Wildcard and type pameter bounds in java

Question

Consider this case:

class A {}

class B<T extends A, E extends T> {
    B<?, A> b;
    B<?, ? extends A> b2;
}

As I understand type bounds, in this case effective upper bounds of both T and E is class A. So the question: why javac doesn't accept class A as argument in declaration of field b, but accepts wildcard ? extends A in declaration of field b2?

Answer 1

With the following classes:

class A {}
class C extends A {}
class B<T extends A, E extends T> {}

Think of it like this:

E extends T extends A
With B<?,A> then T -> ? and E -> A
A extends ? extends A
Where the ? could be any subclass of A, let's say C.
A extends C extends A is clearly invalid.
So that's why it's a compile error.

---

Note for Eclipse users:

Eclipse 4.9.0 compiler disagreed with javac 8u and Intellij and did not emit a compile error for the generic arguments in B<?,A>. I assume this is a bug in the Eclipse compiler but I have not consulted the JLS to confirm this.

class B<T extends A, E extends T> {
    B<?, A> b; // <-- Eclipse does NOT emit a compile error
    B<?, ? extends A> b2;
}

This assumed bug has been reported here.

Answer 2

Your declaration is incorrect. You are missing the point of 'wildcard'-s. They are used for unknown property declaration. For B<?, E> b the E must extend the T and the T must extend the A but you said that the first generic type is ? which is unknown! So you said that an unknown parameter must be extend T and so A. It is incorrect.

Relationship you created is like E -> T -> A. After that you declared as a 1. generic type ? -> E -> T -> A and 2. generic type as A -> E -> T -> ?. So ? must extends ? and A must extends E vs.. It is confusing and unknown for compiler...