Workaround for generic arrays in java

Question

I have an assignment and I need to make my own (simple) generic linked list:

public class Node<T> {

   private int key;
   private T data;
   private Node<T> nextNode;

}

But I need to implement a dictionary with a hash table. I wanted to make a list containing Node(s). In case of a conflict (two objects of type disperse to the same node, I simply link them - linked lists).

I have to implement this by myself, no outside help (already implemented lists or what ever)

How I wanted to do this:

public class GenericDictionary<T> implements GenericDictionary_interface<T> {

    private int capacity;   
    private Node<T> [] slots;

    public GenericDictionary () {   
        this.capacity = 31;
        slots = new Node<T>[capacity];  // the array I need which I disperse to
    }
}

This however is not exactly possible. I did try and read on the subject, tried searching here on SO ... but I didn't get it at all.

My only request is ... don't be lazy on variable / method names, make them easy to understand please.

If you have to implement this by yourself, without outside help, why do you ask? Isn´t that requesting outside help? — TheBlastOne
– TheBlastOne, Commented Oct 24, 2012 at 10:35
@TheBlastOne. Common, there is nothing wrong with getting some help if OP has tried out code. Had he asked to write code for him. That would be wrong. — Rohit Jain
– Rohit Jain, Commented Oct 24, 2012 at 10:37
@RohitJain Agree. I just wonder if OP understands he´s breaking the rule. Surely we don't have a problem with that. But he should. I'd say he should question the rule, since getting (and accepting) hints or help from outside, and accepting it, is a key to learning. — TheBlastOne
– TheBlastOne, Commented Oct 24, 2012 at 10:41
@Kalec all I say is a) don't just break the rule, destroy it! because b) I agree that you are doing nothing wrong when you're asking for the right help. — TheBlastOne
– TheBlastOne, Commented Oct 24, 2012 at 12:47

RokL · Accepted Answer · 2012-10-24 10:23:21Z

5

Here's the best you can do:

    @SuppressWarnings("unchecked")
    Node<T>[] slots = (Node<T>[]) new Node<?>[capacity];

You can't get rid of the warning (aside from suppressing it). When you need an array of a generic class, you need to create the array with unspecified generic type then cast it.

answered Oct 24, 2012 at 10:23

RokL

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6 Comments

Kalec Over a year ago

So I'm typecasting Node<?> as Node<T> ? and thus it would work ? Okay, any way it could backfire ?

RokL Over a year ago

None. You need to understand that unlike C++/C# templates which generate actual implementations, Java generics are just extra syntactic compile time checks and automatically inserted casts. For instance ArrayList<Integer> and ArrayList<String> are the same class, the same object even. Generic type only changes the cast used before returing the element on get(), Object array is used in both cases.

RokL Over a year ago

Another example to help you understand the nature of Java Generics: You can fill List<Object> with 10 strings and 1 integer, cast it to List<String> and the list will work as long as you don't try to get() the integer element (because get() method has code like this return (T)obj_array[idx]; which will cause class cast exception in List<String> when the actual element is integer.

John B Over a year ago

This is called "type erasure": docs.oracle.com/javase/tutorial/java/generics/erasure.html

Kalec Over a year ago

Okay, I understand. I do come from a background of C/C++ so I didn't EXACTLY understand how it works, thank you.

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