Java에서 휘발성이 스레드 안전성을 보장할 수 없는 문제에 대한 설명 예

/**
 * 
 * 类简要描述
 * 
 * <p>
 * 类详细描述
 * </p>
 * 
 * @author think
 * 
 */

public class VolatileThread implements Runnable {

 private volatile int a = 0;

 @Override
 public void run() {
  // TODO Auto-generated method stub
//  synchronized (this) {
   a = a + 1;
   System.out.println(Thread.currentThread().getName() + ":----" + a);
   try {
    Thread.sleep(100);
    a = a + 2;
   } catch (InterruptedException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
   }

   System.out.println(Thread.currentThread().getName() + ":----" + a);
//  }
 }

}

/**
 * 
 * 类简要描述
 * 
 * <p>
 * 类详细描述
 * </p>
 * 
 * @author think
 * 
 */

public class VolatileMain {

 public static void main(String[] args) {

  VolatileThread s = new VolatileThread();

  Thread t1 = new Thread(s);
  Thread t2 = new Thread(s);
  Thread t3 = new Thread(s);
  Thread t4 = new Thread(s);
  t1.start();
  t2.start();
  t3.start();
  t4.start();
  
  
/*  同步的结果
  Thread-2:----1
  Thread-2:----3
  Thread-0:----4
  Thread-0:----6
  Thread-3:----7
  Thread-3:----9
  Thread-1:----10
  Thread-1:----12*/
  
/*  
  去掉同步的结果
  Thread-0:----1
  Thread-1:----2
  Thread-2:----3
  Thread-3:----4
  Thread-0:----8
  Thread-3:----10
  Thread-1:----10
  Thread-2:----12*/
  


 }

}