How does the port opcode work?
| Date | 2017-05-17 22:34 |
| From | Arthur Hunkins <000001e1d761dea2-dmarc-request@LISTSERV.HEANET.IE> |
| Subject | How does the port opcode work? |
My friend Beppi has asked me to pose this general question about port: controllers I have on my GUI. Unfortunately there are some "clicks" that are almost certainly caused by various k-parameters' sudden changes, so I implemented a portamento to any of these k-parameters. But I really don't understand how portamento should be used! The opcode port as I understood halves the distance between the current value and the destination value, at k-cycle. But how does it know what are the current and destination values, as soon as I pass just one of them as parameter? And how is it initialized at the very first run? Does it read the value of the left opcode parameter before writing it? In other words: Is this correct: kVibratoTemp chnget "vibrato" kVibrato port kVibratoTemp, 0.01 or does this make sense instead? kVibrato chnget "vibrato" kVibrato port kVibrato, 0.01 Thank you very much ------------------------------ Art Hunkins |
| Date | 2017-05-18 11:19 |
| From | jpff |
| Subject | Re: How does the port opcode work? |
> But I really don't understand how portamento should be used!
> The opcode port as I understood halves the distance between the current
> value and the destination value, at k-cycle. But how does it know what are
> the current and destination values, as soon as I pass just one of them as
> parameter? And how is it initialized at the very first run? Does it read the
> value of the left opcode parameter before writing it?
>
>
> In other words:
> Is this correct:
>
> kVibratoTemp chnget "vibrato"
> kVibrato port kVibratoTemp, 0.01
>
> or does this make sense instead?
>
> kVibrato chnget "vibrato"
> kVibrato port kVibrato, 0.01
it has internal state so the two versions are really the same
What is the problem>
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| Date | 2017-05-19 00:10 |
| From | Giuseppe Menozzi |
| Subject | Re: How does the port opcode work? |
kVibrato chnget "vibrato" Oh no problem, just trying to understand :) Now I understood. So, that particular "port" opcode in that particular point of the code remembers its previous internal status, now it's all clear, thank you. Beppi. |