Call child method from base method without virtual or references

Question

Consider the code:

#include <iostream>

class A {
public:
    void g() {std::cout << "Call from A" << std::endl;}
    void f() {g();}
};

class B : public A {
public:
    void g() {std::cout << "Call from B" << std::endl;}
};

int main() {
    B b;
    b.f(); // "Call from A" 
}

If I make A::g() virtual, it outputs "Call from B".

I always understood "virtual" useful when we want an interface to allow polymorphism, say A *a = new B(); a->f(), but this example indicates that virtual is also required in this situation. My understanding of this example is that the call g() on A::f is formally equivalent to this->g(), and this is a pointer.

However, this is not what I really wanted: in my real example, I have a general template class A with a general method A::simulate (say A::f), that uses different methods (say A::g()), some of which are overwritten by subclass (say B). What I wanted was to have the method A::simulate in such a way that the code would be compiled in the subclasses of A as if it has been defined on the subclasses so I don't have the same logic coded multiple-times. How can I do this?

This is because my interface will not require polymorphism because the user will always need to compile the code (i.e. it is header only).

Answer 1

You can simply make your base class a CRTP (as you've mentioned it's already a templated class anyway):

#include <iostream>

template<class Derived>
class A {
public:
    void g() {std::cout << "Call from A" << std::endl;}
    void f() { static_cast<Derived*>(this)->g();}
            // ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Call g() from Derived
};

class B : public A<B> {
public:
    void g() {std::cout << "Call from B" << std::endl;}
};

int main() {
    B b;
    b.f(); // "Call from B" 
}