I am trying to notify an event from two different threads. Main thread wakes up on first notify_one() call but not on the second one. Is this the right way to use condition variable?
#include <chrono>
#include <condition_variable>
#include <iostream>
#include <mutex>
#include <thread>
bool is_ready_1(false);
bool is_ready_2(false);
std::mutex m;
std::condition_variable cv;
void test1()
{
std::this_thread::sleep_for(std::chrono::seconds(3));
std::unique_lock<std::mutex> lk(m);
is_ready_1 = true;
cv.notify_one();
}
void test2()
{
std::this_thread::sleep_for(std::chrono::seconds(6));
std::unique_lock<std::mutex> lk(m);
is_ready_2 = true;
cv.notify_one();
}
int main()
{
std::thread t1(test1);
std::thread t2(test2);
std::unique_lock<std::mutex> lk(m);
while (!is_ready_1)
{
cv.wait(lk);
if (is_ready_1)
std::cout << "Spurious wake-1 up!\n";
}
while (!is_ready_2)
{
cv.wait(lk);
if (is_ready_2)
std::cout << "Spurious wake-2 up!\n";
}
t1.join();
t2.join();
system("pause");
}
Try this code. Note that a condition variable accepts a predicate that first checks if the the condition is met, and then takes the lock. When you notify the variable, it runs the predicate (without taking the lock), if the condition is true, the thread will own the lock else the calling thread will keep waiting for the next notification.
#include <chrono>
#include <condition_variable>
#include <iostream>
#include <mutex>
#include <thread>
auto is_ready_1=false;
auto is_ready_2=false;
std::mutex m;
std::condition_variable cv;
void test1()
{
std::this_thread::sleep_for(std::chrono::seconds(3));
std::unique_lock<std::mutex> lk(m);
is_ready_1 = true;
cv.notify_one();
}
void test2()
{
std::this_thread::sleep_for(std::chrono::seconds(6));
std::unique_lock<std::mutex> lk(m);
is_ready_2 = true;
cv.notify_one();
}
int main()
{
std::thread t1(test1);
std::thread t2(test2);
std::unique_lock<std::mutex> lk(m);
cv.wait(lk, [] {
if (!is_ready_1)
{
std::cout << "Spurious 1" << std::endl;
}
return is_ready_1;
});
//preceding line must clear before this is executed
cv.wait(lk, [] {
if (!is_ready_2)
{
std::cout << "Spurious 1" << std::endl;
}
return is_ready_2;
});
t1.join();
t2.join();
system("pause");
}
Also on Windows 10, I see two spurious wakeup events, I suspect these happen when the thread is woken up from sleep.
Related
I was trying to figure out the data race theme, and I made this code. Here we work with the shared element wnd. I thought that by putting lock in the while loop, I would prohibit the th1 thread from working with wnd, but this did not happen and I see an unobstructed output of the th1 thread.
#include <iostream>
#include <thread>
#include <mutex>
#include <chrono>
int main()
{
bool wnd = true;
std::mutex mutex;
std::unique_lock<std::mutex> lock(mutex, std::defer_lock);
std::thread th1([&]() {
int i = 0;
while (true)
{
++i;
if (wnd)
std::cout << i << " WND TRUE" << std::endl;
else
std::cout << i << " WND FALSE" << std::endl;
std::this_thread::sleep_for(std::chrono::milliseconds(100));
}
});
while (true)
{
lock.lock();
std::this_thread::sleep_for(std::chrono::milliseconds(2000));
if (wnd)
wnd = false;
else
wnd = true;
lock.unlock();
std::this_thread::sleep_for(std::chrono::milliseconds(100));
}
th1.join();
return 0;
}
To tell the truth, I was hoping to see that the th1 thread stops printing for 2+-seconds at a time when the main thread is inside the lock section.
You are not using the mutex and specially std::unique_lock properly.
#include <iostream>
#include <thread>
#include <mutex>
#include <chrono>
int main()
{
bool wnd = true;
std::mutex mutex;
std::thread th1{[&]() {
for (int i = 0; i<10000; ++i)
{ std::unique_lock<std::mutex> lock(mutex);
std::cout << i << "\tWND\t " << std::boolalpha << wnd << std::endl;
};
}};
for (int i = 0; i<30; ++i)
{ std::unique_lock<std::mutex> lock(mutex);
std::this_thread::sleep_for(std::chrono::milliseconds(2000));
wnd = !wnd;
};
th1.join();
}
std::unique_lock uses its constructor operand as a resource whose acquisition is lock and release is unlock. It is designed to use RAII as a means of guaranteeing correct lock/unlock sequences on mutexes. a defered lock only means the mutex is not locked at the begining of lifespan of the std::unique_lock and it is not the usual use case. You can manually lock/unlock the mutex, but that generally leads to less maintainable, more error-prone code.
Keep in mind that if the threads involved are not racing over the ownership of the mutex, neither waits for the other; in your original prorgram, the worker thread did not touch the mutex. But in the program above, both threads are competing to lock the mutex; winner gets a chance to continue what he wants, and the loser has to wait until the mutex is unlocked - so that he can take its ownership.
from
#include <iostream>
#include <condition_variable>
#include <thread>
#include <chrono>
using namespace std::chrono_literals;
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 == " << i << '\n';
done = true;
}
void signals()
{
std::this_thread::sleep_for(200ms);
std::cout << "Notifying falsely...\n";
cv.notify_one(); // waiting thread is notified with i == 0.
// cv.wait wakes up, checks i, and goes back to waiting
std::this_thread::sleep_for(300ms); // Add to make sure 'waits' thread wakes up**
std::unique_lock<std::mutex> lk(cv_m);
i = 1;
while (!done)
{
std::cout << "Notifying true change...\n";
lk.unlock();
cv.notify_one(); // waiting thread is notified with i == 1, cv.wait returns**
std::this_thread::sleep_for(300ms);
lk.lock();
}
}
int main()
{
std::thread t1(waits), t2(signals);
t1.join();
t2.join();
}
This is an example I got from cppreference. I have tried to execute the codes and observe the second notify seems not effected. Even I commented it out, the "signals" thread is still done.
Should it be put in the forever loop ??
Also if I add a wait before taking the cv_m mutex, seems it's working as expected
I'm reading this std::condition_variable example:
#include <iostream>
#include <string>
#include <thread>
#include <mutex>
#include <condition_variable>
std::mutex m;
std::condition_variable cv;
std::string data;
bool ready = false;
bool processed = false;
void worker_thread()
{
// Wait until main() sends data
std::unique_lock<std::mutex> lk(m);
cv.wait(lk, []{return ready;});
// after the wait, we own the lock.
std::cout << "Worker thread is processing data\n";
data += " after processing";
// Send data back to main()
processed = true;
std::cout << "Worker thread signals data processing completed\n";
// Manual unlocking is done before notifying, to avoid waking up
// the waiting thread only to block again (see notify_one for details)
lk.unlock();
cv.notify_one();
}
int main()
{
std::thread worker(worker_thread);
data = "Example data";
// send data to the worker thread
{
std::lock_guard<std::mutex> lk(m);
ready = true;
std::cout << "main() signals data ready for processing\n";
}
cv.notify_one();
// wait for the worker
{
std::unique_lock<std::mutex> lk(m);
cv.wait(lk, []{return processed;});
}
std::cout << "Back in main(), data = " << data << '\n';
worker.join();
}
Let's focus on the beggining:
std::thread worker(worker_thread);
here, we start the worked, which will lock the mutex immediately:
std::unique_lock<std::mutex> lk(m);
cv.wait(lk, []{return ready;});
Then, we lock the mutex on main to change the value of ready
{
std::lock_guard<std::mutex> lk(m);
ready = true;
std::cout << "main() signals data ready for processing\n";
}
but how is it possible that we even arrive at the line ready=true? The mutext m is blocked from the worked thread, so the line std::lock_guard<std::mutex> lk(m); will wait until the mutex m is unlocked. As I understand, when a mutex is locked and we try to lock it, we'll wait until it gets unlocked. However, it'll never get unlocked because the worked thread is waiting, therefore not releasing it.
In the link you have attached, note that the following
The wait operations atomically release the mutex and suspend the
execution of the thread.
Hence the line
cv.wait(lk, []{return ready;});
releases the mutex and suspends the execution of the thread until the condition variable is notified, a timeout expires (with wit_for()), or a spurious wake-up occurs, hence the thread is awakened, and the mutex is atomically reacquired
Here's another example that will give you the gist of the mechanism.
#include <cstdio>
#include <thread>
#include <mutex>
#include <condition_variable>
#include <chrono>
std::mutex mtx;
std::condition_variable cv;
bool signal = 0;
void A()
{
while (1)
{
{
std::unique_lock<std::mutex> lock(mtx);
cv.wait(lock, [](){return signal;});
signal = 0;
}
printf("Lock release from th A\n");
}
}
int main()
{
std::thread th(A);
for (unsigned i = 0; i < 100; i++)
{
{
std::lock_guard<std::mutex> lock(mtx);
signal = 1;
}
cv.notify_one();
printf("Sending signal %i\n", i);
std::this_thread::sleep_for(std::chrono::seconds(5));
}
th.join();
return 0;
}
Typical usages of condition variables look like this (see code below): http://en.cppreference.com/w/cpp/thread/condition_variable.
However, it seems that the main thread could potentially call notify_one before the worker thread calls wait, which would result in a deadlock. Am I mistaken? If not, what is the usual workaround for this?
#include <iostream>
#include <string>
#include <thread>
#include <mutex>
#include <condition_variable>
std::mutex m;
std::condition_variable cv;
std::string data;
bool ready = false;
bool processed = false;
void worker_thread()
{
// Wait until main() sends data
std::unique_lock<std::mutex> lk(m);
cv.wait(lk, []{return ready;});
// after the wait, we own the lock.
std::cout << "Worker thread is processing data\n";
data += " after processing";
// Send data back to main()
processed = true;
std::cout << "Worker thread signals data processing completed\n";
// Manual unlocking is done before notifying, to avoid waking up
// the waiting thread only to block again (see notify_one for details)
lk.unlock();
cv.notify_one();
}
int main()
{
std::thread worker(worker_thread);
data = "Example data";
// send data to the worker thread
{
std::lock_guard<std::mutex> lk(m);
ready = true;
std::cout << "main() signals data ready for processing\n";
}
cv.notify_one();
// wait for the worker
{
std::unique_lock<std::mutex> lk(m);
cv.wait(lk, []{return processed;});
}
std::cout << "Back in main(), data = " << data << '\n';
worker.join();
}
Note the definition of wait that uses a condition (the only wait you should ever be using):
while (!pred()) {
wait(lock);
}
if notify was already fired it means the condition is already true (that was sequenced before notify_one in the signalling thread). So when the receiver takes the mutex and looks at pred(), it will be true and it will proceed.
There is example of using condition_variable taken from cppreference.com:
#include <condition_variable>
#include <mutex>
#include <thread>
#include <iostream>
#include <queue>
#include <chrono>
int main()
{
std::queue<int> produced_nums;
std::mutex m;
std::condition_variable cond_var;
bool done = false;
bool notified = false;
std::thread producer([&]() {
for (int i = 0; i < 5; ++i) {
std::this_thread::sleep_for(std::chrono::seconds(1));
std::lock_guard<std::mutex> lock(m);
std::cout << "producing " << i << '\n';
produced_nums.push(i);
notified = true;
cond_var.notify_one();
}
std::lock_guard<std::mutex> lock(m);
notified = true;
done = true;
cond_var.notify_one();
});
std::thread consumer([&]() {
while (!done) {
std::unique_lock<std::mutex> lock(m);
while (!notified) { // loop to avoid spurious wakeups
cond_var.wait(lock);
}
while (!produced_nums.empty()) {
std::cout << "consuming " << produced_nums.front() << '\n';
produced_nums.pop();
}
notified = false;
}
});
producer.join();
consumer.join();
}
If variable done comes true before the consumer thread is started, the consumer thread will not get any message. Indeed, sleep_for(seconds(1)) almost avoids such situation, but could it be possible in theory (or if don't have sleep in code)?
In my opinion correct version should look like this to force running consumer loop at least once:
std::thread consumer([&]() {
std::unique_lock<std::mutex> lock(m);
do {
while (!notified || !done) { // loop to avoid spurious wakeups
cond_var.wait(lock);
}
while (!produced_nums.empty()) {
std::cout << "consuming " << produced_nums.front() << '\n';
produced_nums.pop();
}
notified = false;
} while (!done);
});
Yes, you are absolutely right: there is a (remote) possibility that the consumer thread will not start running until after done has been set. Further, the write to done in the producer thread and the read in the consumer thread produce a race condition, and the behavior is undefined. Same problem in your version. Wrap the mutex around the entire loop in each function. Sorry, don't have the energy to write the correct code.