@mianmogra Ah sorry, i was at work and just took a brief look at the patch in the video, even without listening to the audio, my apologies!

@MeneerJansen Looks like in the video there is just a sample being played...

@MeneerJansen This should do it:

@Jona Interesting, the outlet of [array define] was not there in 0.48.1. There you would have to get the pointer via [send( message.

For [colarray], the dynamic patching was needed, because i wanted to modify a graphical array

Using an array object like [array get] gives a multiply defined error when graphical and non-graphical arrays with the same name are present

@Jona Yes. There is a hack to get the pointer of a graphical array. It is pretty messy, but it works. I used this method in [colarray] (https://forum.pdpatchrepo.info/topic/11184/colarray-a-graphical-array-where-color-and-line-width-can-be-set).

The background is that every array, graphical or not, is in fact a representation of a [struct] that is loaded in the background when Pd starts. The trick i used is to modify this hidden patch that contains the [struct] for all arrays by using dynamic patching, then click on the array by using a mouse message. The struct then sends out the pointer to the specific array. But the situations where this is really necessary should be limited.

You can also use the messages from [array define] like const and resize by just sending them to the graphical array: arraydisplay2.pd

@Jona The graphical array is just an [array] combined with a graphical representation. So you can directly access it with the [array] object without the need of a helper array: arraydisplay.pd

@ManoloPureD Well, there is some stuff going on with data structures and some tricks with a hidden hslider to get an infinite slider. But the main stuff for the audio is this:

Here you have a continuous input from minus infinity to infinity. The [mod 384] makes it circular between 0 and 383. But we divide by 384 so the input range doesn't matter, it only defines the resolution. The output here is between 0 and 1.

From here the actual stuff happens within the expressions. We want the pitch to double for each octave, so we have pow(2, $f1). So now we have a lowest pitch at 0 and the highest pitch at 1. Playing with factor 4 and 110 can give different results.

Now we want the volume to be at maximum in the middle and at zero at 0 and 1 so we can not hear that the pitch switches. This is a little tricky and done by using the absolute function. I think the power function is only there to smooth things out.

I actually had to play around myself quite a bit to get the result, which can still be optimized, i guess. So i would recommend to just deconstruct the expressions and play around with some values yourself, using print, to get a feeling for what is going on.

@Nobody You can try "/usr/local/lib/pd/bin/pd". But simply "pd" worked for me: