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From: Lune Lune on 22 Jan 2010 05:17
Hi Robert

Thanks for your example.

I wanted to play with it to really understand how that works. So I tried
this :
QUEUE = Queue.new
threads = (1..5).map do |a|
Thread.new(QUEUE) do |q|
until ( q == (task = q.deq))
print "thread-", a, " ", task.inspect, "\n"
end
end
end

10.times do |t|
puts "t #{t}"
30.times { |s| puts "s #{s}"; QUEUE.enq(s) }
threads.size.times { QUEUE.enq QUEUE }
threads.each {|t| t.join}
end

In my loop "30 times" 5 * 6 threads are created.
But when I add "10 times" I should see 10 * what I had earlier. This is
not happening. I have my 5 * 6 threads created and that's all. Could you
explain why ?

Thanks
--
Posted via http://www.ruby-forum.com/.

From: Jesús Gabriel y Galán on 22 Jan 2010 05:33
On Fri, Jan 22, 2010 at 11:17 AM, Lune Lune <verduc_m(a)epitech.net> wrote:
> Hi Robert
>
> Thanks for your example.
>
> I wanted to play with it to really understand how that works. So I tried
> this :
> QUEUE = Queue.new
> threads = (1..5).map do |a|
>   Thread.new(QUEUE) do |q|
>      until ( q == (task = q.deq))
>        print "thread-", a, " ", task.inspect, "\n"
>      end
>   end
> end
>
> 10.times do |t|
>  puts "t #{t}"
>  30.times { |s| puts "s #{s}"; QUEUE.enq(s) }
>  threads.size.times { QUEUE.enq QUEUE }
>  threads.each {|t| t.join}
> end
>
> In my loop "30 times" 5 * 6 threads are created.
> But when I add "10 times" I should see 10 * what I had earlier. This is
> not happening. I have my 5 * 6 threads created and that's all. Could you
> explain why ?

The first time around you insert 30 elements and then one end-of-work
token (the QUEUE itself) per thread. Then you join the threads, so the
main thread will wait at that point. The 5 threads will consume the 30
tokens and eventually consume the end-of-work token and exit. After
that, the main thread goes through the next 9 iterations inserting
items in the queue, but there are no more threads to consume the queue
items.

Jesus.

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