classA extends classB, how to have objects from classB in classA

Question

I have this class:

public class Gradient extends Item<GradientRep> {

    /**
     * Initializes a new Gradient
     * @param x X-coordinate of the Item
     * @param y Y-coordinate of the Item
     * @param value The value of the gradient
     */

    public Gradient(int x, int y, int value, String type) {
        super(x, y);
        this.value = value;
        this.type = type;
    }

     ....

And I would like this class to "own" some objects of class Item, for example, I would like to retrieve from class Gradient this object:

Item MyItem = new Item()

Is there an obvious way to do this?

To give some background, for now the class Gradient only represent a fixed point on a map, which is not what we want for a Gradient. So, I want to propagate this gradient, and therefore when I instantiate a Gradient, I want it to provide other Items (with different coordinates on the map, which is dealt by the class Item).

Did you try to add a variable of type Item to Gradient? I don’t really see what the problem is. — Joakim Danielson, Apr 24 '21 at 20:00
Does this answer your question? [ extends Class> and super Class> in Java - why it works this way?](https://stackoverflow.com/questions/35669112/extends-class-and-super-class-in-java-why-it-works-this-way) — Monica Aspiras Labbao, Apr 25 '21 at 06:58

score 1 · Accepted Answer · answered Apr 24 '21 at 20:11

1

I'm not entirely sure what you're after - but you could instanciate items from Gradient like this:

import java.util.ArrayList;

public class Gradient extends Item<GradientRep> {

    private final int value;
    private final String type;
    
    private ArrayList<Item<GradientRep>> items = new ArrayList<>();

    /**
     * Initializes a new Gradient
     *
     * @param x     X-coordinate of the Item
     * @param y     Y-coordinate of the Item
     * @param value The value of the gradient
     */
    public Gradient(int x, int y, int value, String type) {
        super(x, y);
        this.value = value;
        this.type = type;
    }

    Item<GradientRep> getItem() {
        return new Item<>(x, y);
    }
}

The Gradient object will be an item, stores items in the items ArrayList (or your choice of ADT), and can generate new Item objects with the getItem method.

answered Apr 24 '21 at 20:11

Tytrox

483
2
10

Thank you, actually I don't need the Items to be of "type" GradientRep, Can I just do Item getItem() { ... etc. ? – ailauli69 Apr 24 '21 at 20:23
Sorry, I did not specified that the Item class is Abstract, and I would like to create Items objects in Gradient class, based on the values of x and y (for instance create an Item(x+1, y+2), and be able to access it once the Gradient class is instanciated) – ailauli69 Apr 24 '21 at 20:27
If the `Item` class is abstract, you can't instanciate `Item` objects. If you would like objects 'like' items, you will need to either make `Item` not abstract, or create another child of `Item`. Maybe something like: `public class Coordinate extends Item <> {` – Tytrox Apr 24 '21 at 20:40
And yes, you can have a raw use of `Item`, although I would avoid it if you could. – Tytrox Apr 24 '21 at 20:42

score 0 · Answer 2 · answered Apr 25 '21 at 02:17

You can do it through this way as well:

public class Gradient extends Item<GradientRep> {
        Item myItem;
        /**
         * Initializes a new Gradient
         * @param x X-coordinate of the Item
         * @param y Y-coordinate of the Item
         * @param value The value of the gradient
         */
    
        public Gradient(int x, int y, int value, String type) {
            super(x, y);
            this.value = value;
            this.type = type;
            myItem = new Item();
        }

you can access the field as gradient.myItem

If you are making the field static, then use Gradient.myItem

classA extends classB, how to have objects from classB in classA

2 Answers2