c++ - 从线程中获取 native 句柄？

Question

我正在使用 VS2012，我想在运行的线程中设置线程优先级。目标是用最高优先级状态初始化所有线程。为此，我想为线程获取一个HANDLE。

我在访问对应于 thread 对象的指针时遇到一些问题。

这可能吗？

从调用主线程，指针是有效的，从 C++11 线程它被设置为 CCCCCCCC。可以预见的是，取消引用一些无意义的内存位置会导致崩溃。

下面的代码是显示问题的简化版本。

#include "stdafx.h"
#include <Windows.h>
#include <thread>
#include <mutex>
#include <condition_variable>
#include <iostream>
#include <atomic>
using namespace std;
class threadContainer
    {
    thread* mT;
    condition_variable* con;
    void lockMe()
        {
        mutex m;
        unique_lock<std::mutex> lock(m);
        con->wait(lock);//waits for host thread
        cout << mT << endl;//CCCCCCCC
        auto h = mT->native_handle();//causes a crash
        con->wait(lock);//locks forever
        }
    public:
        void run()
            {
            con = new condition_variable();
            mT = new thread(&threadContainer::lockMe,*this);
            cout << mT << endl; //00326420
            con->notify_one();// Without this line everything locks as expected
            mT->join();
            }
    };
int _tmain(int argc, _TCHAR* argv[])
    {
    threadContainer mContainer;
    mContainer.run();
    return 0;
    }

Answer 1

#include <mutex>
#include <condition_variable>
#include <iostream>
#include <atomic>
#include <thread>

class threadContainer {
   std::thread* mT;
  std::mutex m;
  void lockMe() {
    // wait for mT to be assigned:
    {
      std::unique_lock<std::mutex> lock(m);
    }
    std::cout << "lockMe():" << mT << "\n";
    auto h = mT->native_handle();//causes a crash
    std::cout << "Done lockMe!\n";
  }
  public:
    void run() {
      // release lock only after mT assigned:
      {
        std::unique_lock<std::mutex> lock(m);
        mT = new std::thread( [&](){ this->lockMe(); } );
      }
      std::cout << "run():" << mT << "\n"; //00326420
      mT->join();
    }
};

int main() {
  threadContainer mContainer;
  mContainer.run();
  return 0;
}

试试吧。