How does dynamic binding work for private methods in base class in Java?

Question

class Base {
    private void SayHello(){ //PRIVATE
        System.out.println("Hello from Base");
    }
}

class Derived extends Base {
    public void sayHello(){ //PUBLIC
        System.out.println("Hello from Derived");
    }
}

public class TestHello{

    public static void main(String[] args) {
        Derived d = new Derived();
        Base b = d;

        d.sayHello(); //works as expected
        b.sayHello(); //Why does this not work?
    }
}

I want to understand: is the private sayHello from base class visible to the derived class? or is it a redefinition? And why does the call to the derived sayHello from the base pointer does not work? I mean, if it were public (in Base), then the sayHello from the derived class would have been called. So, what I can not understand is that if it has to call the public sayHello from the derived class, then why look at the access modifier from the base class?

Also, if you can point me to some concise resource that will help me understand this in more depth, I'd really appreciate this. Thanks!

When you provide an instance method in a sub-class which overrides the parents implementation, this is the method which is called, not the parents. — Peter Lawrey, Nov 30 '15 at 09:33

score 1 · Accepted Answer · edited May 23 '17 at 12:33

is the private sayHello from base class visible to the derived class?

Of course, no, because it has private access modifier.

More about access modifiers:

Controlling Access to Members of a Class

or is it a redefinition?

As you can see in the accepted answer to this question:

What's the difference between redefining a method and overriding a method?

the term "redefinition" isn't commonly used. We can talk about "overriding" and "overloading", but in your case sayHello from the Derived class is a kind of new method and it's not an overloaded version of sayHello from the Base class.

And why does the call to the derived sayHello from the base pointer does not work?

Simply because you try to call method that doesn't belong to the open class interface.

I mean, if it were public (in Base), then the sayHello from the derived class would have been called.

Of course, it's an expected polymorphic behaviour. In this case, sayHello from the Derived class overrides sayHello from the Base class, so you can call sayHello from the Derived class via the reference to the the Base class.

So, what I can not understand is that if it has to call the public sayHello from the derived class, then why look at the access modifier from the base class?

Because you use reference to the Base class and there's no sayHello method in the interface of the Base class.

I found a good discussion here:

Overriding private methods in Java

May be also useful for you:

Overriding and Hiding Methods

Hope it'll help you.

score 1 · Answer 2 · answered Nov 29 '15 at 15:10

1

Because the line below simply assigns as variable reference of object Derived

Base b = d;

and the private methods are private to the objects and not visible outside of the class.

answered Nov 29 '15 at 15:10

bhantol

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MugenTwo · Answer 3 · 2015-11-29T15:01:11.563

b.sayHello() is not working because when you call the method sayHello() the computer starts looking for the method in the base class, and when it finds the method it will see that it is a private method, so it is going to tell you that you can't use it.

On the otherhand, d.sayHello() is working because the computer will start looking for sayHello() in the subClass derived, and when the computer finds it it will let you use it because it is public.

How does dynamic binding work for private methods in base class in Java?

3 Answers3