C++ SFINAE with CRTP, G++ compile error - c++

Question

I would like to know if the following code are valid.
The original intension is that, I like a base class that dispatch calls to a certain member to either derived class members if it is there or fall back to default behaviors if derived class does not have this member. Another use is that this base class can be used by itself and the Derived template parameter becomes a implementation policy. Anyway, the following MWE compiles and runs correctly with clang++, Intel, icpc and MSVS. However it fails with g++ (from 4.4 to 4.6, any version I had a hand on) with the error message at the end of the question.
If I change the call at point (1), (2), (3) to call_dispatch (which was the sort of thing I did originally), g++ does not complain anymore. I don't think it is a good practice to have the dispatch function and the caller having the same name. I was just curious if it will work, and curiously enough to try it out (I have no idea how does this idea come to me). My rationale behind this is that, at pint (1), call is invoked with one parameter, so the overload resolution will not match its caller, the zero parameter one. It will not match the SFINAE one at point (2) either, since D2 does not have the member, and then it shall match the one at point (3). Just as in the situation when (1)-(3) are named call_dispatch.
But g++ does not agree with me and other compilers. So, is it an incorrect implementation of g++ or the code itself is invalid? Besides the error message is really confusing. Where does the void (B<D2>::*)() and &B<D2>::call come from? Int he called the member pointer was defined as D2's member.
#include <iostream>
#include <functional>
template <typename Derived>
class B
{
public :
void call ()
{
call<Derived>(0); //----------------------------------------- (1)
}
private :
template <typename D, void (D::*)()> class SFINAE {};
template <typename D>
void call (SFINAE<D, &D::call> *) //---------------------------- (2)
{
static_cast<Derived *>(this)->call();
}
template <typename D>
void call (...) //--------------------------------------------- (3)
{
std::cout << "Call B" << std::endl;
}
};
class D1 : public B<D1>
{
public :
void call ()
{
std::cout << "Call D1" << std::endl;
}
};
class D2 : public B<D2> {};
int main ()
{
D1 d1;
D2 d2;
d1.call();
d2.call();
return 0;
}
Error:
foo.cpp: In member function ‘void B<Derived>::call() [with Derived = D2]’:
foo.cpp:48:13: instantiated from here
foo.cpp:11:9: error: ‘&B<D2>::call’ is not a valid template argument for type ‘void (D2::*)()’ because it is of type ‘void (B<D2>::*)()’
foo.cpp:11:9: note: standard conversions are not allowed in this context
Edit
Though I have not fully understand what goes wrong in the above code yet. But I think there is a another way without specifically construct a SFINAE class but archive the same effect.
#include <iostream>
template <typename Derived>
class B
{
public :
void call ()
{
call_dispatch(&Derived::call);
}
template <typename C>
void call_dispatch (void (C::*) ())
{
static_cast<Derived *>(this)->call();
}
void call_dispatch (void (B<Derived>::*) ())
{
std::cout << "Call B" << std::endl;
}
private :
};
class D1 : public B<D1>
{
public :
void call ()
{
std::cout << "Call D1" << std::endl;
}
};
class D2 : public B<D2> {};
int main ()
{
D1 d1;
D2 d2;
d1.call();
d2.call();
return 0;
}
Basically, because D1 and D2 both are derived from B, so the expression &Derived::call will always be resolved. In D1 it resolved to &D1::call, then the template version member is used. In D2, it does not have its own call, so &D2::call is resolved to &B::call, and thanks to
#DavidRodríguez-dribeas, who points out that now &D2::call has the type B::call, therefore the template and the non-template members equally match, but non-template is preferred. So the default call is used.
Can help me see if there is any defect in this new code?

Answer 1

Where does the void (B::*)() and &B::call come from?
The type of a pointer to member is not the type on which you obtained such pointer, but the type on which the member is defined.
struct base { int x; };
struct derived : base {};
int main() {
std::cout << std::is_same< decltype(&derived::x), int (base::*) >::value << std::endl;
}
The above program prints 1. In your case, when you use &D::base, the compiler finds B<D2>::call as a member of the base template, and that is the result of the expression: void (B<D2>::*)().

Answer 2

Your instantiation of SFINAE is expecting the function pointer template argument to be of type void (D2::*)(), but instead is of type void (B<D2>::*)() (because call is not overidden in D2, it uses the one defined in B<D2>).
Edit
It is not the instantiation of template <typename D>
void call (SFINAE<D, &D::call> *) that is failing here. There is no substitution error there. The substitution error occurs with the instantiation of SFINAE<B<D2>, &B<D2>::call>, of which there is no fallback instantiation, hence the error.