% can be different with respect to sign in different implementations of C. fmod() in C is designed to work with floats.

On my system, -10 [mod 3] and -10 [% 3] in Pd work differently. [mod] outputs the positive remainder, which is 2, while % outputs the remainder with the sign of the dividend, which is -1.

[div] and [mod] form a pair. Given two numbers A and B, B*(A [div B])+(A [mod B]) = A. [%] and [/]—[int] should form a similar pair, so -10 [div 3] should yield -4, while int(-10 [/ 3]) should yield -3.

On Sat, May 7, 2016 at 6:44 PM, Miller Puckette <msp@ucsd.edu> wrote:

I _think_ (but am not sure) that "%" works differently on different
CPU architectures.

cheers
Miller

On Sat, May 07, 2016 at 06:27:33PM -0300, Alexandre Torres Porres wrote:
> 2016-05-07 14:53 GMT-03:00 Miller Puckette <msp@ucsd.edu>:
>
> > I put in a sentence to scare users away from "%". Use "mod" instead :)
> >
>
> oh, but I can't see it, so you just did it now, right?
>
> I know they differ for negative values input, never knew why the reason..
>
> expr also has a "%" function that behaves in the same way as the [%]
> object, to make things more confusing, a "fmod" function in expr also
> behaves in the same was as "%", but for float arguments, and not like
> vanilla's [mod]
>
> in max, [%~] (or [modulo~]) will behave the same way as "fmod" in expr,
> that is modulo for float arguments, which is also in agreement to pd
> vanilla's % - only that pd's is for ints.
>
> With all that, what I mean to ask and say is that I can't see what's wrong
> with [%] - the odd one out seems to be [mod].
>
> what do you say?
>
> cheers

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