Java线程join()混淆

Question

我试图按照这个例子来理解 join() 方法:

class PrintDemo {

   public void printCount() {

      try {

         for(int i = 5; i > 0; i--) {
            System.out.println("Counter   ---   "  + i );
         }
      } catch (Exception e) {
         System.out.println("Thread  interrupted.");
      }
   }
}

class ThreadDemo extends Thread {
   private Thread t;
   private String threadName;
   PrintDemo  PD;

   ThreadDemo(String name,  PrintDemo pd) {
      threadName = name;
      PD = pd;
   }

   public void run() {

      synchronized(PD) {
         PD.printCount();
      }
      System.out.println("Thread " +  threadName + " exiting.");
   }

   public void start () {
      System.out.println("Starting " +  threadName );

      if (t == null) {
         t = new Thread (this, threadName);
         t.start ();
      }
   }
}

public class TestThread {

   public static void main(String args[]) {
      PrintDemo PD = new PrintDemo();

      ThreadDemo T1 = new ThreadDemo("Thread - 1 ", PD);
      ThreadDemo T2 = new ThreadDemo("Thread - 2 ", PD);

      T1.start();
      T2.start();

      // wait for threads to end
      try {
         T1.join();
         T2.join();
      } catch (Exception e) {
         System.out.println("Interrupted");
      }
   }
}

我了解到 T1.join() 例如让主线程等待 T1 完成以继续其流程。我在这里正确吗？

所以，我修改了如下代码:

      System.out.println("1");
      T1.start();
      System.out.println("2");
      T2.start();

      // wait for threads to end
      try {
         System.out.println("3");
         T1.join();
         System.out.println("4");
         T2.join();
         System.out.println("5");
      } catch (Exception e) {
         System.out.println("Interrupted");
      }

为了尝试并遵循完整的流程。

这是我每次得到的:

1
Starting Thread - 1 
2
Starting Thread - 2 
3
4
5
Counter   ---   5
Counter   ---   4
Counter   ---   3
Counter   ---   2
Counter   ---   1
Thread Thread - 1  exiting.
Counter   ---   5
Counter   ---   4
Counter   ---   3
Counter   ---   2
Counter   ---   1
Thread Thread - 2  exiting.

我无法解释...

为什么计数器打印在“5”字符之后？？？ 我可能不太了解加入问题...

Answer 1

您启动一个线程 (A)，该线程 (A) 启动另一个线程 (B)。

您等待线程 A 完成 - 这几乎是立即完成的，因为它除了启动线程 B 之外几乎不做任何事情 - 但线程 B 仍在独立运行。