Re: Accessing private member via subclass

From:

Arved Sandstrom <dcest61@hotmail.com>

Newsgroups:

comp.lang.java.programmer

Date:

Mon, 23 Nov 2009 12:39:17 GMT

Message-ID:

<VnvOm.54744$PH1.40660@edtnps82>

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
{
}
Consider the starred line. The field "i" is private to Super and is
being accessed by Super, which seems to me to fit within JLS 6.6.1:

    if the member or constructor is declared private, then access
    is permitted if and only if it occurs within the body of the top
    level class (?7.6) that encloses the declaration of the member
    or constructor.

However, trying to compile these classes leads to:

Super.java:7: i has private access in Super
        s.i = 2;
         ^
1 error

in both 1.4.2_09 and 1.6.0_06.

Obviously, the error can be removed by changing the line to

    ((Super)s).i = 2;

And, just as obviously, the error doesn't actually prevent
encapsulation from being broken. For what it's worth, similar (in
fact, almost identical) C# code compiles with no problems.

Any thoughts about this?

I'll assume that this is a "let's explore the edge cases in the
language" test case. :-) Having said that, I don't read that section of
the JLS as pertaining to what you are doing here; you might as well have
a method in Super that does a similar thing to an instance of a class
that is not related to Super at all, and you certainly wouldn't expect
that to work either. IOW, using Sub in that method is muddying the
waters. The key point is that you're trying to access a private member
variable and you're not actually in the class that owns it because of
the "s.i" syntax.

I'm a bit surprised that C# would permit this. How identical is "almost
identical"? :-)

AHS