Re: MOD operator

[Theo Brauers <th.brauers(at)fz-juelich.de>]

Thanks for your help. In the meantime I found (with help from Franz Rohrer 
and Markus Fuehrer) that there is a difference between 
	a MOD b = a-INT(a/b)*b
and 
	modulo(a,b) = a-FLOOR(a/b)*b

I wrote an function calc_modulo 
------
FUNCTION calc_modulo, a, b
	RETURN, a - FLOOR(a/b)*b
END
------
which works fine for me.
Cheers Theo
> 
> Theo Brauers wrote:
> >
> > I was wondering if the result of the MOD operator in IDL changed
> > from previous versions (4.x, 5.0) to the current version (5.3.1).
> > Now the output is:
> >
> > IDL> PRINT, (FINDGEN(8)-4.) MOD 3
> >      -1.00000     0.000000     -2.00000     -1.00000     0.000000
> >       1.00000      2.00000     0.000000
> >
> > When I programmed a function long ago I used the MOD operator
> > expecting the output
> >
> > IDL> PRINT, (FINDGEN(8)-4.) MOD 3
> >       2.00000     0.000000      1.00000      2.00000     0.000000
> >       1.00000      2.00000     0.000000
> >
> 
> To the best of my knowledge, this is always the way IDL (and most other
> languages providing a MOD operator) has worked.  If you wish to take the
> mod of a negative number , the result will be a negative number between
> -(n+1) and 0.  A positive input yields a positive result.  So when we're
> trying to limit a value to between 0 and 2*pi, we often end up with code
> that looks something like:
> 
> x = ((y mod (2*pi)) + 2*pi) mod (2*pi)
> 
> Hope this helps.
> 
> Phillip

-- 
----------------------------------------------
Dr. Theo Brauers
Institut fuer Atmosphaerische Chemie (ICG-3)
Forschungszentrum Juelich
52425 JUELICH, Germany
Tel. +49-2461-61-6646    Fax. +49-2461-61-5346
http://www.kfa-juelich.de/icg/icg3/MITARBEITER/th.brauers.html