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/*
* Copyright (c) 2014 Nicolas George
*
* This file is part of FFmpeg.
*
* FFmpeg is free software; you can redistribute it and/or
* modify it under the terms of the GNU Lesser General Public License
* as published by the Free Software Foundation; either
* version 2.1 of the License, or (at your option) any later version.
*
* FFmpeg is distributed in the hope that it will be useful,
* but WITHOUT ANY WARRANTY; without even the implied warranty of
* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
* GNU Lesser General Public License for more details.
*
* You should have received a copy of the GNU Lesser General Public License
* along with FFmpeg; if not, write to the Free Software Foundation, Inc.,
* 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301 USA
*/
#include "fifo.h"
#include "threadmessage.h"
#include "thread.h"
struct AVThreadMessageQueue {
#if HAVE_THREADS
AVFifoBuffer *fifo;
pthread_mutex_t lock;
avutil/threadmessage: split the pthread condition in two Fix a dead lock under certain conditions. Let's assume we have a queue of 1 message max, 2 senders, and 1 receiver. Scenario (real record obtained with debug added): [...] SENDER #0: acquired lock SENDER #0: queue is full, wait SENDER #1: acquired lock SENDER #1: queue is full, wait RECEIVER: acquired lock RECEIVER: reading a msg from the queue RECEIVER: signal the cond RECEIVER: acquired lock RECEIVER: queue is empty, wait SENDER #0: writing a msg the queue SENDER #0: signal the cond SENDER #0: acquired lock SENDER #0: queue is full, wait SENDER #1: queue is full, wait Translated: - initially the queue contains 1/1 message with 2 senders blocking on it, waiting to push another message. - Meanwhile the receiver is obtaining the lock, read the message, signal & release the lock. For some reason it is able to acquire the lock again before the signal wakes up one of the sender. Since it just emptied the queue, the reader waits for the queue to fill up again. - The signal finally reaches one of the sender, which writes a message and then signal the condition. Unfortunately, instead of waking up the reader, it actually wakes up the other worker (signal = notify the condition just for 1 waiter), who can't push another message in the queue because it's full. - Meanwhile, the receiver is still waiting. Deadlock. This scenario can be triggered with for example: tests/api/api-threadmessage-test 1 2 100 100 1 1000 1000 One working solution is to make av_thread_message_queue_{send,recv}() call pthread_cond_broadcast() instead of pthread_cond_signal() so both senders and receivers are unlocked when work is done (be it reading or writing). This second solution replaces the condition with two: one to notify the senders, and one to notify the receivers. This prevents senders from notifying other senders instead of a reader, and the other way around. It also avoid broadcasting to everyone like the first solution, and is, as a result in theory more optimized.
9 years ago
pthread_cond_t cond_recv;
pthread_cond_t cond_send;
int err_send;
int err_recv;
unsigned elsize;
void (*free_func)(void *msg);
#else
int dummy;
#endif
};
int av_thread_message_queue_alloc(AVThreadMessageQueue **mq,
unsigned nelem,
unsigned elsize)
{
#if HAVE_THREADS
AVThreadMessageQueue *rmq;
int ret = 0;
if (nelem > INT_MAX / elsize)
return AVERROR(EINVAL);
if (!(rmq = av_mallocz(sizeof(*rmq))))
return AVERROR(ENOMEM);
if ((ret = pthread_mutex_init(&rmq->lock, NULL))) {
av_free(rmq);
return AVERROR(ret);
}
avutil/threadmessage: split the pthread condition in two Fix a dead lock under certain conditions. Let's assume we have a queue of 1 message max, 2 senders, and 1 receiver. Scenario (real record obtained with debug added): [...] SENDER #0: acquired lock SENDER #0: queue is full, wait SENDER #1: acquired lock SENDER #1: queue is full, wait RECEIVER: acquired lock RECEIVER: reading a msg from the queue RECEIVER: signal the cond RECEIVER: acquired lock RECEIVER: queue is empty, wait SENDER #0: writing a msg the queue SENDER #0: signal the cond SENDER #0: acquired lock SENDER #0: queue is full, wait SENDER #1: queue is full, wait Translated: - initially the queue contains 1/1 message with 2 senders blocking on it, waiting to push another message. - Meanwhile the receiver is obtaining the lock, read the message, signal & release the lock. For some reason it is able to acquire the lock again before the signal wakes up one of the sender. Since it just emptied the queue, the reader waits for the queue to fill up again. - The signal finally reaches one of the sender, which writes a message and then signal the condition. Unfortunately, instead of waking up the reader, it actually wakes up the other worker (signal = notify the condition just for 1 waiter), who can't push another message in the queue because it's full. - Meanwhile, the receiver is still waiting. Deadlock. This scenario can be triggered with for example: tests/api/api-threadmessage-test 1 2 100 100 1 1000 1000 One working solution is to make av_thread_message_queue_{send,recv}() call pthread_cond_broadcast() instead of pthread_cond_signal() so both senders and receivers are unlocked when work is done (be it reading or writing). This second solution replaces the condition with two: one to notify the senders, and one to notify the receivers. This prevents senders from notifying other senders instead of a reader, and the other way around. It also avoid broadcasting to everyone like the first solution, and is, as a result in theory more optimized.
9 years ago
if ((ret = pthread_cond_init(&rmq->cond_recv, NULL))) {
pthread_mutex_destroy(&rmq->lock);
av_free(rmq);
return AVERROR(ret);
}
if ((ret = pthread_cond_init(&rmq->cond_send, NULL))) {
pthread_cond_destroy(&rmq->cond_recv);
pthread_mutex_destroy(&rmq->lock);
av_free(rmq);
return AVERROR(ret);
}
if (!(rmq->fifo = av_fifo_alloc(elsize * nelem))) {
avutil/threadmessage: split the pthread condition in two Fix a dead lock under certain conditions. Let's assume we have a queue of 1 message max, 2 senders, and 1 receiver. Scenario (real record obtained with debug added): [...] SENDER #0: acquired lock SENDER #0: queue is full, wait SENDER #1: acquired lock SENDER #1: queue is full, wait RECEIVER: acquired lock RECEIVER: reading a msg from the queue RECEIVER: signal the cond RECEIVER: acquired lock RECEIVER: queue is empty, wait SENDER #0: writing a msg the queue SENDER #0: signal the cond SENDER #0: acquired lock SENDER #0: queue is full, wait SENDER #1: queue is full, wait Translated: - initially the queue contains 1/1 message with 2 senders blocking on it, waiting to push another message. - Meanwhile the receiver is obtaining the lock, read the message, signal & release the lock. For some reason it is able to acquire the lock again before the signal wakes up one of the sender. Since it just emptied the queue, the reader waits for the queue to fill up again. - The signal finally reaches one of the sender, which writes a message and then signal the condition. Unfortunately, instead of waking up the reader, it actually wakes up the other worker (signal = notify the condition just for 1 waiter), who can't push another message in the queue because it's full. - Meanwhile, the receiver is still waiting. Deadlock. This scenario can be triggered with for example: tests/api/api-threadmessage-test 1 2 100 100 1 1000 1000 One working solution is to make av_thread_message_queue_{send,recv}() call pthread_cond_broadcast() instead of pthread_cond_signal() so both senders and receivers are unlocked when work is done (be it reading or writing). This second solution replaces the condition with two: one to notify the senders, and one to notify the receivers. This prevents senders from notifying other senders instead of a reader, and the other way around. It also avoid broadcasting to everyone like the first solution, and is, as a result in theory more optimized.
9 years ago
pthread_cond_destroy(&rmq->cond_send);
pthread_cond_destroy(&rmq->cond_recv);
pthread_mutex_destroy(&rmq->lock);
av_free(rmq);
return AVERROR(ENOMEM);
}
rmq->elsize = elsize;
*mq = rmq;
return 0;
#else
*mq = NULL;
return AVERROR(ENOSYS);
#endif /* HAVE_THREADS */
}
void av_thread_message_queue_set_free_func(AVThreadMessageQueue *mq,
void (*free_func)(void *msg))
{
#if HAVE_THREADS
mq->free_func = free_func;
#endif
}
void av_thread_message_queue_free(AVThreadMessageQueue **mq)
{
#if HAVE_THREADS
if (*mq) {
av_thread_message_flush(*mq);
av_fifo_freep(&(*mq)->fifo);
avutil/threadmessage: split the pthread condition in two Fix a dead lock under certain conditions. Let's assume we have a queue of 1 message max, 2 senders, and 1 receiver. Scenario (real record obtained with debug added): [...] SENDER #0: acquired lock SENDER #0: queue is full, wait SENDER #1: acquired lock SENDER #1: queue is full, wait RECEIVER: acquired lock RECEIVER: reading a msg from the queue RECEIVER: signal the cond RECEIVER: acquired lock RECEIVER: queue is empty, wait SENDER #0: writing a msg the queue SENDER #0: signal the cond SENDER #0: acquired lock SENDER #0: queue is full, wait SENDER #1: queue is full, wait Translated: - initially the queue contains 1/1 message with 2 senders blocking on it, waiting to push another message. - Meanwhile the receiver is obtaining the lock, read the message, signal & release the lock. For some reason it is able to acquire the lock again before the signal wakes up one of the sender. Since it just emptied the queue, the reader waits for the queue to fill up again. - The signal finally reaches one of the sender, which writes a message and then signal the condition. Unfortunately, instead of waking up the reader, it actually wakes up the other worker (signal = notify the condition just for 1 waiter), who can't push another message in the queue because it's full. - Meanwhile, the receiver is still waiting. Deadlock. This scenario can be triggered with for example: tests/api/api-threadmessage-test 1 2 100 100 1 1000 1000 One working solution is to make av_thread_message_queue_{send,recv}() call pthread_cond_broadcast() instead of pthread_cond_signal() so both senders and receivers are unlocked when work is done (be it reading or writing). This second solution replaces the condition with two: one to notify the senders, and one to notify the receivers. This prevents senders from notifying other senders instead of a reader, and the other way around. It also avoid broadcasting to everyone like the first solution, and is, as a result in theory more optimized.
9 years ago
pthread_cond_destroy(&(*mq)->cond_send);
pthread_cond_destroy(&(*mq)->cond_recv);
pthread_mutex_destroy(&(*mq)->lock);
av_freep(mq);
}
#endif
}
int av_thread_message_queue_nb_elems(AVThreadMessageQueue *mq)
{
#if HAVE_THREADS
int ret;
pthread_mutex_lock(&mq->lock);
ret = av_fifo_size(mq->fifo);
pthread_mutex_unlock(&mq->lock);
return ret / mq->elsize;
#else
return AVERROR(ENOSYS);
#endif
}
#if HAVE_THREADS
static int av_thread_message_queue_send_locked(AVThreadMessageQueue *mq,
void *msg,
unsigned flags)
{
while (!mq->err_send && av_fifo_space(mq->fifo) < mq->elsize) {
if ((flags & AV_THREAD_MESSAGE_NONBLOCK))
return AVERROR(EAGAIN);
avutil/threadmessage: split the pthread condition in two Fix a dead lock under certain conditions. Let's assume we have a queue of 1 message max, 2 senders, and 1 receiver. Scenario (real record obtained with debug added): [...] SENDER #0: acquired lock SENDER #0: queue is full, wait SENDER #1: acquired lock SENDER #1: queue is full, wait RECEIVER: acquired lock RECEIVER: reading a msg from the queue RECEIVER: signal the cond RECEIVER: acquired lock RECEIVER: queue is empty, wait SENDER #0: writing a msg the queue SENDER #0: signal the cond SENDER #0: acquired lock SENDER #0: queue is full, wait SENDER #1: queue is full, wait Translated: - initially the queue contains 1/1 message with 2 senders blocking on it, waiting to push another message. - Meanwhile the receiver is obtaining the lock, read the message, signal & release the lock. For some reason it is able to acquire the lock again before the signal wakes up one of the sender. Since it just emptied the queue, the reader waits for the queue to fill up again. - The signal finally reaches one of the sender, which writes a message and then signal the condition. Unfortunately, instead of waking up the reader, it actually wakes up the other worker (signal = notify the condition just for 1 waiter), who can't push another message in the queue because it's full. - Meanwhile, the receiver is still waiting. Deadlock. This scenario can be triggered with for example: tests/api/api-threadmessage-test 1 2 100 100 1 1000 1000 One working solution is to make av_thread_message_queue_{send,recv}() call pthread_cond_broadcast() instead of pthread_cond_signal() so both senders and receivers are unlocked when work is done (be it reading or writing). This second solution replaces the condition with two: one to notify the senders, and one to notify the receivers. This prevents senders from notifying other senders instead of a reader, and the other way around. It also avoid broadcasting to everyone like the first solution, and is, as a result in theory more optimized.
9 years ago
pthread_cond_wait(&mq->cond_send, &mq->lock);
}
if (mq->err_send)
return mq->err_send;
av_fifo_generic_write(mq->fifo, msg, mq->elsize, NULL);
avutil/threadmessage: split the pthread condition in two Fix a dead lock under certain conditions. Let's assume we have a queue of 1 message max, 2 senders, and 1 receiver. Scenario (real record obtained with debug added): [...] SENDER #0: acquired lock SENDER #0: queue is full, wait SENDER #1: acquired lock SENDER #1: queue is full, wait RECEIVER: acquired lock RECEIVER: reading a msg from the queue RECEIVER: signal the cond RECEIVER: acquired lock RECEIVER: queue is empty, wait SENDER #0: writing a msg the queue SENDER #0: signal the cond SENDER #0: acquired lock SENDER #0: queue is full, wait SENDER #1: queue is full, wait Translated: - initially the queue contains 1/1 message with 2 senders blocking on it, waiting to push another message. - Meanwhile the receiver is obtaining the lock, read the message, signal & release the lock. For some reason it is able to acquire the lock again before the signal wakes up one of the sender. Since it just emptied the queue, the reader waits for the queue to fill up again. - The signal finally reaches one of the sender, which writes a message and then signal the condition. Unfortunately, instead of waking up the reader, it actually wakes up the other worker (signal = notify the condition just for 1 waiter), who can't push another message in the queue because it's full. - Meanwhile, the receiver is still waiting. Deadlock. This scenario can be triggered with for example: tests/api/api-threadmessage-test 1 2 100 100 1 1000 1000 One working solution is to make av_thread_message_queue_{send,recv}() call pthread_cond_broadcast() instead of pthread_cond_signal() so both senders and receivers are unlocked when work is done (be it reading or writing). This second solution replaces the condition with two: one to notify the senders, and one to notify the receivers. This prevents senders from notifying other senders instead of a reader, and the other way around. It also avoid broadcasting to everyone like the first solution, and is, as a result in theory more optimized.
9 years ago
/* one message is sent, signal one receiver */
pthread_cond_signal(&mq->cond_recv);
return 0;
}
static int av_thread_message_queue_recv_locked(AVThreadMessageQueue *mq,
void *msg,
unsigned flags)
{
while (!mq->err_recv && av_fifo_size(mq->fifo) < mq->elsize) {
if ((flags & AV_THREAD_MESSAGE_NONBLOCK))
return AVERROR(EAGAIN);
avutil/threadmessage: split the pthread condition in two Fix a dead lock under certain conditions. Let's assume we have a queue of 1 message max, 2 senders, and 1 receiver. Scenario (real record obtained with debug added): [...] SENDER #0: acquired lock SENDER #0: queue is full, wait SENDER #1: acquired lock SENDER #1: queue is full, wait RECEIVER: acquired lock RECEIVER: reading a msg from the queue RECEIVER: signal the cond RECEIVER: acquired lock RECEIVER: queue is empty, wait SENDER #0: writing a msg the queue SENDER #0: signal the cond SENDER #0: acquired lock SENDER #0: queue is full, wait SENDER #1: queue is full, wait Translated: - initially the queue contains 1/1 message with 2 senders blocking on it, waiting to push another message. - Meanwhile the receiver is obtaining the lock, read the message, signal & release the lock. For some reason it is able to acquire the lock again before the signal wakes up one of the sender. Since it just emptied the queue, the reader waits for the queue to fill up again. - The signal finally reaches one of the sender, which writes a message and then signal the condition. Unfortunately, instead of waking up the reader, it actually wakes up the other worker (signal = notify the condition just for 1 waiter), who can't push another message in the queue because it's full. - Meanwhile, the receiver is still waiting. Deadlock. This scenario can be triggered with for example: tests/api/api-threadmessage-test 1 2 100 100 1 1000 1000 One working solution is to make av_thread_message_queue_{send,recv}() call pthread_cond_broadcast() instead of pthread_cond_signal() so both senders and receivers are unlocked when work is done (be it reading or writing). This second solution replaces the condition with two: one to notify the senders, and one to notify the receivers. This prevents senders from notifying other senders instead of a reader, and the other way around. It also avoid broadcasting to everyone like the first solution, and is, as a result in theory more optimized.
9 years ago
pthread_cond_wait(&mq->cond_recv, &mq->lock);
}
if (av_fifo_size(mq->fifo) < mq->elsize)
return mq->err_recv;
av_fifo_generic_read(mq->fifo, msg, mq->elsize, NULL);
avutil/threadmessage: split the pthread condition in two Fix a dead lock under certain conditions. Let's assume we have a queue of 1 message max, 2 senders, and 1 receiver. Scenario (real record obtained with debug added): [...] SENDER #0: acquired lock SENDER #0: queue is full, wait SENDER #1: acquired lock SENDER #1: queue is full, wait RECEIVER: acquired lock RECEIVER: reading a msg from the queue RECEIVER: signal the cond RECEIVER: acquired lock RECEIVER: queue is empty, wait SENDER #0: writing a msg the queue SENDER #0: signal the cond SENDER #0: acquired lock SENDER #0: queue is full, wait SENDER #1: queue is full, wait Translated: - initially the queue contains 1/1 message with 2 senders blocking on it, waiting to push another message. - Meanwhile the receiver is obtaining the lock, read the message, signal & release the lock. For some reason it is able to acquire the lock again before the signal wakes up one of the sender. Since it just emptied the queue, the reader waits for the queue to fill up again. - The signal finally reaches one of the sender, which writes a message and then signal the condition. Unfortunately, instead of waking up the reader, it actually wakes up the other worker (signal = notify the condition just for 1 waiter), who can't push another message in the queue because it's full. - Meanwhile, the receiver is still waiting. Deadlock. This scenario can be triggered with for example: tests/api/api-threadmessage-test 1 2 100 100 1 1000 1000 One working solution is to make av_thread_message_queue_{send,recv}() call pthread_cond_broadcast() instead of pthread_cond_signal() so both senders and receivers are unlocked when work is done (be it reading or writing). This second solution replaces the condition with two: one to notify the senders, and one to notify the receivers. This prevents senders from notifying other senders instead of a reader, and the other way around. It also avoid broadcasting to everyone like the first solution, and is, as a result in theory more optimized.
9 years ago
/* one message space appeared, signal one sender */
pthread_cond_signal(&mq->cond_send);
return 0;
}
#endif /* HAVE_THREADS */
int av_thread_message_queue_send(AVThreadMessageQueue *mq,
void *msg,
unsigned flags)
{
#if HAVE_THREADS
int ret;
pthread_mutex_lock(&mq->lock);
ret = av_thread_message_queue_send_locked(mq, msg, flags);
pthread_mutex_unlock(&mq->lock);
return ret;
#else
return AVERROR(ENOSYS);
#endif /* HAVE_THREADS */
}
int av_thread_message_queue_recv(AVThreadMessageQueue *mq,
void *msg,
unsigned flags)
{
#if HAVE_THREADS
int ret;
pthread_mutex_lock(&mq->lock);
ret = av_thread_message_queue_recv_locked(mq, msg, flags);
pthread_mutex_unlock(&mq->lock);
return ret;
#else
return AVERROR(ENOSYS);
#endif /* HAVE_THREADS */
}
void av_thread_message_queue_set_err_send(AVThreadMessageQueue *mq,
int err)
{
#if HAVE_THREADS
pthread_mutex_lock(&mq->lock);
mq->err_send = err;
avutil/threadmessage: split the pthread condition in two Fix a dead lock under certain conditions. Let's assume we have a queue of 1 message max, 2 senders, and 1 receiver. Scenario (real record obtained with debug added): [...] SENDER #0: acquired lock SENDER #0: queue is full, wait SENDER #1: acquired lock SENDER #1: queue is full, wait RECEIVER: acquired lock RECEIVER: reading a msg from the queue RECEIVER: signal the cond RECEIVER: acquired lock RECEIVER: queue is empty, wait SENDER #0: writing a msg the queue SENDER #0: signal the cond SENDER #0: acquired lock SENDER #0: queue is full, wait SENDER #1: queue is full, wait Translated: - initially the queue contains 1/1 message with 2 senders blocking on it, waiting to push another message. - Meanwhile the receiver is obtaining the lock, read the message, signal & release the lock. For some reason it is able to acquire the lock again before the signal wakes up one of the sender. Since it just emptied the queue, the reader waits for the queue to fill up again. - The signal finally reaches one of the sender, which writes a message and then signal the condition. Unfortunately, instead of waking up the reader, it actually wakes up the other worker (signal = notify the condition just for 1 waiter), who can't push another message in the queue because it's full. - Meanwhile, the receiver is still waiting. Deadlock. This scenario can be triggered with for example: tests/api/api-threadmessage-test 1 2 100 100 1 1000 1000 One working solution is to make av_thread_message_queue_{send,recv}() call pthread_cond_broadcast() instead of pthread_cond_signal() so both senders and receivers are unlocked when work is done (be it reading or writing). This second solution replaces the condition with two: one to notify the senders, and one to notify the receivers. This prevents senders from notifying other senders instead of a reader, and the other way around. It also avoid broadcasting to everyone like the first solution, and is, as a result in theory more optimized.
9 years ago
pthread_cond_broadcast(&mq->cond_send);
pthread_mutex_unlock(&mq->lock);
#endif /* HAVE_THREADS */
}
void av_thread_message_queue_set_err_recv(AVThreadMessageQueue *mq,
int err)
{
#if HAVE_THREADS
pthread_mutex_lock(&mq->lock);
mq->err_recv = err;
avutil/threadmessage: split the pthread condition in two Fix a dead lock under certain conditions. Let's assume we have a queue of 1 message max, 2 senders, and 1 receiver. Scenario (real record obtained with debug added): [...] SENDER #0: acquired lock SENDER #0: queue is full, wait SENDER #1: acquired lock SENDER #1: queue is full, wait RECEIVER: acquired lock RECEIVER: reading a msg from the queue RECEIVER: signal the cond RECEIVER: acquired lock RECEIVER: queue is empty, wait SENDER #0: writing a msg the queue SENDER #0: signal the cond SENDER #0: acquired lock SENDER #0: queue is full, wait SENDER #1: queue is full, wait Translated: - initially the queue contains 1/1 message with 2 senders blocking on it, waiting to push another message. - Meanwhile the receiver is obtaining the lock, read the message, signal & release the lock. For some reason it is able to acquire the lock again before the signal wakes up one of the sender. Since it just emptied the queue, the reader waits for the queue to fill up again. - The signal finally reaches one of the sender, which writes a message and then signal the condition. Unfortunately, instead of waking up the reader, it actually wakes up the other worker (signal = notify the condition just for 1 waiter), who can't push another message in the queue because it's full. - Meanwhile, the receiver is still waiting. Deadlock. This scenario can be triggered with for example: tests/api/api-threadmessage-test 1 2 100 100 1 1000 1000 One working solution is to make av_thread_message_queue_{send,recv}() call pthread_cond_broadcast() instead of pthread_cond_signal() so both senders and receivers are unlocked when work is done (be it reading or writing). This second solution replaces the condition with two: one to notify the senders, and one to notify the receivers. This prevents senders from notifying other senders instead of a reader, and the other way around. It also avoid broadcasting to everyone like the first solution, and is, as a result in theory more optimized.
9 years ago
pthread_cond_broadcast(&mq->cond_recv);
pthread_mutex_unlock(&mq->lock);
#endif /* HAVE_THREADS */
}
#if HAVE_THREADS
static void free_func_wrap(void *arg, void *msg, int size)
{
AVThreadMessageQueue *mq = arg;
mq->free_func(msg);
}
#endif
void av_thread_message_flush(AVThreadMessageQueue *mq)
{
#if HAVE_THREADS
int used, off;
void *free_func = mq->free_func;
pthread_mutex_lock(&mq->lock);
used = av_fifo_size(mq->fifo);
if (free_func)
for (off = 0; off < used; off += mq->elsize)
av_fifo_generic_peek_at(mq->fifo, mq, off, mq->elsize, free_func_wrap);
av_fifo_drain(mq->fifo, used);
avutil/threadmessage: split the pthread condition in two Fix a dead lock under certain conditions. Let's assume we have a queue of 1 message max, 2 senders, and 1 receiver. Scenario (real record obtained with debug added): [...] SENDER #0: acquired lock SENDER #0: queue is full, wait SENDER #1: acquired lock SENDER #1: queue is full, wait RECEIVER: acquired lock RECEIVER: reading a msg from the queue RECEIVER: signal the cond RECEIVER: acquired lock RECEIVER: queue is empty, wait SENDER #0: writing a msg the queue SENDER #0: signal the cond SENDER #0: acquired lock SENDER #0: queue is full, wait SENDER #1: queue is full, wait Translated: - initially the queue contains 1/1 message with 2 senders blocking on it, waiting to push another message. - Meanwhile the receiver is obtaining the lock, read the message, signal & release the lock. For some reason it is able to acquire the lock again before the signal wakes up one of the sender. Since it just emptied the queue, the reader waits for the queue to fill up again. - The signal finally reaches one of the sender, which writes a message and then signal the condition. Unfortunately, instead of waking up the reader, it actually wakes up the other worker (signal = notify the condition just for 1 waiter), who can't push another message in the queue because it's full. - Meanwhile, the receiver is still waiting. Deadlock. This scenario can be triggered with for example: tests/api/api-threadmessage-test 1 2 100 100 1 1000 1000 One working solution is to make av_thread_message_queue_{send,recv}() call pthread_cond_broadcast() instead of pthread_cond_signal() so both senders and receivers are unlocked when work is done (be it reading or writing). This second solution replaces the condition with two: one to notify the senders, and one to notify the receivers. This prevents senders from notifying other senders instead of a reader, and the other way around. It also avoid broadcasting to everyone like the first solution, and is, as a result in theory more optimized.
9 years ago
/* only the senders need to be notified since the queue is empty and there
* is nothing to read */
pthread_cond_broadcast(&mq->cond_send);
pthread_mutex_unlock(&mq->lock);
#endif /* HAVE_THREADS */
}