CODEWITHSUNDEEP
javac++inheritancepolymorphism
Victor S
Victor S

Reputation: 4191

Why does superclass B invoke subclass A's method?

The Test3.java:

public class Test3 {
    public static void main(String[] args) {
        A a = new A();
    }
}

class A extends B {

    A() {
        System.out.println("Now in A()");
    }

    void m() {
        System.out.println("Now in m() of A");
    }
}

class B {
    B() {
        System.out.println("Now in B()");
        m();
    }

    void m() {
        System.out.println("Now in m() of B");
    }
}

The output:

    $ java Test3
Now in B()
Now in m() of A
Now in A()

Why not "Now in m() of B" in the 2nd line of output?

Upvotes: 0

Views: 86

Answers (2)

Sleiman Jneidi
Sleiman Jneidi

Reputation: 23329

Because methods in Java are virtual by default, if you using Java 6 or higher, it is recommended that you use the @Override annotation, it makes things clearer.

The first question is: Why the first line was 

Now in B()

The reason is: When you can extend a class in Java a call to the super constructor will be done automatically for you by the compiler

A() {
    super(); // added by Java !!
    System.out.println("Now in A()");
}

So, now we are in B's constructor 

B() {
    System.out.println("Now in B()"); // print this
    m(); // Ohh hang on, m() is overridden
}

because m() is overriden by the subclass A, and the actual instance is of type A, so we will call A's implementation of m(), and that is a polymorphic call.

Upvotes: 2

Anubian Noob
Anubian Noob

Reputation: 13596

Because the method m is overriden in A. So when it goes to the super constructor, and calls m, the object is still an instance of A, so it calls the method in A.

Upvotes: 3

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