OK. The example here is provided using pthread lib in c.
In textbook I came across the following code:
//for thread 2
pthread_mutex_lock(&lock);
should_wake_up = 1;
pthread_cond_signal(&cond);
pthread_mutex_unlock(&lock);
This code works out pretty fine. I just wonder will the following code also works?
//for thread 2
pthread_mutex_lock(&lock);
should_wake_up = 1;
pthread_mutex_unlock(&lock);
pthread_cond_signal(&cond);//signal the conditional variable but the lock is not held
What's the pros and cons for the following code?
PS. Suppose the cooperating thread has the code:
//for thread 1
pthread_mutex_lock(&lock);
while(!should_wake_up)
pthread_cond_wait(&cond, &lock);
pthead_mutex_unlock(&lock);
PS2. I came across some other question, which points out that if we don't want signal being lost, we must use lock to make sure that the associated predicate (in this case, is should_wake_up) can not be changed when the lock is held in thread 1. In this case, this seems not to be the issue. Link to the post: [1]: signal on condition variable without holding lock. I think his issue is that he forget about locking. But my question is different.
For normal usage, you can unlock the mutex before signalling a condition variable.
The mutex protects the shared state (in this case the should_wake_up flag). Provided the mutex is locked when modifying the shared state, and when checking the shared state, pthread_cond_signal can be called without the mutex locked and everything will work as expected.
Under most circumstances, I would recommend calling pthread_cond_signal after calling pthread_mutex_unlock as a slight performance improvement. If you call pthread_cond_signal before pthread_mutex_unlock then the waiting thread can be woken by the signal before the mutex is unlocked, so the waiting thread then has to go back to sleep, as it blocks on the mutex that is still held by the signalling thread.
Related
when there are waiting semaphores of sem_wait method, I call the sem_destroy method on other thread. But waiting semaphore was not wake up.
In case of mutex, pthread_mutex_destroy was return the value EBUSY when there are some waiting threads.
however sem_destroy return 0 and errno was also set 0.
I want to destroy semaphore after calling sem_destroy to block access as destroyed semaphore and to wake up the waiting thread.
Semaphore handle of Window OS is possible.
please advise me. thank you.
POSIX says this about sem_destroy:
The effect of destroying a semaphore upon which other threads are currently blocked is undefined.
It specifically doesn't say that other threads are woken up. In fact, if sem_t contains a pointer to memory, what it almost certainly does do is free the memory, meaning you then have a use-after-free security problem. (Whether that is the case depends on your libc.)
The general approach of allocation for mutexes and semaphores is that they should be either allocated and freed with their relevant data structure, or they should be allocated before the relevant code needs them and then freed after the entire code is done with them. In C, you cannot safely deallocate data structures (e.g., with sem_destroy) that are in use.
If you want to wake up all users of the semaphore, you must increment it until all users have awoken. You can call sem_getvalue to determine if anyone is waiting on the semaphore and then call sem_post to increment it. Only then can you safely destroy it. Note that this can have a race condition, depending on your code.
However, note that you must be careful that the other code does not continue to use the semaphore after it's destroyed, such as by trying to re-acquire it in a loop. If you are careful to structure your code properly, then you can have confidence that this won't happen.
I would like to know how many threads are waiting on a lock so I would be able to destroy it safely.
The problem is that I can't destroy the lock when someone holds it or someone is waiting on it.
My program can make sure that no new requests are made to acquire the lock, but how can I know when all the threads that waited on it are done with it?
I thought about a conditional variable but I suspect it will create problems..
dlv, could you add some code snippet to your description.
I hope you should be using condition variables,
Each thread will block in pthread_cond_wait() until the other thread signals it to wake up. This will not cause a deadlock. It can easily be extended to many threads, by allocating one int, pthread_cond_t and pthread_mutex_t per thread.
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.
The pthread_cond_signal() routine is used to signal (or wake up) another thread which is waiting on the condition variable. It should be called after mutex is locked, and must unlock mutex in order for pthread_cond_wait() routine to complete.
The pthread_cond_broadcast() routine should be used instead of pthread_cond_signal() if more than one thread is in a blocking wait state.
It is a logical error to call pthread_cond_signal() before calling pthread_cond_wait().
Proper locking and unlocking of the associated mutex variable is essential when using these routines. For example:
Failing to lock the mutex before calling pthread_cond_wait() may cause it NOT to block.
Failing to unlock the mutex after calling pthread_cond_signal() may not allow a matching pthread_cond_wait() routine to complete (it will remain blocked).
If threads that can use the mutex still exist or might be created in the future then don't delete it.
You do know and are tracking what threads are created, right?
If, for some reason, you cannot keep track of the threads using a resource, your only way out is to leak the resource. It can never be safely deleted because you never know when you are done using it.
Say you had a counter that counted the threads using a mutex. That counter would need its own mutex. Then how do you decide when to delete that one?
That way of thinking is the road that leads to hell. You could do what you want with condition variables, but the result would be an extremely weak design.
Assuming you managed to create such a monster, it would basically allow you to kill "safely" any other thread regardless of its internal state. Except for a quick and dirty panic exit (in case of some internal software error), this is the worst possible way of solving synchronization issues.
A design relying on such tricks would have to create implicit synchronizations between tasks to make sure the terminations occur in the proper order. A lot of software are designed that way, and most of them allow mediocre programmers to make a living by maintaining the pile of crap they created in the first place.
Task termination should be an issue solved at global design level, not by a toolbox of wonky objects that allow you to twist synchronization any odd way.
We know that multi-threaded code has the bane of possible deadlocks if the threads acquire mutex locks but before it gets a chance to release it, the thread gets suspended by main thread or pre-empted out by Scheduler?
I am a beginner in using pthread library so please bear with me if my below query/proposed solution might be unfeasible or outright wrong.
void main()
{
thread_create(T1,NULL,thr_function,NULL)
suspend_thread(T1);
acquire_lock(Lock1);<--- //Now here is a possible deadlock if thread_function acquried Lock1 before main and main suspended T1 before its release
//Do something further;
}
void *thr_function(void *val)
{
///do something;
acquire_lock(Lock1);
//do some more things;
//do some more things;
release_lock(Lock1);
}
In this below pseudo code segment above I have, can't the thread run-time/compiler work together to make sure if a thread which has acquired a mutex lock, is suspended/pre-empted then it executes some 'cleanup code' of releasing all locks it has held before it gets out. The compiler/linker can identify the places inside a thread function which acquire , release lock, then when a thread is suspended between those two places(i.e. after acquire but before release) the execution in the thread function should jump via some kind of 'goto label;' inserted by the runtime where at the label: the thread would release the lock and then the thread gets blocked or context switch happens. [ I know if a thread acquires more than 1 locks it might get messy to jump across those points to release those locks...]
But basic idea/question is can the thread function not do the necessary releases of acquired locks for mutexes, semaphores before it gets blocked out or goes out of execution state to wait or some other state?
No. The reason a thread holds a lock is so that it can make data temporarily inconsistent or see a consistent view of that data itself. If some scheme were to automatically release that lock before the thread made the data consistent again, other threads would acquire the lock, see the inconsistent data, and fail. Or when that thread was resumed, it would either not have the lock or have the lock and see inconsistent data itself. This is why you can only reliably suspend a thread with that thread's cooperation.
Consider this logic to add an object to a linked list protected by a mutex:
Acquire the lock protecting a linked list.
Modify the link's head pointer.
Modify the object's next pointer.
Release the lock.
Now imagine if something were to suspend the thread between steps 2 and 3. If the lock were released, other threads would see the link's head pointer pointing to an object that had not been linked to the list. And when the thread resumed, it might set the object to the wrong pointer because the list had changed.
The general consensus is that suspending threads is so evil that even a feeling that you might want to suspend a thread suggests an incorrect application design. There is practically no reason a properly-designed application would ever want to suspend a thread. (If you didn't want that thread to continue doing the work it was doing, why did you code it to continue doing that work in the first place?)
By the way, scheduler pre-emption is not a problem. Eventually, the thread will be scheduled again and release the lock. So long as there are other threads that can make forward progress, no harm is done. And if there are no other threads that can make forward progress, the only thing the system can do is schedule the thread that was pre-empted.
One way to avoid this kind of deadlocks is to have a global, mutexed variable should_stop_thread which eventually gets set to true by the master thread.
The child thread checks the variable regularly and terminates in a controlled manner if it is true. "Controlled" in this sense means that all data (pointers) are valid (again) and mutex locks are released.
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.
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();