[PD] 1 / 0 = 0?
williamahuston at gmail.com
Thu May 24 17:24:08 CEST 2018
I understand that tradeoff when working with real-time audio.
However, I was talking about control logic, not audio.
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On Thu, May 24, 2018 at 10:57 AM, Christof Ressi <christof.ressi at gmx.at>
> Hi William,
> there's no other sane way to handle division by 0 in the audio domain
> since the result must be a number and there are only two options: output 0
> or some ridiculously large number (which would be quite dangerous).
> Pd *could* handle it differently in the control domain, but why?
> as Matt said: just don't divide by 0 :-)
> Gesendet: Donnerstag, 24. Mai 2018 um 16:52 Uhr
> Von: "Matt Davey" <hard.off at gmail.com>
> An: "William Huston" <williamahuston at gmail.com>
> Cc: "pd-list at lists.iem.at" <pd-list at lists.iem.at>
> Betreff: Re: [PD] 1 / 0 = 0?
> Just don’t allow your function to blow up in the first place.
> On Thursday, May 24, 2018, William Huston <williamahuston at gmail.com[
> mailto:williamahuston at gmail.com]> wrote:
> I have a function which blows up at a certain point, moves to infinity.
> In this case, it is basically 1/0.
> I was a little surprised to find that 1/0 = 0 (according to Pd).
> I was expecting NaN, or an overflow condition would could be trapped.
> So what I have to do is examine the *input* for the values
> which will blow up my function, and set a flag. While this
> is easy in this single case, it is awkward in the general case.
> This seems somewhat broken to me.
> Is this a bug?
> William Huston: WilliamAHuston at gmail.com[mailto:WilliamAHuston at gmail.com]
> Binghamton NY
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