Overriding a method whose argument extends “generified” class

Question

I've just encountered an interesting problem related to Java generics. I'd like to use a concrete definition (DerivedPaginator) instead of "generified" definition (Paginator<Derived>). In order to do that, I have to change the method defined in the PaginatorTest. I've tried different combinations, but I still don't know how to do it.

Could you please help me with solving this puzzle?

interface Base {
}

interface Derived extends Base {
}

interface Paginator<T extends Base> {
}

interface DerivedPaginator extends Paginator<Derived> {
}

interface PaginatorTest<T extends Base> {
    // how to define this method so that it would accept DerivedPaginator?
    void check(Paginator<T> it);

    // nice try, but no cigar
    //<Y extends Paginator<T>> void check(Y it);
}

interface DerivedPaginatorTest extends PaginatorTest<Derived> {
    // this works fine
    //@Override
    //void check(Paginator<Derived> it);

    // but this doesn't
    @Override
    void check(DerivedPaginator it);
}

Answer 1

You may declare

interface PaginatorTest<T extends Base, Y extends Paginator<T>> {
    void check(Y it);
}

and use it like

interface DerivedPaginatorTest extends PaginatorTest<Derived, DerivedPaginator> {
    @Override
    void check(DerivedPaginator it);
}

Answer 2

DerivedPaginator extends Paginator<Derived> , but a Paginator<Derived> is not necessarily a DerivedPaginator . I could make some other class that implements Paginator<Derived> but not DerivedPaginator , so you cannot guarantee that that argument to DerivedPaginatorTest .check() is a DerivedPaginator . I'm not even sure why you have the interface DerivedPaginator which does not add anything to Paginator<Derived> . Anyway, if you need to be able to have check() method take something more specific than Paginator<T> , you need to generify over that something too, like Howard shows.