Question

这是一个简单的C ++线程池实现。这是一个改编版本，来自https://github.com/progschj/ThreadPool。

#ifndef __THREAD_POOL_H__
#define __THREAD_POOL_H__

#include <vector>
#include <queue>
#include <memory>
#include <thread>
#include <chrono>
#include <mutex>
#include <condition_variable>
#include <future>
#include <functional>
#include <stdexcept>

namespace ThreadPool {

class FixedThreadPool {
 public:
  FixedThreadPool(size_t);

  template<class F, class... Args>
  auto Submit(F&& f, Args&&... args)
      -> std::future<typename std::result_of<F(Args...)>::type>;

  template<class F, class... Args>
  void Execute(F&& f, Args&&... args);

  ~FixedThreadPool();

  void AwaitTermination();

  void Stop();

 private:
  void ThreadWorker();

  // need to keep track of threads so we can join them
  std::vector<std::thread> workers;

  // the task queue
  std::queue< std::function<void()> > tasks;

  // synchronization
  std::mutex worker_mutex;
  std::mutex queue_mutex;
  std::condition_variable condition;

  // stop flag
  bool stop_;

  // thread size
  int thread_size_;
};

// Constructor does nothing. Threads are created when new task submitted.
FixedThreadPool::FixedThreadPool(size_t num_threads): 
    stop_(false),
    thread_size_(num_threads) {}

// Destructor joins all threads
FixedThreadPool::~FixedThreadPool() {
  //std::this_thread::sleep_for(std::chrono::seconds(5));
  for(std::thread &worker: workers) {
    if (worker.joinable()) {
      worker.join();
    }
  }
}

// Thread worker
void FixedThreadPool::ThreadWorker() {
  std::function<void()> task;
  while (1) {
    {
      std::unique_lock<std::mutex> lock(this->queue_mutex);
      this->condition.wait(lock,
                     [this]() { return this->stop_ || !this->tasks.empty(); });
      printf("wakeeeeeened\n");
      if (this->stop_ && this->tasks.empty()) {
        printf("returning ...\n");
        return;
      }
      task = std::move(this->tasks.front());
      this->tasks.pop();
    }
    task();
  }
}

// Add new work item to the pool
template<class F, class... Args>
auto FixedThreadPool::Submit(F&& f, Args&&... args)
    -> std::future<typename std::result_of<F(Args...)>::type >
{
  {
    std::unique_lock<std::mutex> lock(this->worker_mutex);
    if (workers.size() < thread_size_) {
      workers.emplace_back(std::thread(&FixedThreadPool::ThreadWorker, this));
    }
  }

  using return_type = typename std::result_of<F(Args...)>::type;

  auto task = std::make_shared< std::packaged_task<return_type()> >(
      std::bind(std::forward<F>(f), std::forward<Args>(args)...)
  );

  std::future<return_type> res = task->get_future();
  {
    std::unique_lock<std::mutex> lock(queue_mutex);
    if(stop_) {
      throw std::runtime_error("ThreadPool has been shutdown.");
    }
    tasks.emplace([task]() { (*task)(); });
  }
  condition.notify_one();
  return res;
}

// Execute new task without returning std::future object.
template<class F, class... Args>
void FixedThreadPool::Execute(F&& f, Args&&... args) {
  Submit(std::forward<F>(f), std::forward<Args>(args)...);
}

// Blocks and wait for all previously submitted tasks to be completed.
void FixedThreadPool::AwaitTermination() {
  for(std::thread &worker: workers) {
    if (worker.joinable()) {
      worker.join();
    }
  }
}

// Shut down the threadpool. This method does not wait for previously submitted
// tasks to be completed.
void FixedThreadPool::Stop() {
  printf("Stopping ...\n");
  {
    std::unique_lock<std::mutex> lock(queue_mutex);
    stop_ = true;
  }
}


} // namespace ThreadPool

#endif /* __THREAD_POOL_H__ */

和测试main.cpp：

#include <iostream>
#include <vector>
#include <chrono>
#include <exception>

#include "ThreadPool.h"

int main(int argc, char** argv) {
  ThreadPool::FixedThreadPool pool(3);

  pool.Execute([]() {
      std::cout << "hello world" << std::endl;
    }
  );
  pool.Stop();
  pool.AwaitTermination();
  std::cout << "All tasks complted." << std::endl;

  return 0;
}

我在这个测试程序中有一个错误。只有一个任务提交给线程池，但我得到了两次唤醒工作线程：

>>./test 
Stopping ...
wakeeeeeened
hello world
wakeeeeeened
returning ...
All tasks complted.

我认为问题出在FixedThreadPool :: ThreadWorker（）本身。工作线程持续等待条件变量来获取新任务。函数FixedThreadPool :: Submit（）将新任务添加到队列并调用condition.nofity_one（）以唤醒工作线程。

但我无法弄清楚工作线程如何被唤醒两次。我只提交了一个任务，并且在此测试中只有一个工作线程。

Answer 1

将评论转换为答案：

condition_variable::wait(lock, pred)相当于while(!pred()) wait(lock);。如果pred()返回true，则实际上不会发生等待，并且呼叫会立即返回。

你的第一个叫醒来自notify_one()电话;第二个“唤醒”是因为第二个wait()调用恰好在Stop()调用之后执行，因此您的谓词返回true并且wait()会立即返回而无需等待。

很明显你在这里得到了（非）幸运：如果第二个wait()电话发生在Stop()电话会议之前，那么你的工作线程将永远等待（在没有虚假的唤醒），你的主线也将如此。

另外，摆脱__THREAD_POOL_H__。把那些双下划线烧到地上。

std :: condition_variable - 通知一次但等待线程唤醒两次

1 个答案: