C++ makes possible to reassign reference? [duplicate]

Question

#include <iostream>
using namespace std;

class Base{
    public:
    virtual void f(){
        std::cout << "Base\n";
    }
};

class Derived1 : public Base{
    public:
    void f(){
        std::cout << "Derived1\n";
    }

};

class Derived2 : public Base{
    public:
    void f(){
        std::cout << "Derived2\n";
    }
};




int main() {
    Derived1 d1;
    Derived2 d2;

    Base& ref = d1;
    ref.f();
    ref = d2;
    ref.f();

    return 0;
}

It is said that the reference cannot be reassigned. Here, the compiler accept this code and output is not understandable for me. Output:

Derived1 Derived1

So, what does mean ref = d2; in fact?

Answer 1

What ref = d2 actually did was assign the value of d2 to d1 - however, it's still d1 which is referred to by ref. Because d1 is still referred to, invoking ref.f still prints Derived1.

Suppose you add a data value v to your base class - set it to 1 in d1 and 2 in d2. Change your implementations of f to print v. You get something like:

#include <iostream>
using namespace std;

class Base{
    public:
    int v;

    virtual void f(){
        std::cout << "Base : v=" << v << "\n";
    }
};

class Derived1 : public Base{
    public:
    void f(){
        std::cout << "Derived1 : v=" << v << "\n";
    }

};

class Derived2 : public Base{
    public:
    void f(){
        std::cout << "Derived2 : v=" << v << "\n";
    }
};

int main() {
    Derived1 d1;
    d1.v = 1;
    Derived2 d2;
    d2.v = 2;

    Base& ref = d1;
    ref.f();
    ref = d2;
    ref.f();

    return 0;
}

When run this will print:

Derived1 : v=1
Derived1 : v=2

Hopefully this makes it a bit clearer.

Best of luck.