c++ - std::condition_variable – 通知一次但等待线程被唤醒两次

Question

这是一个简单的 C++ 线程池实现。这是一个修改后的版本，源自 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。只有一个任务被提交到线程池，但我的工作线程被唤醒了两次:

>>./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以获取新任务。函数 FixedThreadPool::Submit() 添加一个新任务到队列并调用 condition.nofity_one() 来唤醒一个工作线程。

但是我想不通工作线程是如何被唤醒两次的。在此测试中，我只提交了一项任务，并且只有一个工作线程。

Answer 1

将评论转化为答案:

condition_variable::wait(lock, pred) 等同于 while(!pred()) wait(lock);。如果 pred() 返回 true，则实际上不会发生等待，调用会立即返回。

您的第一次唤醒来自 notify_one() 调用；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() 立即返回，无需等待。

很明显，您在这里运气不好:如果第二次 wait() 调用发生在 Stop() 调用之前，那么您的工作线程将永远等待(在没有虚假唤醒的情况下)，您的主线程也是如此。

另外，去掉 __THREAD_POOL_H__。把那些双下划线烧到地上。