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*From*: Steve Kargl <sgk at troutmask dot apl dot washington dot edu>*To*: Brooks Moses <bmoses at stanford dot edu>*Cc*: fortran at gcc dot gnu dot org, gcc-patches at gcc dot gnu dot org*Date*: Thu, 19 Oct 2006 15:34:23 -0700*Subject*: Re: [PATCH][4.3] Expand lround inline for x86_64/i?86 SSE math*References*: <Pine.LNX.4.64.0610181606580.8942@zhemvz.fhfr.qr> <EEA92755-FDBD-4FF8-AAAB-ACBB9461A6F7@adacore.com> <9645931.1161211060458.JavaMail.root@dtm1eusosrv72.dtm.ops.eu.uu.net> <4537C969.5080509@moene.indiv.nluug.nl> <4537DC19.5000301@stanford.edu> <84fc9c000610191336m4462380x2394ae0232605628@mail.gmail.com> <4537F091.5090802@stanford.edu>

On Thu, Oct 19, 2006 at 02:39:29PM -0700, Brooks Moses wrote: > (Added a cc: to fortran@, so as to increase the population of people who > will complain if I say something that's incorrect!) > > Richard Guenther wrote: > >On 10/19/06, Brooks Moses <bmoses@stanford.edu> wrote: > >>Toon Moene wrote: > >>>Richard Guenther wrote: > >>>>I wonder if fortran specifies round differently, as the frontend > >>>>explicitly converts NINT(x) = INT(x + ((x > 0) ? 0.5 : -0.5)). > >>> > >>>Yep - sorry, don't have the reference handy. > >> > >>For what it's worth, since I do have the reference handy, that's > >>essentially a direct translation of how the Fortran 95 standard defines > >>the NINT intrinsic. > > > >So, does it define how the x + 0.5 is carried out with respect to > >intermediate rounding before converting to INT? Literaly writing > >the above yields 1.0 for NINT ( 0.5 - epsilon ) assuming the hardware > >rounds to nearest even for the addition. > > It doesn't define this, so far as I am aware -- and, really, I'm not at > all sure whether the standard authors intended the description to be > interpreted as exact math, or as numerically-approximate math. The standard doesn't defined anything about intermediate rounding. It states that NINT(X) = INT(X+0.5) for X > 0. You then find INT(A) = 0 for |A| < 1. A quick scan over the section on numeric binary operators did not reveal the normal weasel words, but I suspect that the result of X+0.5 is a processor-dependent approximation and round-to-nearest meets that criterium. OTOH, I think it is a bug because we can choose a better algorithm to determine if x+0.5 is truly less than 1. See for example round[f] in intrinsics/c99_functions.c -- Steve

**Follow-Ups**:**Re: [PATCH][4.3] Expand lround inline for x86_64/i?86 SSE math***From:*Steve Kargl

**References**:**Re: [PATCH][4.3] Expand lround inline for x86_64/i?86 SSE math***From:*Brooks Moses

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