在类体外定义的成员函数上的sfinae

Question

Kind of a continuation from my previous question . What I've got is a bunch of functions that form a sfinae depencency chain like so (let "A -> B" notation mean that existence of A depends on existence of B):

S::f_base -> S::f -> ns::f_ -> f -> T::f

where T is the template argument. It's implemented like this :

#include <utility>

struct S;

template <typename T>
auto f(S& s, T const& t) -> decltype(t.f(s), void())
{
    t.f(s);
}

namespace ns
{
    template <typename T>
    auto f_(S& s, T const& t) -> decltype(f(s, t), void())
    {
        f(s, t);
    }
} 

struct S
{
    template <typename T>
    auto f(T const& t) -> decltype(ns::f_(std::declval<S&>(), t), void())
    {
        ns::f_(*this, t);
    }

    template <typename T>
    auto f_base(T const* t_ptr) -> decltype(f(*t_ptr), void())
    {
        f(*t_ptr);
    }
};

struct pass
{
   void f(S&) const
   {
   }
};

struct fail
{

};

int main()
{
    S s;
    s.f(pass()); // compiles
    //s.f(fail()); // doesn't compile
    return 0;
}

and works as intended. The problem arises when I attempt to move definitions of S::f and S::f_base outside the class body, like so :

#include <utility>

struct S;

template <typename T>
auto f(S& s, T const& t) -> decltype(t.f(s), void())
{
    t.f(s);
}

namespace ns
{
    template <typename T>
    auto f_(S& s, T const& t) -> decltype(f(s, t), void())
    {
        f(s, t);
    }
} 

struct S
{
    template <typename T>
    auto f(T const& t) -> decltype(ns::f_(std::declval<S&>(), t), void());

    template <typename T>
    auto f_base(T const* t_ptr) -> decltype(f(*t_ptr), void());
};

template <typename T>
auto S::f(T const& t) -> decltype(ns::f_(std::declval<S&>(), t), void())
{
    ns::f_(*this, t);
}

template <typename T>
auto S::f_base(T const* t_ptr) -> decltype(f(*t_ptr), void()) // <---- HERE ---
{
    f(*t_ptr);
}

int main()
{

    return 0;
}

On the line marked by arrow GCC 4.7.1 expresses it's dissatisfaction:

error: prototype for 'decltype ((((S*)0)->S::f((* t_ptr)), void())) S::f_base(const T*)' does not match any in class 'S'
error: candidate is: template decltype ((((S*)this)->S::f((* t_ptr)), void())) S::f_base(const T*)

I tried to explicitly specify which f I'm using in f_base by prepending it (in both declaration and definition) with std::declval<S&>(). , but the error persists.

I know I can modify the dependency graph like so:

S::f_base ->
          -> ns::f_ -> f -> T::f
S::f      ->

to make S::f_base depend on ns::f_ along with S::f , but is there a way to do this with the first dependency graph?

Answer 1

GCC 4.X is not the best for meta programing

I have manage to make it compile in 4.7.3 ( live ) :

#include <utility>

struct S;

template <typename T>
auto f(S& s, T const& t) -> decltype(t.f(s), void())
{
    t.f(s);
}

namespace ns
{
    template <typename T>
    auto f_(S& s, T const& t) -> decltype(f(s, t), void())
    {
        f(s, t);
    }
} 

// some std::void_t like but GCC4.x compatible
template<class T>
struct void_t
{
   using type = typename std::enable_if<std::is_same<T,T>::value >::type;
};

struct S
{
    template <typename T>
    auto f(T const& t) -> decltype(ns::f_(std::declval<S&>(), t), void());

    template <typename T>
    auto f_base(T const* t_ptr) ->  typename void_t< decltype (::f(*std::declval<T const*>()))>::type ;
};


template <typename T>
auto S::f(T const& t) -> decltype(ns::f_(std::declval<S&>(), t), void())
{
    ns::f_(*this, t);
}

// ::f is needed, fail if just 'f'
template <typename T>
auto S::f_base(T const* t_ptr) ->  typename void_t< decltype (::f(*std::declval<T const*>()))>::type
{
    f(*t_ptr);
}
int main()
{

    return 0;
}

Note:

In GCC4.XI have seen some weird stuff like :

template<class T>
struct void_t
{
   using type = void;
};

Where GCC replace void_t<decltype(expr)>::type by void even if expr is not valid. That why I use this implementation of void_t .