I had the same problem, and found a working solution:

 interface Adapter<E> {} interface Adaptulator<I> { void add(Container<?, ? extends I> container); } static final class Container<E, I extends Adapter<E>> { public final Class<E> extl; public final Class<I> intl; public Container(Class<E> extl, Class<I> intl) { this.extl = extl; this.intl = intl; } } 

###Why it's working To understand that, we need to state our requirements:

1. Keep two different generics synchronized on something. In your case, this is the E.
2. One of the two generic needs to have some extra inheritance, here it's I.

Creating an extra class allow to meet these two requirements, by creating a tight context.

3. An extra requirement (but probably the most important) was the need of a generic method (not bound too much to our class).

This is solved by the permissive parameter Container<?, ? extends I>.

Note

It's just a guess, but in this kind of usage, in general, you quickly need a ? super A or ? super I somewhere.

I had the same problem, and found a working solution:

 interface Adapter<E> {} interface Adaptulator<I> { void add(Container<?, ? extends I> container); } static final class Container<E, I extends Adapter<E>> { public final Class<E> extl; public final Class<I> intl; public Container(Class<E> extl, Class<I> intl) { this.extl = extl; this.intl = intl; } } 

Why it's working

To understand that, we need to state our requirements:

1. Keep two different generics synchronized on something. In your case, this is the E.
2. One of the two generic needs to have some extra inheritance, here it's I.

Creating an extra class allow to meet these two requirements, by creating a tight context.

3. An extra requirement (but probably the most important) was the need of a generic method (not bound too much to our class).

This is solved by the permissive parameter Container<?, ? extends I>.

Note

It's just a guess, but in this kind of usage, in general, you quickly need a ? super A or ? super I somewhere.