C++ using declaration with typename in inheriting-constructors

Question

While reading this question, I found a strange point:

template <typename T>
class Subclass : public Baseclass<T>
{
public:
    using typename Baseclass<T>::Baseclass;
    //    ^^^^^^^^
};

Since typename, Baseclass<T>::Baseclass should be injected class name, not a constructor. As far as I know, it's the same case as this:

template <typename T>
class Base
{
public:
    typedef short some_type;
};

template <typename T>
class Sub : public Base<T>
{
public:
    using typename Base<T>::some_type;
};

To make sure, I wrote a test code.

#include <iostream>

template <typename T>
class Base
{
public:
    Base() { std::cout << "A::A()\n"; }
    Base(int) { std::cout << "A::A(int)\n"; }
    Base(const char *) { std::cout << "A::A(const char *)\n"; }
};

template <typename T>
class Sub : public Base<T>
{
    using typename Base<T>::Base;
};

int main()
{
    Sub<char> s1;
    Sub<char> s2(3);
    Sub<char> s3("asdf");
}

However, it runs on gcc 4.8.3.

$ g++ -std=c++1y -Wall -Wextra -Werror -pedantic test.cpp -o test && ./test
A::A()
A::A(int)
A::A(const char *)

It also runs without typename.

$ cat test.cpp
...
    using Base<T>::Base;
...

$ g++ -std=c++1y -Wall -Wextra -Werror -pedantic test.cpp -o test && ./test
A::A()
A::A(int)
A::A(const char *)

Why did I get these results? What did I miss?

Answer 1

The standard is pretty clear about this ([namespace.udecl]/1)

using-declaration:

using typename_opt nested-name-specifier unqualified-id ;

The typename keyword is therefore an optional part of a using declaration that may appear even for using declarations of non-types. The following code should therefore be standard conformant:

template < typename T > class Base {
  protected:
    typedef T Ttype;
    Ttype member;

  public:
    Base() {
        std::cout << "A::A()\n";
    }
    Base(int) {
        std::cout << "A::A(int)\n";
    }
    Base(const char *) {
        std::cout << "A::A(const char *)\n";
    }

  protected:
    void memfunc(void) {
        std::cout << "A::memfunc(void)\n";
    }
};

template< typename T >
struct SubNoTypename : protected Base< T > {
    using Base< T >::Base;
    using Base< T >::member;
    using Base< T >::memfunc;
    using Base< T >::Ttype;  // n.b. no error in clang++
};

template< typename T >
struct SubTypename : protected Base< T > {
    using typename Base< T >::Base;    // error in clang++
    using typename Base< T >::member;  // error in clang++
    using typename Base< T >::memfunc; // error in clang++
    using typename Base< T >::Ttype;
};

Both SubNoTypename and SubTypename are recognized as standard conformant by gcc. On the other hand clang++ complains in SubTypename about malplaced typename keywords. However, this is not even consistent, because it should then complain about a missing typename in using Base< T >::Ttype;. This clearly is a clang bug.

---

Edit The typename keyword is also allowed if the base class is no template class, a place where ordinarily you would never expect this keyword to be valid:

class BaseNoTemplate {
  protected:
    typedef T Ttype;
    Ttype member;

  public:
    BaseNoTemplate() {
        std::cout << "A::A()\n";
    }
    BaseNoTemplate(const char *) {
        std::cout << "A::A(const char *)\n";
    }

    void memfunc(void) {
        std::cout << "A::memfunc(void)\n";
    }
};

struct SubNoTemplateNoTypename : protected BaseNoTemplate {
    using BaseNoTemplate::BaseNoTemplate;
    using BaseNoTemplate::member;
    using BaseNoTemplate::memfunc;
    using BaseNoTemplate::Ttype;
};

struct SubNoTemplateTypename : protected BaseNoTemplate {
    using typename BaseNoTemplate::BaseNoTemplate; // error in clang++
    using typename BaseNoTemplate::member;  // error in clang++
    using typename BaseNoTemplate::memfunc; // error in clang++
    using typename BaseNoTemplate::Ttype;   // n.b. no error in clang++
};

Answer 2

Sorry, why do you want using, why not just typedef?

template <typename T>
class Sub : public Base<T>
{
    typedef Base<T> Base;
};