Implementing the Cloneable Interface

[Tianna Taylor-Albin]

I am trying to implement the cloneable interface of the Product class but I keeping getting an error saying that the clone has protected access in Object. I thought the @Override gave access. Can anyone please tell me what I'm doing wrong? Thanks.

public class ProductCloneApp implements Cloneable
{
    private String code;
    private String description;
    private double price;
    
    // the code for the constructor and methods

    @Override 
    public Object clone() throws CloneNotSupportedException
    {
        return super.clone();
    }

    public static void main(String args[]) throws CloneNotSupportedException
    {
        System.out.println("Welcome to the Product Clone Test\n");

        // create a new product
        Product p1 = new Product();
        p1.setCode("java");
        p1.setDescription("Murach's Beginning Java");
        p1.setPrice(49.50);

        // clone the product
        Product p2 = (Product) p1.clone();

        // change a value in the cloned product
        p2.setPrice(44.50);

        if (p1.getPrice() == p2.getPrice())
        {
            System.out.println("FAILURE: The clone method of the Product class is not cloning data.");
        }
        else if (p1.getPrice() != p2.getPrice())
        {
            System.out.println("SUCCESS: The clone method of the Product class is cloning data.");
        }
        System.out.println();
    }
}
}

[manish ghildiyal]

You are cloning objects of Product class....so you need to have Product class implementing Clone interface
...then create its instance and call clone() on that. Currently you are implenting Clone on ProductCloneApp.

Manish

[Paul Clapham]

No, the @Override annotation just tells the compiler to make sure the method it's annotating actually does override a method from the superclass. Which in this case, it does.

But a method whose body only calls the same method from the superclass is actually pointless. If you didn't write that code, then calling the class's clone() method would still call the clone() method from the superclass, which is Object. That's what inheritance does. So I would suggest just not overriding the clone() method.

Edit: Also, as manish says, make sure you make Product cloneable.

[Ivan Jozsef Balazs]

Paul Clapham wrote:
But a method whose body only calls the same method from the superclass is actually pointless. If you didn't write that code, then calling the class's clone() method would still call the clone() method from the superclass, which is Object. That's what inheritance does.

However Object's clone() is protected, so if you want to provide a public clone method of this type, you have to override Object's one in order to make it public.
But how to implement it? Well, through "super.clone()"

Object's clone's API docu writes:

By convention, the returned object should be obtained by calling super.clone.
If a class and all of its superclasses (except Object) obey this convention, it will be the case that x.clone().getClass() == x.getClass().

[Winston Gutkowski]

Tianna Taylor-Albin wrote:Can anyone please tell me what I'm doing wrong?

I think everyone else has covered most of it; however, just one final thing: You don't have to return an Object, and it's usually best not to. These days, you can use covariant returns (look it up) to return a typed object, viz:

@Override
public Product clone() throws CloneNotSupportedException { ...

That way you don't have to cast every time you use it. Obviously, the above assumes that the method is in your Product class.

HIH

Winston

[Ivan Jozsef Balazs]

// I had a dream...
interface Cloneable<T> {
        public T clone() ;
}

[Winston Gutkowski]

Ivan Jozsef Balazs wrote:

// I had a dream...

Unfortunately, a dream too late...

Winston