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).
Related
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.
In the following code Java, I have created a list nums. I can assign the another list during the declaration. But new items cannot be added except the null. So, does it mean the nums is readonly? Why? Is it possible to add new items in that list?
List<Integer> ints = new ArrayList<Integer>();
ints.add(1);
ints.add(2);
List<? extends Number> nums = ints;
nums.add(3.14); //Generates error
nums.addAll(ints); //Generates error
nums.add(null); //works
System.out.println(nums.get(0)); //works
I have gone through this link. I can't get exact reason.
Is it possible to add new items in that list?
Nope... because that code doesn't know what it's "actually" a list of. Imagine if you could:
List<String> strings = new ArrayList<>();
List<? extends Object> objects = strings; // This is fine
objects.add(new Object()); // Fortunately this *isn't* valid...
System.out.println(strings.get(0).length()); // Or what would this do?
Basically, when you use a wildcard like ? extends T you can only get values out via the API... and when you use a wildcard like ? super T, you can only put the values in via the API - because that's what's safe.
No, it's not read-only... even though that is typically the intention.
Given a List<? extends Number> object, the compiler converts its type to List<X> where X is an unknown subtype of Number. Therefore, the object does have an add(X) method. We can call the method with an X argument... for example, null.
And since get() returns X, we could also call add() with a value from get() .... Directly invoking list.add(list.get(i)) won't work, even though it makes sense. We will need a little helper.
The classic example is Collections.reverse(List<? extends Object> list). This method will modify the list, despite the wildcard.
You can also call mutating methods like clear(), of course, on any list.
That being said, wildcard is indeed mainly for use-site variance, and most often, it conveys the intention from the API designer of whether a type-parameter is intended for in or out. For example, by declaring List<? super/extends Foo>, the API expresses that it intends to inject T in to, or, get T out of, the list.
It is a misconception that wildcard makes read/write-only. But this misconception works in most use cases. And the more people having this misconception, the more it becomes a convention...
see my article on wildcard - http://bayou.io/draft/Capturing_Wildcards.html
It's helps when you think of List<? extends Number> nums as a List of some type of thing that extends Number, but you can't be sure what. As such, everything you do with nums needs to be able to done to such a thing.
Adding null works because null can be cast into absolutely anything. Everything else you try to add will fail, because the compiler can't be 100% certain that the thing you're adding extends the thing the list is made of.
Instead of List<? extends Number> nums do List<Number> nums, because you can still put in anything that extends Number.
? doesn't mean "anything", it is closer to meaning "some specific but unknown".
Generics is compile time only.
So Compiler will decide what is actual type we are going to use.
List<? extends Number>
It means we are not sure what actual type of the object.
So Compiler not make sure what is the actual type that list have.
I am trying to create a generic method which accepts a list of numbers in the format below. I know that I should be able to add any class to the list "o" which is at least a number (floats, integers, doubles) BUT also I should be able to add any object because all classes extend from object. In other words, object is a super class of any classs. So I wonder, why do I get an error on the line o.add(p); ?
public int checkType(List<? super Number> o)
{
Object p = new Object();
//error
o.add(p);
return - 1;
}
I followed the explanation of generics from Difference between <? super T> and <? extends T> in Java which is the accepted answer.
The type-parameter List<? super Number> o specifies that any List which can hold a Number can be passed to it.
Which could be List<Object>, but that could also be List<Number>.
So compiler will allow you to add only Number (or subtype) objects to it.
To test, call your methods with the following arguments.
checkType(new ArrayList<Object>()); // Works fine (and it can hold Object type).
checkType(new ArrayList<Number>()); // Works fine (But it can NOT hold Object type).
So, as you see a List<? super Number> means that you are trying to add a Number to the list (Rule of PECS).
One consideration?
List<Object> myObjectList = Arrays.asList(1, 2, new SuperObject(), 4, 5);
Java comes standard with a utility class Arrays.
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.
I am having a hard time to understand the concept of generics wild cards.
As per my understanding <?> type unknown is introduced to resolve the co-variance not supported in generics and it should fit any type of collection and <?extends T> means that you can have collection of types T or the class which extends T.<?super T> means you can have collection of types T or super(s) of T.
Please correct me, if the above is wrong.
When I try to write it like this:
import java.util.*;
public class Gclass {
static Gclass t;
public void write(List< ?super String > lw){
lw.add("b");
}
public void read(List< ? extends String> lr){
String s=lr.get(2);
System.out.println(s);
}
public static void main(String[] args) {
t=new Gclass();
List<String> l=new ArrayList<String>();
l.add("a");
l.add("");
System.out.println(l);
t.write(l);
System.out.println(l);
t.read(l);
System.out.println(l);
}
}
It works but my places of doubt are:
As per my understanding both (extends and super) includes the type declared, so in this particular case as my List is of type String. I could interchange the extends and super, but I get compilation error?
In case of write ? super Object is not working? It should work as it is super of String?
I did not check for read as String can not be extended, but I think I'm also missing a concept here.
I've read all answers on SO related to this problem, but am still not able to have a good understanding about it.
String is indeed a bit of a bad example as it is a final class, but consider something like Number instead.
If a method takes a parameter of type List<? extends Number> then you can pass it a List<Number> or a List<Integer> or a List<BigDecimal> etc. Within the method body it is therefore fine to take things out of the list (as you know they must be instances of Number) but you can't put anything in because you don't know whether or not it's safe (the compiler can't let you risk putting an Integer into a List<Float>, for example).
Conversely if the method takes List<? super Number> then you can pass it a List<Number> or List<Object> - you can't take anything out of this list because you don't know what type it is*, but you do know that it'll definitely be safe to put a Number in.
* technically you can take things out but the only type you can assign them to is Object
As per my understanding both(extends and super) includes the type declared(String here), so in this particular case as my List is of type String... I could interchange the extends and super but i get compilation error?
You're right that both ? extends String and ? super String includes String. But you are missing the point that, ? super String also includes CharSequence, Object, which is not in bounds of ? extends String. You can add a String to a List<? super String>, b'coz whatever type that list is of, it can definitely refer to a String. But, you cannot add say an Integer to a List<? extends Number>, because the list can be a List<Float> actually.
In case of write ? super Object is not working? It should work as it is super of String?
Object is a super class of String will fit in where you have ? super String, and use Object for that. So, ? super String can capture Object, but ? super Object cannot capture String, as String is not a super type of Object. Think of it like this: "Actual type replaces the ?, and it must satisfy the rules attached to that ?.
List<? super String> means that lw holds a value of List with type argument which is String or it's superclass, so you can add a String value "b" (because it can be casted to list's type argument).
List<? extends String> means that lw holds a value of List with type argument which is String or it's subclasses, so values from lw can be casted to String.