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=-0.3 required=5.0 tests=BAYES_00, REPLYTO_WITHOUT_TO_CC autolearn=no autolearn_force=no version=3.4.4 X-Google-Thread: 103376,703c4f68db81387d X-Google-Thread: 109fba,703c4f68db81387d X-Google-Thread: 115aec,703c4f68db81387d X-Google-Thread: f43e6,703c4f68db81387d X-Google-Attributes: gid103376,gid109fba,gid115aec,gidf43e6,public X-Google-Language: ENGLISH,ASCII-7-bit Path: g2news1.google.com!news3.google.com!news.glorb.com!news.cs.univ-paris8.fr!talisker.lacave.net!lacave.net!feeder.xsnews.nl!feeder.news-service.com!post.news-service.com!news1.surfino.com!not-for-mail Message-Id: <4983741.rg76pI6zgE@linux1.krischik.com> From: Martin Krischik Subject: Re: Teaching new tricks to an old dog (C++ -->Ada) Newsgroups: comp.lang.ada,comp.lang.c++,comp.realtime,comp.software-eng Reply-To: martin@krischik.com Date: Fri, 25 Mar 2005 14:55:40 +0100 References: <4229bad9$0$1019$afc38c87@news.optusnet.com.au> <1110052142.832650@athnrd02> <1110284070.410136.205090@o13g2000cwo.googlegroups.com> <395uqaF5rhu2mU1@individual.net> <1110329098.642196@athnrd02> <1110361741.551255@athnrd02> <422edaec$0$26554$9b4e6d93@newsread4.arcor-online.net> <1111464133.508323@athnrd02> <1111508368.393411.6450@z14g2000cwz.googlegroups.com> <1111522653.164049@athnrd02> <9Gi0e.815$FN4.52@newssvr21.news.prodigy.com> <25875118.D6KecKHI6J@linux1.krischik.com> <1gaaa12mhwkch$.13718sqwfpcq0.dlg@40tude.net> <1159231.e8lUUTcn41@linux1.krischik.com> <1t9ck682jq6m0$.1wtb2w4h2yu4s.dlg@40tude.net> <1134594.trGOGDrpi2@linux1.krischik.com> Organization: None User-Agent: KNode/0.8.0 MIME-Version: 1.0 Content-Type: text/plain; charset=us-ascii Content-Transfer-Encoding: 7Bit X-Complaints-To: abuse@surfino.com NNTP-Posting-Host: 83.169.175.19 (83.169.175.19) NNTP-Posting-Date: Fri, 25 Mar 2005 15:00:18 +0100 X-Trace: 6962a42441972f60c0ab622288 Xref: g2news1.google.com comp.lang.ada:9983 comp.lang.c++:47219 comp.realtime:1718 comp.software-eng:5334 Date: 2005-03-25T14:55:40+01:00 List-Id: Dmitry A. Kazakov wrote: > On Fri, 25 Mar 2005 10:50:46 +0100, Martin Krischik wrote: > >> Dmitry A. Kazakov wrote: >> >>> On Thu, 24 Mar 2005 14:38:39 +0100, Martin Krischik wrote: >>> >>>> Dmitry A. Kazakov wrote: >>>> >>>>> On Thu, 24 Mar 2005 11:47:10 +0100, Martin Krischik wrote: >>>>> >>>>>> Hyman Rosen wrote: >>>>>> >>>>>>> adaworks@sbcglobal.net wrote: >>>>>>>> I wish Ada had a good model of multiple inheritance. >>>>>>> > I don't mean the somewhat haphazard model of C++ >>>>>>> >>>>>>> What's wrong with C++'s model? >>>>>> >>>>>> As allways: the default behaviour. Default is static inheritance. >>>>> >>>>> As it will be with interfaces in Ada 2005, I suppose... >>>> >>>> It make no difference with interfaces as interfaces hold no data. >>> >>> Of course it does. Diamond diagram is as valid for interfaces as for >>> implementations (data). If you allow interfaces to be derived from other >>> interfaces (I don't know if Ada 2005 will), then: >>> >>> C -> A (both are interfaces) >>> C -> B (both are interfaces) >>> >>> Let X implements both A and B. Now: A's C of X = B's C of X? >> >> Where is the problem? There is no data in C, C only consists of its tag >> (or virtual function table). But the tag (or virtual function table) is a >> singelton and A's C and B's C are both the same. > > That's the point, whether they be same or not is to be defined solely by > X, not per language design. > > Forget about data, it is irrelevant. Primitive operations are data and > data are primitive operations. If C has abstract Foo, then may/shall X > implement X.A.C.Foo and X.B.C.Foo differently? Is it the same primitive > operation of X or is it two different operations? Or else, will Foo's > dispatching table split into two branches or not? > > The answer is: it depends on the design. Because both variants are needed. Yes, I see your point and foregot something here (silly me): At least one A ore B need to be an interfaces as well (If we consider the Java approch). And that resolves is: If A is interface than X.A.C.Foo If B is interface then X.B.C.Foo If A and B are interfaces then X.Foo Whatever the situation there is only one Foo. Of course I do see the advantages of full MI as well. Martin -- mailto://krischik@users.sourceforge.net Ada programming at: http://ada.krischik.com