method visibility

[K.S Moon]

public class Test {
private int p = 1;
private void method(){
System.out.println(p);
}
public static void main(String[] args) {
Test t = new Test();
t.method(); //here
}
}
why can i access private method?
i am using jdk1.3.1

Thanks

[Valentin Crettaz]

Because the method is private to the class. Within the scope of the Test class, you can access any member, private or not.

[Garrett Smith]

Becuase this is true, it is also possible to access privates of other objects.

package inheritance;
/** 
 * Object promotion
 *
 * When a method is called and a subclass of the parameter type 
 * is passed in, a cast is affected on the parameter so as 
 * to make it of the parent class.
 */
 
class Parent {
    private void method() {
        System.out.println("Parent.method");
    }
    
    public static void main(String args[]) {
        callMethod(new Parent()); // access private method
        callMethod(new Child());
    }
    
    public static void callMethod(Parent p) {
        p.method();
    }
}
    
class Child extends Parent {
    void method() {
        System.out.println("Child.method");
    }
}

The above program shows that you can access method().
What about the program below?

package inheritance;
class Other {
    public static void main(String args[]) {
        Parent.callMethod(new Parent()); // access private method
        Parent.callMethod(new Child());
    }
}

* Parent.callMethod is public, so that shouldn't be a problem.
* Constructors Parent and Child are both accessible (default access, aka "package-private" access).
∴ This ought to compile.
What happens when Parent.callMethod tries to call method() on the Parent objects passed in? Does it work? Try it.

[K.S Moon]

Thanks a lot! Valentin & Garrett