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Difference between <? super T> and <? extends T> in Java [duplicate]

What is the difference between List<? super T> and List<? extends T> ?

I used to use List<? extends T>, but it does not allow me to add elements to it list.add(e), whereas the List<? super T> does.

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Anand Avatar asked Dec 03 '10 06:12

Anand


People also ask

What is the difference between list <? Super T and list <? Extends T?

super is a lower bound, and extends is an upper bound.

What does <? Super mean in Java?

The super keyword in Java is a reference variable which is used to refer immediate parent class object. Whenever you create the instance of subclass, an instance of parent class is created implicitly which is referred by super reference variable.

What is the difference between Super and extends?

Producer – If you want to only retrieve the elements from a generic collection, use extends . Consumer – If you want to only put elements into a generic collection, use super . If you do both retrieve and put operations with the same collection, you shouldn't use either extends or super .

What does Super T means in Java?

super T denotes an unknown type that is a supertype of T (or T itself; remember that the supertype relation is reflexive). It is the dual of the bounded wildcards we've been using, where we use ? extends T to denote an unknown type that is a subtype of T .


1 Answers

extends

The wildcard declaration of List<? extends Number> foo3 means that any of these are legal assignments:

List<? extends Number> foo3 = new ArrayList<Number>();  // Number "extends" Number (in this context) List<? extends Number> foo3 = new ArrayList<Integer>(); // Integer extends Number List<? extends Number> foo3 = new ArrayList<Double>();  // Double extends Number 
  1. Reading - Given the above possible assignments, what type of object are you guaranteed to read from List foo3:

    • You can read a Number because any of the lists that could be assigned to foo3 contain a Number or a subclass of Number.
    • You can't read an Integer because foo3 could be pointing at a List<Double>.
    • You can't read a Double because foo3 could be pointing at a List<Integer>.
  2. Writing - Given the above possible assignments, what type of object could you add to List foo3 that would be legal for all the above possible ArrayList assignments:

    • You can't add an Integer because foo3 could be pointing at a List<Double>.
    • You can't add a Double because foo3 could be pointing at a List<Integer>.
    • You can't add a Number because foo3 could be pointing at a List<Integer>.

You can't add any object to List<? extends T> because you can't guarantee what kind of List it is really pointing to, so you can't guarantee that the object is allowed in that List. The only "guarantee" is that you can only read from it and you'll get a T or subclass of T.

super

Now consider List <? super T>.

The wildcard declaration of List<? super Integer> foo3 means that any of these are legal assignments:

List<? super Integer> foo3 = new ArrayList<Integer>();  // Integer is a "superclass" of Integer (in this context) List<? super Integer> foo3 = new ArrayList<Number>();   // Number is a superclass of Integer List<? super Integer> foo3 = new ArrayList<Object>();   // Object is a superclass of Integer 
  1. Reading - Given the above possible assignments, what type of object are you guaranteed to receive when you read from List foo3:

    • You aren't guaranteed an Integer because foo3 could be pointing at a List<Number> or List<Object>.
    • You aren't guaranteed a Number because foo3 could be pointing at a List<Object>.
    • The only guarantee is that you will get an instance of an Object or subclass of Object (but you don't know what subclass).
  2. Writing - Given the above possible assignments, what type of object could you add to List foo3 that would be legal for all the above possible ArrayList assignments:

    • You can add an Integer because an Integer is allowed in any of above lists.
    • You can add an instance of a subclass of Integer because an instance of a subclass of Integer is allowed in any of the above lists.
    • You can't add a Double because foo3 could be pointing at an ArrayList<Integer>.
    • You can't add a Number because foo3 could be pointing at an ArrayList<Integer>.
    • You can't add an Object because foo3 could be pointing at an ArrayList<Integer>.

PECS

Remember PECS: "Producer Extends, Consumer Super".

  • "Producer Extends" - If you need a List to produce T values (you want to read Ts from the list), you need to declare it with ? extends T, e.g. List<? extends Integer>. But you cannot add to this list.

  • "Consumer Super" - If you need a List to consume T values (you want to write Ts into the list), you need to declare it with ? super T, e.g. List<? super Integer>. But there are no guarantees what type of object you may read from this list.

  • If you need to both read from and write to a list, you need to declare it exactly with no wildcards, e.g. List<Integer>.

Example

Note this example from the Java Generics FAQ. Note how the source list src (the producing list) uses extends, and the destination list dest (the consuming list) uses super:

public class Collections {    public static <T> void copy(List<? super T> dest, List<? extends T> src) {       for (int i = 0; i < src.size(); i++)          dest.set(i, src.get(i));    }  } 

Also see How can I add to List<? extends Number> data structures?

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Bert F Avatar answered Sep 28 '22 09:09

Bert F