CODEWITHSUNDEEP
supercollider
Rithesh
Rithesh

Reputation: 199

Array of buses in superCollider

I have a Synth generated with a do:

(
SynthDef(\siny, { arg freq, outBus=0; Out.ar( outBus, SinOsc.ar(freq!2,0,0.2) ) } ).send(s);
SynthDef(\filter, { arg cFreq,q=0.8, inBus; Out.ar( 0, BPF.ar(In.ar(inBus), cFreq!2, 1/q ) ) } ).send(s);

)

(
~sourceOut = Bus.audio(s);
~sine_Group = ParGroup.new;
z = [100,500,1000,1500,250];

{
z.do({ arg val; Synth.head(~sine_Group, \siny, [\freq: val, \outBus: ~sourceOut]) });
z.do({ arg val; Synth.after(~sine_Group, \filter, [\inBus: ~sourceOut, \cFreq: 200] ) });

}.play;
)

Right now, my understanding is that, output of multiple instances of Synth \siny get mixed in the bus ~sourceOut, and goes as an input into synth \filter

What i actually want to do is to have a one-to-one connection between the multiple instances of \siny and \filter.. Could I use an array of busses to connect them? If so, how do I do that?

Upvotes: 1

Views: 700

Answers (1)

Dan Stowell
Dan Stowell

Reputation: 4758

Yes you can. Here I've modified your code minimally. First I made ~sourceOut an array of Busses rather than a single Bus. Second, inside the do loops I made use of the fact that the main iteration functions in SuperCollider can provide a second index argument as well as each item itself. Thirdly I use that index argument to select the desired Bus:

(
z = [100,500,1000,1500,250];
~sourceOut = z.collect{ Bus.audio(s) };
~sine_Group = ParGroup.new;

{
z.do({ arg val, index; Synth.head(~sine_Group, \siny, [\freq: val, \outBus: ~sourceOut[index]]) });
z.do({ arg val, index; Synth.after(~sine_Group, \filter, [\inBus: ~sourceOut[index], \cFreq: 200] ) });

}.play;
)

Depending on your needs, you might also like to look at NodeProxy which is useful for prototyping and live coding, and provides some tricks for plugging synths' output into each other.

Upvotes: 1

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