What is syntax?

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 >?

Yishai · Accepted Answer · 2010-05-13 14:33:16Z

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. But what about java.sql.Date? It implements Comparable 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.

I see. So then given that class, would Collection< Object > list = new SortedList< Object>(); be a valid statement? — ShrimpCrackers, May 13 '10 at 14:36
@aloh, no it would not because Object does not implement Comparable. — Yishai, May 13 '10 at 14:37

polygenelubricants · Answer 2 · 2010-05-13 14:49:54Z

It's a lower-bounded wildcard.

JLS 4.5.1 Type Arguments and Wildcards

Wildcards are useful in situations where only partial knowledge about the type parameter is required. [...] An upper bound is signified by the syntax:
? extends B
where B is the upper bound. [...] it is permissible to declare lower bounds on a wildcard, using the syntax:
? super B
where B is a lower bound.

A List, for example, includes List, List, andList.

Wildcards are used to make generics more powerful and flexible; bounds are used to maintain type safety.

It's a bit more complicated than that though - the question isn't about ? super T, it is about T extends Comparable — oxbow_lakes, May 13 '10 at 14:31
Ah, I see. Those all implement Comparable. What about the super keyword? — ShrimpCrackers, May 13 '10 at 14:33
Not really! If you look at docs, you'll realize it's not simply a bounded wildcard. — Epitaph, May 13 '10 at 14:34
Just for clarification, my previous comment was aimed at your initial response which was just 3-4 lines. — Epitaph, May 13 '10 at 16:35

Petar Minchev · Answer 3 · 2010-05-13 14:33:22Z

It means that T must implement Comparable The sense is that because SortedList is sorted, it must know how to compare two classes of its genericsT parameter. That's why T must implement Comparable

John Smith · Answer 4 · 2010-05-13 14:39:02Z

It means that the type T must implement Comparable of T or one of its super classes.

For example, if A extends B, if you want to use SortedList, A must implementComparable or Comparable, or in fact just Comparable.

This allows the list of As to be constructed with any valid comparator.

Epitaph · Answer 5 · 2010-05-13 14:42:28Z

Consider the following example:

Using a type parameter defined in the class declaration

public class ArrayList extends AbstractList ... {
public boolean add(E o) // You can use the "E" here ONLY because it's already been defined as part of the class
Using a type parameter that was NOT defined in the class declaration
```
public <T extends Animal> void takeThing(ArrayList<T> list)
                                            // Here we can use  because we declared "T" earlier in the method declaration                  
```
If the class itself doesn't use a type parameter, you can still specify one for a method, by declaring it in a really unusual (but available) space - before the return type. This method says that T can be "any type of Animal".

NOTE:

public <T extends Animal> void takeThing(ArrayList<T> list)             

is NOT same as              

public void takeThing(ArrayList<Animal> list)

Both are legal, but they are different. The first one indicates that you can pass in a ArrayList object instantiated as Animal or any Animal subtype like ArrayList, ArrayList or ArrayList. But, you can only pass ArrayList in the second, and NOT any of the subtypes.

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What is syntax?

5 Answers

JLS 4.5.1 Type Arguments and Wildcards

See also

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