How a node not in FX-Application-Thread snapshot?

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  I run a thread of the Application of FX for a main thread using Application.launch [described here: {: identifier of the thread = 2530636}]
  
  I try to have the application thread return information on the main thread, but Application.launch returns void. Is there an easy way to communicate between the main thread and the thread of the Application?
  
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  Published by: cr0ck3t on April 30, 2013 21:04
  
  Published by: cr0ck3t on April 30, 2013 21:05

  Is there an easy way to get a reference to these objects to both the main thread and the thread of Application FX - called via Application.launch () since the main thread? Or what I need to use Sockets or MOM?

  Not much to do with the concurrent programming is what I would call easy. It looks easy - but it's not.
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  ---------
  You can find a sample of the various interactions of thread with JavaFX here:
  https://gist.github.com/jewelsea/5500981 "Simulation of dwarf dragons using multiple threads to eat."
  Related code is just demo-ware to try different competitive access facilities and not necessarily a recommended strategy.
  If your curiosity, you can take a look and try to work out what it is, what it does and how it does.
  The main reason followed is that of a blocking queue:
  http://docs.Oracle.com/javase/6/docs/API/Java/util/concurrent/BlockingQueue.html
  ---------
  Note that once you call launch of the main thread, no subsequent statement in the main method will be different until the JavaFX application terminates. If you can't start from the main thread and communicate with the main thread JavaFX application. Instead, you need to spawn another thread (or a set of threads) for communication with the JavaFX application.
  ---------
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  ----------
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