Why Field Can’t Be Overridden?

Category: Classes & Interfaces   November 15, 2012

This article shows the basic object oriented concept in Java - Field Hiding.

1. Can Field Be Overridden in Java?

Let's first take a look at the following example which creates two Sub objects. One is assigned to a Sub reference, the other is assigned to a Super reference.

package oo;
 
class Super {
	String s = "Super";
}
 
class Sub extends Super {
	String s = "Sub";
}
 
public class FieldOverriding {
	public static void main(String[] args) {
		Sub c1 = new Sub();
		System.out.println(c1.s);
 
		Super c2 = new Sub();
		System.out.println(c2.s);
	}
}

What is the output?

Sub
Super

We did create two Sub objects, but why the second one prints out "Super"?

2. Hiding Fields Instead Of Overriding Them

In [1], there is a clear definition of Hiding Fields:

Within a class, a field that has the same name as a field in the superclass hides the superclass's field, even if their types are different. Within the subclass, the field in the superclass cannot be referenced by its simple name. Instead, the field must be accessed through super. Generally speaking, we don't recommend hiding fields as it makes code difficult to read.

From this definition, member fields can not be overridden like methods. When a subclass defines a field with the same name, the subclass just declares a new field. The field in the superclass is hidden. It is NOT overridden, so it can not be accessed polymorphically.

3. Ways to Access Hidden Fields

1). By using parenting reference type, the hidden parent fields can be access, like the example above.
2). By casting you can access the hidden member in the superclass.

System.out.println(((Super)c1).s);

References:
1. Hiding Fields

Related posts:

  1. Inheritance vs. Composition in Java
  2. Notes from “Sun Certified Programmer for Java 6 Study Guide”
  3. Declaration, Initialization and Scoping for Java
  4. How Java Compiler Generate Code for Overloaded and Overridden Methods?
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  • chanaka

    i also want to know about it

  • optimistsimitpo

    As for the first line of this part..
    Super c2 = new Sub();
    System.out.println(c2.s);
    Is it a typo of “Super c2 = new Super();?” If not, why did you do upcasting?

  • Wittahera

    I like this article. Just want to make a comparison, if s is a method like

    String s(){
    return ‘Sub’;
    } in sub class and

    String s(){
    return ‘Super’;
    } in super class

    Then both will return ‘Sub’ because of dynamic linking.

  • Daniel Gurianov

    Name of the article and explanation is quite misleading to me. Fields behave the same way as a methods. As long as your use constructor of the last bottom child in inheritance chain, you can do upcast over the instance and switch to the upper level of abstraction. In this case all fields and methods that were defined on lower level of abstraction will be “forgotten” and all that definitions on the level you just entered, “remebered”. And examples in this article confirms my logic.