Forum: Ruby 1.9.x Thread problem

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Michael M. (Guest)
on 2009-03-09 02:27
(Received via mailing list)
Hi All,

Here's a tricky one for you:

require 'thread'

THREADS = 10
WASTE = 10000

pid = fork() do
   queue = Queue.new
   THREADS.times do |i|
      queue << Thread.new do
         # wasting actual cpu time, rather than sleeping
         Thread.current[:index] = i
         n = rand(WASTE)
         Thread.current[:n] = n
         d = 1.0 + 1/n.to_f
         e = 1.0
         n.times { e = e * d }
      end
   end

   THREADS.times do
      th = queue.pop
      puts "Joining thread #{th[:index]}, n=#{th[:n]}"
      th.join()
   end

end

Process.wait(pid)

If I run this using ruby 1.8.7 (2008-08-11 patchlevel 72) [i486-linux] I
get the output I would expect:
Joining thread 0, n=8767
Joining thread 1, n=2401
Joining thread 2, n=514
Joining thread 3, n=5227
Joining thread 4, n=658
Joining thread 5, n=8618
Joining thread 6, n=2356
Joining thread 7, n=7431
Joining thread 8, n=8426
Joining thread 9, n=588

If I run this using either ruby 1.9.1p0 (2009-01-30 revision 21907)
[i686-linux] or ruby 1.9.2dev (2009-03-05 trunk 22769) [i686-linux] I
get results I would not expect:
Joining thread , n=
Joining thread 1, n=1131
Joining thread 2, n=8233
Joining thread 3, n=8696
Joining thread 4, n=7684
Joining thread 5, n=286
Joining thread 6, n=6618
Joining thread 7, n=2814
Joining thread 8, n=5862
Joining thread 9, n=6307

My best guess is that something similar to the fix for this bug
http://redmine.ruby-lang.org/issues/show/657 is happening, but elsewhere
in the code.  Does anyone have any ideas?

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Nobuyoshi N. (Guest)
on 2009-03-09 03:36
(Received via mailing list)
Hi,

At Mon, 9 Mar 2009 09:25:06 +0900,
Michael M. wrote in [ruby-talk:330697]:
> If I run this using either ruby 1.9.1p0 (2009-01-30 revision 21907)
> [i686-linux] or ruby 1.9.2dev (2009-03-05 trunk 22769) [i686-linux] I
> get results I would not expect:
> Joining thread , n=

It hasn't been defined which thread, the created child thread
or the creating parent thread, runs first at thread creation.
In 1.8, the child runs first by chance.

> My best guess is that something similar to the fix for this bug
> http://redmine.ruby-lang.org/issues/show/657 is happening, but elsewhere
> in the code.  Does anyone have any ideas?

No, it is irrelevant.
Michael M. (Guest)
on 2009-03-09 04:08
(Received via mailing list)
Nobuyoshi N. wrote:
>
> It hasn't been defined which thread, the created child thread
> or the creating parent thread, runs first at thread creation.
> In 1.8, the child runs first by chance.
>
>
Can you then explain why it runs consistently like this or a way in
which I can guarantee I am running in the first created child thread?  I
would have expected to be inside of the child thread because the code
setting up the :index is inside the block passed to Thread.new().  And I
only  put the created child Thread instances in the Queue, so I should
_only_ get child Thread instances when I call Queue#pop.


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Ken B. (Guest)
on 2009-03-09 05:12
(Received via mailing list)
On Sun, 08 Mar 2009 21:05:58 -0500, Michael M. wrote:

>>>
> only  put the created child Thread instances in the Queue, so I should
> _only_ get child Thread instances when I call Queue#pop.

I'm really not sure what you're asking. All you get from Queue#pop is
child threads. The thing is that the operating system decides (and it is
quite out of your control) which thread will have a chance to execute
next. So apparently Ruby 1.8 always chose to let the child threads have
a
chance before you reached the THREADS.times loop. Ruby 1.9, on the other
hand isn't giving the child threads a chance to run any code until you
hit the th.join() call. Thus, when you try to access th[:index] and th
[:n], the child threads haven't had a chance to update them yet. Once
you
run thread 0, that takes long enough that the other threads also get a
chance to execute and set values for th[:index] and th[:n], and that's
why those values are available.

But there's no guarantees about who gets to execute when, and you
shouldn't count on any. This is not a bug.

Unfortunately, I don't have any ideas about a good way to ensure that
the
threads execute far enough to have set th[:index] and th[:n] before you
print them.
Michael M. (Guest)
on 2009-03-09 05:17
(Received via mailing list)
>>
>>

I have read and re-read this about 10 times now and this comment makes
it appear that the contract for Thread.new {block}, where the block is
run exclusively in the newly created thread, is violated.
> Can you then explain why it runs consistently like this or a way in
> which I can guarantee I am running in the first created child thread?
> I would have expected to be inside of the child thread because the
> code setting up the :index is inside the block passed to
> Thread.new().  And I only  put the created child Thread instances in
> the Queue, so I should _only_ get child Thread instances when I call
> Queue#pop.
 require 'thread'
 THREADS = 10

 parent_id = Thread.current.object_id

 queue = Queue.new
 THREADS.times do |i|
    queue << Thread.new do
       th = Thread.current
       th[:index] = i
       raise "Running inside parent" if th.object_id == parent_id
       sleep(2)
    end
 end

 THREADS.times do
    th = queue.pop
    raise "Running inside parent" if th.object_id == parent_id
    puts "Joining thread #{th[:index]}"
    th.join()
 end

I have simplified and changed things.  I am quite certain that the
parent thread is not being joined or indeed added to the queue as no
exceptions are raised (unless this is a bad test) and I still get the
same (strange) output.  Any other ideas?


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Michael M. (Guest)
on 2009-03-09 05:26
(Received via mailing list)
> chance to execute and set values for th[:index] and th[:n], and that's
> why those values are available.
>
> But there's no guarantees about who gets to execute when, and you
> shouldn't count on any. This is not a bug.
>
> Unfortunately, I don't have any ideas about a good way to ensure that the
> threads execute far enough to have set th[:index] and th[:n] before you
> print them.
>
>
Ahhh, now I understand and yes, you're quite right.  Sigh.  If I join
the thread before printing the index, then everything works as
expected.  Back to square one with my other problem.... But I haven't
come up with something simple that displays the right behaviour, so I'll
give that a bit more thought before I post that question.  Thanks for
your help.

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the subject of legal or other privilege, none of which is waived or
lost by reason of this transmission.
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James M. (Guest)
on 2009-03-09 06:20
Here is a bare solution using a mutex and a counter.  The next thing
would be to use a condition variable to put the main thread to sleep
while the child threads initialize, rather than spinning wheels /
burning rubber while waiting.

Also look into monitor.rb, sync.rb, and producer/consumer strategies.

require 'thread'
THREADS = 10

parent_id = Thread.current.object_id

mutex = Mutex.new
num_threads_ready = 0

queue = Queue.new
THREADS.times do |i|
   queue << Thread.new do
      th = Thread.current
      th[:index] = i
      mutex.synchronize { num_threads_ready += 1 }  # <-----
      raise "Running inside parent" if th.object_id == parent_id
      sleep(2)
   end
end

# wait until threads have initialized
until mutex.synchronize { num_threads_ready == THREADS }
   Thread.pass
end

THREADS.times do
   th = queue.pop
   raise "Running inside parent" if th.object_id == parent_id
   puts "Joining thread #{th[:index]}"
   th.join()
end
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