Add elements to a List<? extends Number>

Question

Why it is not possible to add elements to a list of type List<? extends Number>?

for example :

public class GenClass{
  public static void main(String args[]){
    List<? extends Number> ml=new Vector<Integer>();

    ml.get(0); //ok
    ml.add(new Integer(7)); // erro de compilação
    ml.add(new Double(7d));// erro de compilação
    ml.add(new Number(7));// erro de compilação
    ml.add(new Object());// erro de compilação
  }
}

Answer 1

Adding elements to this type of list is not possible as the wildcard declaration of List<? extends Number> ml means that the variable ml may contain any family value Number, instead of any value of a specific type. See the equivalent assignments below that would be allowed:

List<? extends Number> foo3 = new ArrayList<Number>;  // Number "extends" Number
List<? extends Number> foo3 = new ArrayList<Integer>; // Integer extends Number
List<? extends Number> foo3 = new ArrayList<Double>;  // Double extends Number

That said, the type of object you can add to the List foo3 which would be valid after any of the above ArrayList would bump into the problems below:

• You can’t add one Integerbecause foo3 could be pointing to a List<Double>.
• You can’t add one Doublebecause foo3 could be pointing to a List<Integer>.
• You can’t add one Numberbecause foo3 could be pointing to a List<Integer>.

Therefore:

You cannot add any object to List<? extends T> because you can’t guarantee that List is actually pointing to this type, so you cannot guarantee that the object to be added is allowed in that list. The only "guarantee" you have is that you can only read this list and you will get one T subclass of T.

The reverse logic applies to the use of super, according to the assignments below, which are also valid:

List<? super Number> foo3 = new ArrayList<Number>; // Number é "superclasse" de Number
List<? super Number> foo3 = new ArrayList<Object>; // Object é "superclasse" de Number

About List<? super T>, the following conclusion can be reached:

You can’t read the type T specific (for example Number) from List<? super T> because you can’t guarantee what kind of List is really pointing to it. The only "guarantee" you have is that you are able to add a type value T(or any subclass of T) without violating the integrity of the list being pointed out.

The statement to the left(List<? extends Number> foo3) sets the data type that can be assigned to your Arraylists on the right(= new ArrayList<Integer>();), and not the data type that can be added to the created list.

_{Translated and adapted from this reply from Soen}

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References for reading:

• What are covariance and countervariance?

• What is the purpose of the super command when used in the parameter declaration of a method?