Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
599 views
in Technique[技术] by (71.8m points)

c++ - What happens when calling the destructor of a thread object that has a condition variable waiting?

I am using a SynchronisedQueue to communicate between threads. I found that destroying the thread object when the attaching thread is waiting on a condition variable would cause the program crash. This can be corrected by calling detach() before the thread destruction. But I am wondering what happens exactly when a thread waiting a conditional variable got terminated. Is there another way to use condition variable to avoid this?

#include <queue>
#include <thread>
#include <mutex>
#include <condition_variable>

template <typename Type> class SynchronisedQueue {
 public:
  void Enqueue(Type const & data) {
    std::unique_lock<std::mutex> lock(mutex_);
    queue_.push(data);
    condition_.notify_one();
  }
  Type Dequeue() {
    std::unique_lock<std::mutex> lock(mutex_);
    while (queue_.empty())
      condition_.wait(lock);
    Type result = queue_.front();
    queue_.pop();
    return result; 
  }
 private:
  std::queue<Type> queue_;
  std::mutex mutex_;
  std::condition_variable condition_; 
};

class Worker {
public:
  Worker(SynchronisedQueue<int> * queue) : queue_(queue) {}
  void operator()() {
    queue_->Dequeue();    // <-- The thread waits here.
  }
private:
  SynchronisedQueue<int> * queue_;
};

int main() {
  auto queue = new SynchronisedQueue<int>();
  Worker worker(queue);
  std::thread worker_thread(worker);
  worker_thread.~thread();  // <-- Crashes the program.
  return 0;
}
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

From the C++11 spec:

30.3.1.3 thread destructor [thread.thread.destr] ~thread();

If joinable(), calls std::terminate(). Otherwise, has no effects.

[ Note: Either implicitly detaching or joining a joinable() thread in its destructor could result in difficult to debug correctness (for detach) or performance (for join) bugs encountered only when an exception is raised. Thus the pro grammer must ensure that the destructor is never executed while the thread is still joinable. — end note ]

So calling a thread destructor without first calling join (to wait for it to finish) or detach is guarenteed to immediately call std::terminate and end the program.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...