From mboxrd@z Thu Jan 1 00:00:00 1970 X-Spam-Checker-Version: SpamAssassin 3.4.4 (2020-01-24) on polar.synack.me X-Spam-Level: X-Spam-Status: No, score=-1.3 required=5.0 tests=BAYES_00,INVALID_MSGID autolearn=no autolearn_force=no version=3.4.4 X-Google-Language: ENGLISH,ASCII-7-bit X-Google-Thread: 103376,6a11907952c890f5 X-Google-Attributes: gid103376,public From: jsa@alexandria (Jon S Anthony) Subject: Re: Visible or hidden derivation for controlled types ? Date: 1997/03/17 Message-ID: #1/1 X-Deja-AN: 226275591 Distribution: world References: <3329621A.3E9D@elca-matrix.ch> <332B1F05.31A7@bix.com> Organization: PSI Public Usenet Link Newsgroups: comp.lang.ada Date: 1997-03-17T00:00:00+00:00 List-Id: In article <332B1F05.31A7@bix.com> Tom Moran writes: > I know of at least two Ada 95 compilers that don't take > type T is private; > procedure Initialize(X : in out T); > ... > private > type T is new Ada.Finalization.Controlled ... > to mean that the Initialize specified is the Controlled Initialzie for > T. If this is wrong, Hmmm, I would have thought that this is correct as the fact that T is such a derived type is private. > the meantime it's certainly safer and more likely portable to admit in > the public part that T is a controlled type. Why would you want that > fact to be private, BTW? ????? I would also think that the "correct" thing to do is place the Initialize in the private part. /Jon -- Jon Anthony Organon Motives, Inc. Belmont, MA 02178 617.484.3383 jsa@organon.com