继承时无法访问受保护的内部类 [英] Can't access protected inner class while inheriting

问题描述

阅读思考Java我坚持参加ex：6 of Inner Classes章节。

Reading through "Thinking in Java" i stuck in ex:6 of Inner Classes chapter.

练习6：（2）在自己的包中创建一个至少包含一个方法的接口。在单独的包中创建
类。添加一个实现该接口的受保护内部类。在
第三个包中，继承自您的类，并在方法内返回
protected内部类的对象，在返回期间向上转换为接口。

Exercise 6: (2) Create an interface with at least one method, in its own package. Create a class in a separate package. Add a protected inner class that implements the interface. In a third package, inherit from your class and, inside a method, return an object of the protected inner class, upcasting to the interface during the return.

这是我的代码：

interface

interface

package intfpack;
public interface IOne{
        void    f();
}

---

COne.java

具有实现接口的受保护内部类的类

---

COne.java

Class with protected inner class that implements the interface

package classpack;
import intfpack.*;
public class COne{
        protected class Inner implements IOne{
                public void f(){System.out.println("Inner class of COne");}
        } 
}

---

CTwo.java

继承自受保护内部类的类

---

CTwo.java

Inheriting from class with protected inner class

package thirdpack;
import classpack.*;
import intfpack.*;

public class CTwo extends COne{
        public IOne getInner(){
                IOne io = new Inner(); 
                return io;
        }
        public static void main(String[] args){
                CTwo ct = new CTwo();
                ct.getInner();
        }
}

Copmiler说下一个：

Copmiler says next:

javac CTwo.java
CTwo.java:9: Inner() has protected access in classpack.COne.Inner
                IOne io = new Inner(); 
                          ^
1 error

但书中说我可以访问受保护的内部派生类中的类。错误在哪里？

But the book says that i can access protected inner classes in derived class. Where is mistake?

推荐答案

错误消息是抱怨受保护的构造函数，而不是类。但是您没有在发布的代码中明确定义构造函数。在这种情况下，根据JLS ，默认构造函数将受到保护（与类相同）。

The error message is complaining about the constructor being protected, not the class. But you haven't explicitly defined a constructor in the code you posted. In this case, according to the JLS, the default constructor will be protected (the same as the class).

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