What is the difference between 'super' and 'extends' in Java Generics

JavaGenerics

Java Problem Overview


I am trying to learn Java generics. I am not clear when you will use <T extends Foo> and when you will use <T super Foo>. What do each one of these things mean about T? Lets say I have <T extends Comparable> and <T super Comparable>, what do each of these mean?

I've read several tutorials at sun.com but I'm still lost. Can somebody illustrate with examples?

Java Solutions


Solution 1 - Java

It depends which way on the inheritance hierarchy it allows. Assume you have a class "Child" which inherits from "Parent" which inherits from "Grandparent".

<T extends Parent> accepts either Parent or Child while <T super Parent> accepts either Parent or Grandparent.

Solution 2 - Java

There are three types of wildcards:

  • ? extends Type: Denotes a family of subtypes of type Type. This is the most useful wildcard.
  • ? super Type: Denotes a family of supertypes of type Type.
  • ?: Denotes the set of all types or any.

Solution 3 - Java

See Effective Java 2nd Edition, Item 28:

PECS

Producer extends, Consumer super

If your parameter is a producer, it should be <? extends T>, if it's a consumer it has to be <? super T>.

Take a look at the Google Collections, they know how to use it, because they got Bloch ;)

Solution 4 - Java

For me the best answer came from @BSingh, when i read the article of Uncle Bob. I resume here, the conclusion of article.

Use List< T super Suit> whenever you are going to write into the list.

> When you put an Object to the List, all you care about is that the object is of a type that is compatible with type held by the list. So you want the list to take the type of that object or any of the superclasses of that object.

Use List< T extends Suit> whenever you are going to read from a list. > On the other hand, when you read from a list, you want the type you are reading to be the type contained byt the list, or a derivative from that type.

Solution 5 - Java

If you are asking about type parameters, then there is no <T super X> construct in Java. Bounded parameter can only extend, but it can extend more than one type. E.g

public class MyClass< T extends Closeable & Runnable >
{
  // Closeable and Runnable are chosen for demonstration purposes only
}

In this case if you see MyClass< ConcreteT > then ConcreteT must be declared as

public class ConcreteT
  implements Closeable, Runnable
{
   ...
}

For bounded wildcards, read this article. Read section on get-put principle. Basically, super corresponds to write semantics, and extends corresponds to read semantics.

Solution 6 - Java

Remember PECS - Producer Extends Consumer Support. Also, uncle Bob discusses it well in his craftsman series. Check out http://objectmentor.com/resources/articles/The_Craftsman_44__Brown_Bag_I.pdf

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionbenhsuView Question on Stackoverflow
Solution 1 - JavaR Samuel KlatchkoView Answer on Stackoverflow
Solution 2 - JavaIvo BostickyView Answer on Stackoverflow
Solution 3 - JavawhiskeysierraView Answer on Stackoverflow
Solution 4 - JavaStathis AlexopoulosView Answer on Stackoverflow
Solution 5 - JavaAlexander PogrebnyakView Answer on Stackoverflow
Solution 6 - JavaBSinghView Answer on Stackoverflow