C++ – Can a class member function template be virtual

When an object written in C# is constructed, what happens is that the initializers run in order from the most derived class to the base class, and then constructors run in order from the base class to the most derived class (see Eric Lippert's blog for details as to why this is).

Also in .NET objects do not change type as they are constructed, but start out as the most derived type, with the method table being for the most derived type. This means that virtual method calls always run on the most derived type.

When you combine these two facts you are left with the problem that if you make a virtual method call in a constructor, and it is not the most derived type in its inheritance hierarchy, that it will be called on a class whose constructor has not been run, and therefore may not be in a suitable state to have that method called.

This problem is, of course, mitigated if you mark your class as sealed to ensure that it is the most derived type in the inheritance hierarchy - in which case it is perfectly safe to call the virtual method.

Yes, with SFINAE you can check if a given class does provide a certain method. Here's the working code:

#include <iostream>

struct Hello
{
    int helloworld() { return 0; }
};

struct Generic {};    

// SFINAE test
template <typename T>
class has_helloworld
{
    typedef char one;
    struct two { char x[2]; };

    template <typename C> static one test( decltype(&C::helloworld) ) ;
    template <typename C> static two test(...);    

public:
    enum { value = sizeof(test<T>(0)) == sizeof(char) };
};
    
int main(int argc, char *argv[])
{
    std::cout << has_helloworld<Hello>::value << std::endl;
    std::cout << has_helloworld<Generic>::value << std::endl;
    return 0;
}

I've just tested it with Linux and gcc 4.1/4.3. I don't know if it's portable to other platforms running different compilers.