多线程示例输出说明

Question

请任何人向我解释一下下面代码的输出，我很困惑线程是如何执行连接命令的，而不让主线程打印Hellos语句。

    // example for thread::join
    #include <iostream>       // std::cout
    #include <thread>         // std::thread, std::this_thread::sleep_for
    #include <chrono>         // std::chrono::seconds

    void pause_thread(int n) 
    {
      std::this_thread::sleep_for (std::chrono::seconds(n));
      std::cout << "pause of " << n << " seconds ended\n";
    }

    int main() 
    {
      std::cout << "Spawning 3 threads...\n";
      std::thread t1 (pause_thread,10);
      std::thread t2 (pause_thread,5);
      std::thread t3 (pause_thread,3);
      std::cout << "Done spawning threads. Now waiting for them to join:\n";
      t1.join();
      std::cout << "Hello 1!\n";
      t2.join();
      std::cout << "Hello 2!\n";
      t3.join();
      std::cout << "Hello 3!\n";
      std::cout << "All threads joined!\n";

      return 0;
    }

产出：

    *Spawning 3 threads...

    Done spawning threads. Now waiting for them to join:

    pause of 3 seconds ended

    pause of 5 seconds ended

    pause of 10 seconds ended

    Hello 1!

    Hello 2!

    Hello 3!

    All threads joined!*

非常感谢。

Answer 1

join()阻塞直到线程完成。如果线程在调用join()之前完成，则随后的join()将立即返回。因此，当您将多个join()语句放在一个序列中时，您将在那里阻塞，直到所有线程完成为止，无论它们实际执行的顺序是什么。

请注意，join()阻止您从其中调用它的线程，而不是您调用它的线程。也就是说，在代码片段中，main()线程将等待，但t1、t2和t3将一直持续到它们完成。