Declaring a member function with a typedef coming from a metafunction

Question

Consider the following code:

template <class>
struct Foo_s {
    using type = void();
};

template <class>
using Foo_u = void();

template <class T>
struct Bar {
             Foo_u<void>       foo1; // OK
    typename Foo_s<void>::type foo2; // OK
             Foo_u<T>          foo3; // OK
    typename Foo_s<T>::type    foo4; // Boom.
};

template struct Bar<void>;

The declaration of foo4 fails on GCC 7.2, Clang 5.0.0 and MSVC 19.10.25017.

GCC:

<source>: In instantiation of 'struct Bar<void>':
18 : <source>:18:17:   required from here
15 : <source>:15:29: error: field 'Bar<void>::foo4' invalidly declared function type
     typename Foo_s<T>::type foo4;
                             ^~~~

Clang:

15 : <source>:15:29: error: data member instantiated with function type 'typename Foo_s<void>::type' (aka 'void ()')
    typename Foo_s<T>::type foo4;
                            ^
18 : <source>:18:17: note: in instantiation of template class 'Bar<void>' requested here
template struct Bar<void>;
                ^

MSVC:

15 : <source>(15): error C2207: 'Bar<T>::foo4': a member of a class template cannot acquire a function type
18 : <source>(18): note: see reference to class template instantiation 'Bar<void>' being compiled

All of them seem to think that I am trying to declare a data member with a function type. As you can see, this only happens when the type is nested (not a using template), and is dependent on the parameter of the class. This looks like a bug, but the fact that for once all of these three compilers agree has me doubting.

Is this standard behaviour? If so, is there a rationale behind disallowing this, and is there a way to declare a member function whose type is computed with a metaprogram?

Yes, it's standard. See https://timsong-cpp.github.io/cppwp/temp#spec-8. Don't know the rationale or why `Foo_u` is accepted though. — Rakete1111, Nov 24 '17 at 13:23
Is it possible to define such a function? For example, how would you define `foo3`? — geza, Nov 24 '17 at 13:39
@geza you can only use this for declarations, definitions need the usual syntax.. — Quentin, Nov 24 '17 at 13:43
@geza [declaring pure virtual functions with a generated signature](https://stackoverflow.com/a/47460334/3233393) :) — Quentin, Nov 24 '17 at 13:54

Columbo · Accepted Answer · 2017-11-24T17:48:11.570

[temp.spec]/8:

If a function declaration acquired its function type through a dependent type without using the syntactic form of a function declarator, the program is ill-formed.

This makes your last line decidedly ill-formed. Which makes sense, since, as in other cases of dependent constructs, we don't want the meaning of a phrase to depend too strongly on template arguments.

However, [dcl.fct]/13:

A typedef of function type may be used to declare a function but shall not be used to define a function.

This makes your first three lines well-formed---the first two directly, and for the third one, note that

When a template-id refers to the specialization of an alias template, it is equivalent to the associated type obtained by substitution of its template-arguments for the template-parameters in the type-id of the alias template.

by [temp.alias].

Declaring a member function with a typedef coming from a metafunction

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