any scope, including in class definition.

According to CWG 727 , [temp.expl.spec] paragraph 2 was changed from

(emphasis mine)

An explicit specialization shall be declared in a namespace enclosing the specialized template. An explicit specialization whose declarator-id or class-head-name is not qualified shall be declared in the nearest enclosing namespace of the template, or, if the namespace is inline (10.3.1 [namespace.def]), any namespace from its enclosing namespace set. Such a declaration may also be a definition. If the declaration is not a definition, the specialization may be defined later (10.3.1.2 [namespace.memdef]).

to

(emphasis mine)

An explicit specialization may be declared in any scope in which the corresponding primary template may be defined (10.3.1.2 [namespace.memdef], 12.2 [class.mem], 17.6.2 [temp.mem]).

It seems GCC fails to follow this.

EDIT

I have reported the issue as Bug 85282 .

Answer 2

If somebody is looking for a workaround till this is fixed in gcc:

It is possible to use std::is_same_v and if constexpr godbolt

template<typename T>
struct Widget
{
    template<typename U>
    void foo(U)
    {
        if constexpr (std::is_same_v<U, int*>) {
            std::cout << "int*\n";
        }
        std::cout << "U\n";
    }

};