Advanced Java Generics question: why do we need to specify redundant information
I've got some generic class for my JPA model POJO that goes like this:
public interface Identifiable<PK extends Serializable> {
PK getUniqueId();
}
public interface GenericDao<T extends Identifiable<PK>> {
public T findById(PK id);
}
This code won't compile. For this to work, I need to specify
public interface GenericDao<T extends Identifiable<PK>, PK extends Serializable>
But that's redundant information !! The fact that T extends Identifiable imply that a PK type will be specified for the Identifiable instance and that this is the type to use for the DAO's PK.
How can I make this work without redundant information ?
Thanks, Fred
Edit: Simplified example
Asked by: Abigail168 | Posted: 28-01-2022
Answer 1
Have you tried:
public interface WatchableDao<T extends Watchable<?>>
(i.e. it's a Watchable<Something> but I don't care what Something is)
I haven't tried it, but it's worth a go.
EDIT: Post question edit, it seems that you really do need PK as a type parameter to the interface. In that case, I believe you have to effectively repeat the constraint as you are doing. Yes, it's redundant, but I think it's simpler than the language having to specify what effective constraints would apply to PK based on its use as a type argument elsewhere. If it's any consolation, the same is true in C#.
It also makes the constraints on PK clear from just the interface itself, rather than having to look at another interface to see what's feasible.
Answered by: First Name803 | Posted: 01-03-2022Similar questions
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