Do I have to acquire lock before calling condition_variable.notify_one()?

Question

I am a bit confused about the use of std::condition_variable. I understand I have to create a unique_lock on a mutex before calling condition_variable.wait(). What I cannot find is whether I should also acquire a unique lock before calling notify_one() or notify_all().

Examples on cppreference.com are conflicting. For example, the notify_one page gives this example:

#include <iostream> #include <condition_variable> #include <thread> #include <chrono>  std::condition_variable cv; std::mutex cv_m; int i = 0; bool done = false;  void waits() {     std::unique_lock<std::mutex> lk(cv_m);     std::cout << "Waiting... \n";     cv.wait(lk, []{return i == 1;});     std::cout << "...finished waiting. i == 1\n";     done = true; }  void signals() {     std::this_thread::sleep_for(std::chrono::seconds(1));     std::cout << "Notifying...\n";     cv.notify_one();      std::unique_lock<std::mutex> lk(cv_m);     i = 1;     while (!done) {         lk.unlock();         std::this_thread::sleep_for(std::chrono::seconds(1));         lk.lock();         std::cerr << "Notifying again...\n";         cv.notify_one();     } }  int main() {     std::thread t1(waits), t2(signals);     t1.join(); t2.join(); } 

Here the lock is not acquired for the first notify_one(), but is acquired for the second notify_one(). Looking though other pages with examples I see different things, mostly not acquiring the lock.

• Can I choose myself to lock the mutex before calling notify_one(), and why would I choose to lock it?
• In the example given, why is there no lock for the first notify_one(), but there is for subsequent calls. Is this example wrong or is there some rationale?

Answer 1

You do not need to be holding a lock when calling condition_variable::notify_one(), but it's not wrong in the sense that it's still well defined behavior and not an error.

However, it might be a "pessimization" since whatever waiting thread is made runnable (if any) will immediately try to acquire the lock that the notifying thread holds. I think it's a good rule of thumb to avoid holding the lock associated with a condition variable while calling notify_one() or notify_all(). See Pthread Mutex: pthread_mutex_unlock() consumes lots of time for an example where releasing a lock before calling the pthread equivalent of notify_one() improved performance measurably.

Keep in mind that the lock() call in the while loop is necessary at some point, because the lock needs to be held during the while (!done) loop condition check. But it doesn't need to be held for the call to notify_one().

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2016-02-27: Large update to address some questions in the comments about whether there's a race condition is the lock isn't help for the notify_one() call. I know this update is late because the question was asked almost two years ago, but I'd like to address @Cookie's question about a possible race condition if the producer (signals() in this example) calls notify_one() just before the consumer (waits() in this example) is able to call wait().

The key is what happens to i - that's the object that actually indicates whether or not the consumer has "work" to do. The condition_variable is just a mechanism to let the consumer efficiently wait for a change to i.

The producer needs to hold the lock when updating i, and the consumer must hold the lock while checking i and calling condition_variable::wait() (if it needs to wait at all). In this case, the key is that it must be the same instance of holding the lock (often called a critical section) when the consumer does this check-and-wait. Since the critical section is held when the producer updates i and when the consumer checks-and-waits on i, there is no opportunity for i to change between when the consumer checks i and when it calls condition_variable::wait(). This is the crux for a proper use of condition variables.

The C++ standard says that condition_variable::wait() behaves like the following when called with a predicate (as in this case):

while (!pred())     wait(lock); 

There are two situations that can occur when the consumer checks i:

• if i is 0 then the consumer calls cv.wait(), then i will still be 0 when the wait(lock) part of the implementation is called - the proper use of the locks ensures that. In this case the producer has no opportunity to call the condition_variable::notify_one() in its while loop until after the consumer has called cv.wait(lk, []{return i == 1;}) (and the wait() call has done everything it needs to do to properly 'catch' a notify - wait() won't release the lock until it has done that). So in this case, the consumer cannot miss the notification.

• if i is already 1 when the consumer calls cv.wait(), the wait(lock) part of the implementation will never be called because the while (!pred()) test will cause the internal loop to terminate. In this situation it doesn't matter when the call to notify_one() occurs - the consumer will not block.

The example here does have the additional complexity of using the done variable to signal back to the producer thread that the consumer has recognized that i == 1, but I don't think this changes the analysis at all because all of the access to done (for both reading and modifying) are done while in the same critical sections that involve i and the condition_variable.

If you look at the question that @eh9 pointed to, Sync is unreliable using std::atomic and std::condition_variable, you will see a race condition. However, the code posted in that question violates one of the fundamental rules of using a condition variable: It does not hold a single critical section when performing a check-and-wait.

In that example, the code looks like:

if (--f->counter == 0)      // (1)     // we have zeroed this fence's counter, wake up everyone that waits     f->resume.notify_all(); // (2) else {     unique_lock<mutex> lock(f->resume_mutex);     f->resume.wait(lock);   // (3) } 

You will notice that the wait() at #3 is performed while holding f->resume_mutex. But the check for whether or not the wait() is necessary at step #1 is not done while holding that lock at all (much less continuously for the check-and-wait), which is a requirement for proper use of condition variables). I believe that the person who has the problem with that code snippet thought that since f->counter was a std::atomic type this would fulfill the requirement. However, the atomicity provided by std::atomic doesn't extend to the subsequent call to f->resume.wait(lock). In this example, there is a race between when f->counter is checked (step #1) and when the wait() is called (step #3).

That race does not exist in this question's example.