Reputation: 929
I have a two threads, one which works in a tight loop, and the other which occasionally needs to perform a synchronization with the first:
// thread 1
while(1)
{
lock(work);
// perform work
unlock(work);
}
// thread 2
while(1)
{
// unrelated work that takes a while
lock(work);
// synchronizing step
unlock(work);
}
My intention is that thread 2 can, by taking the lock, effectively pause thread 1 and perform the necessary synchronization. Thread 1 can also offer to pause, by unlocking, and if thread 2 is not waiting on lock, re-lock and return to work.
The problem I have encountered is that mutexes are not fair, so thread 1 quickly re-locks the mutex and starves thread 2. I have attempted to use pthread_yield
, and so far it seems to run okay, but I am not sure it will work for all systems / number of cores. Is there a way to guarantee that thread 1 will always yield to thread 2, even on multi-core systems?
What is the most effective way of handling this synchronization process?
Upvotes: 14
Views: 8493
Reputation: 674
Here's a simple solution which will work for your case (two threads). If you're using std::mutex
then this class is a drop-in replacement. Change your mutex to this type and you are guaranteed that if one thread holds the lock and the other is waiting on it, once the first thread unlocks, the second thread will grab the lock before the first thread can lock it again.
If more than two threads happen to use the mutex simultaneously it will still function but there are no guarantees on fairness.
If you're using plain pthread_mutex_t
you can easily change your locking code according to this example (unlock remains unchanged).
#include <mutex>
// Behaves the same as std::mutex but guarantees fairness as long as
// up to two threads are using (holding/waiting on) it.
// When one thread unlocks the mutex while another is waiting on it,
// the other is guaranteed to run before the first thread can lock it again.
class FairDualMutex : public std::mutex {
public:
void lock() {
_fairness_mutex.lock();
std::mutex::lock();
_fairness_mutex.unlock();
}
private:
std::mutex _fairness_mutex;
};
Upvotes: 1
Reputation: 2283
Ticket lock above looks like the best. However, to insure your pthread_yield works, you could have a bool waiting, which is set and reset by thread2. thread1 yields as long as bool waiting is set.
Upvotes: 0
Reputation: 239041
You can build a FIFO "ticket lock" on top of pthreads mutexes, along these lines:
#include <pthread.h>
typedef struct ticket_lock {
pthread_cond_t cond;
pthread_mutex_t mutex;
unsigned long queue_head, queue_tail;
} ticket_lock_t;
#define TICKET_LOCK_INITIALIZER { PTHREAD_COND_INITIALIZER, PTHREAD_MUTEX_INITIALIZER }
void ticket_lock(ticket_lock_t *ticket)
{
unsigned long queue_me;
pthread_mutex_lock(&ticket->mutex);
queue_me = ticket->queue_tail++;
while (queue_me != ticket->queue_head)
{
pthread_cond_wait(&ticket->cond, &ticket->mutex);
}
pthread_mutex_unlock(&ticket->mutex);
}
void ticket_unlock(ticket_lock_t *ticket)
{
pthread_mutex_lock(&ticket->mutex);
ticket->queue_head++;
pthread_cond_broadcast(&ticket->cond);
pthread_mutex_unlock(&ticket->mutex);
}
Under this kind of scheme, no low-level pthreads mutex is held while a thread is within the ticketlock protected critical section, allowing other threads to join the queue.
Upvotes: 9
Reputation: 12963
pthread
offers a notion of thread priority in its API. When two threads are competing over a mutex, the scheduling policy determines which one will get it. The function pthread_attr_setschedpolicy
lets you set that, and pthread_attr_getschedpolicy
permits retrieving the information.
Now the bad news:
SCHED_FIFO
, SCHED_RR
, SCHED_OTHER
and SCHED_SPORADIC
), but in this question, it has been answered that only SCHED_OTHER
was supported on linux)So I would give it a shot if I were you, but not expect too much. pthread_yield
seems more promising to me. More information available here.
Upvotes: 4
Reputation: 5110
In your case it is better to use condition variable to notify second thread when it is required to awake and perform all required operations.
Upvotes: 5