CODEWITHSUNDEEP
c++virtual
Mr.Cat
Mr.Cat

Reputation: 330

Virtual function taking parameters of base class type

what's wrong with the polymorphism here? 

class A
{
public:
    int x;
    virtual void dosomething(const A& ob) = 0;
};

class B : public A
{
public:
    int y;
    void dosomething(const A& ob) { // I assume this will work (base class accepting a derived class by reference)
        x += ob.x;
        y += ob.y; // error: const class A has no memeber named 'y'
        cout<<x<<"    "<<y;
    }
};


int main()
{
    B s, m;
    s.x = 3;
    s.y = 9;
    m.x = 3;
    m.y = 9;

    s.dosomething(m);
}

i tried using pointer instead of reference, and still not working
should i override the function?

Upvotes: 2

Views: 87

Answers (1)

Remy Lebeau
Remy Lebeau

Reputation: 595477

dosomething() accepts a reference to the A portion of an object. A does not have a member named y, just like the compiler error says. Only B has a y member.

Imagine there is another class C that also derived from A but not B, where C does not have a y member either. What do you think would happen if a C object were passed to B::dosomething()?

class C : public A
{
public:
    int z;
    void dosomething(const A& ob) { ... }
};


B b;
C c;
b.dosomething(c);

It would not work. B::dosomething() can accept a C object as input, but there is still no y member for B::dosomething() to read from.

So, you must use a type-cast inside of B::dosomething() to check if the input obj is actually derived from B before accessing its y member:

class B : public A
{
public:
    int y;
    void dosomething(const A& ob) {
        x += ob.x;
        const B *b = dynamic_cast<const B*>(&ob);
        if (b) y += b->y;
        cout << x << "    " << y;
    }
};

Upvotes: 4

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