冗余通用参数 [英] Redundant generic parameters

问题描述

我有这两个接口和类：

I have this two interfaces and classes:

public interface Identifiable<T> {
    T getId();
}

public interface GenericRepository<T extends Identifiable<K>, K> {
    T get(K id);
}

public class MyEntity implements Identifiable<Long> {

    private Long id;

    public Long getId() {
        return id;
    }
}

public class MyService {
    private GenericRepository<MyEntity, Long> myEntityRepository;
}

这一切都按需要运作。但在我看来，GenericRepository（K）中的第二个泛型参数是多余的。因为我知道MyEntity是一个可识别的，我认为如果我最终能够像这样使用它会很棒：

It all works as desired. But in my opinion second generic parameter in GenericRepository (K) is redundant. Because I know that MyEntity is an Identifiable, I think it would be great if I can finally use it like this:

public class MyService {
    private GenericRepository<MyEntity> myEntityRepository;
}

但是，我尝试了各种不成功的事情。可能吗？如果没有，为什么不呢？

But I'm trying different things without succeeding. Is it possible? If not, why not?

更新：回答一些回复。我想编译器知道哪些类型是MyEntity中的通用类型。例如：

UPDATE: Answering some responses. I think compiler knows something about which type is the generic in MyEntity. For example:

public class MyEntityGenericRepository implements GenericRepository<MyEntity, Long> {
    // compiles...
}

public class MyEntityGenericRepository implements GenericRepository<MyEntity, String> {
    // compiler says: "Bound mismatch: The type MyEntity is not a valid substitute for the bounded parameter <T extends Identifiable<K>> of the type GenericRepository<T,K>"
}

推荐答案

我不认为你可以省略它。使用 T扩展可识别的< K> ，您想要说通用类型参数必须是可识别。由于 Identifiable 是一个泛型类，所以您也需要提及它的泛型类型参数（如果您想按照以下规则进行操作 - 如果您忽略它，则会失去所有泛型由于向后兼容性规则，对 GenericRepository >类型安全性）。另请注意， K 实际上用作 GenericRepository.get 的参数类型。由于该类型可能与 T 不同，因此您需要通过将其声明为另一个通用类型参数 GenericRepository 。否则，编译器无法知道 K 是什么。

I don't think you can omit it. With T extends Identifiable<K> you want to say that the generic type parameter must be an Identifiable. Since Identifiable is a generic class, you need to mention its generic type parameter too (if you want to play by the rules that is - if you omit it, you lose all generic type safety for GenericRepository, due to backward compatibility rules). Note also that K is actually used as the parameter type of GenericRepository.get. And since that type may be different from T, you need to satisfy the compiler by declaring it as another generic type parameter of GenericRepository. Otherwise the compiler has no way of knowing what K is.

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