How to stop an std::thread from running, without terminating the program

问题

I am trying to learn std::threads from C++11 to make a threading system.

I was wondering if there is a way to stop a thread from running (Not sleeping, but really destructing the thread or so to speak) without terminating the whole program.

I know std::join exists, but that forces a thread to wait till all threads return.

Is there another way to handle this? (For example for making a ThreadPool class without having to block a thread?)

回答1:

The C++ std::thread class is really just a minimal interface layered on top of some more complete implementation-defined threading package. As such, it only defines a tiny amount of functionality -- creating new threads, detach, and join. That's pretty much it -- there's no standard way of managing, scheduling, stopping/starting/killing threads or doing much else.

There is a native_handle method that returns an implementation-defined type which can probably be used to do what you want, depending on the implementation.

回答2:

Maybe this thread_pool will help you:

#include <boost/thread.hpp>
#include <boost/phoenix.hpp>
#include <boost/optional.hpp>

using namespace boost;
using namespace boost::phoenix::arg_names;

boost::atomic_size_t counter(0ul);

class thread_pool
{
  private:
      mutex mx;
      condition_variable cv;

      typedef function<void()> job_t;
      std::deque<job_t> _queue;

      thread_group pool;

      boost::atomic_bool shutdown;
      static void worker_thread(thread_pool& q)
      {
          while (optional<job_t> job = q.dequeue())
              (*job)();
      }

  public:
      thread_pool() : shutdown(false) {
          for (unsigned i = 0; i < boost::thread::hardware_concurrency(); ++i)
              pool.create_thread(bind(worker_thread, ref(*this)));
      }

      void enqueue(job_t job) 
      {
          lock_guard<mutex> lk(mx);
          _queue.push_back(job);

          cv.notify_one();
      }

      optional<job_t> dequeue() 
      {
          unique_lock<mutex> lk(mx);
          namespace phx = boost::phoenix;

          cv.wait(lk, phx::ref(shutdown) || !phx::empty(phx::ref(_queue)));

          if (_queue.empty())
              return none;

          job_t job = _queue.front();
          _queue.pop_front();

          return job;
      }

      ~thread_pool()
      {
          shutdown = true;
          {
              lock_guard<mutex> lk(mx);
              cv.notify_all();
          }

          pool.join_all();
      }
};

Example of use: live On Coliru

回答3:

There is no way to non-cooperatively stop a thread from running with standard C++. That doesn't mean it's impossible, but you may need to go back to your systems native handle.

For a standard conforming way you can use synchronization primitives (e.g. an std::atomic<bool>) to set a kill flag from the outside and read it within the thread. But it still has to be the thread that finishes on its own.

来源：https://stackoverflow.com/questions/23117354/how-to-stop-an-stdthread-from-running-without-terminating-the-program

标签

c++

multithreading

c++11

stdthread