From: "Kroeger, Simon (ext)" <simon.kroeger.ext@...>
Date: 2006-03-24T19:35:27+09:00
Subject: Re: Stupid Problem

Hi,

your code is fine, it blocks at sThread.join (as you would expect it to
do).
put a 

$stdout.sync = true

at the top if you would like to see the output before the program
terminates.

cheers

Simon


> -----Original Message-----
> From: trevershick@gmail.com [mailto:trevershick@gmail.com] 
> Sent: Friday, March 24, 2006 1:14 AM
> To: ruby-talk ML
> Subject: Stupid Problem
> 
> I'm a total newb as you'll find out. But regardless, i have a 
> question.
>  I run the code below:
> 
> It locks up (doesn't do anything).
> If i remove the while 1 and 'end' but leave the code between it works
> perfectly.
> It must be in how i'm envisioning how the thread works. I'm a java
> programmer and this seems correct.  Create the thread, define it's
> works then start the thread. The call to start should return and not
> block. I'm not sure what it's doing when i run the code with the
> 'while' statement.
> 
> Thanks for your help.
> 
> #load "ListenerList.rb"
> require "socket"
> 
> class MovingPartsServer
> 
> 	def initialize()
> 		# do nothing yet
> 		#@listeners = ListenerList.new();
> 		# initialize the UDP socket
> 		puts " Open Socket"
> 		@socket = UDPSocket.open
> 		puts "Socket Created : "
> 		puts @socket
> 		@socket.bind(nil, 10003)
> 		puts "bound"
> 
> 	end
> 
> 	def run()
> 		puts  "run"
> 		@sThread = Thread.new(self) { |server|
> 			while 1
> 				puts "in thread"
> 				rcv = @socket.recvfrom(64)
> 				puts rcv
> #				@listeners.AddListener rcv[0]
> #				puts @listeners
> 			end
> 		}
> 		return @sThread
> 	end
> 
> end
> 
> puts "Start"
> server = MovingPartsServer.new;
> puts "Start Server Thread"
> sThread  = server.run();
> 
> 
> puts "start client"
> UDPSocket.open.send("ad hoc", 0, 'localhost', 10003)
> puts "end client"
> sThread.join
> 
> 
>