Re: Accessing private member via subclass

From:

Lew <noone@lewscanon.com>

Newsgroups:

comp.lang.java.programmer

Date:

Tue, 24 Nov 2009 09:13:57 -0500

Message-ID:

<hegpn7$b4b$1@news.albasani.net>

Michal Kleczek wrote:

Arved Sandstrom wrote:

Michal Kleczek wrote:

In Java things get strange sometimes due to such irregularities. Look at
the following:

public class Super {

  private int i;

  <T extends Super> void m(T s) {
    s.i = 5;
  }

}

This compiles fine in Java - should it?

I believe it should. When using "extends", that reference "s" must
behave as an instance of Super; we can't use or guess at any extended
behaviour of a specific T. For example, if you had a List<? extends
Number>, reading from (accessing) that List gives you a Number.

In this case, being in the body of S, getting or setting a private
member field of S is OK.

But that basically means:
"we can access 'i' here because any value passed to m() will have a type
that is a subtype of Super and i is a member of Super".
On the other hand in the original example we have:
"we cannot access 'i' here because any proper subtype of Super does not
inherit 'i'"

Don't you think it is a contradiction?

No more so than that casting to 'Super' allows access. In other words, no -
it's entirely consistent. You can think of 'T extends Super' as casting to
'Super' whatever 'T' really is.

It would be a contradiction if ((Super) T).i were legal but the <T extends
Super> example were not.

--
Lew