How to implement a generic recursive interface?

Question

Another case of java generics confusion...

I am defining a hierarchy of interfaces which must include a method to return a list of that interface type. This is a standard use case for the recursive generic pattern (or whatever the name for this is):

interface SuperInter<T extends SuperInter<T>> {
    List<T> getEffects();
}

Now when I extends this interface:

interface SubInter extends SuperInter<SubInter> {}

I can implement the sub-interface and have the correct method to implement:

class SubImpl implements SubInter {

    @Override
    public List<SubInter> getEffects() {
        return null;
    }
}

and similarly any other interface which uses itself as the generic type will have its implementing class contain a method that returns a list of that interface.

However, I can't implement the super interface type correctly:

class SuperImpl implements SuperInter<SuperInter> {

    @Override
    public List<SuperInter> getEffects() {
        return null;
    }
}

Besides raw types warning, I get:

Bound mismatch: The type SuperInter is not a valid substitute for the bounded parameter <T extends SuperInter<T>> of the type SuperInter<T>

I guess because the class does not extends itself. How can I achieve this?

i might be wrong, but due to the recursion the only right way to write is is the following: class SuperImpl implements SuperInter<SuperInter<SuperInter<SuperInter<SuperInter<[to infinity]>>>> { — Lino
– Lino, Commented Jul 28, 2017 at 6:00

Eran · Accepted Answer · 2017-07-28 05:59:45Z

2

You can declare it as follows:

class SuperImpl implements SuperInter<SuperImpl> {

    @Override
    public List<SuperImpl> getEffects() {
        return null;
    }
}

answered Jul 28, 2017 at 5:59

Eran

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6 Comments

Mark Over a year ago

So I can't have the interface returned like in the subclass/interface case, only the implementing class?

mattmess1221 Over a year ago

Try having it return List<? super T>

Stuart Marks Over a year ago

(In your deleted answer to this other question you can use the two-arg overload of Comparator.comparing.)

Mark Over a year ago

@StuartMarks i don't understand what this relates to. Also do you know the answer to my comment?

Stuart Marks Over a year ago

@Mark Re your question/comment, offhand no, but I'll post if I think of anything.

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