CODEWITHSUNDEEP
javaoopgenericsinheritancearraylist
bcsta
bcsta

Reputation: 2327

Java How can a List hold both supertype and subtype at the same time?

I have a class, say Animal and a subclass Dog. I want to have a List named allAnimals that can hold both Animal objects and Dog objects at the same time. Currently, it is set up as List<Animal> allAnimals. I want to be able to perform two tasks:

  1. Add supertypes and subtypes. 

    List<Dogs> dogs = new List<>();
...
allAnimals.addAll(dogs);

  2. I want to be able to pass any subtype or supertype to a list as a parameter in a method.

    public void foo(List<Animal> animals){...}
...
foo(allAnimals);
or
foo(dogs);

Is the above possible? wildcard <? extends Animal> is giving an error in this case since foo is both putting things in the list and taking them out. Do I need to make Animal class abstract in order to achieve this?

Upvotes: 0

Views: 345

Answers (3)

D.Kerner
D.Kerner

Reputation: 1

You can try something like this:

public class Test {
    public void accept(List<? extends Animal> animals) {
        List<? super Animal> container = new ArrayList<>();
        container.add(animals.get(0));
        System.out.println(animals);
    }
    public void test() {
        List<Animal> animals = new ArrayList<>();
        accept(animals);

        List<Dog> dogs = new ArrayList<>();
        accept(dogs);
    }
}

Upvotes: 0

Joop Eggen
Joop Eggen

Reputation: 109613

In List is it specified: addAll(Collection<? extends E> c) so it works:

animals.addAll(dogs);

The same for your method:

void foo(List<? extends Animal> animals) {
}

A supertype of the specified is not possible (foo(new Object()) as parameter.

Upvotes: 0

Lecagy
Lecagy

Reputation: 489

List<? super Animal> animalList = new ArrayList<>();
animalList.add(new Animal());
animalList.add(new Dog("Snoopy"));
animalList.addAll(dogs);

so this list is mutable and can hold all Animals and its sub-classes like dog. Be aware that List<Dog> is not an instance of List<Animal>.

for your second task method overloading might be nesseccary, depending on what you want to do with the list

Upvotes: 0

Related Questions