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Re: [Qemu-devel] [PATCH] semaphore: fix a hangup problem under load on N


From: Laszlo Ersek
Subject: Re: [Qemu-devel] [PATCH] semaphore: fix a hangup problem under load on NetBSD hosts.
Date: Mon, 01 Jul 2013 11:54:41 +0200
User-agent: Mozilla/5.0 (X11; Linux x86_64; rv:17.0) Gecko/20130513 Thunderbird/17.0.6

On 06/29/13 12:22, Izumi Tsutsui wrote:
> Fix following bugs in "fallback implementation of counting semaphores
> with mutex+condvar" added in c166cb72f1676855816340666c3b618beef4b976:
>  - waiting threads are not restarted properly if more than one threads
>    are waiting unblock signals in qemu_sem_timedwait()
>  - possible missing pthread_cond_signal(3) calls when waiting threads
>    are returned by ETIMEDOUT
>  - fix an uninitialized variable
> 
> The problem is analyzed by and fix is provided by Noriyuki Soda.
> 
> Signed-off-by: Izumi Tsutsui <address@hidden>
> ---
>  util/qemu-thread-posix.c | 17 +++++++++--------
>  1 file changed, 9 insertions(+), 8 deletions(-)
> 
> diff --git a/util/qemu-thread-posix.c b/util/qemu-thread-posix.c
> index 4489abf..db7a15b 100644
> --- a/util/qemu-thread-posix.c
> +++ b/util/qemu-thread-posix.c
> @@ -172,10 +172,9 @@ void qemu_sem_post(QemuSemaphore *sem)
>      pthread_mutex_lock(&sem->lock);
>      if (sem->count == INT_MAX) {
>          rc = EINVAL;
> -    } else if (sem->count++ < 0) {
> -        rc = pthread_cond_signal(&sem->cond);
>      } else {
> -        rc = 0;
> +        sem->count++;
> +        rc = pthread_cond_signal(&sem->cond);
>      }
>      pthread_mutex_unlock(&sem->lock);
>      if (rc != 0) {
> @@ -207,19 +206,21 @@ int qemu_sem_timedwait(QemuSemaphore *sem, int ms)
>      struct timespec ts;
>  
>  #if defined(__APPLE__) || defined(__NetBSD__)
> +    rc = 0;
>      compute_abs_deadline(&ts, ms);
>      pthread_mutex_lock(&sem->lock);
> -    --sem->count;
> -    while (sem->count < 0) {
> +    while (sem->count <= 0) {
>          rc = pthread_cond_timedwait(&sem->cond, &sem->lock, &ts);
>          if (rc == ETIMEDOUT) {
> -            ++sem->count;
>              break;
>          }
>          if (rc != 0) {
>              error_exit(rc, __func__);
>          }
>      }
> +    if (rc != ETIMEDOUT) {
> +        --sem->count;
> +    }
>      pthread_mutex_unlock(&sem->lock);
>      return (rc == ETIMEDOUT ? -1 : 0);
>  #else
> @@ -251,10 +252,10 @@ void qemu_sem_wait(QemuSemaphore *sem)
>  {
>  #if defined(__APPLE__) || defined(__NetBSD__)
>      pthread_mutex_lock(&sem->lock);
> -    --sem->count;
> -    while (sem->count < 0) {
> +    while (sem->count <= 0) {
>          pthread_cond_wait(&sem->cond, &sem->lock);
>      }
> +    --sem->count;
>      pthread_mutex_unlock(&sem->lock);
>  #else
>      int rc;
> 

I agree with this patch, but I'd propose something more intrusive (feel
free to ignore it anyway): "QemuSemaphore.count" has no business with
negative values; it should be an unsigned int.

The condition on which consumers block is exactly (count == 0).
Conversely, the only time we need to send a signal is the 0->1 count
transition (*). Checks for negative values should be eliminated in
parallel with the int->unsigned type change.

Also I'd feel safer if pthread_cond_*() and pthread_mutex_*() were
retval-checked consistently, but that's tangential.

Reviewed-by: Laszlo Ersek <address@hidden>

(*) 100% tangential: this reminds me of when I made an attempt to
dissect condvars & co on reddit [1]. I considered pthread_cond_signal()
vs. pthread_cond_broadcast() too; alas my two conclusions there against
the former were wrong. See [2] why -- in short when a wakeup signal is
delivered, the victim thread is removed from the set of potential
victims. In other words, pthread_cond_signal() itself (vs. broadcast)
*is* correct here.

I also like that the signal is sent with the mutex held [3] [4].

[1] 
http://www.reddit.com/r/programming/comments/9ynxv/utter_verbiage_how_to_design_condition_variables/
[2] 
http://thread.gmane.org/gmane.comp.standards.posix.austin.general/4844/focus=4850
[3] 
http://thread.gmane.org/gmane.comp.standards.posix.austin.general/1822/focus=1823
[4] 
http://www.domaigne.com/blog/computing/condvars-signal-with-mutex-locked-or-not/

Thanks,
Laszlo



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