SIMPLE PROBLEM: Send a bang to one of many bpatchers

    Feb 10 2013 | 1:10 pm
    here's the idea.
    I have a live.knob with a range of 0-127. I also have 128 bpatchers. I want to send a bang to the bpatcher selected by the knob.
    preferably without having to connect 128 different objects maybe using 'send'/'receive' or 'forward'.

    • Feb 10 2013 | 3:28 pm
      thinking also about how to do that some elegant way, but maybe worth mentioning that connecting route object to even 128 object can be done in a minute or two using the [thispatcher] scripting / "script connect" message + a counter and uzi...
    • Feb 10 2013 | 6:56 pm
      I would send the value of the knob, and select for it in the bpatcher. This assumes that one of the arguments to the bpatcher is what instance it is.
    • Feb 10 2013 | 8:24 pm
      One possible solution:
    • Feb 11 2013 | 12:17 am
      Django that is pretty nice and succinct coding! before I start manually connecting 128 bpatchers to it possible to create one bpatcher with 'receiver#' (not sure of syntax) embedded with code so that as I duplicate it in the parent, max increments it with a unique identifier?
      I think i've seen this before in other patches but i'm uneducated with syntax...
      You're totally getting credit for this bit of code! Thanks!
    • Feb 11 2013 | 12:31 am
      @Chris Muir ...not sure how to do that easily for 128 bpatchers.
      @Dnk777 ... can you post an example?
    • Feb 11 2013 | 1:44 am
      django: nice!
      mssngmrblz: ok, it took me bit more than 1-2 minutes, because messages won't replace $1 with number, if the $1 is closed in []. so i had to use combine object. but maybe 3-4 minutes and when you have it built, you can reuse. I think this is a must, if you work on a patch with 128 bpatchers...
      (please note one stupid thing i did in this example was giving the scripting names starting with [1]. that was stupid, because the inlets/outlets are numbers from 0 and you usually want to connect first of the many objects to first of the many inlets/outlets. so you better start with [0] on end of the scripting name, and then just duplicating or copy/pasting will create [1], [2], ...[n])
    • Feb 11 2013 | 9:41 am
      I'd do this with pattrforward.
      Give each of your bpatchers a scripting name containing some numeric/index element, e.g. myBpatcher[3]
      Use the value from the live.dial to set which bpatcher the pattrforward is currently routing to
      Send a bang to the pattrforward object (which will forward the message to the currently targeted bpatcher), using an 'in0' prefix to ensure your bang goes the correct inlet of th bpatcher.
      Example below
    • Feb 11 2013 | 1:56 pm
      or use Nathanael Decaude's max toolbox.
    • Feb 11 2013 | 11:34 pm
      +1 for the toolbox...connecting 128 objects can be done with one mouse selection and a keystroke.
      When you originally create the bpatchers, give them the array name: my_bpatcher[1] ...then when you duplicate, they'll auto-name with [2] [3] etc. Problem: How to give each one an argument? Arguments won't auto-increment with this method. Possible solution: [thispatcher] to [sprintf script sendbox my_patcher[%u] argument %u]
      Can't test it now, but can [thispatcher] send a message to [bpatcher] to set its arguments? If so, you're done with a [route #1] in the bpatcher.
    • Feb 12 2013 | 3:34 am
      seejayjames: i don't think there is a way to add argument(s) to existing object using thispatcher, but you could use it to create new 128 objects with arguments...
    • Feb 12 2013 | 4:44 am
      The patcher scripting name does auto-increment, so you could get that name from within the bpatcher, and then use that number to select the number to bang. Here's an example: