On 2/20/07, Lang Martin <lang-gb@coptix.com> wrote:
I have a (possibly irrational) bias against mutexes, and attempted to
solve a threading problem with just message passing. The code works,
but only part of the time.

I haven't run your code myself, but I think I see a few things wrong with it.
 

This code:

(let ((th (thread-start!
            (make-thread
             (lambda ()
               (let lp ()
                 (let ((p (thread-receive 0 #f)))
                   (if p
                       (p 40)
                       (begin
                         (thread-yield!)
                         (lp))))))))))
        (+ 2 (call/cc
         (lambda (ret)
           (thread-send th ret)
           (thread-sleep! 4)))))


Occasionally fails by attempting to add 2 to #!void.

Let's call the original thread "thread #1" and the newly created one "thread #2". 

Thread #1 will start thread #2, capture its own continuation and send it to thread #2.  Then it adds the result of (thread-sleep! 4) [which is #!void] to 2, which cause your error after 4 seconds.

In thread #2, calling (p 40) will make that thread invoke thread #1's continuation, resulting in a final result of 42.  Note that thread #2 will not loop if it gets the continuation as a message, because it discards its own continuation when invoking (p 40).  It basically becomes thread #1 at that point.

Other notes: you should just wait in thread #2 for a message instead of waiting in a spin lock (ie using 'thread-receive' with a timeout of 0).  Also, calling 'thread-yield!' isn't necessary since Gambit's thread system is preemptive.

Finally, for debugging it might help to use Gambit's command-line options so that threads other than the primordial thread will also start a repl on crashes, using for example:
% gsi -:dar
(for more details see Gambit's documentation).


Hope this helps and that I didn't misinterpret the code.  Personally, I don't think you will have to use mutexes if you design your program correctly.


Guillaume