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().

---

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.

Answer 2

Situation

Using vc10 and Boost 1.56 I implemented a concurrent queue pretty much like this blog post suggests. The author unlocks the mutex to minimize contention, i.e., notify_one() is called with the mutex unlocked:

void push(const T& item)
{
  std::unique_lock<std::mutex> mlock(mutex_);
  queue_.push(item);
  mlock.unlock();     // unlock before notificiation to minimize mutex contention
  cond_.notify_one(); // notify one waiting thread
}

Unlocking the mutex is backed by an example in the Boost documentation:

void prepare_data_for_processing()
{
    retrieve_data();
    prepare_data();
    {
        boost::lock_guard<boost::mutex> lock(mut);
        data_ready=true;
    }
    cond.notify_one();
}

Problem

Still this led to the following erratic behaviour:

• while notify_one() has not been called yet cond_.wait() can still be interrupted via boost::thread::interrupt()
• once notify_one() was called for the first time cond_.wait() deadlocks; the wait cannot be ended by boost::thread::interrupt() or boost::condition_variable::notify_*() anymore.

Solution

Removing the line mlock.unlock() made the code work as expected (notifications and interrupts end the wait). Note that notify_one() is called with the mutex still locked, it is unlocked right afterwards when leaving the scope:

void push(const T& item)
{
  std::lock_guard<std::mutex> mlock(mutex_);
  queue_.push(item);
  cond_.notify_one(); // notify one waiting thread
}

That means that at least with my particular thread implementation the mutex must not be unlocked before calling boost::condition_variable::notify_one(), although both ways seem correct.

Answer 3

As others have pointed out, you do not need to be holding the lock when calling notify_one(), in terms of race conditions and threading-related issues. However, in some cases, holding the lock may be required to prevent the condition_variable from getting destroyed before notify_one() is called. Consider the following example:

thread t;

void foo() {
    std::mutex m;
    std::condition_variable cv;
    bool done = false;

    t = std::thread([&]() {
        {
            std::lock_guard<std::mutex> l(m);  // (1)
            done = true;  // (2)
        }  // (3)
        cv.notify_one();  // (4)
    });  // (5)

    std::unique_lock<std::mutex> lock(m);  // (6)
    cv.wait(lock, [&done]() { return done; });  // (7)
}

void main() {
    foo();  // (8)
    t.join();  // (9)
}

Assume there is a context switch to the newly created thread t after we created it but before we start waiting on the condition variable (somewhere between (5) and (6)). The thread t acquires the lock (1), sets the predicate variable (2) and then releases the lock (3). Assume there is another context switch right at this point before notify_one() (4) is executed. The main thread acquires the lock (6) and executes line (7), at which point the predicate returns true and there is no reason to wait, so it releases the lock and continues. foo returns (8) and the variables in its scope (including cv) are destroyed. Before thread t could join the main thread (9), it has to finish its execution, so it continues from where it left off to execute cv.notify_one() (4), at which point cv is already destroyed!

The possible fix in this case is to keep holding the lock when calling notify_one (i.e. remove the scope ending in line (3)). By doing so, we ensure that thread t calls notify_one before cv.wait can check the newly set predicate variable and continue, since it would need to acquire the lock, which t is currently holding, to do the check. So, we ensure that cv is not accessed by thread t after foo returns.

To summarize, the problem in this specific case is not really about threading, but about the lifetimes of the variables captured by reference. cv is captured by reference via thread t, hence you have to make sure cv stays alive for the duration of the thread's execution. The other examples presented here do not suffer from this issue, because condition_variable and mutex objects are defined in the global scope, hence they are guaranteed to be kept alive until the program exits.

Answer 4

@Michael Burr is correct. condition_variable::notify_one does not require a lock on the variable. Nothing prevents you to use a lock in that situation though, as the example illustrates it.

In the given example, the lock is motivated by the concurrent use of the variable i. Because the signals thread modifies the variable, it needs to ensure that no other thread is access it during that time.

Locks are used for any situation requiring synchronization, I don't think we can state it in a more general way.

Answer 5

Just adding this answer because I think the accepted answer might be misleading. In all cases you will need to lock the mutex, prior to calling notify_one() somewhere for your code to be thread-safe, although you might unlock it again before actually calling notify_*().

To clarify, you MUST take the lock before entering wait(lk) because wait() unlocks lk and it would be Undefined Behavior if the lock wasn't locked. This is not the case with notify_one(), but you need to make sure you won't call notify_*() before entering wait() and having that call unlock the mutex; which obviously only can be done by locking that same mutex before you call notify_*().

For example, consider the following case:

std::atomic_int count;
std::mutex cancel_mutex;
std::condition_variable cancel_cv;

void stop()
{
  if (count.fetch_sub(1) == -999) // Reached -1000 ?
    cv.notify_one();
}

bool start()
{
  if (count.fetch_add(1) >= 0)
    return true;
  // Failure.
  stop();
  return false;
}

void cancel()
{
  if (count.fetch_sub(1000) == 0)  // Reached -1000?
    return;
  // Wait till count reached -1000.
  std::unique_lock<std::mutex> lk(cancel_mutex);
  cancel_cv.wait(lk);
}

Warning: this code contains a bug.

The idea is the following: threads call start() and stop() in pairs, but only as long as start() returned true. For example:

if (start())
{
  // Do stuff
  stop();
}

One (other) thread at some point will call cancel() and after returning from cancel() will destroy objects that are needed at 'Do stuff'. However, cancel() is supposed not to return while there are threads between start() and stop(), and once cancel() executed its first line, start() will always return false, so no new threads will enter the 'Do stuff' area.

Works right?

The reasoning is as follows:

1) If any thread successfully executes the first line of start() (and therefore will return true) then no thread did execute the first line of cancel() yet (we assume that the total number of threads is much smaller than 1000 by the way).

2) Also, while a thread successfully executed the first line of start(), but not yet the first line of stop() then it is impossible that any thread will successfully execute the first line of cancel() (note that only one thread ever calls cancel()): the value returned by fetch_sub(1000) will be larger than 0.

3) Once a thread executed the first line of cancel(), the first line of start() will always return false and a thread calling start() will not enter the 'Do stuff' area anymore.

4) The number of calls to start() and stop() are always balanced, so after the first line of cancel() is unsuccessfully executed, there will always be a moment where a (the last) call to stop() causes count to reach -1000 and therefore notify_one() to be called. Note that can only ever happen when the first line of cancel resulted in that thread to fall through.

Apart from a starvation problem where so many threads are calling start()/stop() that count never reaches -1000 and cancel() never returns, which one might accept as "unlikely and never lasting long", there is another bug:

It is possible that there is one thread inside the 'Do stuff' area, lets say it is just calling stop(); at that moment a thread executes the first line of cancel() reading the value 1 with the fetch_sub(1000) and falling through. But before it takes the mutex and/or does the call to wait(lk), the first thread executes the first line of stop(), reads -999 and calls cv.notify_one()!

Then this call to notify_one() is done BEFORE we are wait()-ing on the condition variable! And the program would indefinitely dead-lock.

For this reason we should not be able to call notify_one() until we called wait(). Note that the power of a condition variable lies there in that it is able to atomically unlock the mutex, check if a call to notify_one() happened and go to sleep or not. You can't fool it, but you do need to keep the mutex locked whenever you make changes to variables that might change the condition from false to true and keep it locked while calling notify_one() because of race conditions like described here.

In this example there is no condition however. Why didn't I use as condition 'count == -1000'? Because that isn't interesting at all here: as soon as -1000 is reached at all, we are sure that no new thread will enter the 'Do stuff' area. Moreover, threads can still call start() and will increment count (to -999 and -998 etc) but we don't care about that. The only thing that matters is that -1000 was reached - so that we know for sure that there are no threads anymore in the 'Do stuff' area. We are sure that this is the case when notify_one() is being called, but how to make sure we don't call notify_one() before cancel() locked its mutex? Just locking cancel_mutex shortly prior to notify_one() isn't going to help of course.

The problem is that, despite that we're not waiting for a condition, there still is a condition, and we need to lock the mutex

1) before that condition is reached 2) before we call notify_one.

The correct code therefore becomes:

void stop()
{
  if (count.fetch_sub(1) == -999) // Reached -1000 ?
  {
    cancel_mutex.lock();
    cancel_mutex.unlock();
    cv.notify_one();
  }
}

[...same start()...]

void cancel()
{
  std::unique_lock<std::mutex> lk(cancel_mutex);
  if (count.fetch_sub(1000) == 0)
    return;
  cancel_cv.wait(lk);
}

Of course this is just one example but other cases are very much alike; in almost all cases where you use a conditional variable you will need to have that mutex locked (shortly) before calling notify_one(), or else it is possible that you call it before calling wait().

Note that I unlocked the mutex prior to calling notify_one() in this case, because otherwise there is the (small) chance that the call to notify_one() wakes up the thread waiting for the condition variable which then will try to take the mutex and block, before we release the mutex again. That's just slightly slower than needed.

This example was kinda special in that the line that changes the condition is executed by the same thread that calls wait().

More usual is the case where one thread simply wait's for a condition to become true and another thread takes the lock before changing the variables involved in that condition (causing it to possibly become true). In that case the mutex is locked immediately before (and after) the condition became true - so it is totally ok to just unlock the mutex before calling notify_*() in that case.

Answer 6

In some case, when the cv may be occupied(locked) by other threads. You needs to get lock and release it before notify_*().
If not, the notify_*() maybe not executed at all.