Re: Truncating a fixed-point value

[Adam Beneschan <adam@irvine.com>]

On Jul 9, 1:34 pm, 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 >= 0.0, and I need to
> > compute Floor(X) as an Integer.  'Truncation and 'Floor aren't defined
> > for fixed-point types; the type conversion Integer(X) rounds; and
> > Integer(X-0.5) fails if X=0.0.
>
> > How do others do this?  Or 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 >= 0.0 and 0.5 is a model number
of the fixed-point type (I think I'm using the right term).  Those
conditions do hold for the code I need it for.  It isn't useful for
negative numbers---it yields -2 if X=-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.

                         -- Adam