Re: Pthread-win32 races?

["Sergey Fokin" <green.nsk@gmail.com>]

Hello.

> The library is working correctly since sem_destroy() is returning the
> error EBUSY as required and documented at:
>
> http://sourceware.org/pthreads-win32/manual/sem_init.html
>
> This is also in accordance with the Single Unix Specification. If it was
> hanging your program rather than returning the error then that would be
> a problem.

The sem_destroy function sets errno to the following error code on error:
EBUSY if some threads are currently blocked waiting on the semaphore.

But there's obviously no threads waiting on semaphore, is there?

> By the way, in your sample code you don't check the return code from the
> sem_post(), but the semaphore could already be destroyed at that point.

It couldn't (shouldn't, because actually it does). Because semaphore
is destroyed only after sem_wait(), but sem_wait() returns (should
return) only after sem_post() succeeds. Did I understood right?

> It would be better in this and similar cases to call sem_destroy() after
> the call to pthread_join(), or at least after you can guarantee that the
> semaphore is no longer required by any child threads.

In this example I can destroy semaphore after pthread_join(). But in
my program logic is more complicated and sem_post()'ing thread doesn't
finish after sem_post(). And again the same question: Does sem_post()
perform atomic access to the semaphore or I should perform additional
synchronisation to access the semaphore? Synchronizing access to
semaphore looks strange, don't you think so?

This quotation is from linux sem_post manual:

       !sem_post!  atomically  increases the count of the semaphore pointed to
       by |sem|. This function never blocks and can safely be  used  in  asyn-
       chronous signal handlers.

So, I think supplied code must be correct according to manual.

> A sem_t "handle" is not required to be unique in time, so it's possible
> to destroy a semaphore and init a new one having another purpose
> altogether, which then by chance occupies the same physical memory
> location, i.e. has the same "handle" (in pthreads-win32 this is just the
> pointer to the struct in memory), so a sema op somewhere may not fail
> even though, logically, it is no longer accessing the semaphore it
> should be, and the application may now be mysteriously badly behaved and
> difficult to debug.

Yes, I understand this. And there's no chance to accidentally access
destroyed semaphore.

-- 
eof