Re: [Caml-list] Thread behaviour

[Benedikt Grundmann <bgrundmann@janestreet.com>]

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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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