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Thanks for the advice,
Is there any example about how to do this?
Does it need Lua coding?
Regards
F7hOctober 23, 2019 at 5:10 pm in reply to: Receive MidiMessage in a modulator function Lua script? #116421F0h
msepsis
Thanks!!
That’s what I was looking for.
F7hHello seq303,
Have you find a solution to your LFO with CTRLR?
CheersF0h
In this example I sent CC33 to CC44 with random values.
--send CC33 ... CC44 ---- for i = 1 , size do panel:sendMidiMessageNow(CtrlrMidiMessage(string.format('B0 %.2x %.2x', 32 + i,Table))) end --end
- This reply was modified 4 years, 6 months ago by F0h.
Random_table = function(modulator, value) math.randomseed(Time.getMillisecondCounterHiRes()) size = 12 Min = 1 Max = 20 Table ={} if math.abs((Max-Min)-1)<=size then console('Error table size') do return end end --Fill the table with -1 --- for i = 1, size do Table = -1 end --End fill the table--- ---Number generator with no repetition--- for i = 1 , size do flag = 1 --suppose number exits in table while (flag == 1) do flag = 0 r = math.random(Min , Max) for j = 1 , i do if Table[j] == r then flag = 1 break end end end Table = r end ---END Number generator with no repetition--- --Print the table--- for i = 1, size do console(string.format('Table[%d] = %d', i, Table)) end ---end table printing-- end
- This reply was modified 4 years, 6 months ago by F0h.
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