An example explanation of the problem that volatile cannot guarantee thread safety in Java

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Today I coded and studied whether Java’s volatile keyword can guarantee thread safety. After practice, volatile cannot guarantee thread safety, it only guarantees data. The visibility will not be cached again. Each thread reads data from the main memory, not from the cache. The code is attached as follows. When synchronized is removed, the result of each thread is It's messy, but the result is correct when you add it.

/**
 * 
 * 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*/
  


 }

}

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