Kroeger, Simon (ext)
3/24/2006 10:35:00 AM
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
>
>
>