Naive Inheritance Problem - Java

Question

G'day people,

I am feeling embarrass by asking such a naive question. But I can't understand one thing, I have Inheritance structure like this,

enter image description here

B extends A, code I have wrote is as below,

Class A

public class A{
    private int pos = 0;
    public A(){
        this.pos = 12;
    }
    public int getPos(){
        return this.pos;
    }
}

Class B

public class B extends A{
    int spec = 15;
    public B(){
        super();
    }
    public int getSpec(){
        return this.spec;
    }
}

And I have one more class to test, Which will get us to my question.

Class Test

import java.util.*;
public class Test{
    public static void main(String[] args){
        B a = new B();
        ArrayList<A> c = new ArrayList<A>();
        c.add(a);
        System.out.println(c.get(0).getPos());
        System.out.println(c.get(0).getSpec());
    }
}

Question : Now I am creating an instance of B, Which means I can access to my parent class's method getPos() and B's own method getSpec(). But if I create ArrayList with type A(...B is type A too, as it extends A...) and add my B's instance it losses it's ability to access it's own method. What am I doing wrong? Does ArrayList implementation is casting my B to A internally?

Note : My basic understanding of inheritance is parent cannot access child's method except they are protected. But Child can access their parent class's method.

Hovercraft Full Of Eels · Accepted Answer · 2011-09-10 02:35:26Z

4

There's no casting involved. What you're doing is no different from this:

A bAsA = new B():

While the object referred by bAsA is truly a B object, it is held by an A variable and thus only A methods are available (unless you explicitly cast it as a B variable).

Since your ArrayList is an ArrayList of A, each item in the ArrayList is treated as an A variable and only A methods are available.

answered Sep 10, 2011 at 2:35

Hovercraft Full Of Eels

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Comments

Matt Ball · Accepted Answer · 2011-09-10 02:36:05Z

4

Does ArrayList implementation is casting my B to A internally?

No. There is no "internal casting." You, the programmer, have told the compiler it's a list of A.

You have declared the List as List<A>, which you can read as "a list of A". Since all B are A, you can add any B to a List<A>. On retrieval, however, you're only guaranteed to get back an A, not a B — because it's a List<A>, remember — so the compiler treats everything that comes out of the list as an A, even if (at runtime) it's an instance of B.

answered Sep 10, 2011 at 2:36

Matt Ball

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5 Comments

TeaCupApp Over a year ago

Thanks Matt, Now this makes sense :). Btw What if I want collection of child in one ArrayList? can I do something like ArrayList<Obejct>...; and then cast it each time? or is there any work around?

Hovercraft Full Of Eels Over a year ago

@Skeet: heavy casting gives a program a bad smell and suggests that one should look carefully at the program design because there's a good chance that it's broken and can be improved in a much more object-oriented fashion.

Matt Ball Over a year ago

@Skeet I would not recommend casting. If you want to be able to call the same method on As and B stop worrying about the type and just make A and B implement the same interface. Or, put a stub method in A that B can override.

Matt Ball Over a year ago

@Skeet also it's generally good practice to declare List<Foo> or even looser constraints like Collection<Foo> or Iterable<Foo>. As I said in my previous comment: don't worry about an object's type. Start thinking in terms of what you can do with it.

TeaCupApp Over a year ago

+1 Thanks, Matt and Hovercraft, I will keep that in my mind, Nice answer

Community · Accepted Answer · 2017-05-23 12:30:36Z

3

In addition to the answers provided by @Matt Ball and @Hovercraft Full Of Eels, you can avoid having to explicitly cast by declaring methods implemented by the subclass as abstract methods in the superclass.

public abstract class A{
    .
    .
    public abstract int getSpec();
}

EDIT-

As mentioned by @Kublai Khan, it is necessary to then make the superclass an abstract class.

edited May 23, 2017 at 12:30

CommunityBot

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answered Sep 10, 2011 at 2:53

mre

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1 Comment

Paul Bellora Over a year ago

+1 and don't forget that A now needs to be an abstract class.

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