我正在开发一个使用线程池的应用程序,向其提交任务并同步它们。主线程必须等到单循环迭代完成所有提交的任务,然后再提交另一堆任务(因为下一次迭代的任务对相同的数据进行操作,并且它们将相互依赖)。 p>
我的问题是,最好的方法是什么?
到目前为止,我提出的是每个线程在完成一个任务后,递增一个原子无符号整数。当整数等于提交的任务数时,主线程继续工作并提交另一轮任务。
这是我的第一个多线程应用程序。 这是处理这个问题的最佳和明智的方法。
我正在使用从一本优秀的书“C ++ Concurrency in Action:Anthony Williams。”中复制的线程池类。
以下是课程:
class thread_pool
{
std::atomic_bool done;
thread_safe_queue<std::function<void()> > work_queue;
std::vector<std::thread> threads;
join_threads joiner;
void worker_thread()
{
while(!done)
{
std::function<void()> task;
if(work_queue.try_pop(task))
{
task();
}
else
{
std::this_thread::yield();
}
}
}
public:
thread_pool():
done(false),joiner(threads)
{
unsigned const thread_count=std::thread::hardware_concurrency();
try
{
for(unsigned i=0;i<thread_count;++i)
{
threads.push_back(
std::thread(&thread_pool::worker_thread,this));
}
}
catch(...)
{
done=true;
throw;
}
}
~thread_pool()
{
done=true;
}
template<typename FunctionType>
void submit(FunctionType f)
{
work_queue.push(std::function<void()>(f));
}
};
template<typename T>
class threadsafe_queue
{
private:
mutable std::mutex mut;
std::queue<T> data_queue;
std::condition_variable data_cond;
public:
threadsafe_queue()
{}
void push(T new_value)
{
std::lock_guard<std::mutex> lk(mut);
data_queue.push(std::move(new_value));
data_cond.notify_one();
}
void wait_and_pop(T& value)
{
std::unique_lock<std::mutex> lk(mut);
data_cond.wait(lk, [this]{return !data_queue.empty(); });
value = std::move(data_queue.front());
data_queue.pop();
}
std::shared_ptr<T> wait_and_pop()
{
std::unique_lock<std::mutex> lk(mut);
data_cond.wait(lk, [this]{return !data_queue.empty(); });
std::shared_ptr<T> res(
std::make_shared<T>(std::move(data_queue.front())));
data_queue.pop();
return res;
}
bool try_pop(T& value)
{
std::lock_guard<std::mutex> lk(mut);
if (data_queue.empty())
return false;
value = std::move(data_queue.front());
data_queue.pop();
}
std::shared_ptr<T> try_pop()
{
std::lock_guard<std::mutex> lk(mut);
if (data_queue.empty())
return std::shared_ptr<T>();
std::shared_ptr<T> res(
std::make_shared<T>(std::move(data_queue.front())));
data_queue.pop();
return res;
}
bool empty() const
{
std::lock_guard<std::mutex> lk(mut);
return data_queue.empty();
}
};
main()函数:
std::condition_variable waitForThreads;
std::mutex mut;
std::atomic<unsigned> doneCount = 0;
unsigned threadCount = 4; // sample concurrent thread count that I use for testing
void synchronizeWork()
{
doneCount++;
if (doneCount.load() == threadCount)
{
doneCount = 0;
std::lock_guard<std::mutex> lock(mut);
waitForThreads.notify_one();
}
}
void Task_A()
{
std::cout << "Task A, thread id: " << std::this_thread::get_id() << std::endl;
std::this_thread::sleep_for(std::chrono::milliseconds(3000));
synchronizeWork();
}
int main()
{
unsigned const thread_count = std::thread::hardware_concurrency();
thread_pool threadPool;
for (int i = 0; i < 1000; ++i)
{
for (unsigned j = 0; j < thread_count; j++)
threadPool.submit(Task_A);
// Below is my way of synchronizing the tasks
{
std::unique_lock<std::mutex> lock(mut);
waitForThreads.wait(lock);
}
}
答案 0 :(得分:0)
我不熟悉你正在使用的线程池类。
不使用这样的类,通常的方法是这样的:
std::cout << "Spawning 3 threads...\n";
std::thread t1 (pause_thread,1);
std::thread t2 (pause_thread,2);
std::thread t3 (pause_thread,3);
std::cout << "Done spawning threads. Now waiting for them to join:\n";
t1.join();
t2.join();
t3.join();
std::cout << "All threads joined!\n";
我认为任何体面的线程池类都可以让你做同样的事情,更简单的说,就是给你一个阻止方法直到所有的线程都完成了。我建议你仔细查看文档。