Here's an experiment using the thread C++ class.
Initial conditions (ICs):
Thread A has a condition variable that is waiting on a lock (on a mutex).
Thread B has the mutex locked.
Thread C hasn't done anything.
Now thread C calls m.lock() (when creating a lock). Afterwards, thread B notifies the condition variable. Does the fact that thread A was waiting on a condition variable that was waiting on a lock on that mutex make it any more or less likely that it will lock the mutex first, or is thread C just as likely to do so?
Here's an example of what I mean:
#include <condition_variable>
#include <mutex>
#include <thread>
std::condition_variable cv;
std::mutex m;
void funcB()
{
std::unique_lock<std::mutex> B_lk(m);
sleep(2); // allow thread C to attempt lock; IC-2
cv.notify_one();
B_lk.unlock();
}
void funcC()
{
sleep(1); // allow thread B to lock; IC-3
std::unique_lock<std::mutex> C_lk(m);
/* Perform task C */
}
int main (int argc, char* argv[]) // thread A
{
std::unique_lock<std::mutex> A_lk(m);
std::thread threadB(funcB);
std::thread threadC(funcC);
cv.wait(A_lk); // IC-1
/* Perform task A */
/* Clean up and return */
}
I think threads A and C are (theoretically, anyway) equally likely to lock the mutex after thread B unlocks it because I didn't see any mention of priority in the C++ Standard. I read through many other questions about locking priority here on SO, but I couldn't find any that addressed this particular question.
It's deliberately unspecified by the standard to allow freedom of implementation. Specifically, C++11 §30.5.1 Class condition variable [thread.condition.condvar] states:
void notify_one() noexcept;
7 Effects: If any threads are blocked waiting for *this, unblocks one of those threads.
void notify_all() noexcept;
8 Effects: Unblocks all threads that are blocked waiting for *this.
There is no claim made about preference/fairness/priority to any thread(s), simply "unblocks one" or "unblocks all."
Related
i am looking at this piece of code:
#include <chrono>
#include <iostream>
#include <map>
#include <mutex>
#include <shared_mutex>
#include <string>
#include <thread>
bool flag;
std::mutex m;
void wait_for_flag() {
// std::cout << &m << std::endl;
// return;
std::unique_lock<std::mutex> lk(m);
while (!flag) {
lk.unlock();
std::cout << "unlocked....." << std::endl;
std::this_thread::sleep_for(std::chrono::milliseconds(100));
std::cout << "sleeping....." << std::endl;
lk.lock();
std::cout << "locked by " << std::this_thread::get_id() << "....."
<< std::endl;
}
}
int main(int argc, char const *argv[]) {
std::thread t(wait_for_flag);
std::thread t2(wait_for_flag);
std::thread t3(wait_for_flag);
std::thread t4(wait_for_flag);
std::thread t5(wait_for_flag);
t.join();
t2.join();
t3.join();
t4.join();
t5.join();
return 0;
}
I am new to this, and I thought mutex can only be acquired by one thread. I got two questions:
why there is no deadlock among those threads, e.g. if thread A runs lk.unlock(), then thread B runs lk.lock() and then thread A runs lk.lock().
what does it mean we define a new unique_lock in every thread associating to the same mutex lock (which is called m in here)
Thanks
Because right after acquiring a lock on the mutex each thread calls lk.unlock(); and now other thread can acquire a lock on the mutex. Only if a thread tries to lock an already locked mutex (by a different thread) it has to wait for the mutex to be free. As any thread in your code eventually calls lk.unlock(); there is always a chance for a different thread to get a lock on the mutex and there is no deadlock.
A deadlock would occur for example if you have two mutexes and two threads try to lock them in different order:
// thread A
std::unique_lock<std::mutex> lk1(mutex1);
std::unique_lock<std::mutex> lk2(mutex2); // X
// thread B
std::unique_lock<std::mutex> lk2(mutex2);
std::unique_lock<std::mutex> lk1(mutex1); // X
Here it can happen that thread A locks mutex1, thread B locks mutex2 and then both wait in X for the other thread to release the other mutex, but this will never happen. Its a deadlock.
2.
A lock is merely a slim RAII type. Its only purpose is to call lock on the mutex when created and unlock when destroyed. You can write the same code without the lock, by manually locking / unlocking the mutex, but when there is an exception while a mutex is locked it will never be unlocked.
#SolomonSlow my question is, if we use unique_lock to wrap the mutex in different threads, why there is no deadlock...?
"Deadlock" means that there is some set of threads in which none of the threads can proceed until one of the other members of the set does something. In the simplest possible deadlock, there are just two threads, and there are two mutexes:
Thread A has placed a unique_lock on mutex 1, and it is blocked, waiting to place a lock on mutex 2.
Thread B has placed a lock on mutex 2, and it is blocked, waiting to place a lock on mutex 1.
Thread A can't do anything until thread B does something first, and thread B can't do anything until thread A does something first. Neither thread will ever be able to do anything again. Deadlock.
You can't have a deadlock without at least two different things (e.g., two different mutexes) that the threads wait for. If there's only one mutex, then whichever thread has it locked, that thread will be able to proceed. It's only a deadlock when no thread is able to proceed.
In your example, each of the five threads settles in to a loop:
unlock the mutex,
print, sleep, print,
lock the mutex,
print,
go back to the top of the loop.
Whenever one of your threads locks the mutex, there's nothing to stop it from printing and then going back to the top and unlocking the mutex again so that some other thread can run. There's no deadlock.
This is not an answer. It's just an illustration. I turned your one example into three different examples that all achieve the same result. I hope it may help you to better understand what unique_lock does.
The first way doesn't use unique_lock at all. It only uses the mutex. This is the old-school way—the way we used to do things before RAII was discovered.
std::mutex m;
{
...
while (...) {
do_work_outside_critical_section();
m.lock(); // explicitly put a "lock" on the mutex.
do_work_inside_critical_section();
m.unlock(); // explicitly remove the "lock."
}
}
The old-school way is risky because if do_work_inside_critical_section() throws an exception, it will leave the mutex in a locked state, and any thread that tries to lock it again probably will hang forever.
The second way uses unique_lock, which is an embodiment of RAII.
The RAII pattern ensures that there's no way out of this code block that leaves a lock on mutex m. The unique_lock destructor always will be called, no matter what, and the destructor removes the lock.
std::mutex m;
{
...
while (...) {
do_work_outside_critical_section();
std::unique_lock<std::mutex> lk(m); // constructor puts a "lock" on the mutex.
do_work_inside_critical_section();
} // destructor implicitly removes the "lock."
}
Notice that in this version, a unique_lock is constructed and destructed every time around the loop. That might sound costly, but it really isn't. unique_lock is meant to be used in this way.
The last way is what you did in your example. It only creates and destroys the unique_lock one time, but then it repeatedly locks and unlocks it within the loop. This works, but it's more code lines than the version above, which makes it a little bit harder to read and understand.
std::mutex m;
{
...
std::unique_lock<std::mutex> lk(m); // constructor puts a "lock" on the mutex.
while (...) {
lk.unlock(); // explicitly remove the "lock" from the mutex.
do_work_outside_critical_section();
lk.lock(); // explicitly put a "lock" back on the mutex.
do_work_inside_critical_section();
}
} // destructor implicitly removes the "lock."
Consider the following quote about std::condition_variable from cppreference:
The condition_variable class is a synchronization primitive that can be used to block a thread, or multiple threads at the same time, until another thread both modifies a shared variable (the condition), and notifies the condition_variable.
The thread that intends to modify the variable has to
acquire a std::mutex (typically via std::lock_guard)
perform the modification while the lock is held
execute notify_one or notify_all on the std::condition_variable (the lock does not need to be held for notification)
Even if the shared variable is atomic, it must be modified under the mutex in order to correctly publish the modification to the waiting thread.
An exemplary typical scenario of this approach is as follows:
// shared (e.d., global) variables:
bool proceed = false;
std::mutex m;
std::condition_variable cv;
// thread #1:
{
std::unique_lock<std::mutex> l(m);
while (!proceed) cv.wait(l); // or, cv.wait(l, []{ return proceed; });
}
// thread #2:
{
std::lock_guard<std::mutex> l(m);
proceed = true;
}
cv.notify_one();
However, #Tim in this question came up with (kind-of an academic) alternative:
std::atomic<bool> proceed {false}; // atomic to avoid data race
std::mutex m;
std::condition_variable cv;
// thread #1:
{
std::unique_lock<std::mutex> l(m);
while (!proceed) cv.wait(l);
}
// thread #2:
proceed = true; // not protected with mutex
{ std::lock_guard<std::mutex> l(m); }
cv.notify_one();
It obviously does not meet the requirements of the above cppreference quote since proceed shared variable (atomic in this case) is not modified under the mutex.
The question is if this code is correct. In my opinion, it is, since:
First option is that !proceed in while (!proceed) is evaluated as false. In that case, cv.wait(l); is not invoked at all.
Second option is that !proceed in while (!proceed) is evaluated as true. In that case, cv.notify_one() cannot happen until thread #1 enters cv.wait(l);.
Am I missing something or is cppreference wrong in this regard? (For instance, are some reordering issues involved?)
And, what if proceed = true; would be changed to proceed.store(true, std::memory_order_relaxed);?
For simplicity, let's assume that we have only one conditional variable to match a single condition that is reflected by a boolean.
1) Why does std::condition_variable::wait(...) locks the mutex again after a "notify" has been sent to un-sleep it?
2) Seeing the behaviour in "1)", does that mean that when you do std::condition_variable::notify_all it only makes it so that all of the waiting threads are unblocked/woken up... but in order instead of all at once? If so, what can be done to do it all at once?
3) If I only care about threads sleeping until a condition is met and not care a single bit for any mutex acquisition, what can I do? Is there an alternative or should current std::condition_variable::wait(...) approach(es) be hacked around this?
If "hackery" is to be used, will this function work for unblocking all waiting threads on a condition and can it be called from any(per thread) threads:
//declared somehwere and modified before sending "notify"(ies)
std::atomic<bool> global_shared_condition_atomic_bool;
//the single(for simplicity in our case) condition variable matched with the above boolean result
std::condition_variable global_shared_condition_variable;
static void MyClass:wait()
{
std::mutex mutex;
std::unique_lock<std::mutex> lock(mutex);
while (!global_shared_condition_atomic_bool) global_shared_condition_variable.wait(lock);
}
it would have been called from random "waiting" threads like so:
void random_thread_run()
{
while(someLoopControlValue)
{
//random code...
MyClass:wait(); //wait for whatever condition the class+method is for.
//more random code...
}
}
Edit:
Gate class
#ifndef Gate_Header
#define Gate_Header
#include <mutex>
#include <condition_variable>
class Gate
{
public:
Gate()
{
gate_open = false;
}
void open()
{
m.lock();
gate_open = true;
m.unlock();
cv.notify_all();
}
void wait()
{
std::unique_lock<std::mutex> lock(m);
while (!gate_open) cv.wait(lock);
}
void close()
{
m.lock();
gate_open = false;
m.unlock();
}
private:
std::mutex m;
std::condition_variable cv;
bool gate_open;
};
#endif
Condition variables wake things up spuriously.
You must have a mutex and it must guard a message of some kind for them to work, or you have zero guarantee that any such wakeup occurred.
This was done, presumably, because efficient implementations of a non-spurious version end up being implemeneted in terms of such a spurious version anyhow.
If you fail to guard the message editing with a mutex (ie, no synchronization on it, the state of the message is undefined behavior. This can cause compilers to optimize the read from memory to skip it after the first read.
Even excluding that undefined behavior (imagine you use atomics), there are race conditions where a message is set, a notification occurs, and nobody waiting on the notification sees the message being set if you fail to have the mutex acquired in the time between the variable being set and the condition variable being notified.
Barring extreme cases, you usually want to use the lambda version of wait.
Auditing condition variable code is not possible unless you audit both the notification code and the wait code.
struct gate {
bool gate_open = false;
mutable std::condition_variable cv;
mutable std::mutex m;
void open_gate() {
std::unique_lock<std::mutex> lock(m);
gate_open=true;
cv.notify_all();
}
void wait_at_gate() const {
std::unique_lock<std::mutex> lock(m);
cv.wait( lock, [this]{ return gate_open; } );
}
};
or
void open_gate() {
{
std::unique_lock<std::mutex> lock(m);
gate_open=true;
}
cv.notify_all();
}
No, your code will not work.
The mutex protects modifications to the shared variable. As such, all of the waiting threads and the signaling thread must lock that specific mutex instance. With what you've written, each thread has its own mutex instance.
The main reason for all of this mutex stuff is due to the concept of spurious wakeup, an unfortunate aspect of OS implementations of condition variables. Threads waiting on them sometimes just start running even though the condition hasn't been satisfied yet.
The mutex-bound check of the actual variable allows the thread to test whether it was spuriously awoken or not.
wait atomically releases the mutex and starts waiting on the condition. When wait exits, the mutex is atomically reacquired as part of the wakeup process. Now, consider a race between a spurious wakeup and the notifying thread. The notifying thread can be in one of 2 states: about to modify the variable, or after modifying it and about to notify everyone to wake up.
If the spurious wakeup happens when the notifying thread is about to modify the varaible, then one of them will get to the mutex first. So the spuriously awoken thread will either see the old value or the new value. If it sees the new, then it has been notified and will go do its business. If it sees the old, then it will wait on the condition again. But if it saw the old, then it blocked the notifying thread from modifying that variable, so it had to wait until the spurious thread went back to sleep.
Why does std::condition_variable::wait(...) locks the mutex again after a "notify" has been sent to un-sleep it?
Because the mutex locks access to the condition variable. And the first thing you have to do after waking up from a wait call is to check the condition variable. As such, that must be done under the protection of the mutex.
The signalling thread must be prevented from modifying the variable while other threads are reading it. That's what the mutex is for.
Seeing the behaviour in "1)", does that mean that when you do std::condition_variable::notify_all it only makes it so that all of the waiting threads are unblocked/woken up... but in order instead of all at once?
The order they wake up in is not specified. However, by the time notify_all returns, all threads are guaranteed to have been unblocked.
If I only care about threads sleeping until a condition is met and not care a single bit for any mutex acquisition, what can I do?
Nothing. condition_variable requires that access to the actual variable you're checking is controlled via a mutex.
I started using std::mutexes to stop a thread and wait for another thread to resume it. It works like this:
Thread 1
// Ensures the mutex will be locked
while(myWaitMutex.try_lock());
// Locks it again to pause this thread
myWaitMutex.lock();
Thread 2
// Executed when thread 1 should resume processing:
myWaitMutex.unlock();
However I am not sure if this is correct and will work without problems on all platforms. If this is not correct, what is the correct way to implement this in C++11?
The problems with the code
// Ensures the mutex will be locked
while(myWaitMutex.try_lock());
.try_lock() tries to aquire the lock and returns true if successful, i.e., the code says "if we aquire the lock then retry to lock it again and again until we fail". We can never "fail" as we currently own the lock ourselves that we are waiting on, and so this will be an infinite loop. Also, attempting to lock using a std::mutex that the caller have already aquired a lock on is UB, so this is guaranteed to be UB. If not successful, .try_lock() will return false and the while loop will be exited. In other words, this will not ensure that the mutex will be locked.
The correct way to ensure the mutex will be locked is simply:
myWaitMutex.lock();
This will cause the current thread to block (indefinitely) until it can aquire the lock.
Next, the other thread tries to unlock a mutex it does not have a lock on.
// Executed when thread 1 should resume processing:
myWaitMutex.unlock();
This won't work as it's UB to .unlock() on a std::mutex that you don't already have a lock on.
Using locks
When using mutex locks, it's easier to use a RAII ownership-wrapper object such as std::lock_guard. The usage pattern of std::mutex is always: "Lock -> do something in critical section -> unlock". A std::lock_guard will lock the mutex in its constructor, and unlock it in its destructor. No need to worry about when to lock and unlock and such low-level stuff.
std::mutex m;
{
std::lock_guard<std::mutex> lk{m};
/* We have the lock until we exit scope. */
} // Here 'lk' is destroyed and will release lock.
A simple lock might not be the best tool for the job
If what you want is to be able to signal a thread to wake up, then there's the wait and notify structure using std::condition_variable. The std::condition_variable allows any caller to send a signal to waiting threads without holding any locks.
#include <atomic>
#include <condition_variable>
#include <iostream>
#include <mutex>
#include <thread>
using namespace std::literals;
int main() {
std::mutex m;
std::condition_variable cond;
std::thread t{[&] {
std::cout << "Entering sleep..." << std::endl;
std::unique_lock<std::mutex> lk{m};
cond.wait(lk); // Will block until 'cond' is notified.
std::cout << "Thread is awake!" << std::endl;
}};
std::this_thread::sleep_for(3s);
cond.notify_all(); // Notify all waiting threads.
t.join(); // Remember to join thread before exit.
}
However, to further complicate things there's this thing called spurious wakeups that mean that any waiting threads may wake up at any time for unknown reasons. This is a fact on most systems and has to do with the inner workings of thread scheduling. Also, we probably need to check that waiting is really needed as we're dealing with concurrency. If, for example, the notifying thread happens to notify before we start waiting, then we might wait forever unless we have a way to first check this.
To handle this we need to add a while loop and a predicate that tells when we need to wait and when we're done waiting.
int main() {
std::mutex m;
std::condition_variable cond;
bool done = false; // Flag for indicating when done waiting.
std::thread t{[&] {
std::cout << "Entering sleep..." << std::endl;
std::unique_lock<std::mutex> lk{m};
while (!done) { // Wait inside loop to handle spurious wakeups etc.
cond.wait(lk);
}
std::cout << "Thread is awake!" << std::endl;
}};
std::this_thread::sleep_for(3s);
{ // Aquire lock to avoid data race on 'done'.
std::lock_guard<std::mutex> lk{m};
done = true; // Set 'done' to true before notifying.
}
cond.notify_all();
t.join();
}
There are additional reasons why it's a good idea to wait inside a loop and use a predicate such as "stolen wakeups" as mentioned in the comments by #David Schwartz.
It sounds to me that you are looking for condition variable. In the end there should always be a way to make it work through mutexes, but condition variable is the current C++ idiomatic way to handle the `block and wait until something happens' scenario.
The behavior of a mutex when a thread that holds it attempts to lock it is undefined. The behavior of a mutex when a thread that doesn't hold it attempts to unlock it is undefined. So your code might do anything at all on various platforms.
Instead, use a mutex together with a condition variable and a predicate boolean. In pseudo-code:
To block:
Acquire the mutex.
While the predicate is false, block on the condition variable.
If you want to re-arm here, set the predicate to false.
Release the mutex.
To release:
Acquire the mutex.
Set the predicate to true.
Signal the condition variable.
Release the mutex.
To rearm:
Acquire the mutex.
Set the predicate to false.
Release the mutex.
Please check this code....
std::mutex m_mutex;
std::condition_variable m_cond_var;
void threadOne(){
std::unique_lock<std::mutex> lck(mtx);
while (!ready){
m_cond_var.wait(lck);
}
m_cond_var.notify_all();
}
void threadTwo(){
std::unique_lock<std::mutex> lck(mtx);
read = true;
m_cond_var.notify_all();
}
I hope you will get the solution. And it is very proper!!
For simplicity, let's assume that we have only one conditional variable to match a single condition that is reflected by a boolean.
1) Why does std::condition_variable::wait(...) locks the mutex again after a "notify" has been sent to un-sleep it?
2) Seeing the behaviour in "1)", does that mean that when you do std::condition_variable::notify_all it only makes it so that all of the waiting threads are unblocked/woken up... but in order instead of all at once? If so, what can be done to do it all at once?
3) If I only care about threads sleeping until a condition is met and not care a single bit for any mutex acquisition, what can I do? Is there an alternative or should current std::condition_variable::wait(...) approach(es) be hacked around this?
If "hackery" is to be used, will this function work for unblocking all waiting threads on a condition and can it be called from any(per thread) threads:
//declared somehwere and modified before sending "notify"(ies)
std::atomic<bool> global_shared_condition_atomic_bool;
//the single(for simplicity in our case) condition variable matched with the above boolean result
std::condition_variable global_shared_condition_variable;
static void MyClass:wait()
{
std::mutex mutex;
std::unique_lock<std::mutex> lock(mutex);
while (!global_shared_condition_atomic_bool) global_shared_condition_variable.wait(lock);
}
it would have been called from random "waiting" threads like so:
void random_thread_run()
{
while(someLoopControlValue)
{
//random code...
MyClass:wait(); //wait for whatever condition the class+method is for.
//more random code...
}
}
Edit:
Gate class
#ifndef Gate_Header
#define Gate_Header
#include <mutex>
#include <condition_variable>
class Gate
{
public:
Gate()
{
gate_open = false;
}
void open()
{
m.lock();
gate_open = true;
m.unlock();
cv.notify_all();
}
void wait()
{
std::unique_lock<std::mutex> lock(m);
while (!gate_open) cv.wait(lock);
}
void close()
{
m.lock();
gate_open = false;
m.unlock();
}
private:
std::mutex m;
std::condition_variable cv;
bool gate_open;
};
#endif
Condition variables wake things up spuriously.
You must have a mutex and it must guard a message of some kind for them to work, or you have zero guarantee that any such wakeup occurred.
This was done, presumably, because efficient implementations of a non-spurious version end up being implemeneted in terms of such a spurious version anyhow.
If you fail to guard the message editing with a mutex (ie, no synchronization on it, the state of the message is undefined behavior. This can cause compilers to optimize the read from memory to skip it after the first read.
Even excluding that undefined behavior (imagine you use atomics), there are race conditions where a message is set, a notification occurs, and nobody waiting on the notification sees the message being set if you fail to have the mutex acquired in the time between the variable being set and the condition variable being notified.
Barring extreme cases, you usually want to use the lambda version of wait.
Auditing condition variable code is not possible unless you audit both the notification code and the wait code.
struct gate {
bool gate_open = false;
mutable std::condition_variable cv;
mutable std::mutex m;
void open_gate() {
std::unique_lock<std::mutex> lock(m);
gate_open=true;
cv.notify_all();
}
void wait_at_gate() const {
std::unique_lock<std::mutex> lock(m);
cv.wait( lock, [this]{ return gate_open; } );
}
};
or
void open_gate() {
{
std::unique_lock<std::mutex> lock(m);
gate_open=true;
}
cv.notify_all();
}
No, your code will not work.
The mutex protects modifications to the shared variable. As such, all of the waiting threads and the signaling thread must lock that specific mutex instance. With what you've written, each thread has its own mutex instance.
The main reason for all of this mutex stuff is due to the concept of spurious wakeup, an unfortunate aspect of OS implementations of condition variables. Threads waiting on them sometimes just start running even though the condition hasn't been satisfied yet.
The mutex-bound check of the actual variable allows the thread to test whether it was spuriously awoken or not.
wait atomically releases the mutex and starts waiting on the condition. When wait exits, the mutex is atomically reacquired as part of the wakeup process. Now, consider a race between a spurious wakeup and the notifying thread. The notifying thread can be in one of 2 states: about to modify the variable, or after modifying it and about to notify everyone to wake up.
If the spurious wakeup happens when the notifying thread is about to modify the varaible, then one of them will get to the mutex first. So the spuriously awoken thread will either see the old value or the new value. If it sees the new, then it has been notified and will go do its business. If it sees the old, then it will wait on the condition again. But if it saw the old, then it blocked the notifying thread from modifying that variable, so it had to wait until the spurious thread went back to sleep.
Why does std::condition_variable::wait(...) locks the mutex again after a "notify" has been sent to un-sleep it?
Because the mutex locks access to the condition variable. And the first thing you have to do after waking up from a wait call is to check the condition variable. As such, that must be done under the protection of the mutex.
The signalling thread must be prevented from modifying the variable while other threads are reading it. That's what the mutex is for.
Seeing the behaviour in "1)", does that mean that when you do std::condition_variable::notify_all it only makes it so that all of the waiting threads are unblocked/woken up... but in order instead of all at once?
The order they wake up in is not specified. However, by the time notify_all returns, all threads are guaranteed to have been unblocked.
If I only care about threads sleeping until a condition is met and not care a single bit for any mutex acquisition, what can I do?
Nothing. condition_variable requires that access to the actual variable you're checking is controlled via a mutex.