You're calling the public function print, that's why it's fine. You cannot directly access x's value.
Joined: Jul 23, 2008
It can be seen but not directly referenced.For example, you couldn't say:
..neither could you call it statically, unless it was a class variable (static)
In your case as you are calling the method in the first occasion on an instance of the Test1 class and the method is public than the method works fine.
In the second method you are calling the print method inherited by the Test2 class which again because its public in the parent class its visible.
Visibility in this case,limits the direct reference of a member variable ie,to x. However the public access modifier on the print method means the print method is visible and works, AND ensures a degree of encapsulation (accessing x only through a public interface). Thus your examples are the same as the getter and setters conventions (accessor and mutators)
Steve [ October 03, 2008: Message edited by: Stephen Davies ]
be a well encapsulated person, don't expose your privates, unless you public void getWife()!