# Re: MOD operator

```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

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