Polymorphism in C++

Question

Okay, so I know there are a lot of polymorphism threads flying around but I have yet to encounter this situation.

class Base {
public:
    virtual void method1() {
        cout << "BaseMethod1" << endl;
    }

    void method2() {
        cout << "BaseMethod2" << endl;
    }
};

class Derive: public Base {
public:
    void method1() {
        cout << "DeriveMethod1" << endl;
        method2();
    }

    void method2() {
        cout << "DeriveMethod2" << endl;
    }
};

int main() {
    Base* p = new Derive();
    p->method1();
}

What's tripping me up is method1 in the derived class calls a method2. So which method2 would it be since the method2 in the Base class wasn't declared as virtual?

Thanks ahead of time!

@Pubby trying things won't necessarily lead to enlightenment in C++. In fact it can lead to a dangerous false sense of security in the case of undefined behavior that happens to work. — Antimony
– Antimony, Commented Mar 1, 2013 at 4:56
The key phrase you want to look up to answer this question is "name hiding". — In silico
– In silico, Commented Mar 1, 2013 at 4:56
-1. Do a basic research first since this is a very basic question, ask the compiler, then perhaps ask a more specific question. — Lie Ryan
– Lie Ryan, Commented Mar 1, 2013 at 4:59
@Antimony well if you prefer to read the standard then have it your way. All I care is that people do a little research before asking questions. — Pubby
– Pubby, Commented Mar 1, 2013 at 4:59

Alok Save · Accepted Answer · 2013-03-01 05:01:18Z

4

The code in question will not compile because Base::method1() is declared private. It needs to be public to be able to be called from main.

Derived::method1() is implicitly virtual even if not marked as virtual. So the this in Derived::method1() points to a Dervied object and in this scope compiler can only see Derived::method2(). Hence Derived::method2() will be called. The method in derived class hides the same named method in Base class.

Good Read:

What's the meaning of, Warning: Derived::f(char) hides Base::f(double)?

edited Mar 1, 2013 at 5:01

answered Mar 1, 2013 at 4:55

Alok Save

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Comments

Matthew Sanders · Accepted Answer · 2013-03-01 04:59:48Z

2

The main reason you do not see this pattern very often is that it is in fact an anti-pattern.

The Derived method will be invoked because you are calling it with a Derived object reference (this). If you were calling it with a Base class reference you will get the Base method.

If you redeclare a non-virtual method you are hiding the base method and breaking polymorphism.

answered Mar 1, 2013 at 4:59

Matthew Sanders

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2 Comments

Tony Delroy Over a year ago

+1 for several good points (especially significance of this), but regarding "redeclare a non-virtual method [is] breaking polymorphism" - that's a bit melodramatic: the object may continue to work polymorphically through a Base class reference/pointer perfectly well - as per Liskov Substitution Principle - or it may even be necessary not to have some functions virtual to avoid breaking the principle (e.g. a Derived class serialisation method that wasn't reparsable as per Base method documentation). Not broken, but fragile (e.g. harder to switch to compile-time polymorphism).

Matthew Sanders Over a year ago

Good point. There are almost always exceptions to the rule. I tend to stick with this particular rule pretty heavily as it reduces the likelihood of error in large code bases. This is a contradiction to what one would expect. I guess you could say it inflates polymorphism because now you are making an object that takes many forms based on how you refer to it :P.

c-smile · Accepted Answer · 2013-03-01 04:59:05Z

1

It will use Derive::method2() as name look-up (in body of Derive::method1()) starts from the class itself.

answered Mar 1, 2013 at 4:59

c-smile

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