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java – Initialization in polymorphism of variables-Exceptionshub

Posted by: admin February 25, 2020 Leave a comment

Questions:

Suppose you have the following code

class A {
    int i = 4;

    A() { 
        print();
    }

    void print () {
        System.out.println("A");
    }
}

class B extends A {
    int i = 2;              //"this line"

    public static void main(String[] args){
        A a = new B();
        a.print();
    }

    void print () {
        System.out.println(i);
    }
}

this will print 0 2

Now, if you remove line labeled “this line”
the code will print 4 4

  • I understand that if there was no int i=2; line,

A a = new B(); will call class A, initializes i as 4, call constructor,
which gives control over to print() method in class B, and finally prints 4.

a.print() will call print() method in class B because the methods will bind at runtime, which will also use the value defined at class A, 4.

(Of course if there is any mistake in my reasoning, let me know)

  • However, what i don’t understand is if there is int i=2.

why is it that if you insert the code, the first part (creating object) will all of sudden print 0 instead of 4? Why does it not initialize the variable as i=4, but instead assigns default value?

How to&Answers:

It is a combination of several behaviors in Java.

  1. Method overriding
  2. Instance variable shadowing
  3. order of constructors

I will simply go through what happened in your code, and see if you understand.

Your code conceptually looks like this (skipping main()):

class A {
    int i = 0; // default value

    A() { 
        A::i = 4;  // originally in initialization statement
        print();
    }

    void print () {
        System.out.println("A");
    }
}

class B extends A {
    int i = 0;              // Remember this shadows A::i

    public B() {
        super();
        B::i = 2;
    }

    void print () {
        System.out.println(i);
    }
}

So, when in your original main(), you called A a = new B();, it is constructing a B, for which this happens:

  • A::i and B::i are all in default value 0
  • super(), which means A’s constructor is called
    • A::i is set to 4
    • print() is called. Due to late-binding, it is bound to B::print()
    • B::print() is trying to print out B::i, which which is still 0
  • B::i is set to 2

Then when you call a.print() in your main(), it is bounded to B::print() which is printing out B::i (which is 2 at this moment).

Hence the result you see

Answer:

All the instance variables in the new object, including those declared in superclasses, are initialized to their default values – JLS 12.5

Therefore, your variable B::i will be initialized to 0. The constructor in B will be like:

B() {
    super();
    i = 2;
}

So when you call

A a = new B();

The constructor in A will call the print method in B, which will prints the i in class B, which is 0.

Answer:

In your case the class B, the declaration of “i” hides the declaration of “i” in A, and all references to “i” in the child class refer to the B.i not the A.i.

And so what you see in A.i is the default value of any int attribute in java which is zero.

Java instance variables cannot be overridden in a subclass.

You want to try this for more clarification.

class B extends A {
    int i = 2;              //"this line"

    public static void main(String[] args){
        B b = new B();
        A a = b;
        System.out.println("a.i is " + a.i);
    System.out.println("b.i is " + b.i);
    }

    void print () {
        System.out.println(i);
    }
}

Ouput:

a.i is 4
b.i is 2