From 4973622212445677058 X-Google-Language: ENGLISH,ASCII-7-bit X-Google-Thread: f78e5,e94695845965b249 X-Google-Attributes: gidf78e5,public X-Google-ArrivalTime: 1992-12-11 07:48:16 PST Newsgroups: comp.std.c++ Path: sparky!uunet!munnari.oz.au!metro!extro.ucc.su.OZ.AU!maxtal From: maxtal@extro.ucc.su.OZ.AU (John MAX Skaller) Subject: Re: Zero-length structures and pointer comparisons Message-ID: <1992Dec11.154104.5690@ucc.su.OZ.AU> Sender: news@ucc.su.OZ.AU Nntp-Posting-Host: extro.ucc.su.oz.au Organization: MAXTAL P/L C/- University Computing Centre, Sydney References: <1992Dec8.143504.5590@jyu.fi> <5343@holden.lulea.trab.se> <1992Dec10.051649.10282@jyu.fi> Date: Fri, 11 Dec 1992 15:41:04 GMT Lines: 30 In article <1992Dec10.051649.10282@jyu.fi> sakkinen@jyu.fi (Markku Sakkinen) writes: >In article <5343@holden.lulea.trab.se> jbn@lulea.trab.se (Johan Bengtsson) writes: >>Markku Sakkinen (sakkinen@jyu.fi) wrote: > >You must have misunderstood the idea somehow in two ways. >First, no matter how an object identifier is represented, it can always >be interpreted as an integer or bit string for the purpose of >ordering: "violating the total ordering property" is impossible. > I think you missed something here. It may well be possible to have < > <= >= and the operator @ satifying a @ b means !(ab) form a total order using bitwise comparisons (provided the pointers are void *). However the operator @ need not be the operator ==, it is quite possible for a < b && a == b to be true, where < is bitwise and == the proper pointer comparison. This will be immediately true on the 8086 unless all far pointers are normalised. (== must normalise them, < need not). -- ;---------------------------------------------------------------------- JOHN (MAX) SKALLER, maxtal@extro.ucc.su.oz.au Maxtal Pty Ltd, 6 MacKay St ASHFIELD, NSW 2131, AUSTRALIA ;--------------- SCIENTIFIC AND ENGINEERING SOFTWARE ------------------