Overriding member variables in Java ( Variable Hiding)

Question

I am studying overriding member functions in JAVA and thought about experimenting with overriding member variables.

So, I defined classes

public class A{
    public int intVal = 1;
    public void identifyClass()
    {
        System.out.println("I am class A");
    }
}

public class B extends A
{
    public int intVal = 2;
    public void identifyClass()
    {
        System.out.println("I am class B");
    }
}

public class mainClass
{
    public static void main(String [] args)
    {
        A a = new A();
        B b = new B();
        A aRef;
        aRef = a;
        System.out.println(aRef.intVal);
        aRef.identifyClass();
        aRef = b;
        System.out.println(aRef.intVal);
        aRef.identifyClass();
    }
}

The output is:

1
I am class A
1
I am class B

I am not able to understand why when aRef is set to b intVal is still of class A?

Answer 1

When you make a variable of the same name in a subclass, that's called hiding. The resulting subclass will now actually have both properties. You can access the one from the superclass with super.var or ((SuperClass)this).var. The variables don't even have to be of the same type; they are just two variables sharing a name, much like two overloaded methods.

Answer 2

Variables are not polymorphic in Java; they do not override one another.

Answer 3

There is no polymorphism for fields in Java.

Variables decision happens at a compile time so always Base Class variables (not child’s inherited variables) will be accessed.

So whenever upcasting happens always remember

1) Base Class variables will be accessed.

2) Sub Class methods(overridden methods if overriding happened else inherited methods as it is from parent) will be called.

Answer 4

Variables are resolved compile-time, methods run-time. The aRef is of type A, therefore aRef.Intvalue is compile-time resolved to 1.

Answer 5

OverRiding Concept in Java Functions will override depends on object type and variables will accessed on reference type.

1. Override Function: In this case suppose a parent and child class both have same name of function with own definition. But which function will execute it depends on object type not on reference type on run time.

For e.g.:

Parent parent=new Child();
parent.behaviour();

Here parent is a reference of Parent class but holds an object of Child Class so that's why Child class function will be called in that case.

Child child=new Child();
child.behaviour();

Here child holds an object of Child Class, so the Child class function will be called.

Parent parent=new Parent();
parent.behaviour();

Here parent holds the object of Parent Class, so the Parent class function will be called.

2. Override Variable: Java supports overloaded variables. But actually these are two different variables with same name, one in the parent class and one in the child class. And both variables can be either of the same datatype or different.

When you trying to access the variable, it depends on the reference type object, not the object type.

For e.g.:

Parent parent=new Child();
System.out.println(parent.state);

The reference type is Parent so the Parent class variable is accessed, not the Child class variable.

Child child=new Child();
System.out.println(child.state);

Here the reference type is Child, so the Child class variable is accessed not the Parent class variable.

Parent parent=new Parent();
System.out.println(parent.state);

Here the reference type is Parent, so Parent class variable is accessed.

Answer 6

From JLS Java SE 7 Edition §15.11.1:

This lack of dynamic lookup for field accesses allows programs to be run efficiently with straightforward implementations. The power of late binding and overriding is available, but only when instance methods are used.

Answers from Oliver Charlesworth and Marko Topolnik are correct, I would like to elaborate a little bit more on the why part of the question:

In Java class members are accessed according the type of the reference and not the type of the actual object. For the same reason, if you had a someOtherMethodInB() in class B, you wouldn't be able to access it from aRef after aRef = b is run. Identifiers (ie class, variable, etc names) are resolved at compile time and thus the compiler relies on the reference type to do this.

Now in your example, when running System.out.println(aRef.intVal); it prints the value of intVal defined in A because this is the type of the reference you use to access it. The compiler sees that aRef is of type A and that's the intVal it will access. Don't forget that you have both fields in the instances of B. JLS also has an example similar to yours, "15.11.1-1. Static Binding for Field Access" if you want to take a look.

But why do methods behave differently? The answer is that for methods, Java uses late binding. That means that at compile time, it finds the most suitable method to search for during the runtime. The search involves the case of the method being overridden in some class.

Answer 7

This is called variable hiding. When you assign aRef = b; , aRef has two intVal, 1 is named just intVal another is hidden under A.intVal (see debugger screenshot), Because your variable is of type class A , even when you print just intVal java intelligently picks up A.intVal.

Answer 1: One way of accessing child class's intVal is System.out.println((B)aRef.intVal);

Answer 2: Another way of doing it is Java Reflection because when you use reflection java cant intelligently pickup hidden A.intVal based on Class type, it has to pick up the variable name given as string -

import java.lang.reflect.Field;

class A{
    public int intVal = 1;
    public void identifyClass()
    {
        System.out.println("I am class A");
    }
}

class B extends A
{
    public int intVal = 2;
    public void identifyClass()
    {
        System.out.println("I am class B");
    }
}

public class Main
{
    public static void main(String [] args) throws Exception
    {
        A a = new A();
        B b = new B();
        A aRef;
        aRef = a;
        System.out.println(aRef.intVal);
        aRef.identifyClass();
        aRef = b;
        Field xField = aRef.getClass().getField("intVal");
        System.out.println(xField.get(aRef));
        aRef.identifyClass();
    }
}

Output -

1
I am class A
2
I am class B

Answer 8

I hope this can help:

public class B extends A {
//  public int intVal = 2;

    public B() {
        super();
        super.intVal = 2;
    }

    public void identifyClass() {
        System.out.println("I am class B");
    }
}

So overriding variable of base class is not possible, but base class variable value can be set (changed) from constructor of inherited class.