How to interrupt a thread which is waiting for std::condition_variable_any in C++? - multithreading

I'm reading C++ concurrency in action.
It introduces how to implement interrupting thread using std::condition_variable_any.
I try to understand the code more than a week, but I couldn't.
Below is the code and explanation in the book.
#include <condition_variable>
#include <future>
#include <iostream>
#include <thread>
class thread_interrupted : public std::exception {};
class interrupt_flag {
std::atomic<bool> flag;
std::condition_variable* thread_cond;
std::condition_variable_any* thread_cond_any;
std::mutex set_clear_mutex;
public:
interrupt_flag() : thread_cond(0), thread_cond_any(0) {}
void set() {
flag.store(true, std::memory_order_relaxed);
std::lock_guard<std::mutex> lk(set_clear_mutex);
if (thread_cond) {
thread_cond->notify_all();
} else if (thread_cond_any) {
thread_cond_any->notify_all();
}
}
bool is_set() const { return flag.load(std::memory_order_relaxed); }
template <typename Lockable>
void wait(std::condition_variable_any& cv, Lockable& lk);
};
thread_local static interrupt_flag this_thread_interrupt_flag;
void interruption_point() {
if (this_thread_interrupt_flag.is_set()) {
throw thread_interrupted();
}
}
template <typename Lockable>
void interrupt_flag::wait(std::condition_variable_any& cv, Lockable& lk) {
struct custom_lock {
interrupt_flag* self;
// (1) What is this lk for? Why is lk should be already locked when it is used in costume_lock constructor?
Lockable& lk;
custom_lock(interrupt_flag* self_, std::condition_variable_any& cond,
Lockable& lk_)
: self(self_), lk(lk_) {
self->set_clear_mutex.lock();
self->thread_cond_any = &cond;
}
void unlock() {
lk.unlock();
self->set_clear_mutex.unlock();
}
void lock() { std::lock(self->set_clear_mutex, lk); }
~custom_lock() {
self->thread_cond_any = 0;
self->set_clear_mutex.unlock();
}
};
custom_lock cl(this, cv, lk);
interruption_point();
cv.wait(cl);
interruption_point();
}
class interruptible_thread {
std::thread internal_thread;
interrupt_flag* flag;
public:
template <typename FunctionType>
interruptible_thread(FunctionType f) {
std::promise<interrupt_flag*> p;
internal_thread = std::thread([f, &p] {
p.set_value(&this_thread_interrupt_flag);
f();
});
flag = p.get_future().get();
}
void interrupt() {
if (flag) {
flag->set();
}
};
void join() { internal_thread.join(); };
void detach();
bool joinable() const;
};
template <typename Lockable>
void interruptible_wait(std::condition_variable_any& cv, Lockable& lk) {
this_thread_interrupt_flag.wait(cv, lk);
}
void foo() {
// (2) This is my implementation of how to use interruptible wait. Is it correct?
std::condition_variable_any cv;
std::mutex m;
std::unique_lock<std::mutex> lk(m);
try {
interruptible_wait(cv, lk);
} catch (...) {
std::cout << "interrupted" << std::endl;
}
}
int main() {
std::cout << "Hello" << std::endl;
interruptible_thread th(foo);
th.interrupt();
th.join();
}
Your custom lock type acquires the lock on the internal
set_clear_mutex when it’s constructed 1, and then sets the
thread_cond_any pointer to refer to the std:: condition_variable_any
passed in to the constructor 2.
The Lockable reference is stored for later; this must already be
locked. You can now check for an interruption without worrying about
races. If the interrupt flag is set at this point, it was set before
you acquired the lock on set_clear_mutex. When the condition variable
calls your unlock() function inside wait(), you unlock the Lockable
object and the internal set_clear_mutex 3.
This allows threads that are trying to interrupt you to acquire the
lock on set_clear_mutex and check the thread_cond_any pointer once
you’re inside the wait() call but not before. This is exactly what you
were after (but couldn’t manage) with std::condition_variable.
Once wait() has finished waiting (either because it was notified or
because of a spurious wake), it will call your lock() function, which
again acquires the lock on the internal set_clear_mutex and the lock
on the Lockable object 4. You can now check again for interruptions
that happened during the wait() call before clearing the
thread_cond_any pointer in your custom_lock destructor 5, where you
also unlock the set_clear_mutex.
First, I couldn't understand what is the purpose of Lockabel& lk in mark (1) and why it is already locked in constructor of custom_lock. (It could be locked in the very custom_lock constructor. )
Second there is no example in this book of how to use interruptible wait, so foo() {} in mark (2) is my guess implementation of how to use it. Is it correct way of using it ?

You need a mutex-like object (lk in your foo function) to call the interruptiple waiting just as you would need it for the plain std::condition_variable::wait function.
What's problematic (I also read the book and I have doubts about this example) is that the flag member points to a memory location inside the other thread which could finish right before calling flag->set(). In this specific example the thread only exists after we set the flag so that is okay, but otherwise this approach is limited in my opinion (correct me if I am wrong).

Related

Overridden virtual function not called from thread

I am writing a base class to manage threads. The idea is to allow the thread function to be overridden in child class while the base class manages thread life cycle. I ran into a strange behavior which I don't understand - it seems that the virtual function mechanism does not work when the call is made from a thread. To illustrate my problem, I reduced my code to the following:
#include <iostream>
#include <thread>
using namespace std;
struct B
{
thread t;
void thread_func_non_virt()
{
thread_func();
}
virtual void thread_func()
{
cout << "B::thread_func\n";
}
B(): t(thread(&B::thread_func_non_virt, this)) { }
void join() { t.join(); }
};
struct C : B
{
virtual void thread_func() override
{
cout << "C::thread_func\n";
}
};
int main()
{
C c; // output is "B::thread_func" but "C::thread_func" is expected
c.join();
c.thread_func_non_virt(); // output "C::thread_func" as expected
}
I tried with both Visual studio 2017 and g++ 5.4 (Ubuntu 16) and found the behavior is consistent. Can someone point out where I got wrong?
== UPDATE ==
Based on Igor's answer, I moved the thread creation out of the constructor into a separate method and calling that method after the constructor and got the desired behavior.
Your program exhibits undefined behavior. There's a race on *this between thread_func and C's (implicitly defined) constructor.
#include <iostream>
#include <thread>
using namespace std;
struct B
{
thread t;
void thread_func_non_virt()
{
thread_func();
}
virtual void thread_func()
{
cout << "B::thread_func\n";
}
B(B*ptr): t(thread(&B::thread_func_non_virt, ptr))
{
}
void join() { t.join(); }
};
struct C:public B
{
C():B(this){}
virtual void thread_func() override
{
cout << "C::thread_func\n";
}
};
int main()
{
C c; // "C::thread_func" is expected as expected
c.join();
c.thread_func_non_virt(); // output "C::thread_func" as expected
}

thread sync using mutex and condition variable

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;
}

How many mutex(es) should be used in one thread

I am working on a c++ (11) project and on the main thread, I need to check the value of two variables. The value of the two variables will be set by other threads through two different callbacks. I am using two condition variables to notify changes of those two variables. Because in c++, locks are needed for condition variables, I am not sure if I should use the same mutex for the two condition variables or I should use two mutex's to minimize exclusive execution. Somehow, I feel one mutex should be sufficient because on one thread(the main thread in this case) the code will be executed sequentially anyway. The code on the main thread that checks (wait for) the value of the two variables wont be interleaved anyway. Let me know if you need me to write code to illustrate the problem. I can prepare that. Thanks.
Update, add code:
#include <mutex>
class SomeEventObserver {
public:
virtual void handleEventA() = 0;
virtual void handleEventB() = 0;
};
class Client : public SomeEventObserver {
public:
Client() {
m_shouldQuit = false;
m_hasEventAHappened = false;
m_hasEventBHappened = false;
}
// will be callbed by some other thread (for exampe, thread 10)
virtual void handleEventA() override {
{
std::lock_guard<std::mutex> lock(m_mutexForA);
m_hasEventAHappened = true;
}
m_condVarEventForA.notify_all();
}
// will be called by some other thread (for exampe, thread 11)
virtual void handleEventB() override {
{
std::lock_guard<std::mutex> lock(m_mutexForB);
m_hasEventBHappened = true;
}
m_condVarEventForB.notify_all();
}
// here waitForA and waitForB are in the main thread, they are executed sequentially
// so I am wondering if I can use just one mutex to simplify the code
void run() {
waitForA();
waitForB();
}
void doShutDown() {
m_shouldQuit = true;
}
private:
void waitForA() {
std::unique_lock<std::mutex> lock(m_mutexForA);
m_condVarEventForA.wait(lock, [this]{ return m_hasEventAHappened; });
}
void waitForB() {
std::unique_lock<std::mutex> lock(m_mutexForB);
m_condVarEventForB.wait(lock, [this]{ return m_hasEventBHappened; });
}
// I am wondering if I can use just one mutex
std::condition_variable m_condVarEventForA;
std::condition_variable m_condVarEventForB;
std::mutex m_mutexForA;
std::mutex m_mutexForB;
bool m_hasEventAHappened;
bool m_hasEventBHappened;
};
int main(int argc, char* argv[]) {
Client client;
client.run();
}

properly ending an infinite std::thread

I have a reusable class that starts up an infinite thread. this thread can only be killed by calling a stop function that sets a kill switch variable. When looking around, there is quite a bit of argument over volatile vs atomic variables.
The following is my code:
program.cpp
int main()
{
ThreadClass threadClass;
threadClass.Start();
Sleep(1000);
threadClass.Stop();
Sleep(50);
threaClass.Stop();
}
ThreadClass.h
#pragma once
#include <atomic>
#include <thread>
class::ThreadClass
{
public:
ThreadClass(void);
~ThreadClass(void);
void Start();
void Stop();
private:
void myThread();
std::atomic<bool> runThread;
std::thread theThread;
};
ThreadClass.cpp
#include "ThreadClass.h"
ThreadClass::ThreadClass(void)
{
runThread = false;
}
ThreadClass::~ThreadClass(void)
{
}
void ThreadClass::Start()
{
runThread = true;
the_thread = std::thread(&mythread, this);
}
void ThreadClass::Stop()
{
if(runThread)
{
runThread = false;
if (the_thread.joinable())
{
the_thread.join();
}
}
}
void ThreadClass::mythread()
{
while(runThread)
{
//dostuff
Sleep(100); //or chrono
}
}
The code that i am representing here mirrors an issue that our legacy code had in place. We call the stop function 2 times, which will try to join the thread 2 times. This results in an invalid handle exception. I have coded the Stop() function in order to work around that issue, but my question is why would the the join fail the second time if the thread has completed and joined? Is there a better way programmatically to assume that the thread is valid before trying to join?

How to use clang thread annotations with a RAII style try-lock?

I would like to wrap the following code with clang thread annotations:
std::mutex mutex;
int counter = 0; // should be accessed while the mutex is locked
std::unique_lock<std::mutex> lock(mutex, std::try_to_lock);
if (lock) {
++counter;
}
I want both to use a RAII lock (std::unique_guard) and I also would like to add thread annotations to this code.
class __attribute__((capability("mutex"))) Mutex {
public:
Mutex() = default;
std::mutex &getInternalMutex() { return m_mutex; }
private:
std::mutex m_mutex{};
};
class __attribute__((scoped_lockable)) TryLock {
public:
TryLock(Mutex &mutex)
// PROBLEM HERE: __attribute__((acquire_capability(mutex)))
// is not suitable (lock may fail) but
// __attribute__((try_acquire_capability(true, mutex)))
// cannot be used neither (requires to be used in a method
// returning whether the lock succeed or not)
: m_try_lock(mutex.getInternalMutex(), std::try_to_lock) {}
~TryLock() __attribute__((release_capability)) = default;
bool isLocked() const {
return !!m_try_lock;
}
private:
std::unique_lock<std::mutex> m_try_lock;
};
clang only provides try_acquire_capability which is not suitable here (as it should be used for a function returning a boolean indicating if the lock succeeded or not).
What would be the correct way to annotate this lock?

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