java多线程Hello World

Question

使用两个线程，您应该打印“helloworld helloworld helloworld helloworld helloworld”

在两个线程中，一个线程应该打印“Hello:和另一个线程“World”

我可以用不同的类来实现这一点，比如hello和world各一个类 我也可以用一个内部类来实现它

有没有这样一种方法：只有一个主类而没有内部类

Answer 1

这看起来怎么样？没有线程对特定的单词有特定的责任，但是通过使用两个Atomic可以很容易地确保线程处于锁定状态

这个算法并不依赖于只有两个线程——正如你所看到的，它仍然适用于任意数量的线程，在本例中是42个。只要2个，甚至1个，它仍然可以正常工作

public class HelloWorld implements Runnable {
  // The words.
  private final String[] words;
  // Which word to print next.
  private final AtomicInteger whichWord;
  // Cycles remaining.
  private final AtomicInteger cycles;

  private HelloWorld(String[] words, AtomicInteger whichWord, AtomicInteger cycles) {
    // The words to print.
    this.words = words;
    // The Atomic holding the next word to print.
    this.whichWord = whichWord;
    // How many times around we've gone.
    this.cycles = cycles;
  }

  @Override
  public void run() {
    // Until cycles are complete.
    while ( cycles.get() > 0 ) {
      // Must transit from this word
      int thisWord = whichWord.get();
      // to the next word.
      int nextWord = thisWord + 1;
      // Are we cycling?
      boolean cycled = false;
      if ( nextWord >= words.length ) {
        // We cycled!
        cycled = true;
        // Back to zero.
        nextWord = 0;
      }
      // Grab hold of System.out to ensure no race there either.
      synchronized ( System.out ) {
        // Atomically step the word number - must still be at thisWord for the step calculations to still be correct.
        if ( whichWord.compareAndSet(thisWord, nextWord)) {
          // Success!! We are the priveliged one!
          System.out.print(words[thisWord]);
          // Count the cycles.
          if ( cycled ) {
            // Just step it down.
            cycles.decrementAndGet();
          }
        }
      }
    }
  }

  public static void test() throws InterruptedException {
    // The words to print.
    String [] words = {"Hello ", "world. "};
    // Which word to print next (start at 0 obviously).
    AtomicInteger whichWord = new AtomicInteger(0);
    // How many cycles to print - 6 as specified.
    AtomicInteger cycles = new AtomicInteger(6);
    // My threads - as many as I like.
    Thread [] threads = new Thread[/*2*/42];
    for ( int i = 0; i < threads.length; i++ ) {
      // Make each thread.
      threads[i] = new Thread(new HelloWorld(words, whichWord, cycles));
      // Start it.
      threads[i].start();
    }
    // Wait for them to finish.
    for ( int i = 0; i < threads.length; i++ ) {
      // Wait for completion.
      threads[i].join();
    }
  }

  public static void main(String args[]) throws InterruptedException {
    System.out.println("HelloWorld:Test");
    test();
  }

}

Answer 2

Is there a way such that there's only one main class and and no inner class?

当然。你可以把字符串传给你的主类。当然，棘手的部分是协调线程，以便它们实际打印出"HelloWorld"，而不是"WorldHello"或其他排列。线程将并行运行，不保证顺序。这就是线程程序异步运行的全部意义。试图强制输出特定单词会否定使用线程的目的

<；咆哮>；对我来说，这有点像是一个设计拙劣的计算机科学作业。使用线程编写的全部意义在于，它们独立地并行运行。协调通常发生在每个线程从工作队列中提取结果，然后将结果放入结果队列或其他东西时。任何时候，当你有一个线程程序需要这么多的协调时，你都不应该使用线程</咆哮>

但是，由于每个人都反对我之前的答案，可能是因为它不能完美地解决他们的家庭作业问题，我将添加一些逻辑来协调这两个线程，并说出“Hello World…”

这两个线程需要能够锁定某些内容，相互发出信号，并知道何时应该等待或打印。因此，我将添加一个boolean printHello，并锁定传入的公共锁定对象：

public class HelloWorld implements Runnable {

    private static boolean printHello = true;

    private final String toPrint;
    private final boolean iPrintHello;
    private final Object lock;

    public static void main(String[] args) {
        final Object lock = new Object();
        // first thread prints Hello
        new Thread(new HelloWorld("Hello ", true, lock)).start();
        // second one prints World
        new Thread(new HelloWorld("World ", false, lock)).start();
    }

    public HelloWorld(String toPrint, boolean iPrintHello, Object lock) {
        this.toPrint = toPrint;
        this.iPrintHello = iPrintHello;
        this.lock = lock;
    }

    @Override
    public void run() {
        // don't let it run forever
        for (int i = 0; i < 1000 && !Thread.currentThread().isInterrupted(); ) {
            // they need to synchronize on a common, constant object
            synchronized (lock) {
                // am I printing or waiting?
                if (printHello == iPrintHello) {
                    System.out.print(toPrint);
                    // switch the print-hello to the other value
                    printHello = !printHello;
                    // notify the other class that it can run now
                    lock.notify();
                    i++;
                } else {
                    try {
                        // wait until we get notified that we can print
                        lock.wait();
                    } catch (InterruptedException e) {
                        // if thread is interrupted, _immediately_ re-interrupt it
                        Thread.currentThread().interrupt();
                        return;
                    }
                }
            }
        }
    }
}

Answer 3

1. 如果你想让T1打印“Hello”，T2打印“World”，你的预期结果是“Hello World Hello World Hello World Hello World Hello World”

   T1必须先启动，T2由T1调用，否则可以将“World Hello World”作为输出

   我建议定制读写器或生产者/消费者结构，使用notify()或notifyAll()方法唤醒其他线程

2. 如果您不关心输出的格式，请使用Runnable接口和首选实现