Can't add value to the Java collection with wildcard generic type

Question

Why this code does not compile (Parent is an interface)?

List list = ...
Parent p = factory.get();   // returns concrete         


        

Answer 1

Here's my understanding.

Suppose we have a generic type with 2 methods

type L<T>
    T get();
    void set(T);

Suppose we have a super type P, and it has sub types C1, C2 ... Cn. (for convenience we say P is a subtype of itself, and is actually one of the Ci)

Now we also got n concrete types L<C1>, L<C2> ... L<Cn>, as if we have manually written n types:

type L_Ci_
    Ci get();
    void set(Ci);

We didn't have to manually write them, that's the point. There are no relations among these types

L<Ci> oi = ...;
L<Cj> oj = oi; // doesn't compile. L<Ci> and L<Cj> are not compatible types. 

For C++ template, that's the end of story. It's basically macro expansion - based on one "template" class, it generates many concrete classes, with no type relations among them.

For Java, there's more. We also got a type L<? extends P>, it is a super type of any L<Ci>

L<Ci> oi = ...;
L<? extends P> o = oi; // ok, assign subtype to supertype

What kind of method should exist in L<? extends P>? As a super type, any of its methods must be hornored by its subtypes. This method would work:

type L<? extends P>
    P get();

because in any of its subtype L<Ci>, there's a method Ci get(), which is compatible with P get() - the overriding method has the same signature and covariant return type.

This can't work for set() though - we cannot find a type X, so that void set(X) can be overridden by void set(Ci) for any Ci. Therefore set() method doesn't exist in L<? extends P>.

Also there's a L<? super P> which goes the other way. It has set(P), but no get(). If Si is a super type of P, L<? super P> is a super type of L<Si>.

type L<? super P>
    void set(P);

type L<Si>
    Si get();
    void set(Si);

set(Si) "overrides" set(P) not in the usual sense, but compiler can see that any valid invocation on set(P) is a valid invocation on set(Si)

Answer 2

It's doing that for the sake of safety. Imagine if it worked:

List<Child> childList = new ArrayList<Child>();
childList.add(new Child());

List<? extends Parent> parentList = childList;
parentList.set(0, new Parent());

Child child = childList.get(0); // No! It's not a child! Type safety is broken...

The meaning of List<? extends Parent> is "The is a list of some type which extends Parent. We don't know which type - it could be a List<Parent>, a List<Child>, or a List<GrandChild>." That makes it safe to fetch any items out of the List<T> API and convert from T to Parent, but it's not safe to call in to the List<T> API converting from Parent to T... because that conversion may be invalid.

Answer 3

List<? super Parent>

PECS - "Producer - Extends, Consumer - Super". Your List is a consumer of Parent objects.

Answer 4

This is because of "capture conversion" that happens here.

Every time the compiler will see a wildcard type - it will replace that by a "capture" (seen in compiler errors as CAP#1), thus:

List<? extends Parent> list

will become List<CAP#1> where CAP#1 <: Parent, where the notation <: means subtype of Parent (also Parent <: Parent).

java-12 compiler, when you do something like below, shows this in action:

List<? extends Parent> list = new ArrayList<>();
list.add(new Parent());

Among the error message you will see:

.....
CAP#1 extends Parent from capture of ? extends Parent
.....

When you retrieve something from list, you can only assign that to a Parent. If, theoretically, java language would allow to declare this CAP#1, you could assign list.get(0) to that, but that is not allowed. Because CAP#1 is a subtype of Parent, assigning a virtual CAP#1, that list produces, to a Parent (the super type) is more that OK. It's like doing:

String s = "s";
CharSequence s = s; // assign to the super type

Now, why you can't do list.set(0, p)? Your list, remember, is of type CAP#1 and you are trying to add a Parent to a List<CAP#1>; that is you are trying to add super type to a List of subtypes, that can't work.