Java dynamic binding when using inheritance

Question

For the following code:

class A { 
    public void ex(Object o) {
        System.out.println("A");
    }
}

class B extends A { 
    public void ex(B o) {
        System.out.println("B");
    }
}

class C extends B {
    public void ex(A o) {
        System.out.println("C");
    }
}

public class Main {
    public static void main(String argv[]) {
        C xx;
        xx = new C();
        xx.ex(xx); // This prints B
    }
}

Why is the result of calling the ex method "B"? Both the reference and class type are C, still the method executed is the one from the superclass.

andresp · Accepted Answer · 2020-02-02 21:59:40Z

2

Due to the class chain you created C instances will have all 3 methods available:

A.ex(Object)
B.ex(B)
C.ex(A)

All of them could be used for your xx.ex(xx) invocation. At this point, according to the Java spec, the most specific method is invoked.

If more than one member method is both accessible and applicable to a method invocation, it is necessary to choose one to provide the descriptor for the run-time method dispatch. The Java programming language uses the rule that the most specific method is chosen.

As you don't have an ex(C) method, the most specific method that matches your invocation is B.ex(B) as B is a subclass of A, which is a subclass of Object.

answered Feb 2, 2020 at 21:59

andresp

1,66420 silver badges32 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

Java dynamic binding when using inheritance

1 Answer 1

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Your Answer

Sign up or log in

Post as a guest

Related