Question

Let’s say you have:

public interface A {}

public class B implements A {}

public class C {
  void foo (List<A>) {}
}

public class Test {
  //Declaration one
  List<A> x = new List<A>();

  //Declaration two
  List<A> x = new List<B>();

  B b = new B();
  x.add(b);


  new C().foo(x);
}

Now obviously declaration one is the correct way to do this, and you receive a compile error on declaration two. I would like to know though why Java chooses to enforce type safety in this specific manner; if a list of Cats is still a list of Animals, why does a method expecting a list of animals balk at receiving a bunch of cats?

Curiousity, more than anything else – and a chance to better refine my knowledge.

Cheers,
Dave.

Answer 1

Java generics are not covariant. If you could do this:

ArrayList<Animal> x = new ArrayList<Cat>();

then you would be able to do:

x.add(new Dog());

which violates the concept that an ArrayList<Cat> can only hold Cat objects (or subclass objects).

Read this for for more details: Java theory and practice: Generics gotchas.