Scenario:
I have a condition_variable based wait and signal mechanism. This works! But I need a little more than just the classic wait and signal mechanism. I need to be able to do a timed wait as well as an infinite wait "on the same condition_variable". Hence, I created a wrapper class around a condition_variable which takes care of the spurious wake up issue as well. Following is the code for that:
Code:
// CondVarWrapper.hpp
#pragma once
#include <mutex>
#include <chrono>
#include <condition_variable>
class CondVarWrapper {
public:
void Signal() {
std::unique_lock<std::mutex> unique_lock(cv_mutex);
cond_var_signalled = true;
timed_out = false;
unique_lock.unlock();
cond_var.notify_one();
}
bool WaitFor(const std::chrono::seconds timeout) {
std::unique_lock<std::mutex> unique_lock(cv_mutex);
timed_out = true;
cond_var.wait_for(unique_lock, timeout, [this] {
return cond_var_signalled;
});
cond_var_signalled = false;
return (timed_out == false);
}
bool Wait() {
std::unique_lock<std::mutex> unique_lock(cv_mutex);
timed_out = true;
cond_var.wait(unique_lock, [this] {
return cond_var_signalled;
});
cond_var_signalled = false;
return (timed_out == false);
}
private:
bool cond_var_signalled = false;
bool timed_out = false;
std::mutex cv_mutex;
std::condition_variable cond_var;
};
// main.cpp
#include "CondVarWrapper.hpp"
#include <iostream>
#include <string>
#include <thread>
int main() {
CondVarWrapper cond_var_wrapper;
std::thread my_thread = std::thread([&cond_var_wrapper]{
std::cout << "Thread started" << std::endl;
if (cond_var_wrapper.WaitFor(std::chrono::seconds(10))) {
std::cout << "Thread stopped by signal from main" << std::endl;
} else {
std::cout << "ERROR: Thread stopping because of timeout" << std::endl;
}
});
std::this_thread::sleep_for(std::chrono::seconds(3));
// Uncomment following line to see the timeout working
cond_var_wrapper.Signal();
my_thread.join();
}
Question:
Above code is good but I think there is one problem? Would I really be able to do a wait as as well do a wait_for on the same condition_variable? What if a thread has acquired cv_mutex by calling CondVarWrapper::Wait() and this one never returned for some reason. And then another thread comes in calling CondVarWrapper::WaitFor(std::chrono::seconds(3)) expecting to return out if it does not succeed in 3 seconds. Now, this second thread would not be able to return out of WaitFor after 3 seconds isnt it? In fact it wouldn't ever return. Because the condition_variable wait is a timed wait but not the lock on cv_mutex. Am I correct or Am I wrong in understanding here?
If I am correct above then I need to replace std::mutex cv_mutex with a std::timed_mutex cv_mutex and do a timed_wait in CondVarWrapper::WaitFor and do a infinite wait on CondVarWrapper::Wait? Or are there any better/easier ways of handling it?
The mutex is released when calling std::condition::wait on the condition variable cond_var. Thus, when you call CondVarWrapper::Wait from one thread, it releases the mutex when calling std::condition::wait and it hangs in there forever, the second thread can still call CondVarWrapper::WaitFor and successfully lock the mutex cv_mutex.
I'm trying to implement an multi-thread job, a producer and a consumer, and basically what I want to do is, when consumer finishes the data, it notifies the producer so that producer provides new data.
The tricky part is, in my current impl, producer and consumer both notifies each other and waits for each other, I don't know how to implement this part correctly.
For example, see the code below,
mutex m;
condition_variable cv;
vector<int> Q; // this is the queue the consumer will consume
vector<int> Q_buf; // this is a buffer Q into which producer will fill new data directly
// consumer
void consume() {
while (1) {
if (Q.size() == 0) { // when consumer finishes data
unique_lock<mutex> lk(m);
// how to notify producer to fill up the Q?
...
cv.wait(lk);
}
// for-loop to process the elems in Q
...
}
}
// producer
void produce() {
while (1) {
// for-loop to fill up Q_buf
...
// once Q_buf is fully filled, wait until consumer asks to give it a full Q
unique_lock<mutex> lk(m);
cv.wait(lk);
Q.swap(Q_buf); // replace the empty Q with the full Q_buf
cv.notify_one();
}
}
I'm not sure this the above code using mutex and condition_variable is the right way to implement my idea,
please give me some advice!
The code incorrectly assumes that vector<int>::size() and vector<int>::swap() are atomic. They are not.
Also, spurious wakeups must be handled by a while loop (or another cv::wait overload).
Fixes:
mutex m;
condition_variable cv;
vector<int> Q;
// consumer
void consume() {
while(1) {
// Get the new elements.
vector<int> new_elements;
{
unique_lock<mutex> lk(m);
while(Q.empty())
cv.wait(lk);
new_elements.swap(Q);
}
// for-loop to process the elems in new_elements
}
}
// producer
void produce() {
while(1) {
vector<int> new_elements;
// for-loop to fill up new_elements
// publish new_elements
{
unique_lock<mutex> lk(m);
Q.insert(Q.end(), new_elements.begin(), new_elements.end());
cv.notify_one();
}
}
}
Maybe that is close to what you want to achive. I used 2 conditional variables to notify producers and consumers between each other and introduced variable denoting which turn is now:
#include <ctime>
#include <condition_variable>
#include <iostream>
#include <mutex>
#include <queue>
#include <thread>
template<typename T>
class ReaderWriter {
private:
std::vector<std::thread> readers;
std::vector<std::thread> writers;
std::condition_variable readerCv, writerCv;
std::queue<T> data;
std::mutex readerMutex, writerMutex;
size_t noReaders, noWriters;
enum class Turn { WRITER_TURN, READER_TURN };
Turn turn;
void reader() {
while (1) {
{
std::unique_lock<std::mutex> lk(readerMutex);
while (turn != Turn::READER_TURN) {
readerCv.wait(lk);
}
std::cout << "Thread : " << std::this_thread::get_id() << " consumed " << data.front() << std::endl;
data.pop();
if (data.empty()) {
turn = Turn::WRITER_TURN;
writerCv.notify_one();
}
}
}
}
void writer() {
while (1) {
{
std::unique_lock<std::mutex> lk(writerMutex);
while (turn != Turn::WRITER_TURN) {
writerCv.wait(lk);
}
srand(time(NULL));
int random_number = std::rand();
data.push(random_number);
std::cout << "Thread : " << std::this_thread::get_id() << " produced " << random_number << std::endl;
turn = Turn::READER_TURN;
}
readerCv.notify_one();
}
}
public:
ReaderWriter(size_t noReadersArg, size_t noWritersArg) : noReaders(noReadersArg), noWriters(noWritersArg), turn(ReaderWriter::Turn::WRITER_TURN) {
}
void run() {
int noReadersArg = noReaders, noWritersArg = noWriters;
while (noReadersArg--) {
readers.emplace_back(&ReaderWriter::reader, this);
}
while (noWritersArg--) {
writers.emplace_back(&ReaderWriter::writer, this);
}
}
~ReaderWriter() {
for (auto& r : readers) {
r.join();
}
for (auto& w : writers) {
w.join();
}
}
};
int main() {
ReaderWriter<int> rw(5, 5);
rw.run();
}
Here's a code snippet. Since the worker treads are already synchronized, requirement of two buffers is ruled out. So a simple queue is used to simulate the scenario:
#include "conio.h"
#include <iostream>
#include <thread>
#include <mutex>
#include <queue>
#include <atomic>
#include <condition_variable>
using namespace std;
enum state_t{ READ = 0, WRITE = 1 };
mutex mu;
condition_variable cv;
atomic<bool> running;
queue<int> buffer;
atomic<state_t> state;
void generate_test_data()
{
const int times = 5;
static int data = 0;
for (int i = 0; i < times; i++) {
data = (data++) % 100;
buffer.push(data);
}
}
void ProducerThread() {
while (running) {
unique_lock<mutex> lock(mu);
cv.wait(lock, []() { return !running || state == WRITE; });
if (!running) return;
generate_test_data(); //producing here
lock.unlock();
//notify consumer to start consuming
state = READ;
cv.notify_one();
}
}
void ConsumerThread() {
while (running) {
unique_lock<mutex> lock(mu);
cv.wait(lock, []() { return !running || state == READ; });
if (!running) return;
while (!buffer.empty()) {
auto data = buffer.front(); //consuming here
buffer.pop();
cout << data << " \n";
}
//notify producer to start producing
if (buffer.empty()) {
state = WRITE;
cv.notify_one();
}
}
}
int main(){
running = true;
thread producer = thread([]() { ProducerThread(); });
thread consumer = thread([]() { ConsumerThread(); });
//simulating gui thread
while (!getch()){
}
running = false;
producer.join();
consumer.join();
}
Not a complete answer, though I think two condition variables could be helpful, one named buffer_empty that the producer thread will wait on, and another named buffer_filled that the consumer thread will wait on. Number of mutexes, how to loop, and so on I cannot comment on, since I'm not sure about the details myself.
Accesses to shared variables should only be done while holding the
mutex that protects it
condition_variable::wait should check a condition.
The condition should be a shared variable protected by the mutex that you pass to condition_variable::wait.
The way to check the condition is to wrap the call to wait in a while loop or use the 2-argument overload of wait (which is
equivalent to the while-loop version)
Note: These rules aren't strictly necessary if you truly understand what the hardware is doing. However, these problems get complicated quickly when with simple data structures, and it will be easier to prove that your algorithm is working correctly if you follow them.
Your Q and Q_buf are shared variables. Due to Rule 1, I would prefer to have them as local variables declared in the function that uses them (consume() and produce(), respectively). There will be 1 shared buffer that will be protected by a mutex. The producer will add to its local buffer. When that buffer is full, it acquires the mutex and pushes the local buffer to the shared buffer. It then waits for the consumer to accept this buffer before producing more data.
The consumer waits for this shared buffer to "arrive", then it acquires the mutex and replaces its empty local buffer with the shared buffer. Then it signals to the producer that the buffer has been accepted so it knows to start producing again.
Semantically, I don't see a reason to use swap over move, since in every case one of the containers is empty anyway. Maybe you want to use swap because you know something about the underlying memory. You can use whichever you want and it will be fast and work the same (at least algorithmically).
This problem can be done with 1 condition variable, but it may be a little easier to think about if you use 2.
Here's what I came up with. Tested on Visual Studio 2017 (15.6.7) and GCC 5.4.0. I don't need to be credited or anything (it's such a simple piece), but legally I have to say that I offer no warranties whatsoever.
#include <thread>
#include <vector>
#include <mutex>
#include <condition_variable>
#include <chrono>
std::vector<int> g_deliveryBuffer;
bool g_quit = false;
std::mutex g_mutex; // protects g_deliveryBuffer and g_quit
std::condition_variable g_producerDeliver;
std::condition_variable g_consumerAccepted;
// consumer
void consume()
{
// local buffer
std::vector<int> consumerBuffer;
while (true)
{
if (consumerBuffer.empty())
{
std::unique_lock<std::mutex> lock(g_mutex);
while (g_deliveryBuffer.empty() && !g_quit) // if we beat the producer, wait for them to push to the deliverybuffer
g_producerDeliver.wait(lock);
if (g_quit)
break;
consumerBuffer = std::move(g_deliveryBuffer); // get the buffer
}
g_consumerAccepted.notify_one(); // notify the producer that the buffer has been accepted
// for-loop to process the elems in Q
// ...
consumerBuffer.clear();
// ...
}
}
// producer
void produce()
{
std::vector<int> producerBuffer;
while (true)
{
// for-loop to fill up Q_buf
// ...
producerBuffer = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
// ...
// once Q_buf is fully filled, wait until consumer asks to give it a full Q
{ // scope is for lock
std::unique_lock<std::mutex> lock(g_mutex);
g_deliveryBuffer = std::move(producerBuffer); // ok to push to deliverybuffer. it is guaranteed to be empty
g_producerDeliver.notify_one();
while (!g_deliveryBuffer.empty() && !g_quit)
g_consumerAccepted.wait(lock); // wait for consumer to signal for more data
if (g_quit)
break;
// We will never reach this point if the buffer is not empty.
}
}
}
int main()
{
// spawn threads
std::thread consumerThread(consume);
std::thread producerThread(produce);
// for for 5 seconds
std::this_thread::sleep_for(std::chrono::seconds(5));
// signal that it's time to quit
{
std::lock_guard<std::mutex> lock(g_mutex);
g_quit = true;
}
// one of the threads may be sleeping
g_consumerAccepted.notify_one();
g_producerDeliver.notify_one();
consumerThread.join();
producerThread.join();
return 0;
}
I am quite new to the C++11 feature std::async and I fail to grasp why the code below never prints bar.
Could someone shed some light on this for me?
class Thready {
public:
Thready() {
std::async(std::launch::async, &Thready::foo, this);
}
void foo() {
while (true) {
std::cout << "foo" << std::endl;
}
}
void bar() {
while (true) {
std::cout << "bar" << std::endl;
}
}
};
int main() {
Thready t;
t.bar();
}
See "Notes" section on this page: http://en.cppreference.com/w/cpp/thread/async
The implementation may extend the behavior of the first overload of
std::async by enabling additional (implementation-defined) bits in the
default launch policy. Examples of implementation-defined launch
policies are the sync policy (execute immediately, within the async
call) and the task policy (similar to async, but thread-locals are not
cleared) If the std::future obtained from std::async is not moved from
or bound to a reference, the destructor of the std::future will block
at the end of the full expression until the asynchronous operation
completes, essentially making code such as the following synchronous:
std::async(std::launch::async, []{ f(); }); // temporary's dtor waits for f()
std::async(std::launch::async, []{ g(); }); // does not start until f() completes
(note that the destructors of std::futures
obtained by means other than a call to std::async never block)
TL;DR:
try to save the returned value of std::async call into some variable:
auto handle = std::async(std::launch::async, &Thready::foo, this);
EDIT:
the following code should work as you expect.
#include <future>
#include <iostream>
class Thready {
public:
Thready() {
handle = std::async(std::launch::async, &Thready::foo, this);
}
void foo() {
while (true) {
std::cout << "foo" << std::endl;
}
}
void bar() {
while (true) {
std::cout << "bar" << std::endl;
}
}
std::future<void> handle;
};
int main() {
Thready t;
t.bar();
}
I am trying to teach myself C++11 threading, and I would like to start a background producer thread at the beginning of the application, and have it run until application exit. I would also like to have consumer thread (which also runs for the life of the application).
A real-world example would be a producer thread listening on a Com port for incoming GPS data. Once a full message had been accumulated, it could be parsed to see if it was a message of interest, then converted into a string (say), and 'delivered back' to be consumed (update current location, for example).
My issue is I haven't been able to figure out how to do this without blocking the rest of the application when I 'join()' on the consumer thread.
Here is my very simplified example that hopefully shows my issues:
#include <QCoreApplication>
#include <QDebug>
#include <thread>
#include <atomic>
#include <iostream>
#include <queue>
#include <mutex>
#include <chrono>
#include "threadsafequeuetwo.h"
ThreadSafeQueueTwo<int> goods;
std::mutex mainMutex;
std::atomic<bool> isApplicationRunning = false;
void theProducer ()
{
std::atomic<int> itr = 0;
while(isApplicationRunning)
{
// Simulate this taking some time...
std::this_thread::sleep_for(std::chrono::milliseconds(60));
// Push the "produced" value onto the queue...
goods.push(++itr);
// Diagnostic printout only...
if ((itr % 10) == 0)
{
std::unique_lock<std::mutex> lock(mainMutex);
std::cout << "PUSH " << itr << " on thread ID: "
<< std::this_thread::get_id() << std::endl;
}
// Thread ending logic.
if (itr > 100) isApplicationRunning = false;
}
}
void theConsumer ()
{
while(isApplicationRunning || !goods.empty())
{
int val;
// Wait on new values, and 'pop' when available...
goods.waitAndPop(val);
// Here, we would 'do something' with the new values...
// Simulate this taking some time...
std::this_thread::sleep_for(std::chrono::milliseconds(10));
// Diagnostic printout only...
if ((val % 10) == 0)
{
std::unique_lock<std::mutex> lock(mainMutex);
std::cout << "POP " << val << " on thread ID: "
<< std::this_thread::get_id() << std::endl;
}
}
}
int main(int argc, char *argv[])
{
std::cout << "MAIN running on thread ID: "
<< std::this_thread::get_id() << std::endl;
// This varaiable gets set to true at startup, and,
// would only get set to false when the application
// wants to exit.
isApplicationRunning = true;
std::thread producerThread (theProducer);
std::thread consumerThread (theConsumer);
producerThread.detach();
consumerThread.join(); // BLOCKS!!! - how to get around this???
std::cout << "MAIN ending on thread ID: "
<< std::this_thread::get_id() << std::endl;
}
The ThreadSafeQueueTwo class is the thread safe queue implementation taken almost exactly as is from the "C++ Concurrency In Action" book. This seems to work just fine. Here it is if anybody is interested:
#ifndef THREADSAFEQUEUETWO_H
#define THREADSAFEQUEUETWO_H
#include <queue>
#include <memory>
#include <mutex>
#include <condition_variable>
template<typename T>
class ThreadSafeQueueTwo
{
public:
ThreadSafeQueueTwo()
{}
ThreadSafeQueueTwo(ThreadSafeQueueTwo const& rhs)
{
std::lock_guard<std::mutex> lock(myMutex);
myQueue = rhs.myQueue;
}
void push(T newValue)
{
std::lock_guard<std::mutex> lock(myMutex);
myQueue.push(newValue);
myCondVar.notify_one();
}
void waitAndPop(T& value)
{
std::unique_lock<std::mutex> lock(myMutex);
myCondVar.wait(lock, [this]{return !myQueue.empty(); });
value = myQueue.front();
myQueue.pop();
}
std::shared_ptr<T> waitAndPop()
{
std::unique_lock<std::mutex> lock(myMutex);
myCondVar.wait(lock, [this]{return !myQueue.empty(); });
std::shared_ptr<T> sharedPtrToT (std::make_shared<T>(myQueue.front()));
myQueue.pop();
return sharedPtrToT;
}
bool tryPop(T& value)
{
std::lock_guard<std::mutex> lock(myMutex);
if (myQueue.empty())
return false;
value = myQueue.front();
myQueue.pop();
return true;
}
std::shared_ptr<T> tryPop()
{
std::lock_guard<std::mutex> lock(myMutex);
if (myQueue.empty())
return std::shared_ptr<T>();
std::shared_ptr<T> sharedPtrToT (std::make_shared<T>(myQueue.front()));
myQueue.pop();
return sharedPtrToT;
}
bool empty()
{
std::lock_guard<std::mutex> lock(myMutex);
return myQueue.empty();
}
private:
mutable std::mutex myMutex;
std::queue<T> myQueue;
std::condition_variable myCondVar;
};
#endif // THREADSAFEQUEUETWO_H
Here's the output:
I know there are obvious issues with my example, but my main question is how would I run something like this in the background, without blocking the main thread?
Perhaps an even better way of trying to solve this is, is there a way that every time the producer has 'produced' some new data, could I simply call a method in the main thread, passing in the new data? This would be similar to queued signal/slots it Qt, which I am big fan of.
I tried yesterday to use std::thread correctly, but it's very dark for me.
My program implementation with pthread works well I don't have any problem with it. I would like to have the same solution with std::thread (if possible).
Solution with pthread:
void *MyShell(void *data) {
std::string str;
while(1) {
std::cin >> str;
std::cout << str << std::endl;
}
}
void mainloop() {
pthread_t thread;
pthread_create(&thread, NULL, aed::map::shell::Shell, this);
...
pthread_cancel(thread);
}
And now the solution which doesn't work everytime, with std::thread:
class ShellThreadInterrupFlag {
public:
void interrupt() {
throw std::string("Thread interruption test\n");
}
};
class ShellThread {
public:
template<typename FunctionType, typename ParamsType>
ShellThread(FunctionType f, ParamsType params) {
std::promise<ShellThreadInterrupFlag *> p[3];
_internal_thread = new std::thread(f, p, params);
_flag = p[0].get_future().get();
_internal_thread->detach();
p[1].set_value(_flag); // tell the thread that we detached it
p[2].get_future().get(); // wait until the thread validates the constructor could end (else p[3] is freed)
}
~ShellThread() {
delete _internal_thread;
}
void interrupt() {
_flag->interrupt();
}
private:
std::thread *_internal_thread;
ShellThreadInterrupFlag *_flag;
};
void Shell(std::promise<ShellThreadInterrupFlag *> promises[3],
aed::map::MapEditor *me)
{
ShellThreadInterrupFlag flag;
promises[0].set_value(&flag); // give the ShellThread instance the flag adress
promises[1].get_future().get(); // wait for detaching
promises[2].set_value(&flag); // tell ShellThread() it is able to finish
while(1) {
std::cin >> str;
std::cout << str << std::endl;
}
}
void mainloop()
{
ShellThread *shell_thread;
shell_thread = new ShellThread(Shell, this);
... // mainloop with opengl for drawing, events gestion etc...
shell_thread->interrupt();
}
Sometimes, when I launch the program, the std::cin >> str is called and the mainloop is blocked.
Does anyone know why the thread is blocking my mainloop ? And how could I avoid this problem ?