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: 103376,338371dbbe7075d X-Google-Attributes: gid103376,public From: stt@houdini.camb.inmet.com (Tucker Taft) Subject: Re: [Q] Portability of <= and >= with real operands Date: 1996/12/03 Message-ID: #1/1 X-Deja-AN: 202099161 sender: news@inmet.camb.inmet.com (USENET news) x-nntp-posting-host: houdini.camb.inmet.com references: organization: Intermetrics, Inc. newsgroups: comp.lang.ada Date: 1996-12-03T00:00:00+00:00 List-Id: Keith Thompson (kst@aonix.com) wrote: : In dewar@merv.cs.nyu.edu (Robert Dewar) writes: : [...] : > The idea that the use of <= is more portable than either < or = : > is particular rubbish, I see no possible justification for such a : > statement, and, unlike the old rule about avoiding equality, I cannot : > even guess the thought behind this misunderstanding. : Suppose you've computed two quantities, X and Y, such that X is : mathematically known to be less than Y, but it may be arbitrarily close. : Since floating-point is of finite precision, the representations of X : and Y may be equal. (For example, X = 0.0, Y = some tiny value which : underflows to 0.0). Then X <= Y may reflect the relationship more : accurately than X < Y. Are you also saying that X <= Y is somehow "better" than not (X > Y)? That is truly mysterious, if so. : Of course, in real life you should analyze the particular case, and you : may well reach the opposite conclusion, but I'd bet that the original : rationale for the statement was something like the above. : Keith Thompson (The_Other_Keith) kst@aonix.com <*> -Tucker Taft stt@inmet.com http://www.inmet.com/~stt/ Intermetrics, Inc. Cambridge, MA USA