gpt4 book ai didi

c++ - 使用 condition_variable::notify_all 通知多个线程

转载 作者:搜寻专家 更新时间:2023-10-31 02:07:53 30 4
gpt4 key购买 nike

我一直在尝试编写餐饮哲学家的代码,以此作为更好地使用多线程编程的一种方式。在我的代码中,我有一个 condition_variable 停止线程,直到创建所有线程。但是,似乎当我调用 condition_variable::notify_all 通知所有线程都已创建并开始“进食”时,只有一个线程被通知。例如:

我有一个具有这些成员变量的 Philosophers 类:

static std::condition_variable start;
static std::mutex start_mutex;

还有这些成员函数。

static void start_eating() {
start.notify_all();
}

void dine() {
signal(SIGINT, ctrl_c_catch);

std::unique_lock lk{ start_mutex };
start.wait(lk);

std::cout << id << "started\n";

// see end for complete class...

每个线程都等待 condition_variable 开始,直到我调用 start_eating() 时才会继续。问题是当我调用 start.notify_all(); 时,只有一个线程得到通知并继续。但是,当我更改代码以在等待后解锁互斥锁时,一切运行正常(所有线程继续):

    std::unique_lock lk{ start_mutex };
start.wait(lk);
lk.unlock();

我不明白这里发生了什么。为什么我需要解锁互斥量?

完整代码:

#include <chrono>
#include <mutex>
#include <vector>
#include <thread>
#include <condition_variable>
#include <atomic>
#include <signal.h>
#include <iostream>
#include <shared_mutex>
#include <ctime>


namespace clk = std::chrono;

const auto EAT_SLEEP_TIME = clk::milliseconds{1}; // 5 seconds
const auto NUM_SEATS = 5U;

using Fork = std::mutex; // is the fork being used or not

std::mutex cout_mutex;

void ctrl_c_catch(int dummy);

class Philosopher {
Fork& left;
Fork& right;
unsigned id;
unsigned times_eaten;

static std::condition_variable start;
static std::mutex start_mutex;

static std::atomic_bool end;

public:
Philosopher(Fork& l, Fork& r, unsigned i) : left{ l }, right{ r }, id{ i }, times_eaten{} {}

static void start_eating() {
start.notify_all();
}

static void stop_eating() {
end = true;
}

void dine() {
signal(SIGINT, ctrl_c_catch);

std::unique_lock lk{ start_mutex };
start.wait(lk);
// lk.unlock(); // uncommenting this fixes the issue

std::cout << id << " started\n";

while (!end) {
if (&right < &left) {
right.lock();
left.lock();
} else {
left.lock();
right.lock();
}

cout_mutex.lock();
std::clog << id << " got both forks, eating\n";
cout_mutex.unlock();

++times_eaten;

std::this_thread::sleep_for(EAT_SLEEP_TIME * (rand() % 50));

right.unlock();
left.unlock();

std::this_thread::sleep_for(EAT_SLEEP_TIME * (rand() % 50));
}

cout_mutex.lock();
std::cout << id << " stopped, terminating thread. Eaten " << times_eaten << "\n";
cout_mutex.unlock();

delete this;
}


};

std::atomic_bool Philosopher::end = false;
std::condition_variable Philosopher::start{};
std::mutex Philosopher::start_mutex{};

template <size_t N, typename T = unsigned>
constexpr std::array<T, N> range(T b = 0, T s = 1) {
std::array<T, N> ret{};

for (auto& i : ret) {
i = b;
b += s;
}

return ret;
}

void ctrl_c_catch(int dummy) {
std::cout << "Caught ctrl-c or stop\nStoping Philosophers\n";
Philosopher::stop_eating();
std::this_thread::sleep_for(clk::seconds{5});
exit(0);
}

int main() {
srand(time(NULL));

signal(SIGINT, ctrl_c_catch);

std::vector<Fork> forks{ NUM_SEATS }; // 5 forks
std::vector<std::thread> phil; // vector of philosophers

for (unsigned i : range<NUM_SEATS - 1>()) {
auto p = new Philosopher{forks[i], forks[i + 1], i};
phil.emplace_back(&Philosopher::dine, p);
}
auto p = new Philosopher{forks[NUM_SEATS - 1], forks[0], NUM_SEATS - 1};
phil.emplace_back(&Philosopher::dine, p);

std::clog << "Waiting for 5 seconds\n";
std::this_thread::sleep_for(clk::seconds{10});

std::clog << "Starting Philosophers\n Type 'stop' to stop\n";
Philosopher::start_eating();

for (auto& t : phil)
t.detach();

std::this_thread::sleep_for(clk::seconds{15});
ctrl_c_catch(0);

std::string dummy;
std::cin >> dummy;

if (dummy == "stop")
ctrl_c_catch(0);

return 0;
}

最佳答案

如解释here ,调用std::condition_variable::wait释放锁,等待,唤醒后重新获取锁。所以你需要手动解锁它(或者使用 RAII 自动解锁)以允许其他线程锁定它。 C++ 中的条件变量与非阻塞监视器具有相似的语义,因此您可以阅读它以获得更好的直观理解。此外,由于虚假解锁无法避免,您应该使用该函数的另一个版本,即使用谓词的版本(更多信息在上面的链接中)。

关于c++ - 使用 condition_variable::notify_all 通知多个线程,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48271230/

30 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com