Question

I have a couple of classes that share a common base class, with the exception that they differ in the way their methods work. So in the example below, Adder and Multiplier are the same except for the way in which their calculation is performed.

Is there a way to change “a” to a Multiplier on the fly? Do I need to implement methods which convert derived classes to each other? e.g. something like

a = a.asMultiplier()?

As you can see in the code below I tried reinterpret_cast to a Multiplier, with no luck, it still acts like an Adder. (gcc OS X v4.2.1)

#include <iostream>


class Base {
protected:
    int a,b;    
public:
    Base(int a, int b) {
        this->a = a;
        this->b = b;
    }  
    virtual ~Base() { }    
    virtual int calculate() = 0;
};


class Adder : public Base {    
public:
    Adder(int a, int b) : Base(a, b) {

    }      
    int calculate() {
        return this->a + this->b;
    }
};


class Multiplier : public Base {    
public:
    Multiplier(int a, int b) : Base(a, b) {    
    }  

    int calculate() {
        return this->a * this->b;
    }
};



int main() {

    Base* a = new Adder(3,4);
    Base* m = new Multiplier(3,4);

    std::cout << "Adder gives " << a->calculate() << std::endl;
    std::cout << "Multiplier gives " << m->calculate() << std::endl;

    a = reinterpret_cast<Multiplier*>(a);

    std::cout << "Now adder gives " << a->calculate() << std::endl;

    delete a;
    delete m;
    return 0;
}

Answer 1

The best thing that comes up to me to solve this, is implementing a copy constructor, taking the base class:

class Multiplier : public Base {    
public:
    Multiplier(int a, int b) : Base(a, b) {    
    }  

    explicit Multiplier(const Base& iBase) : Base(iBase.a, iBase.b) {    
    }  

    int calculate() {
        return this->a * this->b;
    }
};

But since I am not the most advanced c++ developer here, it might not be correct or other people might have a better idea, just trying 🙂