Set vs. Set<?>

Two facts about Set<?>

There are two facts about Set<?>:

• Item 1: Since the question mark ? stands for any type. Set<?> is capable of holding any type of elements.
• Item 2: Because we don't know the type of ?, we can't put any element into Set<?>

So a Set<?> can hold any type of element(Item 1), but we can't put any element into it(Item 2). Do the two statements conflict to each other? Of course they are not. This can be clearly illustrated by the following two examples:

Item 1 means the following situation:

//Legal Code
public static void main(String[] args) {
    HashSet<Integer> s1 = new HashSet<Integer>(Arrays.asList(1, 2, 3));
    printSet(s1);

    HashSet<String> s2 = new HashSet<String>(Arrays.asList("a", "b", "c"));
    printSet(s2);
}

public static void printSet(Set<?> s) {
    for (Object o : s) {
        System.out.println(o);
    }
} 

Since Set<?> can hold any type of elements, we simply use Object in the loop.

Item 2 means the following situation which is illegal:

//Illegal Code
public static void printSet(Set<?> s) {
    s.add(10);//this line is illegal 
    for (Object o : s) {
        System.out.println(o);
    }
}

Because we don't know the type of <?> exactly, we can not add any thing to it other than null. For the same reason, we can not initialize a set with Set<?>. The following is illegal:

//Illegal Code
Set<?> set = new HashSet<?>();
Set vs. Set<?>

What's the difference between raw type Set and unbounded wildcard Set<?>?

This method declaration is fine:

public static void printSet(Set s) {
    s.add("2");
    for (Object o : s) {
        System.out.println(o);
    }
}

because raw type has no restrictions. However, this will easily corrupt the invariant of collection.

In brief, wildcard type is safe and the raw type is not. We can not put any element into a Set<?>.

When Set<?> is useful?

When you want to use a generic type, but you don't know or care what the actual type the parameter is, you can use <?>[1]. It can only be used as parameters for a method.

For example:

public static void main(String[] args) {
    HashSet<Integer> s1 = new HashSet<Integer>(Arrays.asList(1,2,3));
    HashSet<Integer> s2 = new HashSet<Integer>(Arrays.asList(4,2,3));
 
    System.out.println(getUnion(s1, s2));
}
 
public static int getUnion(Set<?> s1, Set<?> s2){
    int count = s1.size();
    for(Object o : s2){
        if(!s1.contains(o)){
            count++;
        }
    }
    return count;       
}

Reference:

1. Bloch, Joshua. Effective java. Addison-Wesley Professional, 2008.