Wildcard and mutator methods

Question

I was totally confused, when saw this snippet:

class Animal {}

class Dog extends Animal {}

public class Test {

    public static void main(String[] args) {
        List<? super Animal> list = new ArrayList<>();
        list.add(new Dog());  //it's OK
        list.add(new Animal()); //and this is OK too
    }
}

Why such things are allowed? When i changed my list to List<? super Dog> list = new ArrayList<>();compile-time error occurs in list.add(new Animal()); With extends wildcard all combinations cause errors. Who can tell the exact reason of this behaviour? Thanks in advance.

Think of it this way: a dog is an animal, but an animal is not a dog. — tnw, Feb 26 '15 at 16:46

score 1 · Accepted Answer · edited May 23 '17 at 11:50

1

    List<? super Animal> list = new ArrayList<>();
    list.add(new Dog());  //it's OK
    list.add(new Animal()); //and this is OK too

The above code is allowed as it should be : because A dog is also an animal.

    List<? super Dog> list = new ArrayList<>();
    list.add(new Dog());  //it's OK
    list.add(new Animal()); //error

The above code is an error as it should be again, because not every animal is a dog.

inheritance is as simple as this. :)

NOTE: To complete the answer I refer to this super good answer in :

[Difference between <? super T> and <? extends T> in Java 1

edited May 23 '17 at 11:50

Community

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answered Feb 26 '15 at 16:50

nafas

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Of course! Thanks. But what about `extends`? Why we cannot put `Dog` in `List extends Dog>`? – Dmytro Feb 26 '15 at 16:54
@Jskyi I think this is what you are asking abt mate [http://stackoverflow.com/questions/4343202/difference-between-super-t-and-extends-t-in-java] – nafas Feb 26 '15 at 17:01

Wildcard and mutator methods

1 Answers1