Generics : List<? extends Animal> is same as List<Animal>?
List<Dog>
is a subtype of List<? extends Animal>
, but not a subtype of List<Animal>
.
Why is List<Dog>
not a subtype of List<Animal>
? Consider the following example:
void mySub(List<Animal> myList) {
myList.add(new Cat());
}
If you were allowed to pass a List<Dog>
to this function, you would get a run-time error.
EDIT: Now, if we use List<? extends Animal>
instead, the following will happen:
void mySub(List<? extends Animal> myList) {
myList.add(new Cat()); // compile error here
Animal a = myList.get(0); // works fine
}
You could pass a List<Dog>
to this function, but the compiler realizes that adding something to the list could get you into trouble. If you use super
instead of extends
(allowing you to pass a List<LifeForm>
), it's the other way around.
void mySub(List<? super Animal> myList) {
myList.add(new Cat()); // works fine
Animal a = myList.get(0); // compile error here, since the list entry could be a Plant
}
The theory behind this is Co- and Contravariance.
With List<Animal>
, you know what you have is definitely a list of animals. It's not necessary for all of them to actually be exactly 'Animal's - they could also be derived types. For example, if you have a List of Animals, it makes sense that a couple could be Goats, and some of them Cats, etc - right?
For example this is totally valid:
List<Animal> aL= new List<Animal>();
aL.add(new Goat());
aL.add(new Cat());
Animal a = aL.peek();
a.walk(); // assuming walk is a method within Animal
Of course, the following would not be valid:
aL.peek().meow(); // we can't do this, as it's not guaranteed that aL.peek() will be a Cat
With List<? extends Animal>
, you're making a statement about the type of list you're dealing with.
For example:
List<? extends Animal> L;
This is actually not a declaration of the type of object L can hold. It's a statement about what kinds of lists L can reference.
For example, we could do this:
L = aL; // remember aL is a List of Animals
But now all the compiler knows about L is that it is a List of [either Animal or a subtype of Animal]s
So now the following is not valid:
L.add(new Animal()); // throws a compiletime error
Because for all we know, L could be referencing a list of Goats - to which we cannot add an Animal.
Here's why:
List<Goat> gL = new List<Goat>(); // fine
gL.add(new Goat()); // fine
gL.add(new Animal()); // compiletime error
In the above, we're trying to cast an Animal as a Goat. That doesn't work, because what if after doing that we tried to make that Animal do a 'headbutt', like a goat would? We don't necessarily know that the Animal can do that.
It is not. List<Animal>
says that the value which is assigned to this variable must be of "type" List<Animal>
. This however doesn't mean that there must only be Animal
objects, there can be subclasses too.
List<Number> l = new ArrayList<Number>();
l.add(4); // autoboxing to Integer
l.add(6.7); // autoboxing to Double
You use the List<? extends Number>
construct if you are interest in an list which got Number
objects, but the List object itself doesn't need to be of type List<Number>
but can any other list of subclasses (like List<Integer>
).
This is sometime use for method arguments to say "I want a list of Numbers
, but I don't care if it is just List<Number>
, it can be a List<Double>
too". This avoid some weird down casts if you have a list of some subclasses, but the method expects a list of the baseclass.
public void doSomethingWith(List<Number> l) {
...
}
List<Double> d = new ArrayList<Double>();
doSomethingWith(d); // not working
This is not working as you expecting List<Number>
, not a List<Double>
. But if you wrote List<? extends Number>
you can pass List<Double>
objects even as they aren't List<Number>
objects.
public void doSomethingWith(List<? extends Number> l) {
...
}
List<Double> d = new ArrayList<Double>();
doSomethingWith(d); // works
Note: This whole stuff is unrelated to inheritance of the objects in the list itself. You still can add Double
and Integer
objects in a List<Number>
list, with or without ? extends
stuff.