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General tree type question

I've been learning java's Generics and, as far as I understand, a Tree with generic type is internally a (dynamic) subclass of either an ArrayTree, ArrayList, or LinkedList.  
So this works:
class MyClass 
    private ArrayList list;
   ...


Is that true?

A:

There are no arbitrary subclasses of ArrayList.  The only type that can be the List of T is java.util.List.  (Note that the only subclass of ArrayList that can be List is ArrayList itself.)  The only subclass of ArrayList that can be List is java.util.ArrayList.
If you don't specify a List, the compiler uses the default.  In Java, all of the interfaces in java.util are also "default implementation", so both List and ArrayList implement List.  This explains why you can do:
ArrayList list;

and the compiler will not complain.  The behaviour you see if you did do something like this:
List list;

The compiler would complain that list is a raw type and so you can't add elements to it.  I think what you really want is:
class MyClass 
    private ArrayList list;
   ...


and then this should compile without any problems.
Update:
If you can't modify MyClass to use an ArrayList, you can rewrite it to use a LinkedList.  While you will then have two implementations of List, you can simply remove the one in java.util with no ill effects.  Java is designed to be modular, so you can switch out an entire library's content without a recompile.

A:

The type parameters T can only be constraints of a type for which there is already a declared superclass or interface. In your example, this is not true as there's no definition of a type for which there is such a superclass. So, T cannot be a type parameter in this case ee730c9e81
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Jared Spongebob Man Boy Avigolkes

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