c++ - 多个线程可以加入同一个boost::thread吗？

Question

pthreads如果多个线程尝试加入同一个线程，则具有未定义的行为:

If multiple threads simultaneously try to join with the same thread, the results are undefined.

boost::thread也是如此秒？文档似乎没有具体说明这一点。

如果它未定义，那么多线程等待一个线程完成的干净方法是什么？

Answer 1

If it is undefined, then what would be a clean way for multiple threads to wait on one thread completing?

干净的方法是让一个线程通知其他线程它已完成。一个 packaged_task 包含一个可以等待的 future，这对我们有帮助。

这是一种方法。我使用了 std::thread 和 std::packaged_task，但您也可以使用 boost 等价物。

#include <thread>
#include <mutex>
#include <future>
#include <vector>
#include <iostream>

void emit(const char* msg) {
    static std::mutex m;
    std::lock_guard<std::mutex> l(m);
    std::cout << msg << std::endl;
    std::cout.flush();
}

int main()
{
    using namespace std;

    auto one_task = std::packaged_task<void()>([]{
        emit("waiting...");
        std::this_thread::sleep_for(std::chrono::microseconds(500));
        emit("wait over!");
    });

    // note: convert future to a shared_future so we can pass it
    // to two subordinate threads simultaneously
    auto one_done = std::shared_future<void>(one_task.get_future());
    auto one = std::thread(std::move(one_task));

    std::vector<std::thread> many;
    many.emplace_back([one_done] {
        one_done.wait();
        // do my thing here
        emit("starting thread 1");
    });

    many.emplace_back([one_done] {
        one_done.wait();
        // do my thing here
        emit("starting thread 2");
    });

    one.join();
    for (auto& t : many) {
        t.join();
    }

    cout << "Hello, World" << endl;
    return 0;
}

预期输出:

waiting...
wait over!
starting thread 2
starting thread 1
Hello, World