Question

How to call base class method if it is not abstract.

class WithAbstMethod {
public:
    virtual void do() = 0;
}

class WithImplMethod : public WithAbstMethod {
public:
    virtual void do() {
        // do something
    }
}

template<typename BaseT>
class DerivedClass : BaseT {
public:
    virtual void do() {
        BaseT::do(); // here is a question. How to modify code, so that do() is called if it is not abstract?
        // do something
    }
}

void main() {
    DerivedClass<WithAbstMethod> d1;
    d1.do(); // only DerivedClass::do() should be called
    DerivedClass<WithImplMethod> d2;
    d2.do(); // both WithImplMethod::do() and DerivedClass::do() should be called
}

Is it possible to do this using templates in compile-time without much code (instantiate DerivedClass::do() method with BaseT::do() call and without depending on BaseT type)?
Obviously, provide implementation in WithAbstMethod class is not an option. Code above is pseudo-code so may contain minor errors.

Answer 1

Actually, providing an implementation for WithAbstMethod::do() might be an option. Abstract functions are allowed to have an implementation.

void WithAbstMethod::do()
{
    // do nothing...
}