Question

Let say I want to create an array of length 5 through the default constructor

public class A <E extends Comparable<? super E>> implements B<E>
{
     private E[] myArray;


     public A()
     {
         myArray = (E[]) new Object[5]; 
     }    

}

Is this the right way of doing it? I’m confused that whether I have to
state “Comparable” before []. so

 private  Comparable[] myArray;

Answer 1

The problem with the code you posted is that the erasure of E[] is Comparable[] (because E‘s upper bound is Comparable), so casting an object whose real type is Object[] to Comparable[] will fail.

You could simply change it to the following and it won’t throw the exception:

public class A <E extends Comparable<? super E>> implements B<E>
{
     private E[] myArray;


     public A()
     {
         myArray = (E[]) new Comparable[5]; 
     }    

}

However, there is more subtlety here: myArray is not really of type E[]. Inside the class, E[] is erased to Comparable[], so it is okay. But you have to make sure never to expose myArray to outside of the class. Since it is private, you just have to make sure that no method returns it or something.

The benefit of doing it this way, keeping myArray of type E[], over Jatin’s answer of using Comparable[], is that you don’t have to cast every time you get something out of it, so the code is nicer.