c ++条件变量通知未按预期工作

Question

我正在尝试在以前的worker_thread 中的工作开始 后立即启动新线程，但可能结束或没有结束。我已经用时间延迟替换了开始和结束的工作。我的代码是：

#include <iostream>
#include <string>
#include <mutex>
#include <condition_variable>
#include <future>
#include <atomic>
#include <chrono>
#include <thread>

std::mutex m;
std::condition_variable cv;
bool started = false;

void worker_thread()
{
    std::unique_lock<std::mutex> lk(m);

    static std::atomic<int> count(1);
    std::this_thread::sleep_for(std::chrono::milliseconds{(count % 5) * 100});
    std::cerr << "Start Worker thread: " << count << "\n";

    started = true;
    lk.unlock();
    cv.notify_one();

    std::this_thread::sleep_for(std::chrono::milliseconds{3000});
    std::cerr << "Exit Worker thread: " << count << "\n";
    ++count;
}

int main()
{
    while(1) {
        std::async(std::launch::async, worker_thread);
        std::unique_lock<std::mutex> lk(m);
        cv.wait(lk, []{return started;});
        started = false;
    }
}

输出如下所示：

Start Worker thread: 1
Exit Worker thread: 1
Start Worker thread: 2
Exit Worker thread: 2
Start Worker thread: 3
Exit Worker thread: 3
Start Worker thread: 4
Exit Worker thread: 4
Start Worker thread: 5
Exit Worker thread: 5

这不是我想要的行为。我想要的是（不完全是）这样的：

Start Worker thread: 1
Start Worker thread: 2
Start Worker thread: 3
Start Worker thread: 4
Exit Worker thread: 1
Exit Worker thread: 3
Exit Worker thread: 4
Exit Worker thread: 2
Start Worker thread: 5
Exit Worker thread: 5

目前下一个线程只有在前一个线程的工作完成时才会启动。但我想在前一个线程开始工作后立即启动下一个线程，而不是等待它结束，只等待开始。

Answer 1

std::async 返回一个std::future 持有函数执行的结果。在您的情况下，它是一个被立即销毁的临时对象。 std::future 的文档说：

these actions will not block for the shared state to become ready, except that it may block if all of the following are true:

✔ 共享状态是通过调用 std::async 创建的

✔ 共享状态尚未准备好

✔ 这是对共享状态的最后一次引用

所有这些都是正确的，因此 future 的销毁将阻塞，直到工作函数完成执行。

你可以创建分离线程来避免这个问题：

std::thread(worker_thread).detach();