Re: [Caml-list] Thread behaviour

[Tom Ridge <tom.j.ridge+caml@googlemail.com>]

Would it be fair to say that OCaml does not currently support
pre-emptively scheduled threads?

I have read the lecture from Xavier archived here:

http://alan.petitepomme.net/cwn/2002.11.26.html#8

I would like to implement a library to handle messaging between
possibly-distributed OCaml processes. Alas, my design naively requires
pre-emptively scheduled threads (although it may be possible to change
the design e.g. to work with Lwt) - each message queue is accompanied
by a thread which reinitializes connections when connections go down
etc., hiding this complexity from the user.

Quoting Xavier:

"Scheduling I/O and computation concurrently, and managing process
stacks, is the job of the operating system."

But what if you want to implement a messaging library in OCaml? It
seems unlikely that all operating systems would fix on a standard
implementation of distributed message passing (or, even more funky,
distributed persistent message queues).


On 27 September 2013 11:51, Benedikt Grundmann
<bgrundmann@janestreet.com> wrote:
> The ticker thread will cause yields which will be honored on the next
> allocation of the thread that currently has the caml lock.  That said we
> have seen that sometimes the lock is reacquired by the same thread again.
> So there are some fairness issues.
>
>
> On Fri, Sep 27, 2013 at 11:27 AM, Romain Bardou <romain.bardou@inria.fr>
> wrote:
>>
>> Le 27/09/2013 12:10, Tom Ridge a écrit :
>> > Dear caml-list,
>> >
>> > I have a little program which creates a thread, and then sits in a loop:
>> >
>> > --
>> >
>> > let f () =
>> >   let _ = ignore (print_endline "3") in
>> >   let _ = ignore (print_endline "hello") in
>> >   let _ = ignore (print_endline "4") in
>> >   ()
>> >
>> > let main () =
>> >   let _ = ignore (print_endline "1") in
>> >   let t = Thread.create f () in
>> >   (* let _ = Thread.join t in *)
>> >   let _ = ignore (print_endline "2") in
>> >   while true do
>> >     flush stdout;
>> >   done
>> >
>> > let _ = main ()
>> >
>> > --
>> >
>> > I compile the program with the following Makefile clause:
>> >
>> > test.byte: test.ml FORCE
>> > ocamlc -o $@ -thread unix.cma threads.cma $<
>> >
>> > When I run the program I get the output:
>> >
>> > 1
>> > 2
>> >
>> > and the program then sits in the loop. I was expecting the output from
>> > f to show up as well. If you wait a while, it does. But you have to
>> > wait quite a while.
>> >
>> > What am I doing wrong here? I notice that if I put Thread.yield in the
>> > while loop then f's output gets printed pretty quickly. But why should
>> > the while loop affect scheduling of f's thread?
>> >
>> > Thanks
>> >
>>
>> OCaml's thread, unfortunately, are kind of cooperative: you need to
>> yield explicitly. Note that you will obtain an even different (worse)
>> result with a native program. I observed this myself without looking at
>> the thread code itself so maybe there is actually a way to
>> "automatically yield" but as far as I know there is no way to obtain the
>> behavior you want without using either yields or processes instead of
>> threads. This is the reason for the Procord library I am developing
>> (first version to be released before the next OUPS meeting).
>>
>> Also, you don't need to ignore the result of print_endline, as
>> print_endline returns unit. And using let _ = ... in is the same as
>> using ignore, so using both is not needed.
>>
>> Cheers,
>>
>> --
>> Romain Bardou
>>
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>
>