The mechanics of extension via free functions or member functions

Question

Loads of C++ libraries, the standard included, allow you to adapt your objects for use in the libraries. The choice is often between a member function or a free function in the

Answer 1

If I understood correctly your problem is simply solved using (maybe multiple) inheritance. You have somewhere a namespace free function:

namespace NS {
void DoSomething()
{
    std::cout << "NS::DoSomething()" << std::endl;
}
} // namespace NS

Use a base class which forwards the same function:

struct SomethingBase
{
    void DoSomething()
    {
        return NS::DoSomething();
    }
};

If some class A deriving from SomethingBase does not implement DoSomething() calling it will call SomethingBase::DoSomething() -> NS::DoSomething():

struct A : public SomethingBase // probably other bases
{
    void DoSomethingElse()
    {
        std::cout << "A::DoSomethingElse()" << std::endl;
    }
};

If another class B deriving from SomethingBase implement DoSomething() calling it will call B::DoSomething():

struct B : public SomethingBase // probably other bases

{
    void DoSomething()
    {
        std::cout << "B::DoSomething()" << std::endl;
    }
};

So calling DoSomething() on an object deriving from SomethingBase will execute the member if existing, or the free function otherwise. Note that there is nothing to throw, you get a compile error if there is no match to your call.

int main()
{
    A a;
    B b;
    a.DoSomething(); // "NS::DoSomething()"
    b.DoSomething(); // "B::DoSomething()"
    a.DoSomethingElse(); // "A::DoSomethingElse()"
    b.DoSomethingElse(); // error 'DoSomethingElse' : is not a member of 'B'
}