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>)
Related
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.
I was going through generics in Java and I'm having trouble trying to understand where I would use the following two.
I understand that the first myList would ensure that the list only contains elements of type Integer and all it's superclasses. Now I'm trying to figure out where myList2 would fit in here.
List<? super Integer> myList;
List<Class<? super Integer>> myList2;
Edit: It's not a duplicate of the question being linked...since this is clearly regarding the use of ? vs Class<? whereas the other question is about super vs extend
The first one, 'myList' may contain integer values. Example: myList.add(200);
The second one, 'myList2' may contain classes of type integer. Example: myList2.add(Integer.class);
Update:
As correctly pointed out in the comments, due to the "super" keyword being applied, in addition to objects/ classes of type Integer, all superclasses of the type Integer (that is: Number and Object) can be contained as well.
The first list contains objects of Integer and its superclasses.
The second one contains class objects (or simply classes) of Integer and its superclasses, i.e:
Integer.class;
Number.class;
You can read more about the differences here: The difference between Classes, Objects, and Instances
I was reading an interesting dzone article on covariance in java which is pretty easy to follow but there is one thing bugging me which doesnt make sense, the article is here https://dzone.com/articles/covariance-and-contravariance
I am quoting examples from the article here where it is explaining why a collection cannot be added to:
With covariance we can read items from a structure, but we cannot write anything into it. All these are valid covariant declarations.
List<? extends Number> myNums = new ArrayList<Integer>();
Because we can be sure that whatever the actual list contains, it can be upcasted to a Number (after all anything that extends Number is a Number, right?)
However, we are not allowed to put anything into a covariant structure.
myNumst.add(45); //compiler error
This would not be allowed because the compiler cannot determine what is the actual type of the object in the generic structure. It can be anything that extends Number (like Integer, Double, Long), but the compiler cannot be sure what
The paragraph above is what doesn't make sense to me, the compiler knows that the list contains Number or anything that extends it, and the ArrayList is typed to Integer. And the compiler knows about the literal int that is inserted.
So why does it enforce this as it seems to me like it can determine the type?
When the compiler comes across:
List<? extendsNumber> myNums = new ArrayList<Integer>();
it checks that the types on left and right hand side fit together.
But the compiler does not "remember" the specific type used for the instantiation on the right hand side.
One could say that the compiler remembers this about myNums:
it has been initialized
it is of type List<? extendsNumber>
Yes compilers can do constant folding; and data flow analysis; and it might be possible for a compiler to track that "instantiation type" information as well - but alas: the java compiler doesn't do that.
It only knows that myNums is an initialized List<? extends Numbers> - nothing more.
You are missing two points:
You are thinking in terms of local variables only:
public void myMethod() {
List<? extends Number> list = new ArrayList<Integer>();
list.add(25);
}
A compiler could easily detect the actual value of ?, here but I know of nobody who would write such a code; if you are going to use an integer list you just declare the variable as List<Integer>.
Covariance and contravariance are most useful when dealing with parameters and/or results; this example is more realistic:
public List<Integer> convert(List<? extends Number> source) {
List<Integer> target = new ArrayList<>();
for (Number number : source) {
target.add(number.intValue());
}
return target;
}
How is a compiler expected to know which is the type used to parametrize the list? Even if at compile time all the calls only pass instances of ArrayList<Integer>, at a later time code can use the method with a diferent parameter without the class being recompiled.
The paragraph above is what doesn't make sense to me, the compiler knows that the list contains Number or anything that extends it, and the ArrayList is typed to Integer. And the compiler knows about the literal int that is inserted.
No, what the compiler knows is that the list contains something that extends Number (including Number). It cannot tell if it is a List<Number> (in which case you may insert any instance of Number) or a List<Integer> (in which case you may only insert Integer instances). But it does know that everything you retrieve using get will be an instance of Number (even if it is not sure about the specific class).
The paragraph above is what doesn't make sense to me, the compiler knows that the list contains Number or anything that extends it, and the ArrayList is typed to Integer. And the compiler knows about the literal int that is inserted.
Consider a slightly different but related example:
Object obj = "";
By the argument above, the compiler should also be able to know that obj is actually a String, and so you'd be able to invoke String-specific methods on it:
obj.substring(0);
which anybody with even a little Java experience knows you can't.
You (and only you) (or, at least, the person who wrote the code) has the type information, and a wilful decision has been made to discard it: there is no reason to have declared the variable type to be Object, so why should the compiler have to put in work to try to recover that information? (*)
By the same token, if you want the compiler to know that the value of the variable
List<? extends Number> myNums = new ArrayList<Integer>();
is an ArrayList<Integer>, declare the variable to be of that type. Otherwise, it's just much simpler for the compiler to assume "this can be absolutely anything within the type bounds".
(*) I read this argument somewhere in another answer somewhere on SO. I can't remember where it was, or I'd give appropriate attribution.
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.
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>.