java – 超类引用的访问修饰符和方法[复制]

参见英文答案 > Understanding java’s protected modifier????????????????????????????????????6个
>???????????? Why can’t I call a protected method from an inheriting class in another package in Java?????????????????????????????????????2个
为什么当我在子类中创建超类的引用时,只能从引用中调用public方法而不是受保护的方法. (这些类在不同的包中)

package pet;

public class Dog {
    protected void bark(){};
    void jump(){};  
    public void lick(){};
}


package other;
import pet.*;

public class Husky extends Dog {
    public static void main(String[] args){ 
        Husky h = new Husky();
        h.bark();     //COMPILES (Husky is a subclass of Dog - Protected method)
        h.jump();     //DOES NOT COMPILE (Different packages - package-private access method)

        Dog d = new Dog();
        d.bark();   //DOES NOT COMPILE WHY?
        d.jump();   //DOES NOT COMPILE (Different packages - package-private access method)
        d.lick();   //COMPILES (Method is public)
    }
}

我的问题是为什么不编译d.bark()？ bark方法有一个protected的访问修饰符,允许从同一个包或子类中的类访问它.发生什么了？

如果哈士奇引用能够访问bark方法,则通过相同的逻辑,dog引用也应该能够访问bark方法.

所以我只能假设Dog引用一定存在问题？

Dog d = new Dog();
    d.bark();