What is <? super T> syntax? [duplicate]

Question

I'm having trouble understanding the following syntax:

public class SortedList< T extends Comparable< ? super T> > extends LinkedList< T > 

I see that class SortedList extends LinkedList. I just don't know what

T extends Comparable< ? super T> 

means.

My understanding of it so far is that type T must be a type that implements Comparable...but what is < ? super T >?

Answer 1

super in Generics is the opposite of extends. Instead of saying the comparable's generic type has to be a subclass of T, it is saying it has to be a superclass of T. The distinction is important because extends tells you what you can get out of a class (you get at least this, perhaps a subclass). super tells you what you can put into the class (at most this, perhaps a superclass).

In this specific case, what it is saying is that the type has to implement comparable of itself or its superclass. So consider java.util.Date. It implements Comparable<Date>. But what about java.sql.Date? It implements Comparable<java.util.Date> as well.

Without the super signature, SortedList would not be able accept the type of java.sql.Date, because it doesn't implement a Comparable of itself, but rather of a super class of itself.