[PD] more help file suggestions for pd 0.47
brbrofsvl at gmail.com
Sun May 8 01:20:01 CEST 2016
% 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 at ucsd.edu> wrote:
> I _think_ (but am not sure) that "%" works differently on different
> CPU architectures.
> 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 at 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
> > with [%] - the odd one out seems to be [mod].
> > what do you say?
> > cheers
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