There is the unmodifiableList method in the Collections class. It returns a List<T> but in the parameters it accepts List<? extends T>.
So if I understand this right then this method accepts a List which has T objects or objects that inherited from T class.
But I don't understand why the <? extends T> part in the parameter. I mean the parameter determines the return type of the method. If the method is called with List<Animal> then the return will be List<Animal> so it is going to be the same. Why we need this <? extends T>? Why don't we need just List<T> in the parameter list? I don't understand this.
If you could provide an example I would be happy about it.
Thank you in advance!
So if I understand this right then this method accepts a List which has T objects or objects that inherited from T class
This is oversimplifying matters.
a List<Number> can contain solely integers. The point of a List<Number> is that it is constrained to contain only number instances. Not that each instance MUST be Number and specifically Number - that would be impossible, as Number is abstract (no instances of Number itself can possibly exist, only instances of some subclass of it). The only difference between a List<Integer> containing only integers and a List<Number> containing only integers, is that you can add, say, a Double instance to that List<Number>, but you can't add it to a List<Integer>.
Because a method that takes, as argument, a List<Something> could invoke .add(), the type system needs to worry about the difference. It needs to do so even if this method does not, in fact, call add at all, and never intends to. The type system doesn't know that, and you're free to change the body of a method in some future version without that breaking backwards compatibility, thus, it matters. Even if you feel like it does not.
That then also explains your question:
If you have some method that demands a List<Number>, then you may provide a List<Integer> to Collections.unmodifiableList, and that's fine - Collections.uL will give you that List<Number>. This would ordinarily be broken, in that you can invoke .add(someDouble) which you should not be doing to a list of integers, but .add doesn't work on an unmodifiable, so it's fine.
List<? extends T> means it also accepts lists of subtypes.
Imagine having a class Plant and a class Tree that extends Plant.
You could create an unmodifiable List of Plants using a List<Tree> because Collections.unmodifiableList accepts lists of any subtype of T.
If it was List<T>, it would accept a List<Plant> as a parameter in order to create an unmodifiable List<Plant> but you could not create an unmodifiable List<Plant> using a List<Tree> as parameter.
Related
List<? extends Number> l = new ArrayList<Integer>();
I have the above line of code.
The object 'l' can refer any objects which can be Number any of its sub types.
I know that we can not add anyting to the above list.
Then what is the use of "? extends any_object" since we can not add anything to it ?
In what kind of context it is used ?
Can anyone please explain.
Thanks!!
It’s most useful for method arguments. Consider this:
public void printFormatted(Collection<? extends Number> values) {
NumberFormat format = new DecimalFormat("0.000");
for (Number value : values) {
System.out.println(format.format(value));
}
}
If I declare the method with Collection<Number>, callers must pass a Collection (or List or Set) whose generic type is known to be Number. They cannot pass Collection<Integer>, because Collection<Integer> is not a subclass of Collection<Number> and in fact is not polymorphically compatible with Collection<Number>.
By declaring the method with Collection<? extends Number>, I am saying that callers can pass Collection<Number>, or Collection<Integer>, or Collection<Double>, etc. List<Integer> or Set<Integer> is also acceptable, since the class is polymorphic, even though the generic type is not.
One of possible examples:
When you pass such method as argument, you can iterate over elements of such such list and use methods appropriate to Number class. You need to be aware that you cannot add any elements to such list. You are able to use list.get() method and that way, you're sure that you will get an instanceof Number class.
Such list is then called a producer. If you would need to add elements to list, instead of retrieving them, you would use <? super Number>. That rule is called PECS (Producer Extends Consumer Super).
I understand that one reason Lower-bounded wildcards exist is so that a collection is not immutable when adding new elements.
E.g.
List<? extends Number> obj = new ArrayList<>();//Now this list is immutable
obj.add(new Integer(5));//Does not compile
List<? super Number> objTwo = new ArrayList<>();//This list is mutable
objTwo.add(new Integer(5));//Compiles
The following does not compile because I tried to get the long value of numbers.
Q1: What methods would I be able to use? Only Objects methods?:
public void testLowerBounds(List<? super Number> numbers){
if (!numbers.isEmpty()){
System.out.println(numbers.get(0).longValue());//Does not compile
}
}
How my question came about:
I am currently learning about streams and the book specifies the following stream method:
Optional<T> min(Comparator<? super T> comparator)
And implements it as follows:
Stream<String> s = Stream.of("monkey", "ape", "bonobo");
Optional<String> min = s.min((s1, s2) -> s1.length()—s2.length());
Q2: How is the comparator allowed to use string methods when is used?
If I had to answer Q2: I would say that optional is specifying "You have to pass me an implementation of Comparator that has a generic type "String" or something that implements "String". Would I be correct in saying this?
Looking forward to your response.
First of all, you should not confuse wildcard type parameters with mutability. Having a wildcard in a List’s element type does not prevent modifications, it only imposes a few practical restrictions to what you can do with the list.
Having a list declared like List<? extends Number> implies that the referenced list has an actual element type of Number or a subclass of Number, e.g. it could be a List<Integer> or List<Double>. So you can’t add an arbitrary Number instance as you can’t know whether it is compatible to the actual element type.
But you can still add null, as the null reference is known to be compatible with all reference types. Further, you can always remove elements from a list, e.g. call remove or clear without problems. You can also call methods like Collections.swap(list, index1, index2), which is interesting as it wouldn’t be legal to call list.set(index1, list.get(index2)) due to formal rules regarding wildcard types, but passing the list to another method that might use a non-wildcard type variable to represent the list’s element type works. It’s obviously correct, as it only sets elements stemming from the same list, which must be compatible.
Likewise, if you have a Comparator<Number>, you can call Collections.sort(list, comparator), as a comparator which can handle arbitrary numbers, will be able to handle whatever numbers are actually stored in the list.
To sum it up, having ? extends in a collection’s element type does not prevent modifications.
As said, you can’t insert arbitrary new elements into a list whose actual element type might be an unknown subclass of the bound, like with List<? extends Number>. But you are guaranteed to get a Number instance when retrieving an element, as every instance of subtype of Number is also an instance of Number. When you declare a List<? super Number>, its actual element type might be Number or a super type of Number, e.g. Object or Serializable. You can insert arbitrary Number instances, as you know it will be compatible to whatever actual element type the list has, as it is a super type of number. When you retrieve an instance, you only know that it is an instance of Object, as that’s the super type of all instances. To compare with the ? extends case, having a ? super declaration does not prevent reading, it only imposes some practical limitations. And likewise, you can still pass it to Collections.swap, because, regardless of how little we known about the actual type, inserting what we just retrieved from the same list, works.
In your second question, you are confusing the sides. You are now not looking at the implementation of min, but at the caller. The declaration of min(Comparator<? super T> c) allows the caller to pass any comparator being parameterized with T or a super type of T. So when you have a Stream<String>, it is valid to pass a Comparator<String> to the min method, which is exactly, what you are implementing via the (s1, s2) -> s1.length()—s2.length() lambda expression (though, I’d prefer Comparator.comparingInt(String::length)).
Within the implementation of min, there is indeed no knowledge about what either, T or the actual type argument of the Comparator, is. But it’s sufficient to know that any stream element that is of type T can be passed to the comparator’s compare method, which might expect T or a super type of T.
I'm having difficulty understanding wildcards in Java generics. Specifically I have the following questions:
If we have a LinkedList<?>, why can we not add an Object to it? I understand that it doesn't know the type of the list, but wouldn't adding an Object to the list have us covered in any situation?
Similar to the question above, if we have LinkedList<? extends Number>, why can we not add a Number to it?
Finally, if we have LinkedList<? super Number>, why can we add an Integer to the list, shouldn't we only be able to add things that are a superclass of Number?
I guess I'm trying to understand how wildcards work in general, I've read the Oracle tutorials on them, and a few other things, but I don't understand why they work I suppose.
You're confusing objects and types.
Unlike simple generic parameters, wildcards describe the type of the generic parameter.
A List<? super Number> isn't a list of superclasses of Number; it's a list of some unknown type, where that type is a superclass of number.
A LinkedList<?> might be a LinkedList<Car>.
Since an Object is not a Car, you can't add an Object to it.
In fact, since you don't know anything about what type the list contains, you can't add anything to it. (except null)
Similarly, a LinkedList<? extends Number> might be a List<Long>, so you can't add an Integer to it. (since an Integer is not a Long)
On the other hand, a List<? super Number> is definitely allowed to contain Number or any derived class, since it can only be a list of one of Number's superclasses (eg, List<Object>)
Here is the problem that I have been being tried to find the solution.
We have two class definitions. One of two extends other one.
class T{}
class TT extends T{}
The requirement is that there should be a list keeps object extends T
List<? extends T> list = new ArrayList<>();
But the problem occures when I try to put a TT object ( barely seems it is a subclass of T )
into the list.
list.add(new TT());
Compilation Error Message
The method add(capture#2-of ? extends Cell) in the type List is not applicable for the arguments (Cell)
You can create a List<T> list = new ArrayList<T>(); directly, this can allow all subtypes of T into the list. This is actually little difficult to understand. when you declare it as
List<? extends T> list = ...
It means that it can allow any unknown subtypes of T into the list. But, from that declaration we cannot ensure which is the exact sub-type of T. so, we can only add null into it
List<? extends T> indicates that anything can comes out of it can be cast to T, so the true list could be any of the following:
List<T>
List<T2>
List<TT>
etc
You can see that even a new T cannot safely be added to such a collection because it could be a List<T2> which T cannot be put into. As such, such List cannot have non null entries added to them.
In this case you may simply want List<T>
So why would you ever use this?!
This contravariance can be useful for method parameters or returns, in which a collection will be read, rather than added to. A use for this could be to create a method that accepts any collection that holds items that are T, or extend T.
public static void processList(Collection<? extends Vector3d> list){
for(Vector3d vector:list){
//do something
}
}
This method could accept any collection of objects that extends Vector3d, so ArrayList<MyExtendedVector3d> would be acceptable.
Equally a method could return such a collection. An example of a use case is described in Returning a Collection<ChildType> from a method that specifies that it returns Collection<ParentType>.
The requirement is that there should be a list keeps object extends T
If you just want a List where you can store objects of any class that extend from T, then just create a List like this:
List<T> list = new ArrayList<T>();
The way you've created a list currently, will not allow you to add anything except null to it.
There are boundary rules defined for Java Generics when using WildCards
**extends Wildcard Boundary**
List means a List of objects that are instances of the class T, or subclasses of T (e.g. TT). This means a Read is fine , but insertion would fail as you dont know whether the class is Typed to T
**super Wildcard Boundary**
When you know that the list is typed to either T, or a superclass of T, it is safe to insert instances of T or subclasses of T (e.g.TT ) into the list.
In your example , you should use "super"
An addition to the other answers posted here, I would simply add that I only use wild cards for method parameters and return types. They're intended for method signatures, not implementations. When I put a wildcard into a variable declaration, I always get into trouble.
A book I am reading on Java tells me that the following two pieces of code are equivalent:
public <T extends Animal> void takeThing(ArrayList<T> list)
public void takeThing(ArrayList<? extends Animal> list);
On the opposite page, I am informed that the latter piece of code uses the '?' as a wildcard, meaning that nothing can be added to the list.
Does this mean that if I ever have a list (or other collection types?) that I can't make them simultaneously accept polymorphic arguments AND be re-sizable? Or have I simply misunderstood something?
All help/comments appreciated, even if they go slightly off topic. Thanks.
Does this mean that if I ever have a list (or other collection types?) that I can't make them simultaneously accept polymorphic arguments AND be re-sizable?
No.
The two pieces of code are not completely equivalent. In the first line, the method takeThing has a type parameter T. In the second line, you're using a wildcard.
When you would use the first version, you would specify what concrete type would be used for T. Because the concrete type is then known, there's no problem to add to the list.
In the second version, you're just saying "list is an ArrayList that contains objects of some unknown type that extends Animal". What exactly that type is, isn't known. You can't add objects to such a list because the compiler doesn't have enough information (it doesn't know what the actual type is) to check if what you're adding to the list should be allowed.
Usually, if adding to a list is involved inside a method that accepts just the list without the thing to add, you'll have somewhere else something that is an Animal and you'll want to add it to the list. In this case your method must be declared so that all the list types it accepts allow adding an Animal into them. This will have to be a List<Animal> or a list of some supertype of Animal. It can't possibly be a List<Dog>— the element you are adding could be any Animal.
This is where the concept of the lower bound, and the keyword super, come in. The type declaration List<? super Animal> matches all the acceptable types as described above. On the other hand, you won't be able to get elements out of such a list in a typesafe way because they can in general be of any type at all. If a method wants to both add and get elements of declared type Animal, the only valid type it can accept is a List<Animal>.