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>>
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> 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> 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>>();
I'm not very familiar with Java syntax but it seems that your issue is this:
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, which is why the code above does not compile.
Your statement does not compile because 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> (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.
List<List<? extends Number>> aList = new ArrayList<List<? extends Number>>();
aList.add(new ArrayList<Integer>());
> on the LHS would better match the poster's intent.
– erickson Apr 14 '09 at 15:29>` gives the compilation error `incompatible types: ArrayList
– skomisa Mar 11 '20 at 17:34> cannot be converted to List
>`. It's counter intuitive (to me), but you can't use List> on the LHS when `new ArrayList
>();` is on the RHS. Things get tricky with multi-level wildcards. See [What is the difference between extends Base> and?](https://stackoverflow.com/q/60536437/2985643) for a recent related question.