Saturday, March 26, 2011

Java Generics Tutorial - Part III - Wildcards


The previous posts introduced us to the basics of Java generics y their subtyping relations. In this posts we'll introduce wildcards and how can covariant and contravariant subtyping relations be established with generics.

Wildcards

As we've seen in the previous post, the subtyping relation of generic types is invariant. Sometimes, though, we'd like to use generic types in the same way we can use ordinary types:
  • Narrowing a reference (covariance).
  • Widening a reference (contravariance

Covariance

Let's suppose, for example, that we've got a set of boxes, each one of a different kind of fruit. We'd like to be able to write methods that could accept a any of them. More formally, given a subtype A of a type B, we'd like to find a way to use a reference (or a method parameter) of type C<B> that could accept instances of C<A>.

To accomplish this task we can use a wildcard with extends, such as in the following example:

List<Apple> apples = new ArrayList<Apple>();
List<? extends Fruit> fruits = apples;

? extends reintroduces covariant subtyping for generics types: Apple is a subtype of Fruit and List<Apple> is a subtype of List<? extends Fruit>.

Contravariance

Let's now introduce another wildcard: ? super. Given a supertype B of a type A, then C<B> is a subtype of C<? super A>:

List<Fruit> fruits = new ArrayList<Fruit>();
List<? super Apple> = fruits;

How Can Wildcards Be Used?

Enough theory for now: how can we take advantage of these new constructs?

? extends
Let's go back to the example we used in Part II when introducing Java array covariance:

Apple[] apples = new Apple[1];
Fruit[] fruits = apples;
fruits[0] = new Strawberry();

As we saw, this code compiles but results in a runtime exception when trying to add a Strawberry to an Apple array through a reference to a Fruit array.

Now we can use wildcards to translate this code to its generic counterpart: since Apple is a subtype of Fruit, we will use the ? extends wildcard to be able to assign a reference of a List<Apple> to a reference of a List<? extends Fruit> :

List<Apple> apples = new ArrayList<Apple>();
List<? extends Fruit> fruits = apples;
fruits.add(new Strawberry());

This time, the code won't compile! The Java compiler now prevents us to add a strawberry to a list of fruits. We will detect the error at compile time and we won't even need any runtime check (such as in the case of array stores) to ensure that we're adding to the list a compatible type. The code won't compile even if we try to add a Fruit instance into the list:

fruits.add(new Fruit());

No way. It comes out that, indeed, you can't put anything into a structure whose type uses the ? extends wildcard.

The reason is pretty simple, if we think about it: the ? extends T wildcard tells the compiler that we're dealing with a subtype of the type T, but we cannot know which one. Since there's no way to tell, and we need to guarantee type safety, you won't be allowed to put anything inside such a structure. On the other hand, since we know that whichever type it might be, it will be a subtype of T, we can get data out of the structure with the guarantee that it will be a T instance:

Fruit get = fruits.get(0);

? super
What's the behavior of a type that's using the ? super wildcard? Let's start with this:

List<Fruit> fruits = new ArrayList<Fruit>();
List<? super Apple> = fruits;

We know that fruits is a reference to a List of something that is a supertype of Apple. Again, we cannot know which supertype it is, but we know that Apple and any of its subtypes will be assignment compatible with it. Indeed, since such an unknown type will be both an Apple and a GreenApple supertype, we can write:

fruits.add(new Apple());
fruits.add(new GreenApple());

If we try to add whichever Apple supertype, the compiler will complain:

fruits.add(new Fruit());
fruits.add(new Object());

Since we cannot know which supertype it is, we aren't allowed to add instances of any.

What about getting data out of such a type? It turns out that you the only thing you can get out of it will be Object instances: since we cannot know which supertype it is, the compiler can only guarantee that it will be a reference to an Object, since Object is the supertype of any Java type.

The Get and Put Principle or the PECS Rule

Summarizing the behavior of the ? extends and the ? super wildcards, we draw the following conclusion:


Use the ? extends wildcard if you need to retrieve object from a data structure.
Use the ? super wildcard if you need to put objects in a data structure.
If you need to do both things, don't use any wildcard.

This is what Maurice Naftalin calls The Get and Put Principle in his Java Generics and Collections and what Joshua Bloch calls The PECS Rule in his Effective Java.

Bloch's mnemonic, PECS, comes from "Producer Extends, Consumer Super" and is probably easier to remember and use.

Next Steps

In the next post (coming soon), we will put all together in some examples to clarify how generics can be used to help us write cleaner, clearer and more type safe code.

Part I - The Basics
Part II - Subtyping
Part III - Wildcards
Part IV - Bounded Type Variables

2 comments:

sckonln said...

You might want to update this stmt:
List = fruits;

Binh Thanh Nguyen said...

Thanks, nice post