Java generics - 带有接口绑定问题的通配符

Question

Suppose that we have class X and generic class Y :

class X {}

class Y<T extends X> {}

Why is the following declaration legal?

Y<? extends Serializable> obj; //I understand this as any type extends X and Serializable

What I understand is that ? means any type that satisfies <T extends X> , including X itself. So, subclasses of X that implement Serializable fits here. However, X itself does not implement Serializable, so why does the previous declaration work?

Answer 1

This is valid because in ? extends Serializable ? extends Serializable ? is not determined yet & ? class must be implementing Serializable while it must be extending X per this definition. So, as soon as you provide assignment to this reference variable a type check would be done for that object (generics are to detect type error at compile time). The object must inherit Serializable as well as X else error will be shown.

Answer 2

What I understand is that ? means any type that satisfies <T extends X> , including X itself.

Yes, and you've added the additional restriction for the variable obj that the unknown type which ? stands for, must implement Serializable .

So, subclasses of X that implement Serializable fits here.

That's right.

However, X itself does not implement Serializable , so why does the previous declaration work?

It doesn't matter that X does not implement Serializable - why should this matter?

The extends Serializable is just an additional restriction that you added to the type particularly for the variable obj .

Note that to the variable obj you can only assign a value that is an instance of a class that extends X and that also implements Serializable .

// Error! X does not implement Serializable, so this is not allowed
Y<? extends Serializable> obj = new Y<X>();

class Z extends X implements Serializable { ... }

// No problem, since Z extends X and also implements Serializable
Y<? extends Serializable> obj = new Y<Z>();