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From: Gene <gene.ressler@gmail.com>
Newsgroups: comp.lang.ada
Subject: Re: Truncating a fixed-point value
Date: Fri, 9 Jul 2010 20:13:00 -0700 (PDT)
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On Jul 9, 7:21=A0pm, Adam Beneschan <a...@irvine.com> wrote:
> On Jul 9, 1:34=A0pm, Simon Wright <si...@pushface.org> wrote:
>
> > Adam Beneschan <a...@irvine.com> writes:
> > > I must be suffering from an embarrassing mental block, because I'm
> > > sure there's a simple solution, but I can't see what it is.
>
> > > I have a value X of a fixed-point type, X >=3D 0.0, and I need to
> > > compute Floor(X) as an Integer. =A0'Truncation and 'Floor aren't defi=
ned
> > > for fixed-point types; the type conversion Integer(X) rounds; and
> > > Integer(X-0.5) fails if X=3D0.0.
>
> > > How do others do this? =A0Or do I have to resort to an IF statement (=
or
> > > conditional expression in Ada 2012) to handle the different cases?
>
> > Looks as though Integer(X+0.5)-1 might do the trick ..
>
> Yes, it looks like it, as long as X >=3D 0.0 and 0.5 is a model number
> of the fixed-point type (I think I'm using the right term). =A0Those
> conditions do hold for the code I need it for. =A0It isn't useful for
> negative numbers---it yields -2 if X=3D-1.0, which is not useful whether
> you want a truncation or a floor operation---but I did stipulate
> nonnegative X in my original question.
>
> =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0-- Adam

Shouldn't there be a way to convert a fixed point type to a
corresponding modular or integer type with one integer value per
floating point value, use rem to compute a floor, and convert back?