Re: Accessing private member via subclass

From:

Patricia Shanahan <pats@acm.org>

Newsgroups:

comp.lang.java.programmer

Date:

Mon, 23 Nov 2009 07:02:24 -0800

Message-ID:

<z6KdnQlt-diZOZfWnZ2dnUVZ_qmdnZ2d@earthlink.com>

Michal Kleczek wrote:

Lew wrote:

Mike Schilling wrote:

I thought I understood Java's access control rules pretty well, but
this case puzzles me.

public abstract class Super
{
    private int i;

    void method(Sub s)
    {
        s.i = 2; // (*)
    }
}

public class Sub extends Super
{
}

[snip]

'private' members even to subclasses. Since 'Sub' is not an inner class,
it
does not have access to the private members of 'Super'. So 's.i' is
illegal.

But 's.i' is in the body of 'Super'. So it is legal (or there is another
rule I'm not aware of that makes it illegal)

The rule that makes it illegal is the i is not even a member of Sub, and
membership in Sub is needed to make the s.i notation valid.

"Members are either declared in the type, or inherited because they are
accessible members of a superclass or superinterface which are neither
private nor hidden nor overridden (?8.4.8)."

http://java.sun.com/docs/books/jls/third_edition/html/names.html#6.4.3

"If the identifier does not name an accessible member field of type T,
then the field access is undefined and a compile-time error occurs."

http://java.sun.com/docs/books/jls/third_edition/html/expressions.html#15.11.1

Patricia