Re: Accessing private member via subclass

From:

Arved Sandstrom <dcest61@hotmail.com>

Newsgroups:

comp.lang.java.programmer

Date:

Tue, 24 Nov 2009 13:24:12 GMT

Message-ID:

<08ROm.54881$PH1.17550@edtnps82>

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?

I don't think it's a contradiction as things are defined now - generics
have their own rules - but I'm not fond of pushing the envelope like
this. Since the intent of the method is to set the private member field
of the base class to 5, I'd just code it as

i = 5;

and it becomes unambiguous. I think your general objection is valid - we
ought not have to ask questions like this.

My personal belief is that when we run across situations like this is
that (1) sometimes we should not be allowed to do it, and (2) we
shouldn't try to do it anyway. So in this particular case perhaps the
"s.i" should be illegal.

AHS