I am trying to use mutex to arrange the output between two threads to print the message from Thread 1 then print output from thread 2.
but I am getting the messages to be printed randomly so it seems like I am not using mutex correctly.
std::mutex mu;
void share_print(string msg, int id)
{
mu.lock();
cout << msg << id << endl;
mu.unlock();
}
void func1()
{
for (int i = 0; i > -50; i--)
{
share_print(string("From Func 1: "), i);
}
}
int main()
{
std::thread t1(func1);
for (int i = 0; i < 50; i++)
{
share_print(string("From Main: "), i);
}
t1.join();
return 0;
}
the output is:
Your usage of mutexes is 100% correct. It's your expectation of mutex behavior, and execution thread behavior, that misses the mark. For example, C++ execution threads give you no guarantees whatsoever that any line in func1 will be executed before main() completely finishes executing its for loop.
As far as mutexes are concerned, your only guarantees, that matter here are:
Only one execution thread can lock a given std::mutex at the same time.
If a std::mutex is not locked, one of two things will happen when an execution thread attempts to lock it, either: a) it will lock it b) if another thread already has it locked or manages to lock it first it will block until the mutex is no longer locked, and then it will attempt to lock the mutex again.
It is very important to understand all the implications of these rules. Even if your execution thread has a mutex locked, then proceeds to unlock it, and then lock it again, it may end up re-locking the mutex immediately even if another execution thread is also waiting to lock the mutex. Mutexes do not impose any kind of a queueing, a locking order, or a priority between different execution threads that are trying to lock it. It's a free-for-all.
Even if mutexes worked the way you expected them to work, that still gives you no guarantees whatsoever:
std::thread t1 (func1 );
Your only guarantee here is that func1 will be called by a new execution thread at some point on or after this std::thread object's construction finishes.
for (int i = 0; i < 50; i++)
{
share_print(string("From Main: "), i);
}
This entire for loop can finish even before a single line from func1 gets executed. It'll lock and unlock the mutex 50 times and call it a day, before func1 wakes up and does the same.
Or, alternatively, it's possible for func1 to run to completion before main enters the for loop.
You have no expectations of any order of execution of multiple execution threads, unless explicit syncronization takes place.
In order to achieve your interleaving output a lot more work is needed. In addition to just a mutex there will need to be some kind of a condition variable, and a separate variable that indicates whose "turn" it is. Each execution thread, both main and func1, will not only need to lock the mutex, but block on the condition variable until the shared variable indicates that it's turn is up, then do its printing, set the shared variable to indicate that it's the other thread's turn, signal the condition variable, and only then unlock the mutex (or, always keep the mutex locked and always spin on the condition variable).
Related
I have the following code, which is from https://en.cppreference.com/w/cpp/thread/unique_lock. However, upon, printing the output, I see some unexpected result and would like some explaination.
The code is:
#include <mutex>
#include <thread>
#include <chrono>
#include <iostream>
struct Box {
explicit Box(int num) : num_things{num} {}
int num_things;
std::mutex m;
};
void transfer(Box &from, Box &to, int anotherNumber)
{
// don't actually take the locks yet
std::unique_lock<std::mutex> lock1(from.m, std::defer_lock);
std::unique_lock<std::mutex> lock2(to.m, std::defer_lock);
// lock both unique_locks without deadlock
std::lock(lock1, lock2);
from.num_things += anotherNumber;
to.num_things += anotherNumber;
std::cout<<std::this_thread::get_id()<<" "<<from.num_things<<"\n";
std::cout<<std::this_thread::get_id()<<" "<<to.num_things<<"\n";
// 'from.m' and 'to.m' mutexes unlocked in 'unique_lock' dtors
}
int main()
{
Box acc1(100); //initialized acc1.num_things = 100
Box acc2(50); //initialized acc2.num_things = 50
std::thread t1(transfer, std::ref(acc1), std::ref(acc2), 10);
std::thread t2(transfer, std::ref(acc2), std::ref(acc1), 5);
t1.join();
t2.join();
}
My expectation:
acc1 will be initialized with num_things=100, and acc2 with num_things=50.
say thread t1 runs first, it acquire mutex m, with 2 locks. Once the locks are locked, and can assign num_things to num=10
upon completion, it will print from.num_things = 110 and to.numthings = 60 in order. "from" first, then "to" later.
thread1 finishes the critical section of the code, and wrapper unique_lock calls its destructor which basically unlock the mutex.
Here is what I don't understand.
I expected the lock1 fill be unlocked first, and lock2 later.
Thread t2 then acquire the mutex in the same order and lock the lock1 first, then lock2. It will also runs the critical code sequentially up to cout.
Thread t2 will take the global acc1.num_things = 110 and acc2.num_things = 60 from t1.
I expect that t2 will print from.num_things = 115 first, then to.numthings = 65.
However, upon countless trial, I always get the reverse order. And that is my confusion.
I expected the lock1 fill be unlocked first, and lock2 later.
No, the reverse is true. In your function lock1 gets constructed first, then lock2. Therefore, when the function returns lock2 gets destroyed first, then lock1, so lock2's destructor releases its lock before lock1's destructor.
The actual order in which std::lock manages to acquire the multiple locks has no bearing on how the locks gets destroyed, and release their ownership of their respective mutexes. That still follows normal C++ rules for doing so.
say thread t1 runs first,
You have no guarantee of that, whatsoever. In the above code it's entirely possible that t2 will enter the function first and acquire the locks on the mutexes. And, it is also entirely possible that each time you run this program you'll get different results, with both t1 and t2 winning the race, randomly.
Without getting into technical mumbo-jumbo, the only thing that C++ guarantees you is that std::thread gets fully constructed before the thread function gets invoked in a new execution thread. You have no guarantees whatsoever that, when creating two execution threads one after another, the first one will call its function and run some arbitrary part of the thread function before the second execution thread does the same.
So it's entirely possible that t2 will get the first dibs on the locks occasionally. Or, always. Attempting to control the relative sequence of events across execution threads is much harder than you think.
I've a vector of strings which is a shared resourse.
std::vector<std::string> vecstr;
Have 2 threads which run in parallel:
Thread1: To insert strings to shared resourse.
Thread2: To calculate the size of the shared resourse.
std::mutex mt;
void f1()
{
mt.lock();
while(some_condition())
{
std::string str = getStringFromSomewhere();
vecstr.push_back(str);
}
mt.unlock();
}
size_t f2()
{
mt.lock();
while(string_sending_hasEnded())
{
size_t size = vecstr.size();
}
mt.unlock();
}
int main()
{
std::thread t1(f1);
std::thread t2(f2);
t1.join();
t2.join();
}
My question is : if the t1 thread keeps the vecstr shared resource mutex locked for the entire while loop duration how will the t2 get hold of the shared resource vecstr to calculate it's size ?
Does the execution keep switching between the 2 threads or it depends on who gets hold of mutex 1st. So if T1 got hold of mutex then it will release it only after while loop ends ? Is this true ? Or the execution keeps switching between the 2 threads.
If any one of the thread is going to hijack the execution by not
allowing other thread to be switched in between then how do i handle
such a scenario with while/for loops in each thread but both threads
needs to be continuously executed ? Where I want both the threads to
keep switching their execution. Shall I lock and unlock inside the
while loop, so that each iteration has mutex locked & unlocked ?
You got it. If you want to use mutexes successfully in real life, you will keep a mutex looked only for the smallest amount of time possible. For example just around the push_back() and size() calls.
But really, what you need to do first is figure out what your program is supposed to do, and then use mutexes to make sure to achieve that. At the moment I know that you want to run some threads, but that's not what you want to achieve.
So if T1 got hold of mutex then it will release it only after while loop ends ? Is this true ?
Yes, that's true.
Either of the threads will lock the mt mutex over the whole time these loops are executed.
As for your comment
If that's the case how do i handle such a scenario ? Where I want both the threads to keep switching their execution. Shall I lock and unlock inside the while loop, so that each iteration has mutex locked & unlocked
Yes use more fine grained locking, just for the operations that change/access the vector:
std::mutex mt;
void f1() {
while(some_condition()) {
std::string str = getStringFromSomewhere();
{ std::unique_lock(mt); // -+
vecstr.push_back(str); // | locked
} // -+
}
}
size_t f2() {
while(string_sending_hasEnded()) {
size_t size = 0;
{ std::unique_lock(mt); // -+
size = vecstr.size(); // | locked
} // -+
}
}
I also highly recommend to use a lock-guard (as the std::unique_lock in my example), instead of using lock() unock() yourself manually. So it's safe that the mutex will be unlocked, e.g. in case of exceptions thrown.
In a distributed job system written in C++11 I have implemented a fence (i.e. a thread outside the worker thread pool may ask to block until all currently scheduled jobs are done) using the following structure:
struct fence
{
std::atomic<size_t> counter;
std::mutex resume_mutex;
std::condition_variable resume;
fence(size_t num_threads)
: counter(num_threads)
{}
};
The code implementing the fence looks like this:
void task_pool::fence_impl(void *arg)
{
auto f = (fence *)arg;
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)
}
}
This works very well if threads enter the fence over a period of time. However, if they try to do it almost simultaneously, it seems to sometimes happen that between the atomic decrementation (1) and starting the wait on the conditional var (3), the thread yields CPU time and another thread decrements the counter to zero (1) and fires the cond. var (2). This results in the previous thread waiting forever in (3), because it starts waiting on it after it has already been notified.
A hack to make the thing workable is to put a 10 ms sleep just before (2), but that's unacceptable for obvious reasons.
Any suggestions on how to fix this in a performant way?
Your diagnose is correct, this code is prone to lose condition notifications in the way you described. I.e. after one thread locked the mutex but before waiting on the condition variable another thread may call notify_all() so that the first thread misses that notification.
A simple fix is to lock the mutex before decrementing the counter and while notifying:
void task_pool::fence_impl(void *arg)
{
auto f = static_cast<fence*>(arg);
std::unique_lock<std::mutex> lock(f->resume_mutex);
if (--f->counter == 0) {
f->resume.notify_all();
}
else do {
f->resume.wait(lock);
} while(f->counter);
}
In this case the counter need not be atomic.
An added bonus (or penalty, depending on the point of view) of locking the mutex before notifying is (from here):
The pthread_cond_broadcast() or pthread_cond_signal() functions may be called by a thread whether or not it currently owns the mutex that threads calling pthread_cond_wait() or pthread_cond_timedwait() have associated with the condition variable during their waits; however, if predictable scheduling behavior is required, then that mutex shall be locked by the thread calling pthread_cond_broadcast() or pthread_cond_signal().
Regarding the while loop (from here):
Spurious wakeups from the pthread_cond_timedwait() or pthread_cond_wait() functions may occur. Since the return from pthread_cond_timedwait() or pthread_cond_wait() does not imply anything about the value of this predicate, the predicate should be re-evaluated upon such return.
In order to keep the higher performance of an atomic operation instead of a full mutex, you should change the wait condition into a lock, check and loop.
All condition waits should be done in that way. The condition variable even has a 2nd argument to wait which is a predicate function or lambda.
The code might look like:
void task_pool::fence_impl(void *arg)
{
auto f = (fence *)arg;
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);
while(f->counter) {
f->resume.wait(lock); // (3)
}
}
}
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?
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 if the lock isn't held 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.
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.
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.
#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.
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.
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.
As I understand notify_one calls pthread_cond_signal.
If so, then what do think about this?
For predictable scheduling behavior and to prevent lost wake-ups, the mutex should be held when signaling a condition variable.
https://www.unix.com/man-page/hpux/3T/pthread_cond_signal/
All of the threads waiting on the condition variable are suspended until another thread uses the signal function:
pthread_cond_signal(&myConVar);
In this case the mutex has to be locked before calling the function and unlocked after it.
https://www.i-programmer.info/programming/cc/12288-fundamental-c-condition-variables.html
I personally had cases when notifications were missed because notify_one was called without locking the mutex.
Suppose we have two workers. Each worker has an id of 0 and 1. Also suppose that we have jobs arriving all the time, each job has also an identifier 0 or 1 which specifies which worker will have to do this job.
I would like to create 2 threads that are initially locked, and then when two jobs arrive, unlock them, each of them does their job and then lock them again until other jobs arrive.
I have the following code:
#include <iostream>
#include <thread>
#include <mutex>
using namespace std;
struct job{
thread jobThread;
mutex jobMutex;
};
job jobs[2];
void executeJob(int worker){
while(true){
jobs[worker].jobMutex.lock();
//do some job
}
}
void initialize(){
int i;
for(i=0;i<2;i++){
jobs[i].jobThread = thread(executeJob, i);
}
}
int main(void){
//initialization
initialize();
int buffer[2];
int bufferSize = 0;
while(true){
//jobs arrive here constantly,
//once the buffer becomes full,
//we unlock the threads(workers) and they start working
bufferSize = 2;
if(bufferSize == 2){
for(int i = 0; i<2; i++){
jobs[i].jobMutex.unlock();
}
}
break;
}
}
I started using std::thread a few days ago and I'm not sure why but Visual Studio gives me an error saying abort() has been called. I believe there's something missing however due to my ignorance I can't figure out what.
I would expect this piece of code to actually
Initialize the two threads and then lock them
Inside the main function unlock the two threads, the two threads will do their job(in this case nothing) and then they will become locked again.
But it gives me an error instead. What am I doing wrong?
Thank you in advance!
For this purpose you can use boost's threadpool class.
It's efficient and well tested. opensource library instead of you writing newly and stabilizing it.
http://threadpool.sourceforge.net/
main()
{
pool tp(2); //number of worker threads-currently its 2.
// Add some tasks to the pool.
tp.schedule(&first_task);
tp.schedule(&second_task);
}
void first_task()
{
...
}
void second_task()
{
...
}
Note:
Suggestion for your example:
You don't need to have individual mutex object for each thread. Single mutex object lock itself will does the synchronization between all the threads. You are locking mutex of one thread in executejob function and without unlocking another thread is calling lock with different mutex object leading to deadlock or undefined behaviour.
Also since you are calling mutex.lock() inside whileloop without unlocking , same thread is trying to lock itself with same mutex object infinately leading to undefined behaviour.
If you donot need to execute threads parallel you can have one global mutex object can be used inside executejob function to lock and unlock.
mutex m;
void executeJob(int worker)
{
m.lock();
//do some job
m.unlock();
}
If you want to execute job parallel use boost threadpool as I suggested earlier.
In general you can write an algorithm similar to the following. It works with pthreads. I'm sure it would work with c++ threads as well.
create threads and make them wait on a condition variable, e.g. work_exists.
When work arrives you notify all threads that are waiting on that condition variable. Then in the main thread you start waiting on another condition variable work_done
Upon receiving work_exists notification, worker threads wake up, and grab their assigned work from jobs[worker], they execute it, they send a notification on work_done variable, and then go back to waiting on the work_exists condition variable
When main thread receives work_done notification it checks if all threads are done. If not, it keeps waiting till the notification from last-finishing thread arrives.
From cppreference's page on std::mutex::unlock:
The mutex must be unlocked by all threads that have successfully locked it before being destroyed. Otherwise, the behavior is undefined.
Your approach of having one thread unlock a mutex on behalf of another thread is incorrect.
The behavior you're attempting would normally be done using std::condition_variable. There are examples if you look at the links to the member functions.