Java generics – ArrayList initialization

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You can’t assign a List<Number> to a reference of type List<Integer> because List<Number> allows types of numbers other than Integer. If you were allowed to do that, the following would be allowed:

List<Number> numbers = new ArrayList<Number>();
numbers.add(1.1); // add a double
List<Integer> ints = numbers;
Integer fail = ints.get(0); // ClassCastException!

The type List<Integer> is making a guarantee that anything it contains will be an Integer. That’s why you’re allowed to get an Integer out of it without casting. As you can see, if the compiler allowed a List of another type such as Number to be assigned to a List<Integer> that guarantee would be broken.

Assigning a List<Integer> to a reference of a type such as List<?> or List<? extends Number> is legal because the ? means “some unknown subtype of the given type” (where the type is Object in the case of just ? and Number in the case of ? extends Number).

Since ? indicates that you do not know what specific type of object the List will accept, it isn’t legal to add anything but null to it. You are, however, allowed to retrieve any object from it, which is the purpose of using a ? extends X bounded wildcard type. Note that the opposite is true for a ? super X bounded wildcard type… a List<? super Integer> is “a list of some unknown type that is at least a supertype of Integer“. While you don’t know exactly what type of List it is (could be List<Integer>, List<Number>, List<Object>) you do know for sure that whatever it is, an Integer can be added to it.

Finally, new ArrayList<?>() isn’t legal because when you’re creating an instance of a paramterized class like ArrayList, you have to give a specific type parameter. You could really use whatever in your example (Object, Foo, it doesn’t matter) since you’ll never be able to add anything but null to it since you’re assigning it directly to an ArrayList<?> reference.

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