Re: [Caml-list] Thread behaviour

[Romain Bardou <romain.bardou@inria.fr>]

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