Hiding Fields in Java Inheritance

Question

Within a class, a field that has the same name as a field in the superclass hides the superclass's field.

public class Test {

    public static void main(String[] args) {

        Father father = new Son();
        System.out.println(father.i); //why 1?
        System.out.println(father.getI());  //2
        System.out.println(father.j);  //why 10?
        System.out.println(father.getJ()); //why 10?

        System.out.println();

        Son son = new Son();
        System.out.println(son.i);  //2 
        System.out.println(son.getI()); //2
        System.out.println(son.j); //20
        System.out.println(son.getJ()); //why 10?
    }  
}

class Son extends Father {

    int i = 2;
    int j = 20;

    @Override
    public int getI() {
        return i;
    }
}

class Father {

    int i = 1;
    int j = 10;

    public int getI() {
        return i;
    }

    public int getJ() {
        return j;
    }
}

Can someone explain the results for me?

Oracle Tutorial on hiding variables.

alex
– alex

2012-09-03 08:56:39 +00:00
Commented Sep 3, 2012 at 8:56 — alex
– alex, Commented Sep 3, 2012 at 8:56
A son is not a type of father

jsj
– jsj

2012-09-03 09:13:26 +00:00
Commented Sep 3, 2012 at 9:13 — jsj
– jsj, Commented Sep 3, 2012 at 9:13
I don't see how this question is "too localized"... :D

Willi Mentzel
– Willi Mentzel

2016-11-30 16:43:03 +00:00
Commented Nov 30, 2016 at 16:43 — Willi Mentzel
– Willi Mentzel, Commented Nov 30, 2016 at 16:43

Pang · Accepted Answer · 2023-11-08 09:35:03Z

17

In Java, fields are not polymorphic.

Father father = new Son();
System.out.println(father.i); //why 1? Ans : reference is of type father, so 1 (fields are not polymorphic)
System.out.println(father.getI());  //2 : overridden method called
System.out.println(father.j);  //why 10? Ans : reference is of type father, so 2
System.out.println(father.getJ()); //why 10? there is not overridden getJ() method in Son class, so father.getJ() is called

System.out.println();

// same explanation as above for following 
Son son = new Son();
System.out.println(son.i);  //2 
System.out.println(son.getI()); //2
System.out.println(son.j); //20
System.out.println(son.getJ()); //why 10?

edited Nov 8, 2023 at 9:35

Pang

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answered Sep 3, 2012 at 9:07

Nandkumar Tekale

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1 Comment

Asif Mushtaq Over a year ago

Just in java? any other language provide polymorphism on fields?

Willi Mentzel · Accepted Answer · 2017-11-12 16:02:18Z

4

As per Overriding and Hiding Methods

The version of the hidden method that gets invoked depends on whether it is invoked from the superclass or the subclass.

i.e. when you invoke a method which is overridden in subclass via a super class reference the super class method is invoked and it access super class members.

This explains following as the reference used is of superclass:

System.out.println(father.i);  //why 1?
System.out.println(father.j);  //why 10?
System.out.println(father.getJ()); //why 10?

Similarly for the following:

System.out.println(son.getJ()); //why 10?

since getJ() is not defined in Son a Father version is invoked which sees member defined in the Father class.

If you read Hiding Fields; they specifically don't recommend such method of coding as

Generally speaking, we don't recommend hiding fields as it makes code difficult to read.

edited Nov 12, 2017 at 16:02

Willi Mentzel

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answered Sep 3, 2012 at 9:15

Bharat Sinha

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2 Comments

Grateful Over a year ago

I am sorry, but this is not true. The original source has been misquoted. Your quotation is specifically in reference to static methods, not general methods. Please note the original source wording, as follows. (1) "The version of the overridden instance method that gets invoked is the one in the subclass." (2) "The version of the hidden static method that gets invoked depends on whether it is invoked from the superclass or the subclass."

Grateful Over a year ago

So, in essence, if you invoke a method that is overridden in the subclass, via the superclass reference, the SUBCLASS method version is invoked... NOT the superclass method as you implied.

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