Imagine I have a mutex locked. There is unlimited number of other threads waiting to lock the mutex. When I unlock the mutex, one of those threads will be chosen to enter the critical section. However I have no control over which one. What if I want specific thread to enter the critical section?
I am prety sure this cannot be done using POSIX mutex, however, can I emulate the behaviour using different synchronisation object(s)?
You can use a mutex, a condition variable and a thread id to achive that.
Before unlocking the mutex the thread sets the target thread id, broadcasts the condition variable and releases the mutex. The waiting threads wake up, lock the mutex and check whether the target thread id equals to this thread id. If not the thread goes back to wait.
An optimization to this method to avoid waking up all waiting threads just to check the target thread id and then go back to wait would be to use a separate condition variable for each waiting thread. This way the signaling thread would notify the condition variable of the particular target thread.
Another option is to use signals sent to a particular thread. Let's say we use SIGRTMINfor this purpose. First, all threads block this signal at the start, so that the signal becomes pending and doesn't get lost when the thread isn't waiting for it. When a thread wants to lock the mutex it first calls sigwait() which atomically unblocks SIGRTMIN and waits for it or delivers an already pending one. Once the thread received the signal it can proceed and lock the mutex. The signaling thread uses pthread_kill(target_thread_id, SIGRTMIN) to wake up a particular thread.
Related
My thread does not need to be locked. std::unique_lock locks thread on construction. I am simply using cond_var.wait() as a way to avoid busy waiting. I have essentially circumvented the auto-locking by putting the unique_lock within a tiny scope and hence destroying the unique lock after it leaves the tiny scope. Additionally, there is only a single consumer thread if that's relevant.
{
std::unique_lock<std::mutex> dispatch_ul(dispatch_mtx);
pq_cond.wait(dispatch_ul);
}
Is there possibly a better option to avoid the unnecessary auto-lock functionality from the unique_lock? I'm looking for a mutexless option to simply signal the thread, I am aware of std::condition_variable_any but that requires a mutex of sorts which is yet again unnessesary in my case.
You need a lock to prevent this common newbie mistake:
Producer thread produces something,
Producer thread calls some_condition.notify_all(),
Producer thread goes idle for a while,
meanwhile:
Consumer thread calls some_condition.wait(...)
Consumer thread waits,...
And waits,...
And waits.
A condition variable is not a flag. It does not remember that it was notified. If the producer calls notify_one() or notify_all() before the consumer has entered the wait() call, then the notification is "lost."
In order to prevent lost notifications, there must be some shared data that tells the consumer whether or not it needs to wait, and there must be a lock to protect the shared data.
The producer should:
Lock the lock,
update the shared data,
notify the condition variable,
release the lock
The consumer must then:
Lock the lock,
Check the shared data to see if it needs wait,
Wait if needed,
consume whatever,
release the lock.
The consumer needs to pass the lock in to the wait(...) call so that wait(...) can temporarily unlock it, and then re-lock it before returning. If wait(...) did not unlock the lock, then the producer would never be able to reach the notify() call.
I am trying to understand what guarantees I have just after I signalled a condvar.
The basic pattern of usage is, I believe, this (pseudocode):
Consumer Thread:
Mutex.Enter()
while(variable != ready)
Condvar.Wait()
Mutex.Exit()
Producer Thread:
Mutex.Enter()
variable = ready
Condvar.Broadcast()
[Unknown?]
Mutext.Exit()
My question is. What am I guaranteed about the [Unknown] point in the code above? I am still holding the mutex, but what can I know about the state of the consumer?
From the Man page, it is unclear to me what state the producer is in after it broadcasts/signals and before it releases the mutex.
From condition vars:
pthread_cond_wait() blocks the calling thread until the specified condition is signalled. This routine should be called while mutex is locked, and it will automatically release the mutex while it waits. After signal is received and thread is awakened, mutex will be automatically locked for use by the thread. The programmer is then responsible for unlocking mutex when the thread is finished with it.
So, when producer is executing the Unknown section of code, producer is holding the mutex, and the consumer is locked on the mutex until producer releases it.
For example, suppose you are using atomic spinlock on an integer flag to ensure only one thread modifies the wait-queue that the mutex maintains at any given time. When a thread tries to lock the mutex, we want it to enqueue itself and set the flag to zero before it blocks itself and the unlocker to dequeue a thread from the queue and set it to runnable.
Consider only two threads to be present, one locking and the other releasing the mutex at the same time. if the locker was preempted after it added himself to the queue and set the flag to zero (but not blocked itself yet) and the unlocker then tried to dequeue and make the thread runnable, it wouldn't be useful since the thread hasn't blocked itself yet. So the make-runnable call would be waste but more importantly, the locker thread would then block itself after that and would remain blocked forever.
How is this atomicity achieved to ensure correctness? A similar scenario can be imagined in condition variables with the release of mutex and blocking itself.
Singularity - If a thread managed to lock a mutex, it is assured that no other thread will be able to lock the thread until the original thread releases the lock.
Non-Busy Wait - If a thread attempts to lock a thread that was locked by a second thread, the first thread will be suspended (and will not consume any CPU resources) until the lock is freed by the second thread. At this time, the first thread will wake up and continue execution, having the mutex locked by it.
From: Multi-Threaded Programming With POSIX Threads
Question: I thought threads lock the mutex variables. Threads don't lock other threads?
What do the bold statements above mean? How can one thread lock other thread?
Corrections:
If a thread managed to lock a mutex, it is assured that no other thread will be able to lock the mutex until the original thread releases the lock.
Non-Busy Wait - If a thread attempts to lock a mutex that was locked by a second thread, the first thread will be suspended (and will not consume any CPU resources) until the lock is freed by the second thread. At this time, the first thread will wake up and continue execution, having the mutex locked by it.
It's a good thing you don't take for granted whatever you read on the internet, also I give you thumbs up for paying attention to what you read.
pthread_cond_broadcast is used to wake up several threads waiting on a condition variable. But, at the same time it is also said that we should place a mutex before the condition variable to avoid race conditions.
So, if a mutex is there, and one thread already holds it and thus waits on the variable, how can any other thread hold the same mutex (to enter to the waiting part)?
When a thread calls pthread_cond_wait() it needs to hold the associated mutex. The API will release the mutex while it blocks the thread. Once the API decides the thread needs to be released, it will acquire the mutex before returning from the API.
Holding the mutex is required because checking the condition then calling pthread_cond_wait() isn't an atomic operation. The mutex (and the proper use of the mutex) prevents the thread from missing the condition becoming true in the short period between checking it and calling the wait.
But the short answer to the specific question (how can another thread obtain the mutex) is that while the thread is blocked in pthread_cond_wait(), the mutex is not held.