c++ - C++多线程--条件变量以错误的方式提供数据。
我有一个多线程项目,我应该创建一个模拟酒店。
我有一个接待员结构,他们的工作是不断地寻找是否有空房间(用 check_free_rooms
方法。accommodate_guests
是他的线程方法)。) 如果他发现一个 number_to_check_in
, is_a_room_ready
设置为 true
并通知其中一个正在等待房间的客人线程。
每个客人都有一个字段,这个字段是对前台的引用(酒店只有一个),客人在等待前台的条件变量是 receptionist.cv
状况的通知 receptionist.is_a_room_ready
成为事实。那么,如果我没有理解错的话,随机一个客人应该会得到一个房间,其他的客人应该耐心地等待接待员的再次通知。
#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,在终端出现的线条和我想象的不太一样。客人id从0到14,房间id从0到9,但输出中似乎只有房间多少还可以。我不知道为什么客人的id是一个随机的大int,而不是我在对象构造时分配的。
我对多线程和条件变量没有那么丰富的经验,几乎不知道如何解决这个问题,因为我想了很多,但一无所获。我非常感谢大家的帮助。
一些例子输出。
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
【回答】:
替换。
for (Guest guest : guests) {
threadList.emplace_back(std::thread(&Guest::have_holiday, std::ref(guest)));
}
替换为:
for (Guest &guest : guests) {
threadList.emplace_back(std::thread(&Guest::have_holiday, std::ref(guest)));
}
在你的代码中,你在每次迭代中都创建了一个 guest 的副本,而不是使用现有的那个。请注意 Guest guest
[复制]与 Guest &guest
参考]。阅读关于C++中的引用.
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,你在每次迭代中都创建了两次客人对象。在这一节中,你在每次迭代中创建了两次客体对象。emplace_back
你可以只传递构造函数的参数而不创建副本。
for (int i = 0; i < 15; i++) {
guests.emplace_back(i, receptionist, coffee_machine, swimming_pool);
}