通配符和变异器方法

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<>(); 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.

Answer 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