Calling constructor of a generic type

Question

If I have an abstract class like this:

public abstract class Item
{
    private Integer value;
    public Item()
    {
        value=new Integer(0);
    }
    public Item(Integer value)
    {
        this.value=new Integer();
    }
}

And some classes deriving from Item like this:

public class Pencil extends Item
{
    public Pencil()
    {
        super();
    }
    public Pencil(Integer value)
    {
        super(value);
    }
}

I have not understood why I can't call the constructor using a generic:

public class Box <T extends Item>
{
    T item;
    public Box()
    {
        item=new T(); // here I get the error
    }
}

I know that is possible to have a type which hasn't a constructor, but this case is impossible because Pencil has the constructor without parameters, and Item is abstract. But I get this error from eclipse: cannot instanciate the type T
I don't understand why, and how to avoid this?

Answer 1

This is because Java uses erasure to implement generics, see this:

• http://en.wikipedia.org/wiki/Generics_in_Java#Problems_with_type_erasure

To quote the relevant parts from the above Wikipedia article:

Generics are checked at compile-time for type-correctness. The generic type information is then removed in a process called type erasure.

As a result of type erasure, type parameters cannot be determined at run-time.

Consequently, instantiating a Java class of a parameterized type is impossible because instantiation requires a call to a constructor, which is unavailable if the type is unknown.

You can go around this by actually providing the class yourself. This is well explained here:

• Create instance of generic type in Java?