解释的线程-每次im运行代码时都有不同的答案

Question

我试图创建10个线程，每个线程运行100.000次。因此，我做了一个循环，循环每一个线程。它应该每次递增静态x+1。

public class CAOS_thread_demo_5 extends Thread
    {
        public static int x = 0;
    
        public static void main(String[] args)
        {
        CAOS_thread_demo_5 humhum1 = new CAOS_thread_demo_5();
        CAOS_thread_demo_5 humhum2 = new CAOS_thread_demo_5();
        CAOS_thread_demo_5 humhum3 = new CAOS_thread_demo_5();
        CAOS_thread_demo_5 humhum4 = new CAOS_thread_demo_5();
        CAOS_thread_demo_5 humhum5 = new CAOS_thread_demo_5();
        CAOS_thread_demo_5 humhum6 = new CAOS_thread_demo_5();
        CAOS_thread_demo_5 humhum7 = new CAOS_thread_demo_5();
        CAOS_thread_demo_5 humhum8 = new CAOS_thread_demo_5();
        CAOS_thread_demo_5 humhum9 = new CAOS_thread_demo_5();
        CAOS_thread_demo_5 humhum10 = new CAOS_thread_demo_5();


        humhum1.start();
        humhum2.start();
        humhum3.start();
        humhum4.start();
        humhum5.start();
        humhum6.start();
        humhum7.start();
        humhum8.start();
        humhum9.start();
        humhum10.start();
        try
        {
            humhum1.join();
            humhum2.join();
            humhum3.join();
            humhum4.join();
            humhum5.join();
            humhum6.join();
            humhum7.join();
            humhum8.join();
            humhum9.join();
            humhum10.join();
        }
        catch(Exception e)
        {}

        System.out.println();
        System.out.println("*** the end ***");
        System.out.println("#### counter: " + x);
    }
    

    public void run()
    {
        for (int i = 0; i < 100000; i++) {
            x++;
        }
    }
}

Answer 1

首先，x++不是原子的：它是对增量值的读取，然后是对增量值的写入。其他线程可以在这些操作之间交错。

其次，您不能保证x以前更新的可见性：您不能保证看到其他线程编写的更新，因此线程可能只是践踏其他线程编写的值。您可以通过使xvolatile来解决这个问题，但这并不能解决原子性问题。

在x++周围使用同步块（使用公共监视器，例如CAOS_Thread_Demo_5.Class)；或者使用AtomicInteger，它保证了可见性，并提供了原子增量方法(GetAndIncrement()、IncrementAndGet())。