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Wildcards (The Java™ Tutorials > Bonus > Generics)
Trail: Bonus
Lesson: Generics
Home Page > Bonus > Generics
Wildcards
Consider the problem of writing a routine that prints out all the elements in a collection. Here's how you might write it in an older version of the language (i.e., a pre-5.0 release):
void printCollection(Collection c) {
    Iterator i = c.iterator();
    for (k = 0; k < c.size(); k++) {
        System.out.println(i.next());
    }
}
And here is a naive attempt at writing it using generics (and the new for loop syntax):
void printCollection(Collection<Object> c) {
    for (Object e : c) {
        System.out.println(e);
    }
}
The problem is that this new version is much less useful than the old one. Whereas the old code could be called with any kind of collection as a parameter, the new code only takes Collection<Object>, which, as we've just demonstrated, is not a supertype of all kinds of collections!

So what is the supertype of all kinds of collections? It's written Collection<?> (pronounced "collection of unknown"), that is, a collection whose element type matches anything. It's called a wildcard type for obvious reasons. We can write:

void printCollection(Collection<?> c) {
    for (Object e : c) {
        System.out.println(e);
    }
}
and now, we can call it with any type of collection. Notice that inside printCollection(), we can still read elements from c and give them type Object. This is always safe, since whatever the actual type of the collection, it does contain objects. It isn't safe to add arbitrary objects to it however:
Collection<?> c = new ArrayList<String>();
c.add(new Object()); // Compile time error
Since we don't know what the element type of c stands for, we cannot add objects to it. The add() method takes arguments of type E, the element type of the collection. When the actual type parameter is ?, it stands for some unknown type. Any parameter we pass to add would have to be a subtype of this unknown type. Since we don't know what type that is, we cannot pass anything in. The sole exception is null, which is a member of every type.

On the other hand, given a List<?>, we can call get() and make use of the result. The result type is an unknown type, but we always know that it is an object. It is therefore safe to assign the result of get() to a variable of type Object or pass it as a parameter where the type Object is expected.

Bounded Wildcards

Consider a simple drawing application that can draw shapes such as rectangles and circles. To represent these shapes within the program, you could define a class hierarchy such as this:
public abstract class Shape {
    public abstract void draw(Canvas c);
}

public class Circle extends Shape {
    private int x, y, radius;
    public void draw(Canvas c) {
        ...
    }
}

public class Rectangle extends Shape {
    private int x, y, width, height;
    public void draw(Canvas c) {
        ...
    }
}
These classes can be drawn on a canvas:
public class Canvas {
    public void draw(Shape s) {
        s.draw(this);
   }
}
Any drawing will typically contain a number of shapes. Assuming that they are represented as a list, it would be convenient to have a method in Canvas that draws them all:
public void drawAll(List<Shape> shapes) {
    for (Shape s: shapes) {
        s.draw(this);
   }
}
Now, the type rules say that drawAll() can only be called on lists of exactly Shape: it cannot, for instance, be called on a List<Circle>. That is unfortunate, since all the method does is read shapes from the list, so it could just as well be called on a List<Circle>. What we really want is for the method to accept a list of any kind of shape:
public void drawAll(List<? extends Shape> shapes) {
    ...
}
There is a small but very important difference here: we have replaced the type List<Shape> with List<? extends Shape>. Now drawAll() will accept lists of any subclass of Shape, so we can now call it on a List<Circle> if we want.

List<? extends Shape> is an example of a bounded wildcard. The ? stands for an unknown type, just like the wildcards we saw earlier. However, in this case, we know that this unknown type is in fact a subtype of Shape. (Note: It could be Shape itself, or some subclass; it need not literally extend Shape.) We say that Shape is the upper bound of the wildcard.

There is, as usual, a price to be paid for the flexibility of using wildcards. That price is that it is now illegal to write into shapes in the body of the method. For instance, this is not allowed:

public void addRectangle(List<? extends Shape> shapes) {
    shapes.add(0, new Rectangle()); // Compile-time error!
}
You should be able to figure out why the code above is disallowed. The type of the second parameter to shapes.add() is ? extends Shape-- an unknown subtype of Shape. Since we don't know what type it is, we don't know if it is a supertype of Rectangle; it might or might not be such a supertype, so it isn't safe to pass a Rectangle there.

Bounded wildcards are just what one needs to handle the example of the DMV passing its data to the census bureau. Our example assumes that the data is represented by mapping from names (represented as strings) to people (represented by reference types such as Person or its subtypes, such as Driver). Map<K,V> is an example of a generic type that takes two type arguments, representing the keys and values of the map.

Again, note the naming convention for formal type parameters--K for keys and V for values.

public class Census {
    public static void addRegistry(Map<String, ? extends Person> registry) {
}
...

Map<String, Driver> allDrivers = ... ;
Census.addRegistry(allDrivers);
Previous page: Generics and Subtyping
Next page: Generic Methods
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