Generics - wildcard

[Tim Bant]

Hi all,

I feel like I understand generics and its purpose, particularly around upper and lower bounding, I get that.

However, I can't quite work out what  unbounded wildcards  ? are really doing... Is it simply, being translated by the compiler as Object?

Thus, why doesn't the below compile? I can't see the purpose of the unbounded wildcard (?).

List<?> aList = new ArrayList<>();
aList.add("Hello");

Thanks as always

[Jhonson Fernando]

The wildcard character (?) can be used to specify an unbounded wildcard type, as shown in List<?>. This creates a list of unknown type, and can be useful in certain scenarios such as:

1. When writing a method that can be implemented using functionality provided in the Object class.
2. When using methods in a generic class that don't depend on the type parameter, such as List.size or List.clear.

One example of such a method is printList, which can print a list of any type. However, if it is specified as List<Object>, it will only be able to print lists of Object instances and not subtypes like List<Integer>, List<String>, List<Double>, etc. To make printList more generic, it can be rewritten using List<?> to accept any type of list.

// javac --release 21 --enable-preview Test.java
//java  --enable-preview Test
import java.util.*;

public static void printListA(List<Object> list) {
  for (Object elem: list)
    System.out.println(elem + " ");
  System.out.println();
}

public static void printListB(List<? > list) {
  for (Object elem: list)
    System.out.print(elem + " ");
  System.out.println();
}
void main() {
  List<Integer> la = new ArrayList<Integer> ();
  la.add(1);
  la.add(2);
  printListB(la); // compile
  printListA(la); // will not compile
}

Refer: https://docs.oracle.com/javase/tutorial/java/generics/unboundedWildcards.html

[Mike Simmons]

Tim Bant wrote:I feel like I understand generics and its purpose, particularly around upper and lower bounding, I get that.

However, I can't quite work out what  unbounded wildcards  ? are really doing... Is it simply, being translated by the compiler as Object?

No, better to say it's being translated as a List with lower bound of Object, a List<? extends Object>.  But it's still a wildcard.

Tim Bant wrote:Thus, why doesn't the below compile? I can't see the purpose of the unbounded wildcard (?).

List<?> aList = new ArrayList<>();
aList.add("Hello");

Well to answer the easier part first, there is no purpose; it doesn't make any sense here.

But the reason it doesn't compile is, since aList is a List<?>, the compiler treats it as if it might be a List of anything - a List<Object> or List<String> or List<Integer> or List<SomeUnknownClass>.  So when you get to

aList.add("Hello");

that could make sense if it were a List<Object> or List<String>.  But not if it's a List<Integer> or List<SomeUnknownClass>.  Because the string "Hello" is not a subtype of Integer or SomeUnknownClass. So, the compiler can't allow that, and doesn't.  

There is, of course, no reason to declare the type as List<?> if you know you're going to want to add a String to it later.  You can't do that.  You might want to declare a List<?> for other reasons - Jhonson gave a good example of that.  But in that example, he didn't try to add anything to the list, in the code that declared it as List<?>.  Rather, the list was created outside that method, and passed in.  He didn't need to add anything, or use any method calls on the elements of the list, other than an implicit call to toString() which is guaranteed to exist for any object.

[Tim Bant]

Thanks both