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Date: | Thu, 30 Oct 2003 23:00:54 +0200 |
From: | "Eli Zaretskii" <eliz AT elta DOT co DOT il> |
Sender: | halo1 AT zahav DOT net DOT il |
To: | djgpp AT delorie DOT com |
Message-Id: | <6654-Thu30Oct2003230053+0200-eliz@elta.co.il> |
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=?iso-8859-15?Q?Rafael_Garc=EDa?= on Thu, 30 Oct 2003 14:07:52 +0100) | |
Subject: | Re: double with only 13 digit precision? |
References: | <oprxuxfe0j0bb83x AT news DOT mad DOT ttd DOT net> |
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> From: =?iso-8859-15?Q?Rafael_Garc=EDa?= <rafael AT geninfor DOT com> > Newsgroups: comp.os.msdos.djgpp > Date: Thu, 30 Oct 2003 14:07:52 +0100 > > I thought type double had at least 15 decimal digits of precision It does. > but I have found I can get a result with only 13 precise digits. Can > anyone please explain that result? This is expected, since a - b = a*(1 - b/a) and, for the case in point, where a and b are close to each other and thus b/a is on the order of magnitude of one, the precision of this expression is |a|*10^(-16). |a| is 4*10^3, so you get 10^(-13) of absolute pecision. In other words, this happens because the precision of FP arithmetics is 10^(-16) relative to the magnitude of the original operands.
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