出于某种原因,我无法将接口(interface)中的方法声明为仅限包;它自动声明为公共(public)。这是简化的代码:
package com.example.project;
public interface MyInterface {
void foo();
Thing bar(); // The class Thing is in the com.example.project package, but what it does isn't important.
}
package com.example.project;
public abstract class SimpleConcrete implements MyInterface { // Initializes all methods as hooks
protected Thing bar = new Thing();
void foo() {}
Thing bar() { return bar }
}
package com.example.project;
public class ConcreteA extends SimpleConcrete {
void bar() {
// code here...
}
}
package com.example.project;
public class ConcreteB extends SimpleConcrete {
void bar() {
// more code here...
}
}
当我尝试编译它时,出现了这些错误,所有错误都相互关联:
File: C:\ProjectFolder\com\example\project\SimpleConcrete.java [line: 7]
Error: Cannot reduce the visibility of the inherited method from me.mathmaniac.everworlds.Block
File: C:\ProjectFolder\com\example\project\ConcreteA.java [line: 3]
Error: The inherited method com.example.project.SimpleConcrete.foo() cannot hide the public abstract method in com.example.project.MyInterface
File: C:\ProjectFolder\com\example\project\ConcreteB.java [line: 3]
Error: The inherited method com.example.project.SimpleConcrete.foo() cannot hide the public abstract method in com.example.project.MyInterface
有谁知道如何解决这个问题,或者我是否必须将我的接口(interface)对整个 Java 代码公开,而不仅仅是对包开放? 出于安全原因,我想让方法仅对包开放,但这是必要的,我会尝试找到另一种方式解决问题。
最佳答案
Java 接口(interface)中的所有方法都是隐式公共(public)的。请参阅此处了解更多信息:http://docs.oracle.com/javase/tutorial/java/IandI/interfaceDef.html
您可以通过如下声明来使接口(interface)成为包级别:
interface MyInterface {
void foo();
Thing bar();
}
这至少可以将其封装在该包中。
关于Java - 继承的方法com.example.project.ConcreteA无法隐藏com.example.project.MyInterface中的抽象方法,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/28754751/