Is there a way to wake a specific thread in Haskell? There is a function that suspends the current thread. But the waking counterpart doesn't seem to exist.
yield doesn't suspend the current thread - it moves it to the back of the run queue. It's still in the run queue, it just makes sure that other runnable threads (potentially not all runnable threads, if there are multiple execution contexts defined, which makes this a a pretty weak guarantee) have a chance to run before it continues. For the most part, you should ignore yield. The exception is when you understand exactly what it does, and why that matters.
To actually suspend and resume a thread, MVars are the way to go. When a thread waits on an empty MVar, it is removed from the runnable queue. When a value is put into an MVar, a thread waiting on it (I believe in GHC it's always the thread that has been waiting on that MVar longest, but it's not guaranteed) is put back into the runnable queue.
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I am trying to understand forking with multithreading. So what happens in below scenario ?
Application thread has spawned a thread - polling thread
Application thread runs fork
atpthread_fork handler's pre_fork stops the polling thread using a std::condition_variable. It also waits on a different condition variable to resume the polling
atpthread_fork handler's post_fork in child does cv.notify_one for the waiting poll thread and stops the poll thread
atpthread_fork handler's post_fork in parent does cv.notify_one for the waiting poll thread and resumes the poll thread
But what happens is, post_fork in child enters an infinite loop where it keeps on waiting. This also doesn't seem to notified the poll thread cv at all.
Why is this happening ?
I am trying to understand forking with multithreading.
The #1 thing to understand about combining forking with multithreading is don't do it. The combination is highly problematic other than in a handful of special cases.
So what happens in below scenario ?
Application thread has spawned a thread - polling thread
Application thread runs fork
atpthread_fork handler's pre_fork stops the polling thread using a std::condition_variable. It also waits on a different condition
variable to resume the polling
That makes no sense. A condition variable does not have the power to preemptively make any thread stop. And if the polling thread eventually did stop by blocking on the CV, then what role would a different CV have to play in starting it again?
atpthread_fork handler's post_fork in child does cv.notify_one for the waiting poll thread and stops the poll thread
I suppose you meant to say that a post_fork handler registered via pthread_atfork performs a cv.notify_one in the child to resume the poll thread.
Any way around, it is impossible for the child to do anything with the polling thread because it doesn't have one. The child process has only one thread -- a copy of the one that called fork(). This is one of the main reasons why forking and multithreading don't mix.
atpthread_fork handler's post_fork in parent does cv.notify_one for the waiting poll thread and resumes the poll thread
This seems questionable in light of the overall questionable behavior you are attributing to the CVs, but there is nothing inherently wrong with the concept.
But what happens is, post_fork in child enters an infinite loop where
it keeps on waiting.
Something is missing here. Are you doing a timed wait? Is its wait failing? These are the only ways I can think of that the child could be both looping and waiting. There is initially no other thread in the child process to wake the single one that resulted from the fork, so that thread cannot perform a successful return from a wait on any CV, unless spurriously. There is no one to signal it.
This also doesn't seem to notified the poll
thread cv at all.
Do you mean the one in the child that doesn't exist? Or the one in the parent that probably isn't waiting on the CV you think it's waiting on?
Most of the above is moot. There is absolutely no reason to think that your program is exercising one of the special exceptions, so refer to #1: don't combine forking with multithreading. Choose one.
I was studying about multi-threading and came across join().
As I understand right, using join() on the thread makes process wait until 'joined' thread terminates. For example, calling t1.join() in main will make main wait until the job in thread t1 is finished and t1 terminates.
I'm just curious that how the function join() make this possible - how does it make current thread 'blocked' inside the function? Does join() force execution of joined thread first so any other thread should wait until that thread terminates? Or, is there some way to communicate between two threads(the thread who called join() and the thread who is joined)?
I will be waiting for the answer. Thanks a lot!
To be able to join you need to be able to wait on some event. Then join looks like this:
function join(t : Thread)
// do this atomically
if already done
return
wait on termination event of t
end
Waiting can be done in one of two ways:
Looping and periodically checking if the event has happened (busy wait)
Letting the system reclaim the resources of the thread and be woken up on a system event, in that case waking the thread is managed by the scheduler of the OS
It's rather language specific.
Once you create a thread, it starts running.
A join operation is when your main process stops and waits for the thread to exit and capture a return code. It will block until your thread completes - that's rather the point, as it allows for a synchronization to occur - everything in your program is at a 'known state'.
Related is the detach operation, which is effectively saying 'I don't care any more'.
If I had threads as below
void thread(){
while() {
lock.acquire();
if(condition not true)
{
Cond.wait()
}
// blah blah
Cond.Signal();
lock.release();
}
}
Well I guess my main question is that whether the signalling thread continues running for a while after cond.signal() or immediately gives up the CPU?. I would like it in some cases not to release the lock before the woken up thread finishes execution and in some other cases it may be beneficial to release the lock immediately after signalling, without waiting for the other woken thread to finish.
I understand that if there are any threads waiting on the condition then they get woken up on Cond.signal(). But what do you mean by woekn up - put on the ready queue or does the scheduler make sure that it runs immediately?.
and what about the signalling thread.. does it go to sleep on the same condtion upon signalling? .. so then some other thread has to wake it up to make it release the lock?.
This is in large part dependent on your environment (OS, library, language...) and how the synchronisation primitives are implemented. Since you haven't specified any I'll just give a general answer.
When putting a thread to sleep, most environment will choose to remove it from the scheduler's ready queue and the thread will give up its remaining CPU time. When woken up, the thread is simply placed back into the ready queue and will resume execution the next time the scheduler selects it from the queue.
It's also possible that the thread will do some active waiting (spinning) instead of being removed from the scheduler's ready queue. In this case, the thread will resume execution right away. Note that since a thread can still be run out of CPU of time while spinning, it might have to wait to be rescheduled before waking up. This is a useful strategy if your critical sections are very small and you don't want to pay for the scheduling overheads.
A hybrid approach would be to do a small amount of active waiting before removing the thread from the scheduler's ready queue.
As for the signaling thread, unless specified explicitly by your environment (I can't of any reasons but you never know), I wouldn't expect a call to signal() to block in a way that you have to wake it up. Signal() might have to synchronize itself with other threads calling signal() but those are implementation details and you shouldn't have to do anything about it.
I've been reading up on multithreading and shared resources access and one of the many (for me) new concepts is the mutex lock. What I can't seem to find out is what is actually happening to the thread that finds a "critical section" is locked. It says in many places that the thread gets "blocked", but what does that mean? Is it suspended, and will it resume when the lock is lifted? Or will it try again in the next iteration of the "run loop"?
The reason I ask, is because I want to have system supplied events (mouse, keyboard, etc.), which (apparantly) are delivered on the main thread, to be handled in a very specific part in the run loop of my secondary thread. So whatever event is delivered, I queue in my own datastructure. Obviously, the datastructure needs a mutex lock because it's being modified by both threads. The missing puzzle-piece is: what happens when an event gets delivered in a function on the main thread, I want to queue it, but the queue is locked? Will the main thread be suspended, or will it just jump over the locked section and go out of scope (losing the event)?
Blocked means execution gets stuck there; generally, the thread is put to sleep by the system and yields the processor to another thread. When a thread is blocked trying to acquire a mutex, execution resumes when the mutex is released, though the thread might block again if another thread grabs the mutex before it can.
There is generally a try-lock operation that grab the mutex if possible, and if not, will return an error. But you are eventually going to have to move the current event into that queue. Also, if you delay moving the events to the thread where they are handled, the application will become unresponsive regardless.
A queue is actually one case where you can get away with not using a mutex. For example, Mac OS X (and possibly also iOS) provides the OSAtomicEnqueue() and OSAtomicDequeue() functions (see man atomic or <libkern/OSAtomic.h>) that exploit processor-specific atomic operations to avoid using a lock.
But, why not just process the events on the main thread as part of the main run loop?
The simplest way to think of it is that the blocked thread is put in a wait ("sleeping") state until the mutex is released by the thread holding it. At that point the operating system will "wake up" one of the threads waiting on the mutex and let it acquire it and continue. It's as if the OS simply puts the blocked thread on a shelf until it has the thing it needs to continue. Until the OS takes the thread off the shelf, it's not doing anything. The exact implementation -- which thread gets to go next, whether they all get woken up or they're queued -- will depend on your OS and what language/framework you are using.
Too late to answer but I may facilitate the understanding. I am talking more from implementation perspective rather than theoretical texts.
The word "blocking" is kind of technical homonym. People may use it for sleeping or mere waiting. The term has to be understood in context of usage.
Blocking means Waiting - Assume on an SMP system a thread B wants to acquire a spinlock held by some other thread A. One of the mechanisms is to disable preemption and keep spinning on the processor unless B gets it. Another mechanism probably, an efficient one, is to allow other threads to use processor, in case B does not gets it in easy attempts. Therefore we schedule out thread B (as preemption is enabled) and give processor to some other thread C. In this case thread B just waits in the scheduler's queue and comes back with its turn. Understand that B is not sleeping just waiting rather passively instead of busy-wait and burning processor cycles. On BSD and Solaris systems there are data-structures like turnstiles to implement this situation.
Blocking means Sleeping - If the thread B had instead made system call like read() waiting data from network socket, it cannot proceed until it gets it. Therefore, some texts casually use term blocking as "... blocked for I/O" or "... in blocking system call". Actually, thread B is rather sleeping. There are specific data-structures known as sleep queues - much like luxury waiting rooms on air-ports :-). The thread will be woken up when OS detects availability of data, much like an attendant of the waiting room.
Blocking means just that. It is blocked. It will not proceed until able. You don't say which language you're using, but most languages/libraries have lock objects where you can "attempt" to take the lock and then carry on and do something different depending on whether you succeeded or not.
But in, for example, Java synchronized blocks, your thread will stall until it is able to acquire the monitor (mutex, lock). The java.util.concurrent.locks.Lock interface describes lock objects which have more flexibility in terms of lock acquisition.
In Qt, I have a method which contains a mutex lock and unlock. The problem is when the mutex is unlock it sometimes take long before the other thread gets the lock back. In other words it seems the same thread can get the lock back(method called in a loop) even though another thread is waiting for it. What can I do about this? One thread is a qthread and the other thread is the main thread.
You can have your thread that just unlocked the mutex relinquish the processor. On Posix, you do that by calling pthread_yield() and on Windows by calling Sleep(0).
That said, there is no guarantee that the thread waiting on the lock will be scheduled before your thread wakes up again.
It shouldn't be possible to release a lock and then get it back if some other thread is already waiting on it.
Check that you actually releasing the lock when you think you do. Check that waiting thread actually waits (and not spins a loop with a trylock tests and sleeps, I actually done that once and was very puzzled at first :)).
Or if waiting thread really never gets time to even reach locking code, try QThread::yieldCurrentThread(). This will stop current thread and give scheduler a chance to give execution to somebody else. Might cause unnecessary switching depending on tightness of your loop.
If you want to make sure that one thread has priority over the other ones, an option is to use a QReadWriteLock. It's adapted to a typical scenario where n threads are going to read a value in a infinite loop, with only one thread updating it. I think it's the scenario you described.
QReadWriteLock offers two ways to lock: lockForRead() and lockForWrite(). The threads depending on the value will use the latter, while the thread updating the value (typically via the GUI) will use the former (lockForWrite()) and will have top priority. You won't need to sleep or yield or whatever.
Example code
Let's say you have a QReadWrite lock; somewhere.
"Reader" thread
forever {
lock.lockForRead();
if (condition) {
do_stuff();
}
lock.unlock();
}
"Writer" thread
// external input (eg. user) changes the thread
lock.lockForWrite(); // will block as soon as the reader lock ends
update_condition();
lock.unlock();