Suppose I have two threads and one shared c++ 11 condition variable.
What whould happen if thread1 call notify and after that thread2 call wait?
Will thread2 block forever or it will continue it's work due to call of notify by thread1?
Edit:
enum bcLockOperation
{
bcLockOperation_Light = -1,
bcLockOperation_Medium = 50,
bcLockOperation_Heavy = 1
}
class BC_COREDLL_EXP bcCustomMutex
{
private:
bcCustomMutex(const bcCustomMutex&);
bcCustomMutex& operator=(const bcCustomMutex&);
protected:
bcAtomic<int> mFlag;
bcMutex mMutex;
bcConditionVariable mCond;
public:
bcCustomMutex() { bcAtomicOperation::bcAtomicInit(mFlag, 0); };
~bcCustomMutex() {};
void lock(bcLockOperation pLockOperation = bcLockOperation_Medium)
{
bcINT32 lNewLoopCount = static_cast<bcINT32>(pLockOperation);
bcINT32 lLoopCounter = 0;
bcINT32 lExpected = 0;
bcINT32 lLoopCount = bcAtomicOperation::bcAtomicLoad(mFlag, bcMemoryOrder_Relaxed);
while (true)
{
while(bcAtomicOperation::bcAtomicLoad(mFlag, bcMemoryOrder_Relaxed) != 0 && lLoopCounter != lLoopCount)
++lLoopCounter;
bcAtomicOperation::bcAtomicCompareExchangeStrong(
mFlag,
&lExpected,
lNewLoopCount,
bcMemoryOrder_Acquire,
bcMemoryOrder_Relaxed);
if(lExpected == 0)
{
return;
}
else if(lLoopCounter == lLoopCount)
{
bcLockGuard<bcMutex> lGuard(mMutex);
mCond.wait(mMutex);
}
else
{
continue;
}
}
void UnLock()
{
bcAtomicOperation::bcAtomicStore(mFlag, 0, bcMemoryOrder_Relaxed);
bcUniqueLock<bcMutex> lGuard(mMutex);
mCond.notifyOne();
}
bcBOOL TryLock()
{
};
};
I want to write a custom mutex such that each thread can provide an argument that represents the complexity of operations that the current thread wants to execute.
If the complexity of the operation is low other threads will be in a loop like a spin lock but if the complexity of the operation is medium each thread will iterate 50 times and then will sleep by condition variable and if operation is very complex other threads will go to sleep directly.
now assume thread1 locks this mutex and thread2 goes for waiting due to its loopCounter reaching its end and right before locking the condition variable's mutex, thread1 calls notify on the condition variable. Now thread2 will sleep until another thread locks the custom mutex and then calls unlock on it.
I am new to multithreading and I want to learn. I know that my class may contain errors or may be completely wrong, but is there any way to correct this problem or a good algorithm to write such a mutex.
Thread2 will block until someone calls notify. Calls to notify release threads that are waiting at the time of the call. If there are no threads waiting, they do nothing. They aren't saved.
Usually both the code that decides to wait and the code that decides to notify share the same mutex. So thread2 will never "miss" the notify from thread1.
Here's the classic lock-based concurrent queue example:
void push(int x)
{
lock_guard<mutex> guard{queue_mutex};
thequeue.push(x);
not_empty_condition.notify_one();
}
int pop()
{
unique_lock<mutex> guard{queue_mutex};
not_empty_condition.wait(guard, []{ return !thequeue.empty(); } );
int x = thequeue.front();
thequeue.pop();
return x;
}
Assume thread1 and thread2 are running push() and pop() respectively. Only one of them will be in the critical section at a time.
If thread2 has the lock, either it never waits because the queue is not empty (so "losing" a notify is harmless), or it sits there waiting for a notify (which won't be lost).
If thread1 got the lock, it will put an element in the queue; if thread2 was waiting, it will get notified properly; if thread2 was still waiting for the mutex, it will never wait, as there is at least one element on the queue, so losing a notify is harmless.
In this manner, a notify is only lost if it was not needed in the first place.
Now, if you have a different usage for condition variables in mind, where "losing" a notification has any consequence, I believe you either have a race condition, or are using the wrong tool altogether.
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; });
I would like to write a class that wraps around std::thread and behaves like a std::thread but without actually allocating a thread every time I need to process something async. The reason is that I need to use multi threading in a context where I'm not allow to dynamically allocate and I also don't want to have the overhead of creating a std::thread.
Instead, I want a thread to run in a loop and wait until it can start processing. The client calls invoke which wakes up the thread. The Thread locks a mutex, does it's processing and falls asleep again. A function join behaves like std::thread::join by locking until the thread frees the lock (i.e. falls asleep again).
I think I got the class to run but because of a general lack of experience in multi threading, I would like to ask if anybody can spot race conditions or if the approach I used is considered "good style". For example, I'm not sure if temporary locking the mutex is a decent way to "join" the thread.
EDIT
I found another race condition: when calling join directly after invoke, there is no reason the thread already locked the mutex and thus locks the caller of join until the thread goes to sleep. To prevent this, I had to add a check for the invoke counter.
Header
#pragma once
#include <thread>
#include <atomic>
#include <mutex>
class PersistentThread
{
public:
PersistentThread();
~PersistentThread();
// set function to invoke
// locks if thread is currently processing _func
void set(const std::function<void()> &f);
// wakes the thread up to process _func and fall asleep again
// locks if thread is currently processing _func
void invoke();
// mimics std::thread::join
// locks until the thread is finished with it's loop
void join();
private:
// intern thread loop
void loop(bool *initialized);
private:
bool _shutdownRequested{ false };
std::mutex _mutex;
std::unique_ptr<std::thread> _thread;
std::condition_variable _cond;
std::function<void()> _func{ nullptr };
};
Source File
#include "PersistentThread.h"
PersistentThread::PersistentThread()
{
auto lock = std::unique_lock<std::mutex>(_mutex);
bool initialized = false;
_thread = std::make_unique<std::thread>(&PersistentThread::loop, this, &initialized);
// wait until _thread notifies, check bool initialized to prevent spurious wakeups
_cond.wait(lock, [&] {return initialized; });
}
PersistentThread::~PersistentThread()
{
{
std::lock_guard<std::mutex> lock(_mutex);
_func = nullptr;
_shutdownRequested = true;
// wake up and let join
_cond.notify_one();
}
// join thread,
if (_thread->joinable())
{
_thread->join();
}
}
void PersistentThread::set(const std::function<void()>& f)
{
std::lock_guard<std::mutex> lock(_mutex);
this->_func = f;
}
void PersistentThread::invoke()
{
std::lock_guard<std::mutex> lock(_mutex);
_cond.notify_one();
}
void PersistentThread::join()
{
bool joined = false;
while (!joined)
{
std::lock_guard<std::mutex> lock(_mutex);
joined = (_invokeCounter == 0);
}
}
void PersistentThread::loop(bool *initialized)
{
std::unique_lock<std::mutex> lock(_mutex);
*initialized = true;
_cond.notify_one();
while (true)
{
// wait until we get the mutex again
_cond.wait(lock, [this] {return _shutdownRequested || (this->_invokeCounter > 0); });
// shut down if requested
if (_shutdownRequested) return;
// process
if (_func) _func();
_invokeCounter--;
}
}
You are asking about potential race conditions, and I see at least one race condition in the shown code.
After constructing a PersistentThread, there is no guarantee that the new thread will acquire its initial lock in its loop() before the main execution thread returns from the constructor and enters invoke(). It is possible that the main execution thread enters invoke() immediately after the constructor is complete, ends up notifying nobody, since the internal execution thread hasn't locked the mutex yet. As such, this invoke() will not result in any processing taking place.
You need to synchronize the completion of the constructor with the execution thread's initial lock acquisition.
EDIT: your revision looks right; but I also spotted another race condition.
As documented in the description of wait(), wait() may wake up "spuriously". Just because wait() returned, doesn't mean that some other thread has entered invoke().
You need a counter, in addition to everything else, with invoke() incrementing the counter, and the execution thread executing its assigned duties only when the counter is greater than zero, decrementing it. This will guard against spurious wake-ups.
I would also have the execution thread check the counter before entering wait(), and enter wait() only if it is 0. Otherwise, it decrements the counter, executes its function, and loops back.
This should plug up all the potential race conditions in this area.
P.S. The spurious wake-up also applies to the initial notification, in your correction, that the execution thread has entered the loop. You'll need to do something similar for that situation, too.
I don't understand what you're trying to ask exactly. It's a nice style you used.
It would be much safer using bools and check the single routines because void returns nothing so you could be maybe stuck caused by bugs. Check everything you can since the thread runs under the hood. Make sure the calls are running correctly, if the process had really success. Also you could read some stuff about "Thread Pooling".
I have two condition variables:
CondVar1
CondVar2
Used in two threads like this (pseudo-code):
// thread1 starts in 'waiting' mode, and then Thread2 signals
void Thread1()
{
CondVar1->Wait();
CondVar2->Signal();
}
void Thread2()
{
CondVar1->Signal();
CondVar2->Wait();
}
Can this cause a deadlock? meaning, thread1 waits, thread2 signals, and then can thread1 signals before thread2 enters Wait(), meaning thread2 will never return?
Thanks
You don't usually just wait on a condition variable. The common use pattern is holding a lock, checking a variable that determines whether you can proceed or not and if you cannot wait in the condition:
// pseudocode
void push( T data ) {
Guard<Mutex> lock( m_mutex ); // Hold a lock on the queue
while (m_queue.full()) // [1]
m_cond1.wait(lock); // Wait until a consumer leaves a slot for me to write
// insert data
m_cond2.signal_one(); // Signal consumers that might be waiting on an empty queue
}
Some things to note: most libraries allow for spurious wakes in condition variables. While it is possible to implement a condition variable that avoid spurious wakes, the cost of the operations would be higher, so it is considered a lesser evil to require users to recheck the state before continuing (while loop in [1]).
Some libraries, notably C++11, allow you to pass a predicate, and will implement the loop internally: cond.wait(lock, [&queue](){ return !queue.full(); } );
There are two situations that could lead to a deadlock here:
In normal execution, the one you described. It is possible that the variable is signaled before the thread reaches the call to Wait, so the signal is lost.
A spurious wake-up could happen, causing the first thread to leave the call to Wait before actually being signaled, hence signaling Thread 2 who is not yet waiting.
You should design your code as follows when using signaling mechanisms:
bool thread1Waits = true;
bool thread2Waits = true;
void Thread1()
{
while(thread1Waits) CondVar1->Wait();
thread2Waits = false;
CondVar2->Signal();
}
void Thread2()
{
thread1Waits = false;
CondVar1->Signal();
while(thread2Waits) CondVar2->Wait();
}
Of course, this assumes there are locks protecting the condition variables and that additionally thread 1 runs before thread 2.