Confusion using a parameterized collection with a sub class

D

dugrocker

Hi,

I've got a super class A and A subclass of A called B. I want to make a
collection that can be created as either a collection of A or a
subclass of A. That is fine, and seems to compile fine.

However, when I actually make a collection that contains a subclass of
A (B in this example), I cannot add elements of B to the collection.

Does anyone know the solution?

Code snippet below:
<code>

package generics;
import java.util.ArrayList;

public class MyContainer
{
ArrayList<? extends A> bList; //The list should be a list of
something
//that either extends or
is an A. B
//Fullfills this
requirement


MyContainer()
{
bList = new ArrayList<B>();//List of child class B (compiles)
}

public static void main(String[] args)
{
MyContainer myContainer = new MyContainer();


/* *********************************************************
* The problem is here in the next line.
*
* Compiler message:
*
* The method add(capture-of ? extends A) in the type
* ArrayList<capture-of ? extends A> is not applicable for the
* arguments (B)
*
* *********************************************************/
myContainer.bList.add(new B()); //(fails to compile)

}
}

</code>
 
O

Oliver Wong

ArrayList<? extends A> bList; //The list should be a list of
something
//that either extends or
is an A. B
//Fullfills this
requirement
[...]

/* *********************************************************
* The problem is here in the next line.
*
* Compiler message:
*
* The method add(capture-of ? extends A) in the type
* ArrayList<capture-of ? extends A> is not applicable for the
* arguments (B)
*
* *********************************************************/
myContainer.bList.add(new B()); //(fails to compile)

First of all, realize that when a class extends another class, there is
an IS-A relationship formed between the two classes. That is, if B extends
A, then for any given instance of B, you can say that that instance IS-A A.
This more intuitive if you use real class names. Let's say Vehicule and Car.
Car extends Vehicule, and so you can say that a car IS-A Vehicule.

bList is an ArrayList<? extends Vehicule> which in informal terms means
it's a list of something which extends vehicule. Let's give that something a
name: call it S. Now you don't actually know what S is, only that S extends
Vehicule. So S could be Vehicule itself, or it could be Car, or it could be
some other class that extends Vehicule (such as Boat).

You can never add anything to this list. Why? Because there's no way to
guarantee that whatever it is your adding extends S. In the above example,
you're trying to add Car to the list, but it's not clear that the list can
accept Car, because Car might not extend S. (e.g. Car doesn't extend Boat,
so if S is actually Boat, you couldn't add the Car to a list of boats)

My guess is that you don't actually want an ArrayList<? extends
Vehicule>, but rather an ArrayList<Vehicule>. This is saying that you have a
list of vehicules. You can put a car in a list of vehicules, because a car
IS-A vehicule.

- Oliver
 
D

dugrocker

So, what are you saying, if you make the following statement,
"ArrayList<? extends A> aList;" you will be able to instantiate it as
an ArrayList<A> or ArrayList<subclass of A> but you will never be able
to add any objects to it?

should only be used in a method said:
First of all, realize that when a class extends another class, there is
an IS-A relationship formed between the two classes. That is, if B extends
A, then for any given instance of B, you can say that that instance IS-A A.
This more intuitive if you use real class names. Let's say Vehicule and Car.
Car extends Vehicule, and so you can say that a car IS-A Vehicule.

Okie dokie, I'll be sure to write that down right away.



Oliver said:
ArrayList<? extends A> bList; //The list should be a list of
something
//that either extends or
is an A. B
//Fullfills this
requirement
[...]

/* *********************************************************
* The problem is here in the next line.
*
* Compiler message:
*
* The method add(capture-of ? extends A) in the type
* ArrayList<capture-of ? extends A> is not applicable for the
* arguments (B)
*
* *********************************************************/
myContainer.bList.add(new B()); //(fails to compile)

First of all, realize that when a class extends another class, there is
an IS-A relationship formed between the two classes. That is, if B extends
A, then for any given instance of B, you can say that that instance IS-A A.
This more intuitive if you use real class names. Let's say Vehicule and Car.
Car extends Vehicule, and so you can say that a car IS-A Vehicule.

bList is an ArrayList<? extends Vehicule> which in informal terms means
it's a list of something which extends vehicule. Let's give that something a
name: call it S. Now you don't actually know what S is, only that S extends
Vehicule. So S could be Vehicule itself, or it could be Car, or it could be
some other class that extends Vehicule (such as Boat).

You can never add anything to this list. Why? Because there's no way to
guarantee that whatever it is your adding extends S. In the above example,
you're trying to add Car to the list, but it's not clear that the list can
accept Car, because Car might not extend S. (e.g. Car doesn't extend Boat,
so if S is actually Boat, you couldn't add the Car to a list of boats)

My guess is that you don't actually want an ArrayList<? extends
Vehicule>, but rather an ArrayList<Vehicule>. This is saying that you have a
list of vehicules. You can put a car in a list of vehicules, because a car
IS-A vehicule.

- Oliver
 
O

Oliver Wong

dugrocker said:
So, what are you saying, if you make the following statement,
"ArrayList<? extends A> aList;" you will be able to instantiate it as
an ArrayList<A> or ArrayList<subclass of A> but you will never be able
to add any objects to it?

Right.
This syntax, ArrayList<? extends A>, should only be used in a method
signature then, and not in a variable declaration?

I don't know about that. It's mainly useful when you want to support all
different kinds of list, and you don't plan on putting anything into the
list, but only in extracting things from the list.

- Oliver
 

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