How did the private method called by a reference object variable

Question

How did bird able to call the method fly if it's private

public class Bird {
    private static void fly() { 
        System.out.println("Bird is flying"); 
    }
    public static void main(String[] args) {
        fly(); // make sense
        Bird bird = new Bird();
        bird.fly();
    }
}

change `private void fly() { System.out.println("Bird is flying"); } ` to `private static void fly() { System.out.println("Bird is flying"); }` — Morteza Jalambadani, Oct 18 '18 at 06:32
Read this [tutorial](https://docs.oracle.com/javase/tutorial/java/javaOO/accesscontrol.html) and this [Q&A](https://stackoverflow.com/questions/16727414/the-use-of-visibility-modifiers-in-java). That will make it clear. — LuCio, Oct 18 '18 at 06:44

score 1 · Answer 1 · answered Oct 18 '18 at 06:36

1

The access modifier private means that the method can only be seen from within the class. Since you are within the class, this is possible.

answered Oct 18 '18 at 06:36

B.Jahrer

11
3

score 0 · Answer 2 · answered Oct 18 '18 at 06:35

In Java there are 3 modifiers:

-Public: You can access from anywhere

-Private: You can access from the same class

-Protected You can access from the same package and from any class that extends the class in which it is.

Your method is private and you are accessing it in the same class.

How did the private method called by a reference object variable

2 Answers2