Conditional variables use a mutex and the .wait() function unlocks the
mutex so another thread can access the shared data. When the condition
variable is notified it tries to lock the mutex again to use the shared
data.
This pattern is used in the following concurrent_queue example from Anthony Williams:
template<typename Data>
class concurrent_queue
{
private:
boost::condition_variable the_condition_variable;
public:
void wait_for_data()
{
boost::mutex::scoped_lock lock(the_mutex);
while(the_queue.empty())
{
the_condition_variable.wait(lock);
}
}
void push(Data const& data)
{
boost::mutex::scoped_lock lock(the_mutex);
bool const was_empty=the_queue.empty();
the_queue.push(data);
if(was_empty)
{
the_condition_variable.notify_one();
}
}
};
Since the code uses std::queue it's clear that the mutex has to be
locked when accessing the queue.
But let's say instead of std::queue one uses Microsofts
Concurrency::concurrent_queue from PPL. Member functions like empty,
push and try_pop are thread safe. Do I still need to lock a mutex in
this case or can the condition variable be used like this, without
creating any possible race conditions.
My code (that seems to work, but what does that mean in multithreading?) looks like this. I have one producer that pushes items into Microsofts concurrent_queue and one background thread that waits for new items in this queue.
The consumer/background thread:
while(runFlag) //atomic
{
while(the_queue.empty() && runFlag) //wait only when thread should still run
{
boost::mutex mtx; //local mutex thats locked afterwards. awkward.
boost::mutex::scoped_lock lock(mtx);
condition.wait(lock);
}
Data d;
while(!the_queue.empty() && the_queue.try_pop(d))
{
//process data
}
}
The producer/main thread:
const bool was_empty = the_queue.empty();
Data d;
the_queue.push(d);
if(was_empty) cond_var.notify_one();
The shutdown procedure:
bool expected_run_state = true;
if(runLoop.compare_exchange_strong(expected_run_state, false))
{
//atomically set our loop flag to false and
//notify all clients of the queue to wake up and exit
cond_var.notify_all();
}
As said above this code seems to work but that doesn't necessarily mean it's correct. Especially the local mutex that is only used because the condition variable interface forces me to use a mutex, seems like a very bad idea. I wanted to use condition variables since the time between data items added to the queue hard to predict and I would have to create to sleep and wake up periodically like this:
if(the_queue.empty()) Sleep(short_amount_of_time);
Are there any other, maybe OS (in my case: Windows) specific tools, that make a background thread sleep until some condition is met without regularly waking up and checking the condition?
The code is not correct in different scenarios, for example. If the queue has a single element when const bool was_empty = the_queue.empty(); is evaluated, but a thread consumes the element and a different thread tries to consume and waits on the condition, the writer will not notify that thread after inserting the element in the queue.
The key issue is that the fact that all of the operations in an interface are thread safe does not necessarily mean that your use of the interface is safe. If you depend on multiple operations being performed atomically, you need to provide synchronization mechanisms externally.
Are there any other, maybe OS (in my case: Windows) specific tools,
that make a background thread sleep until some condition is met
without regularly waking up and checking the condition?
This is exactly what Events are for
But if you are targeting only Windows platform (Vista+) you should check out
Slim Reader/Writer (SRW) Locks
Related
We have implemented TaskRunner whose functions will be called by different threads to start, stop and post tasks. TaskRunner will internally create a thread and if the queue is not empty, it will pop the task from queue and executes it. Start() will check if the thread is running. If not creates a new thread. Stop() will join the thread. The code is as below.
bool TaskRunnerImpl::PostTask(Task* task) {
tasks_queue_.push_back(task);
return true;
}
void TaskRunnerImpl::Start() {
std::lock_guard<std::mutex> lock(is_running_mutex_);
if(is_running_) {
return;
}
is_running_ = true;
runner_thread_ = std::thread(&TaskRunnerImpl::Run, this);
}
void TaskRunnerImpl::Run() {
while(is_running_) {
if(tasks_queue_.empty()) {
continue;
}
Task* task_to_run = tasks_queue_.front();
task_to_run->Run();
tasks_queue_.pop_front();
delete task_to_run;
}
}
void TaskRunnerImpl::Stop() {
std::lock_guard<std::mutex> lock(is_running_mutex_);
is_running_ = false;
if(runner_thread_.joinable()) {
runner_thread_.join();
}
}
We want to use conditional variables now otherwise the thread will be continuously checking whether the task queue is empty or not. We implemented as below.
Thread function (Run()) will wait on condition variable.
PostTask() will signal if some one posts a task.
Stop() will signal if some one calls stop.
Code is as below.
bool TaskRunnerImpl::PostTask(Task* task) {
std::lock_guard<std::mutex> taskGuard(m_task_mutex);
tasks_queue_.push_back(task);
m_task_cond_var.notify_one();
return true;
}
void TaskRunnerImpl::Start() {
std::lock_guard<std::mutex> lock(is_running_mutex_);
if(is_running_) {
return;
}
is_running_ = true;
runner_thread_ = std::thread(&TaskRunnerImpl::Run, this);
}
void TaskRunnerImpl::Run() {
while(is_running_) {
Task* task_to_run = nullptr;
{
std::unique_lock<std::mutex> mlock(m_task_mutex);
m_task_cond_var.wait(mlock, [this]() {
return !(is_running_ && tasks_queue_.empty());
});
if(!is_running_) {
return;
}
if(!tasks_queue_.empty()) {
task_to_run = tasks_queue_.front();
task_to_run->Run();
tasks_queue_.pop_front();
}
}
if(task_to_run)
delete task_to_run;
}
}
void TaskRunnerImpl::Stop() {
std::lock_guard<std::mutex> lock(is_running_mutex_);
is_running_ = false;
m_task_cond_var.notify_one();
if(runner_thread_.joinable()) {
runner_thread_.join();
}
}
I have couple of questions as below. Can some one please help me to understand these.
Condition variable m_task_cond_var is linked with mutex m_task_mutex. But Stop() already locks mutex is_running_mutex to gaurd 'is_running_'. Do I need to lock m_task_mutex before signaling? Here I am not convinced why to lock m_task_mutex as we are not protecting any thing related to task queue.
In Thread function(Run()), we are reading is_running_ without locking is_running_mutex. Is this correct?
Do I need to lock m_task_mutex before signaling [In Stop]?
When the predicate being tested in condition_variable::wait method depends on something happening in the signaling thread (which is almost always), then you should obtain the mutex before signaling. Consider the following possibility if you are not holding the m_task_mutex:
The watcher thread (TaskRunnerImpl::Run) wakes up (via spurious wakeup or being notified from elsewhere) and obtains the mutex.
The watcher thread checks its predicate and sees that it is false.
The signaler thread (TaskRunnerImpl::Stop) changes the predicate to return true (by setting is_running_ = false;).
The signaler thread signals the condition variable.
The watcher thread waits to be signaled (bad)
the signal has already come and gone
the predicate was false, so the watcher begins waiting, possibly indefinitely.
The worst that can happen if you are holding the mutex when you signal is that, the blocked thread (TaskRunnerImpl::Run) wakes up and is immediately blocked when trying to obtain the mutex. This can have some performance implications.
In [TaskRunnerImpl::Run] , we are reading is_running_ without locking is_running_mutex. Is this correct?
In general no. Even if it's of type bool. Because a boolean is typically implemented as a single byte, it's possible that one thread is writing to the byte while you are reading, resulting in a partial read. In practice, however, it's safe. That said, you should obtain the mutex before you read (and then release immediately afterwards).
In fact, it may be preferable to use std::atomic<bool> instead of a bool + mutex combination (or std::atomic_flag if you want to get fancy) which will have the same effect, but be easier to work with.
Do I need to lock m_task_mutex before signaling [In Stop]?
Yes you do. You must change condition under the same mutex and send signal either after the mutex is locked or unlocked after the change. If you do not use the same mutex, or send signal before that mutex is locked you create race condition that std::condition_variable is created to solve.
Logic is this:
Watching thread locks mutex and checks watched condition. If it did not happen it goes to sleep and unlocks the mutex atomically. So signaling thread lock the mutex, change condition and signal. If signalling thread does that before watching one locks the mutex, then watchiong one would see condition happen and would not go to sleep. If it locks before, it would go to sleep and woken when signalling thread raise the signal.
Note: you can signal condition variable before or after mutex is unlocked, both cases is correct but may affect performance. But it is incorrect to signal before locking the mutex.
Condition variable m_task_cond_var is linked with mutex m_task_mutex. But Stop() already locks mutex is_running_mutex to gaurd 'is_running_'. Do I need to lock m_task_mutex before signaling? Here I am not convinced why to lock m_task_mutex as we are not protecting any thing related to task queue.
You overcomlicated your code and made things worse. You should use only one mutex in this case and it would work as intended.
In Thread function(Run()), we are reading is_running_ without locking is_running_mutex. Is this correct?
On x86 hardware it may "work", but from language point of view this is UB.
Poring through legacy code of old and large project, I had found that there was used some odd method of creating thread-safe queue, something like this:
template < typename _Msg>
class WaitQue: public QWaitCondition
{
public:
typedef _Msg DataType;
void wakeOne(const DataType& msg)
{
QMutexLocker lock_(&mx);
que.push(msg);
QWaitCondition::wakeOne();
}
void wait(DataType& msg)
{
/// wait if empty.
{
QMutex wx; // WHAT?
QMutexLocker cvlock_(&wx);
if (que.empty())
QWaitCondition::wait(&wx);
}
{
QMutexLocker _wlock(&mx);
msg = que.front();
que.pop();
}
}
unsigned long size() {
QMutexLocker lock_(&mx);
return que.size();
}
private:
std::queue<DataType> que;
QMutex mx;
};
wakeOne is used from threads as kind of "posting" function" and wait is called from other threads and waits indefinitely until a message appears in queue. In some cases roles between threads reverse at different stages and using separate queues.
Is this even legal way to use a QMutex by creating local one? I kind of understand why someone could do that to dodge deadlock while reading size of que but how it even works? Is there a simpler and more idiomatic way to achieve this behavior?
Its legal to have a local condition variable. But it normally makes no sense.
As you've worked out in this case is wrong. You should be using the member:
void wait(DataType& msg)
{
QMutexLocker cvlock_(&mx);
while (que.empty())
QWaitCondition::wait(&mx);
msg = que.front();
que.pop();
}
Notice also that you must have while instead of if around the call to QWaitCondition::wait. This is for complex reasons about (possible) spurious wake up - the Qt docs aren't clear here. But more importantly the fact that the wake and the subsequent reacquire of the mutex is not an atomic operation means you must recheck the variable queue for emptiness. It could be this last case where you previously were getting deadlocks/UB.
Consider the scenario of an empty queue and a caller (thread 1) to wait into QWaitCondition::wait. This thread blocks. Then thread 2 comes along and adds an item to the queue and calls wakeOne. Thread 1 gets woken up and tries to reacquire the mutex. However, thread 3 comes along in your implementation of wait, takes the mutex before thread 1, sees the queue isn't empty, processes the single item and moves on, releasing the mutex. Then thread 1 which has been woken up finally acquires the mutex, returns from QWaitCondition::wait and tries to process... an empty queue. Yikes.
Since I have recently started coding multi threaded programs this might be a stupid question. I found out about the awesome mutex and condition variable usage. From as far as I can understand there use is:
Protect sections of code/shared resources from getting corrupted by multiple threads access. Hence lock that portion thus one can control which thread will be accessing.
If a thread is waiting for a resource/condition from another thread one can use cond.wait() instead of polling every msec
Now Consider the following class example:
class Queue {
private:
std::queue<std::string> m_queue;
boost::mutex m_mutex;
boost::condition_variable m_cond;
bool m_exit;
public:
Queue()
: m_queue()
, m_mutex()
, m_cond()
, m_exit(false)
{}
void Enqueue(const std::string& Req)
{
boost::mutex::scoped_lock lock(m_mutex);
m_queue.push(Req);
m_cond.notify_all();
}
std::string Dequeue()
{
boost::mutex::scoped_lock lock(m_mutex);
while(m_queue.empty() && !m_exit)
{
m_cond.wait(lock);
}
if (m_queue.empty() && m_exit) return "";
std::string val = m_queue.front();
m_queue.pop();
return val;
}
void Exit()
{
boost::mutex::scoped_lock lock(m_mutex);
m_exit = true;
m_cond.notify_all();
}
}
In the above example, Exit() can be called and it will notify the threads waiting on Dequeue that it's time to exit without waiting for more data in the queue.
My question is since Dequeue has acquired the lock(m_mutex), how can Exit acquire the same lock(m_mutex)? Isn't unless the Dequeue releases the lock then only Exit can acquire it?
I have seen this pattern in Destructor implementation too, using same class member mutex, Destructor notifies all the threads(class methods) thats it time to terminate their respective loops/functions etc.
As Jarod mentions in the comments, the call
m_cond.wait(lock)
is guaranteed to atomically unlock the mutex, releasing it for the thread, and starts listening to notifications of the condition variable (see e.g. here).
This atomicity also ensures any code in the thread is executed after the listening is set up (so no notify calls will be missed). This assumes of course that the thread first locks the mutex, otherwise all bets are off.
Another important bit to understand is that condition variables may suffer from "spurious wakeups", so it is important to have a second boolean condition (e.g. here, you could check the emptiness of your queue) so that you don't end up awoken with an empty queue. Something like this:
m_cond.wait(lock, [this]() { return !m_queue.empty() || m_exit; });
There are several questions on SO dealing with atomic, and other that deal with std::condition_variable. But my question if my use below is correct?
Three threads, one ctrl thread that does preparation work before unpausing the two other threads. The ctrl thread also is able to pause the worker threads (sender/receiver) while they are in their tight send/receive loops.
The idea with using the atomic is to make the tight loops faster in case the boolean for pausing is not set.
class SomeClass
{
public:
//...
// Disregard that data is public...
std::condition_variable cv; // UDP threads will wait on this cv until allowed
// to run by ctrl thread.
std::mutex cv_m;
std::atomic<bool> pause_test_threads;
};
void do_pause_test_threads(SomeClass *someclass)
{
if (!someclass->pause_test_threads)
{
// Even though we use an atomic, mutex must be held during
// modification. See documentation of condition variable
// notify_all/wait. Mutex does not need to be held for the actual
// notify call.
std::lock_guard<std::mutex> lk(someclass->cv_m);
someclass->pause_test_threads = true;
}
}
void unpause_test_threads(SomeClass *someclass)
{
if (someclass->pause_test_threads)
{
{
// Even though we use an atomic, mutex must be held during
// modification. See documentation of condition variable
// notify_all/wait. Mutex does not need to be held for the actual
// notify call.
std::lock_guard<std::mutex> lk(someclass->cv_m);
someclass->pause_test_threads = false;
}
someclass->cv.notify_all(); // Allow send/receive threads to run.
}
}
void wait_to_start(SomeClass *someclass)
{
std::unique_lock<std::mutex> lk(someclass->cv_m); // RAII, no need for unlock.
auto not_paused = [someclass](){return someclass->pause_test_threads == false;};
someclass->cv.wait(lk, not_paused);
}
void ctrl_thread(SomeClass *someclass)
{
// Do startup work
// ...
unpause_test_threads(someclass);
for (;;)
{
// ... check for end-program etc, if so, break;
if (lost ctrl connection to other endpoint)
{
pause_test_threads();
}
else
{
unpause_test_threads();
}
sleep(SLEEP_INTERVAL);
}
unpause_test_threads(someclass);
}
void sender_thread(SomeClass *someclass)
{
wait_to_start(someclass);
...
for (;;)
{
// ... check for end-program etc, if so, break;
if (someclass->pause_test_threads) wait_to_start(someclass);
...
}
}
void receiver_thread(SomeClass *someclass)
{
wait_to_start(someclass);
...
for (;;)
{
// ... check for end-program etc, if so, break;
if (someclass->pause_test_threads) wait_to_start(someclass);
...
}
I looked through your code manipulating conditional variable and atomic, and it seems that it is correct and won't cause problems.
Why you should protect writes to shared variable even if it is atomic:
There could be problems if write to shared variable happens between checking it in predicate and waiting on condition. Consider following:
Waiting thread wakes spuriously, aquires mutex, checks predicate and evaluates it to false, so it must wait on cv again.
Controlling thread sets shared variable to true.
Controlling thread sends notification, which is not received by anybody, because there is no threads waiting on conditional variable.
Waiting thread waits on conditional variable. Since notification was already sent, it would wait until next spurious wakeup, or next time when controlling thread sends notification. Potentially waiting indefinetly.
Reads from shared atomic variables without locking is generally safe, unless it introduces TOCTOU problems.
In your case you are reading shared variable to avoid unnecessary locking and then checking it again after lock (in conditional wait call). It is a valid optimisation, called double-checked locking and I do not see any potential problems here.
You might want to check if atomic<bool> is lock-free. Otherwise you will have even more locks you would have without it.
In general, you want to treat the fact that variable is atomic independently of how it works with a condition variable.
If all code that interacts with the condition variable follows the usual pattern of locking the mutex before query/modification, and the code interacting with the condition variable does not rely on code that does not interact with the condition variable, it will continue to be correct even if it wraps an atomic mutex.
From a quick read of your pseudo-code, this appears to be correct. However, pseudo-code is often a poor substitute for real code for multi-threaded code.
The "optimization" of only waiting on the condition variable (and locking the mutex) when an atomic read says you might want to may or may not be an optimization. You need to profile throughput.
atomic data doesn't need another synchronization, it's basis of lock-free algorithms and data structures.
void do_pause_test_threads(SomeClass *someclass)
{
if (!someclass->pause_test_threads)
{
/// your pause_test_threads might be changed here by other thread
/// so you have to acquire mutex before checking and changing
/// or use atomic methods - compare_exchange_weak/strong,
/// but not all together
std::lock_guard<std::mutex> lk(someclass->cv_m);
someclass->pause_test_threads = true;
}
}
boost::condition cond;
boost::mutex access;
void listener_thread()
{
boost::mutex::scoped_lock lock(access);
while (true) {
while (!condition_check_var) {
cond.wait(lock);
}
do_some_work();
}
}
/// ... Main thread ...
cond.notify_all();
check_work:
{
boost::mutex::scoped_lock lock(access);
function_relies_on_work_been_done();
}
Is this proper design? Is it safe to assume that once the notify_all() returns, the listener_thread will have already acquired the lock? And that when the check_work block will run (since it's locking the same mutex as the listener_thread()), some "work" will have already been done by the listener_thread()?
If not, what is the preferred way to achieve this kind of behavior?
There is no guarantee that any other thread has acted upon a notification or even, yet, received. In fact, there isn't even a guarantee that there is a thread currently waiting for its reception although in your setup it looks as if it is likely the case that there are threads waiting. If you want to make sure that the receiving threads have done their work you'll need to set up a reverse communication channel, e.g., using another condition variable and a suitable condition.
I realize that your question is about Boost but here is what the standard has to say about this (30.5.1 [thread.condition.condvar] paragraph 8):
void notify_all() noexcept;
Effects: Unblocks all threads that are blocked waiting for *this.
It doesn't give any guarantee about what happens to the threads and/or any involved mutex.
It's generally OK, though the typical way to write it is like this:
while (true)
{
cond.wait(lock, [&]() -> bool { return condition_check_var; });
do_some_work();
}
You can't speak to the simultaneity of calling notify_all() and the return of wait(), since there is no formal causal relationship between the two. All you need to know for synchronisation is that when wait() returns you will have acquired the lock. Since your check_work block also locks the mutex, it is guaranteed to execute only while the other thread is blocking on the condition variable.