实现接口的Java类可以自动继承注释吗？
假设我有一个带有一些注释的接口，例如：
@SpecialClass
public interface IFoo { /* ... */ }
假设我创建一个实现该接口的类：
public class Foo implements IFoo { /* ... */ }
类 Foo 是否有可能以某种方式“继承”或自动复制全部或部分内容来自 IFoo 及其成员的注释（例如自动将 Foo 注释为 @SpecialClass 等）？
这对于实现 Web 服务类（例如，由 JAX-WS“wsimport”工具生成的类）来说非常方便，只需实现其带注释的接口，而无需显式地将接口注释复制到实现类（例如 javax.jws. WebService、javax.xml.ws.RequestWrapper 等）。
Suppose I had an interface with some annotation(s), for example:
@SpecialClass
public interface IFoo { /* ... */ }
And suppose I make a class that implements the interface:
public class Foo implements IFoo { /* ... */ }
Is it possible for class Foo to somehow "inherit" or automatically copy all or some of the annotations from IFoo and its members (e.g. automagically annotate Foo as @SpecialClass, etc.)?
This would be convenient for implementing web service classes (e.g. those generated by the JAX-WS "wsimport" tool) by just implementing their annotated interfaces without explicitly having to copy the interface annotations to the implementing class (e.g. javax.jws.WebService, javax.xml.ws.RequestWrapper, etc).
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编辑：我将这个答案留在这里供一般信息和未来的读者使用，但 Andreas 指出了我错过的 Javadoc 的重要部分：
换句话说，在这种情况下它不会有帮助。当然，只有当您可以控制注释本身时，它才有用。
我怀疑真正的答案是您只需在各处应用注释即可。如果您担心忘记一个，您可能需要编写一个单元测试来查找所有类（我意识到说起来容易做起来难）并检查实现给定接口的所有类是否存在注释。
您是否尝试过应用
继承对SpecialClass注释本身进行注释？这当然听起来正是您想要的。
EDIT: I'm leaving this answer here for general information and future readers, but Andreas pointed out an important bit of the Javadoc which I'd missed:
In other words, it wouldn't help in this situation. Also it's only useful if you have control over the annotation itself, of course.
I suspect the real answer is that you simply have to apply the annotation everywhere. If you're worried about forgetting one, you might want to write a unit test which finds all your classes (easier said than done, I realise) and checks that the annotation is present for all classes implementing the given interface.
Have you tried applying the
Inheritedannotation to theSpecialClassannotation itself?That certainly sounds like exactly what you want.