问题
I have a A class in package1 and B is in package2 which inherits A. A contains method m1 which is protected. Now my doubt is when I create an object of B in another class C which is also package2, the object of B is unable to access method m1 why? Below is my code
package com.package1;
public class A {
protected void m1(){
System.out.println("I'm protectd method of A");
}
}
package com.package2;
import com.package1.A;
public class B extends A {
public static void main(String[] args) {
B b = new B();
b.m1(); // b object able to access m1
}
}
package com.package2;
public class C {
public static void main(String[] args) {
System.out.println("Hi hello");
B b = new B();
b.m1(); //The method m1() from the type A is not visible
}
}
Do protected method of super class become private in subclass?
回答1:
From JLS 6.6.2. Details on protected Access
A protected member or constructor of an object may be accessed from outside the package in which it is declared only by code that is responsible for the implementation of that object.
Let C be the class in which a protected member is declared. Access is permitted only within the body of a subclass S of C.
Means The protected modifier specifies that the member can only be accessed within its own package (as with package-private) and, in addition, by a subclass of its class in another package.
From Java Doc Controlling Access to Members of a Class
So you can access method m1 from class B even its not on same package because it subclass of A.
But you can't access method m1 from class C because neither its in same package as A nor its subclass of A.
So for accessing this method you can make method m1 public or move your class C into same package as class A
来源：https://stackoverflow.com/questions/30207940/what-happens-to-protected-method-of-a-super-class-in-base-class-in-java