2014-02-20 13:14:34 +00:00
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/*
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* Copyright (c) 2014 Nicolas George
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*
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* This file is part of FFmpeg.
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*
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* FFmpeg is free software; you can redistribute it and/or
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* modify it under the terms of the GNU Lesser General Public License
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* as published by the Free Software Foundation; either
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* version 2.1 of the License, or (at your option) any later version.
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*
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* FFmpeg is distributed in the hope that it will be useful,
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* but WITHOUT ANY WARRANTY; without even the implied warranty of
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* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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* GNU Lesser General Public License for more details.
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*
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* You should have received a copy of the GNU Lesser General Public License
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* along with FFmpeg; if not, write to the Free Software Foundation, Inc.,
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* 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301 USA
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*/
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#include "fifo.h"
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#include "threadmessage.h"
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2015-12-07 14:08:51 +00:00
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#include "thread.h"
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2014-02-20 13:14:34 +00:00
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struct AVThreadMessageQueue {
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#if HAVE_THREADS
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AVFifoBuffer *fifo;
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pthread_mutex_t lock;
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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.
2015-12-01 14:54:31 +00:00
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pthread_cond_t cond_recv;
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pthread_cond_t cond_send;
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2014-02-20 13:14:34 +00:00
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int err_send;
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int err_recv;
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unsigned elsize;
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2015-11-30 17:18:42 +00:00
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void (*free_func)(void *msg);
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2014-02-20 13:14:34 +00:00
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#else
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int dummy;
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#endif
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};
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int av_thread_message_queue_alloc(AVThreadMessageQueue **mq,
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unsigned nelem,
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unsigned elsize)
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{
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#if HAVE_THREADS
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AVThreadMessageQueue *rmq;
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int ret = 0;
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if (nelem > INT_MAX / elsize)
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return AVERROR(EINVAL);
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if (!(rmq = av_mallocz(sizeof(*rmq))))
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return AVERROR(ENOMEM);
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if ((ret = pthread_mutex_init(&rmq->lock, NULL))) {
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av_free(rmq);
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return AVERROR(ret);
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}
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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.
2015-12-01 14:54:31 +00:00
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if ((ret = pthread_cond_init(&rmq->cond_recv, NULL))) {
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pthread_mutex_destroy(&rmq->lock);
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av_free(rmq);
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return AVERROR(ret);
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}
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if ((ret = pthread_cond_init(&rmq->cond_send, NULL))) {
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pthread_cond_destroy(&rmq->cond_recv);
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2014-02-20 13:14:34 +00:00
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pthread_mutex_destroy(&rmq->lock);
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av_free(rmq);
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return AVERROR(ret);
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}
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if (!(rmq->fifo = av_fifo_alloc(elsize * nelem))) {
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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.
2015-12-01 14:54:31 +00:00
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pthread_cond_destroy(&rmq->cond_send);
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pthread_cond_destroy(&rmq->cond_recv);
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2014-02-20 13:14:34 +00:00
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pthread_mutex_destroy(&rmq->lock);
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av_free(rmq);
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2017-07-12 18:41:56 +00:00
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return AVERROR(ENOMEM);
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2014-02-20 13:14:34 +00:00
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}
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rmq->elsize = elsize;
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*mq = rmq;
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return 0;
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#else
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*mq = NULL;
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return AVERROR(ENOSYS);
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#endif /* HAVE_THREADS */
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}
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2015-11-30 17:18:42 +00:00
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void av_thread_message_queue_set_free_func(AVThreadMessageQueue *mq,
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void (*free_func)(void *msg))
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{
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2015-12-07 15:39:57 +00:00
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#if HAVE_THREADS
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2015-11-30 17:18:42 +00:00
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mq->free_func = free_func;
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2015-12-07 15:39:57 +00:00
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#endif
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2015-11-30 17:18:42 +00:00
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}
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2014-02-20 13:14:34 +00:00
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void av_thread_message_queue_free(AVThreadMessageQueue **mq)
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{
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#if HAVE_THREADS
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if (*mq) {
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2015-11-30 17:18:42 +00:00
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av_thread_message_flush(*mq);
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2014-02-20 13:14:34 +00:00
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av_fifo_freep(&(*mq)->fifo);
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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.
2015-12-01 14:54:31 +00:00
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pthread_cond_destroy(&(*mq)->cond_send);
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pthread_cond_destroy(&(*mq)->cond_recv);
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2014-02-20 13:14:34 +00:00
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pthread_mutex_destroy(&(*mq)->lock);
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av_freep(mq);
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}
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#endif
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}
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#if HAVE_THREADS
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static int av_thread_message_queue_send_locked(AVThreadMessageQueue *mq,
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void *msg,
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unsigned flags)
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{
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while (!mq->err_send && av_fifo_space(mq->fifo) < mq->elsize) {
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if ((flags & AV_THREAD_MESSAGE_NONBLOCK))
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return AVERROR(EAGAIN);
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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.
2015-12-01 14:54:31 +00:00
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pthread_cond_wait(&mq->cond_send, &mq->lock);
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2014-02-20 13:14:34 +00:00
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}
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if (mq->err_send)
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return mq->err_send;
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av_fifo_generic_write(mq->fifo, msg, mq->elsize, NULL);
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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.
2015-12-01 14:54:31 +00:00
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/* one message is sent, signal one receiver */
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pthread_cond_signal(&mq->cond_recv);
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2014-02-20 13:14:34 +00:00
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return 0;
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}
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static int av_thread_message_queue_recv_locked(AVThreadMessageQueue *mq,
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void *msg,
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unsigned flags)
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{
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while (!mq->err_recv && av_fifo_size(mq->fifo) < mq->elsize) {
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if ((flags & AV_THREAD_MESSAGE_NONBLOCK))
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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.
2015-12-01 14:54:31 +00:00
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pthread_cond_wait(&mq->cond_recv, &mq->lock);
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2014-02-20 13:14:34 +00:00
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}
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if (av_fifo_size(mq->fifo) < mq->elsize)
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return mq->err_recv;
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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.
2015-12-01 14:54:31 +00:00
|
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|
/* one message space appeared, signal one sender */
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pthread_cond_signal(&mq->cond_send);
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2014-02-20 13:14:34 +00:00
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return 0;
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}
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#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.
2015-12-01 14:54:31 +00:00
|
|
|
pthread_cond_broadcast(&mq->cond_send);
|
2014-02-20 13:14:34 +00:00
|
|
|
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.
2015-12-01 14:54:31 +00:00
|
|
|
pthread_cond_broadcast(&mq->cond_recv);
|
2014-02-20 13:14:34 +00:00
|
|
|
pthread_mutex_unlock(&mq->lock);
|
|
|
|
#endif /* HAVE_THREADS */
|
|
|
|
}
|
2015-11-30 17:18:42 +00:00
|
|
|
|
2015-12-07 13:26:39 +00:00
|
|
|
#if HAVE_THREADS
|
2015-11-30 17:18:42 +00:00
|
|
|
static void free_func_wrap(void *arg, void *msg, int size)
|
|
|
|
{
|
|
|
|
AVThreadMessageQueue *mq = arg;
|
|
|
|
mq->free_func(msg);
|
|
|
|
}
|
2015-12-07 13:26:39 +00:00
|
|
|
#endif
|
2015-11-30 17:18:42 +00:00
|
|
|
|
|
|
|
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.
2015-12-01 14:54:31 +00:00
|
|
|
/* only the senders need to be notified since the queue is empty and there
|
|
|
|
* is nothing to read */
|
|
|
|
pthread_cond_broadcast(&mq->cond_send);
|
2015-11-30 17:18:42 +00:00
|
|
|
pthread_mutex_unlock(&mq->lock);
|
|
|
|
#endif /* HAVE_THREADS */
|
|
|
|
}
|