C++ 多线程 - 条件变量以错误的方式提供数据

Question

我有这个多线程项目，我应该在其中创建酒店的模拟。

我有这个接待员结构，他们的工作是不断地寻找是否有空闲房间（在 check_free_rooms 方法中完成，accommodate_guests 是他的线程方法）。如果他找到 number_to_check_in，is_a_room_ready 将设置为 true，并且通知等待房间的访客线程之一。

每位客人都有一个字段，该字段是对接待员的引用（酒店只有一个），客人正在等待接待员的条件变量receptionist.cv以[=的条件通知客人20=] 成为现实。然后，如果我没理解错的话，随机的一位客人应该得到一个房间，其他客人应该耐心等待接待员的另一个通知。

#include <iostream>
#include <vector>
#include <thread>
#include <mutex>
#include <chrono>
#include <algorithm>
#include <experimental/random>
#include <atomic>
#include <ncurses.h>
#include <condition_variable>
#include <memory>

std::mutex mx_writing;

struct Guest;

struct Room
{
    Room() {}
    int id;
    int guest_id;
    std::atomic<bool> is_ready_for_guest{true};

    void guest_arrives(int guest_id)
    {
        this->guest_id = guest_id;
        this->is_ready_for_guest = false;
    }
    void guest_leaves(int guest_id)
    {
        this->guest_id = -1;
    }
};

struct Receptionist
{
    Receptionist(std::vector<Room> &rooms) : rooms(rooms) {}
    std::vector<Room> &rooms;

    std::mutex mx;
    std::condition_variable cv;
    std::atomic<bool> is_a_room_ready{false};
    int number_to_check_in = 0;

    void check_free_rooms()
    {
        std::unique_lock<std::mutex> lock_receptionist(mx);
        do
        {
            this->number_to_check_in = std::experimental::randint(0, (int)rooms.size() - 1); //find an empty room
        } while (!rooms[this->number_to_check_in].is_ready_for_guest);
        is_a_room_ready = true;
        cv.notify_one();
    }

    void accommodate_guests()
    {
        while (true)
        {
            check_free_rooms();
            std::this_thread::sleep_for(std::chrono::milliseconds(100));
        }
    }
};

struct Guest
{
    Guest(int id, Receptionist &receptionist, Coffee_machine &coffee_machine,
          Swimming_pool &swimming_pool) : id(id), receptionist(receptionist),
                                          coffee_machine(coffee_machine), swimming_pool(swimming_pool) {}
    int id;
    int room_id;

    Receptionist &receptionist;
    Coffee_machine &coffee_machine;
    Swimming_pool &swimming_pool;

    void check_in()
    {
        std::unique_lock<std::mutex> lock_receptionist(receptionist.mx);
        while (!receptionist.is_a_room_ready)
        {
            receptionist.cv.wait(lock_receptionist);
        }
        receptionist.is_a_room_ready = false;
        this->room_id = receptionist.number_to_check_in;           //assign room to guest
        receptionist.rooms[this->room_id].guest_arrives(this->id); //assign guest to room && room becomes occupied
        {
            std::lock_guard<std::mutex> writing_lock(mx_writing);
            std::cout << "Guest " << this->id << " accomodated in room " << this->room_id << std::endl;
        }
        std::this_thread::sleep_for(std::chrono::milliseconds(20));
    }

    void have_holiday()
    {
        check_in();
        std::this_thread::sleep_for(std::chrono::milliseconds(std::experimental::randint(500, 700)));
    }
};

int main()
{
    std::vector<Room> rooms(10);
    for (int i = 0; i < 10; i++)
    {
        rooms[i].id = i;
    }

    Receptionist receptionist(rooms);

    std::vector<Guest> guests;
    for (int i = 0; i < 15; i++)
    {
        guests.emplace_back(Guest(i, receptionist, coffee_machine, swimming_pool));
    }

    std::vector<std::thread> threadList;

    threadList.emplace_back(std::thread(&Receptionist::accommodate_guests, std::ref(receptionist)));
    for (Guest guest : guests)
    {
        threadList.emplace_back(std::thread(&Guest::have_holiday, std::ref(guest)));
    }

    for (std::thread &t : threadList)
    {
        t.join();
    }

    return 0;
}

然而，客人拿到房间后终端出现的线路与我想象的不太一样。客人 ID 的范围从 0 到 14，房间 ID 的范围从 0 到 9，但在输出中只有房间似乎或多或少没问题。我不知道为什么来宾 ID 是一个随机的大整数，而不是我在对象构造时分配的那些。

我在多线程和条件变量方面不是很有经验，并且几乎不知道如何解决这个问题，因为我想了很多却一无所获。如果有任何帮助，我将不胜感激。

一些示例输出：

Guest 5 accommodated in room 32657
Guest -624431120 accommodated in room 7
Guest -624431120 accommodated in room 9
Guest -624431120 accommodated in room 8
Guest -624431120 accommodated in room 5
Guest -624431120 accommodated in room 4
Guest -624431120 accommodated in room 3
Guest -624431120 accommodated in room 2
Guest -624431120 accommodated in room 0
Guest -624431120 accommodated in room 6
^C

或

Guest 4 accommodated in room 32539
Guest -497561616 accommodated in room 1
Guest -497561616 accommodated in room 9
Guest -497561616 accommodated in room 5
Guest -497561616 accommodated in room 8
Guest -497561616 accommodated in room 6
Guest -497561616 accommodated in room 7
Guest -497561616 accommodated in room 3
Guest -497561616 accommodated in room 4
Guest -497561616 accommodated in room 0
^C

或

Guest 4 accommodated in room 32746
Guest -1510756368 accommodated in room 8
Guest -1510756368 accommodated in room 1
Guest -1510756368 accommodated in room 7
Guest -1510756368 accommodated in room 4
Guest -1510756368 accommodated in room 2
Guest -1510756368 accommodated in room 5
Guest -1510756368 accommodated in room 9
Guest -1510756368 accommodated in room 6
Guest -1510756368 accommodated in room 3
^C

Answer 1

替换：

for (Guest guest : guests) {
    threadList.emplace_back(std::thread(&Guest::have_holiday, std::ref(guest)));
}

有：

for (Guest &guest : guests) {
    threadList.emplace_back(std::thread(&Guest::have_holiday, std::ref(guest)));
}

在您的代码中，您在每次迭代中都创建了 guest 的副本，而不是使用现有的副本。
注意 Guest guest [copy] vs Guest &guest [reference].
Read about references in C++.

另一个修正：

for (int i = 0; i < 15; i++) {
    guests.emplace_back(Guest(i, receptionist, coffee_machine, swimming_pool));
}

在本节中，您将在每次迭代中创建两次来宾对象。在 emplace_back 中，您可以只传递构造函数的参数而无需创建副本：

for (int i = 0; i < 15; i++) {
    guests.emplace_back(i, receptionist, coffee_machine, swimming_pool);
}

C++ 多线程 - 条件变量以错误的方式提供数据

C++ multithreading - condition variable supplying data in a wrong way

c++

multithreading

mutex

condition-variable

c++11