c++ - 删除锁定的互斥量

Question

我有一个包含多个资源的程序，需要用它们自己的互斥锁来锁定。

在这个程序中，可能会发生这样的情况，当资源A的互斥锁被锁定时，资源A在另一个线程中被删除。

下面的代码试图重现我试图完成的逻辑:

#include <thread>
#include <mutex>
#include <iostream>
#include <map>
int g_i = 0;
struct Resource
{
    std::mutex* m_mutex;
};

std::map<unsigned int, Resource> myResources;

std::mutex g_i_mutex;  // protects g_i

void shutdown()
{
    std::cout << "shutdown -> myMap.size = : " << myResources.size() << std::endl;
    std::lock_guard<std::mutex> lock(*myResources[1].m_mutex);
    ++g_i;
    std::this_thread::sleep_for(std::chrono::milliseconds(1000));
    delete myResources[1].m_mutex;
    myResources[1].m_mutex = NULL;
    myResources.erase(1);
    std::cout << "shutdown -> myMap.size = : " << myResources.size() << std::endl;
    std::cout << "shutdown : " << g_i << '\n';


}

 void onRecognize()
{
    std::cout << "onRecognize -> myMap.size = : " << myResources.size() << std::endl;
    std::lock_guard<std::mutex> lock(*myResources[1].m_mutex);
    std::cout << "onRecognize -> myMap.size = : " << myResources.size() << std::endl;
    ++g_i;

    std::cout <<  "onRecognize : " << g_i << '\n';


}

int main()
{
    std::cout << __func__ << ": " << g_i << '\n';
    Resource myStruct;
    myStruct.m_mutex = new std::mutex();
    myResources[1] = myStruct;
    std::thread t1(shutdown);
    std::thread t2(onRecognize);

    t1.join();
    std::this_thread::sleep_for(std::chrono::milliseconds(500));
    t2.join();

    std::cout << __func__ << ": " << g_i << '\n';
}

我尝试了这段代码并且它有效。但是我想知道 onRecognize 函数中的 lock_guard 会发生什么，因为互斥量在他被锁定时被删除。所以，我的问题可能是:

在他被锁定在别处时删除互斥锁是否危险？

谢谢

Answer 1

不要在锁定时销毁互斥锁​​。

The behavior is undefined if the mutex is owned by any thread or if any thread terminates while holding any ownership of the mutex.

http://en.cppreference.com/w/cpp/thread/mutex/~mutex