Java Generic List<List<? extends Number>>

Question

How come in java we cannot do:

List<List<? extends Number>> aList = new ArrayList<List<Number>>();

Even though this is OK:

List<? extends Number> aList = new ArrayList<Number>();

Compiler error message is:

Type mismatch: cannot convert from ArrayList<List<Number>> to List<List<? extends Number>>

Answer 1

In Java, if Car is a derived class of Vehicle , then we can treat all Cars as Vehicles ; a Car is a Vehicle . However, a List of Cars is not also a List of Vehicles . We say that List<Car> is not covariant with List<Vehicle> .

Java requires you to explicitly tell it when you would like to use covariance and contravariance with wildcards, represented by the ? token. Take a look at where your problem happens:

List<List<? extends Number>> l = new ArrayList<List<Number>>();
//        ----------------                          ------
// 
// "? extends Number" matched by "Number". Success!

The inner List<? extends Number> List<? extends Number> works because Number does indeed extend Number , so it matches " ? extends Number ". So far, so good. What's next?

List<List<? extends Number>> l = new ArrayList<List<Number>>();
//   ----------------------                    ------------
// 
// "List<? extends Number>" not matched by "List<Number>". These are
//   different types and covariance is not specified with a wildcard.
//   Failure.

However, the combined inner type parameter List<? extends Number> List<? extends Number> is not matched by List<Number> ; the types must be exactly identical . Another wildcard will tell Java that this combined type should also be covariant:

List<? extends List<? extends Number>> l = new ArrayList<List<Number>>();

Answer 2

I'm not very familiar with Java syntax but it seems that your issue is this:

Covariance & Contravariance

Answer 3

You should definitely use the ? type wildcard when appropriate, do not avoid it as a general rule. For example:

public void doThingWithList(List<List<? extends Number>> list);

allows you to pass a List<Integer> or a List<Long> .

public void doThingWithList(List<List<Number>> list);

allows you to only pass arguments declared as List<Number> . A small distinction, yes, but using the wildcard is powerful and safe . Contrary to how it may seem, a List<Integer> is not a subclass, or is not assignable, from List<Number> . Nor is List<Integer> a subclass of List<? extends Number List<? extends Number , which is why the code above does not compile.

Answer 4

Your statement does not compile because List<? extends Number> List<? extends Number> is not the same type as List<Number> . The former is a supertype of the latter.

Have you tried this? Here I'm expressing that the List is covariant in its type argument, so it will accept any subtype of List<? extends Number> List<? extends Number> (which includes List<Number> ).

List<? extends List<? extends Number>> aList = new ArrayList<List<Number>>();

Or even this. Here the type parameter for the ArrayList on the right-hand side is the same as the type parameter on the left-hand side, so variance is not an issue.

List<List<? extends Number>> aList = new ArrayList<List<? extends Number>>();

You should be able to just say

List<List<Number>> aList = new ArrayList<List<Number>>();

I tend to avoid the ? type wildcard whenever possible. I find that the expense incurred in type annotation is not worth the benefit.

Answer 5

List<List<? extends Number>> aList = new ArrayList<List<? extends Number>>();
aList.add(new ArrayList<Integer>());