Generics, Type Parameters and Wildcards

问题:

I am trying to understand java generics and they seem extremely difficult to understand. For example, this is fine...

public class Main {

    public static void main(String[] args) {
        List<?> list = null;
        method(list);
    }

    public static <T> void method(List<T> list) { }
}

... as is this...

public class Main {

    public static void main(String[] args) {
        List<List<?>> list = null;
        method(list);
    }

    public static <T> void method(List<T> list) { }
}

... and this ...

public class Main {

    public static void main(String[] args) {
        List<List<List<?>>> list = null;
        method(list);
    }

    public static <T> void method(List<List<T>> list) { }
}

... but this doesn't compile:

public class Main {

    public static void main(String[] args) {
        List<List<?>> list = null;
        method(list);
    }

    public static <T> void method(List<List<T>> list) { }
}

Can someone explain what is going on in simple language?

回答1:

The main thing to understand with generic types, is that they aren't covariant.

So whilst you can do this:

final String string = "string";
final Object object = string;

The following will not compile:

final List<String> strings = ...
final List<Object> objects = strings;

This is to avoid the situations where you circumvent the generic types:

final List<String> strings = ...
final List<Object> objects = strings;
objects.add(1);
final String string = strings.get(0); <-- oops

So, going through your examples one by one

1

Your generic method takes a List<T>, you pass in a List<?>; which is (essentially) a List<Object>. T can be assigned to the Object type and the compiler is happy.

2

Your generic method is the same, you pass in a List<List<?>>. T can be assigned to the List<?> type and the compiler is again happy.

3

This is basically the same as 2 with another level of nesting. T is still the List<?> type.

4

Here is where it goes a little pear shaped, and where my point from above comes in.

Your generic method takes a List<List<T>>. You pass in a List<List<?>>. Now, as generic types are not covariant, List<?> cannot be assigned to a List<T>.

The actual compiler error (Java 8) is:

required: java.util.List<java.util.List<T>> found: java.util.List<java.util.List<?>> reason: cannot infer type-variable(s) T (argument mismatch; java.util.List<java.util.List<?>> cannot be converted to java.util.List<java.util.List<T>>)

Basically the compiler is telling you that it cannot find a T to assign because of having to infer the type of the List<T> nested in the outer list.

Lets look at this in a little more detail:

List<?> is a List of some unknown type - it could be a List<Integer> or a List<String>; we can get from it as Object, but we cannot add. Because otherwise we run into the covariance issue I mentioned.

List<List<?>> is a List of List of some unknown type - it could be a List<List<Integer>> or a List<List<String>>. In case 1 it was possible to assign T to Object and just not allow add operations on wildcard list. In case 4 this cannot be done - primarily because there is not a generics construct to prevent add to the outer List.

If the compiler were to assign T to Object in the second case then something like the following would be possible:

final List<List<Integer>> list = ...
final List<List<?>> wildcard = list;
wildcard.add(Arrays.asList("oops"));

So, due to covariance, it is not possible to assign a List<List<Integer>> to any other generic List safely.