问题
What is the reason for a notified condition variable to re-lock the mutex after being notified.
The following piece of code deadlock if unique_lock is not scoped or if mutex is not explicitely unlocked
#include <future>
#include <mutex>
#include <iostream>
using namespace std;
int main()
{
std::mutex mtx;
std::condition_variable cv;
//simulate another working thread sending notification
auto as = std::async([&cv](){ std::this_thread::sleep_for(std::chrono::seconds(2));
cv.notify_all();});
//uncomment scoping (or unlock below) to prevent deadlock
//{
std::unique_lock<std::mutex> lk(mtx);
//Spurious Wake-Up Prevention not adressed in this short sample
//UNLESS it is part of the answer / reason to lock again
cv.wait(lk);
//}
std::cout << "CV notified\n" << std::flush;
//uncomment unlock (or scoping above) to prevent deadlock
//mtx.unlock();
mtx.lock();
//do something
mtx.unlock();
std::cout << "End may never be reached\n" << std::flush;
return 0;
}
Even re-reading some documentation and examples I still do not find this obvious.
Most examples that can be found over the net are small code samples that have inherent scoping of the unique_lock.
Shall we use different mutex to deal with critical sections (mutex 1) and condition variables wait and notify (mutex 2) ?
Note: Debug shows that after end of the waiting phase, the "internal" "mutex count" (I think field __count of structure __pthread_mutex_s ) goes from 1 to 2. It reaches back 0 after unlock
回答1:
You're trying to lock the mutex twice. Once with the unique_lock and again with the explicit mutex.lock() call. For non-recursive mutex, it will deadlock on a re-lock attempt to let you know you have a bug.
std::unique_lock<std::mutex> lk(mtx); // This locks for the lifetime of the unique_lock object
cv.wait(lk); // this will unlock while waiting, but relock on return
std::cout << "CV notified\n" << std::flush;
mtx.lock(); // This attempts to lock the mutex again, but will deadlock since unique_lock has already invoked mutex.lock() in its constructor.
The fix is pretty close to what you have with those curly braces uncommented. Just make sure you only have one lock active at a time on the mutex.
Also, as you have it, your code is prone to spurious wake-up. Here's some adjustments for you. You should always stay in the wait loop until the condition or state (usually guarded by the mutex itself) has actually occurred. For a simple notification, a bool will do.
int main()
{
std::mutex mtx;
std::condition_variable cv;
bool conditon = false;
//simulate another working thread sending notification
auto as = std::async([&cv, &mtx, &condition](){
std::this_thread::sleep_for(std::chrono::seconds(2));
mtx.lock();
condition = true;
mtx.unlock();
cv.notify_all();});
std::unique_lock<std::mutex> lk(mtx); // acquire the mutex lock
while (!condition)
{
cv.wait(lk);
}
std::cout << "CV notified\n" << std::flush;
//do something - while still under the lock
return 0;
}
回答2:
Because the condition wait might return for reasons besides being notified such as a signal, or just because someone else wrote onto the same 64-byte cache line. Or it might have been notified but the condition is no longer true because another thread handled it.
So the mutex is locked so that your code can check its condition variable while holding the mutex. Maybe that's just a boolean value saying it's ready to go.
Do NOT skip that part. If you do, you will regret it.
回答3:
Let's temporarily imagine that the mutex is not locked on return from wait:
Thread 1:
Locks mutex, checks predicate (whatever that may be), and upon finding the predicate not in an acceptable form, waits for some other thread to put it in an acceptable form. The wait atomically puts thread 1 to sleep and unlocks the mutex. With the mutex unlocked, some other thread will have permission to put the predicate in the acceptable state (the predicate is not naturally thread safe).
Thread 2:
Simultaneously, this thread is trying to lock the mutex and put the predicate in a state that is acceptable for thread 1 to continue past its wait. It must do this with the mutex locked. The mutex protects the predicate from being accessed (either read or written) by more than one thread at a time.
Once thread 2 puts the mutex in an acceptable state, it notifies the condition_variable and unlocks the mutex (the order of these two actions is not relevant to this argument).
Thread 1:
Now thread 1 has been notified and we presume the hypothetical that the mutex isn't locked on return from wait. The first thing thread 1 has to do is check the predicate to see if it is actually acceptable (this could be a spurious wakeup). But it shouldn't check the predicate without the mutex being locked. Otherwise some other thread could change the predicate right after this thread checks it, invalidating the result of that check.
So the very first thing this thread has to do upon waking is lock the mutex, and then check the predicate.
So it is really more of a convenience that the mutex is locked upon return from wait. Otherwise the waiting thread would have to manually lock it 100% of the time.
Let's look again at the events as thread 1 is entering the wait: I said that the sleep and the unlock happen atomically. This is very important. Imagine if thread 1 has to manually unlock the mutex and then call wait: In this hypothetical scenario, thread 1 could unlock the mutex, and then be interrupted while another thread obtains the mutex, changes the predicate, unlocks the mutex and signals the condition_variable, all before thread 1 calls wait. Now thread 1 sleeps forever, because no thread is going to see that the predicate needs changing, and the condition_variable needs signaling.
So it is imperative that the unlock/enter-wait happen atomically. And it makes the API easier to use if the lock/exit-wait also happens atomically.
来源：https://stackoverflow.com/questions/59888653/use-of-mutex-after-condition-variable-has-been-notified