Re: [Caml-list] threads, signals, and timeout

[Gerd Stolpmann <gerd@gerd-stolpmann.de>]

Am Montag, den 26.10.2009, 12:06 -0700 schrieb yoann padioleau:
> On Mon, Oct 26, 2009 at 11:36 AM, Till Varoquaux <till@pps.jussieu.fr> wrote:
> > You'd have the same problem in any other programming language; this is
> > due to the underlying POSIX model.
> > The POSIX standard mentions that alarms are global to the process [1].
> >
> > " There were two possible choices for alarm generation in
> > multi-threaded applications: generation for the calling thread or
> > generation for the process. The first option would not have been
> > particularly useful since the alarm state is maintained on a
> > per-process basis and the alarm that is established by the last
> > invocation of alarm() is the only one that would be active."
> 
> I've seen that that there are other C APIs for timers: setitimer,
> create_timer, ... are they all subject to the same limitations ?
> 
> > As a general matter mixing signals and threads is a thorny issue under
> > unix; you should avoid doing so unless you have a good grasp of what
> > is going to happen. Having a global timer thread that kills timed-out
> > threads would be possible but seems very painful.
> 
> I dont want to kill them. I want to deliver timeout exceptions to those
> threads, and each of those threads should be able to react differently
> to those timeouts.
> 
> > If you give us more
> > information on what you are trying to achieve (eg your computations
> > threads are crunshing numbers in a tight loop or you are performing io
> > etc...) we might be able to guide you.
> 
> I want to write a toy program that uses threads and timeouts in an
> easy way. Those
> threads may make heavy use of CPU or IO, I don't know. I just want
> an easy API for thread and timeout.
> 
> > Here is a quick run down of my
> > flawed and partial understand of the world:
> >
> > For heavy computations threads will not help you. I am sure you are
> > aware of this but there never is more than one ocaml thread running at
> > the same time.
> 
> To be honest, the problem I have to solve is not in OCaml, but I thought
> OCaml programmers would give better answers than C programmers :)
> 
> I find it wierd that using timeout with process is easy, and using
> timeouts with threads
> is not.  Apparently on google they say a usual technique is to have a
> master thread
> which handles all the signals, with all the other threads blocking them. But
> it would still require this master thread  juggling with multiple
> timeouts requests and deliveries, and then to deliver manually some
> signals to the specific thread (with
> pthread_kill ?). I don't want to write my own "scheduler".
> Those things should be handled at kernel level;
> the kernel should know about threads, and should be able to deliver
> signals to the thread that requested them. Maybe the new Linux
> thread library (NPTL) can do that ?

Till gave already a quite good answer. I think your assumption is
already wrong that "using timeouts with processes is easy". It's only
easy as long you don't care about race conditions and subtle problems of
signals (e.g. that two pending signals are merged by the kernel). Most
likely you get a program that works 99% of the time, and fails (hangs,
or loops, or misses signals) 1% of the time. Signals are really
low-level.

Combining threads and signals leads to numerous implementation problems.
Both the threads and signals API define primitives for synchronization
(e.g. mutexes for threads, and masks for signals). As signals can be
generated at any time, there is no control about the state of the
thread-level synchronization objects: When the signal handler is
entered, you don't know whether a thread (either the thread the handler
is running in, or some other) has locked the mutex or not (given that
the signal handler accesses some shared data). Even worse, you cannot
simply wait in the handler until the mutex is free again - this could be
a deadlock. Essentially, you cannot synchronize data access of a signal
handler and of threads - it's conceptually impossible. Think like this:
A single thread is executed sequentially, and the primitives for
synchronizing several threads rely on that. Signals, however, allow it
to spontaneously interrupt the sequential execution, as if one thread
would be many. That breaks assumptions of multi-threading.

That's the background why POSIX chose not to define a per-thread signal
API. There are no per-thread timers, and you need to write your own mini
scheduler.

Gerd

> 
> 
> > If you can fork and collect the resulting data through
> > pipes or shared memory you should be able to get a noticeable
> > performance boost. There have been countless posts on the subject; I
> > am too lazy to google them up.
> >
> > I/O is a very different beast and you are probably better of with an
> > asynchronous framework. Luckily for you Ocaml has a pretty good one
> > already available [2]
> >
> > Till
> >
> > [1]: http://www.opengroup.org/onlinepubs/000095399/functions/alarm.html
> > [2]: http://ocsigen.org/lwt/
> > On Mon, Oct 26, 2009 at 2:08 PM, yoann padioleau <pad.aryx@gmail.com> wrote:
> >> Hi,
> >>
> >> I would like to create different threads where each thread do some
> >> computation and are subject to different
> >> timeout. Without threads I usually use Unix.alarm with a SIGALARM
> >> handler that just raise a Timeout exception
> >> and everything works fine, but when I try to do something similar with
> >> threads it does not work
> >> because apparently the Unix.alarm done in one thread override the
> >> Unix.alarm done in another
> >> thread. I had a look at thread.mli but was not able to find anything
> >> related to timeout.
> >> Is there a way to have multiple timeout and multiple threads at the same time ?
> >>
> >> Here is a program that unforunately get the first timeout, but not the second :(
> >>
> >>
> >> (*
> >> ocamlc -g -thread unix.cma threads.cma signals_and_threads.ml
> >> *)
> >>
> >> exception Timeout
> >>
> >> let mytid () =
> >>  let t = Thread.self () in
> >>  let i = Thread.id t in
> >>  i
> >>
> >> let set_timeout () =
> >>
> >>  Sys.set_signal Sys.sigalrm
> >>    (Sys.Signal_handle (fun _ ->
> >>      prerr_endline "Time is up!";
> >>      print_string (Printf.sprintf "id: %d\n" (mytid()));
> >>      raise Timeout
> >>    ));
> >>
> >>  ignore(Unix.alarm 1);
> >>  ()
> >>
> >>
> >> let main =
> >>  let t1 =
> >>    Thread.create (fun () ->
> >>      set_timeout ();
> >>      print_string (Printf.sprintf "t1 id: %d\n" (mytid()));
> >>
> >>      let xs = [1;2;3] in
> >>      while(true) do
> >>        let _ = List.map (fun x -> x + 1) xs in
> >>        ()
> >>      done;
> >>      ()
> >>    ) ()
> >>  in
> >>
> >>  let t2 =
> >>    Thread.create (fun () ->
> >>      set_timeout ();
> >>      print_string (Printf.sprintf "t2 id: %d\n" (mytid()));
> >>
> >>      let xs = [1;2;3] in
> >>      while(true) do
> >>        let _ = List.map (fun x -> x + 1) xs in
> >>        ()
> >>      done;
> >>      ()
> >>    ) ()
> >>  in
> >>  Thread.join t1;
> >>  Thread.join t2;
> >>  ()
> >>
> >> ------------------
> >>
> >>
> >>
> >>
> >> Here is the output
> >> Time is up!
> >> t2 id: 2
> >> t1 id: 1
> >> id: 1
> >> Thread 1 killed on uncaught exception Signals_and_threads.Timeout
> >> .... <the program loops, meaning the second thread never received its timeout
> >>
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> 
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