为什么我们不需要带StampedLock的volatile?

https://docs.oracle.com/javase/8/docs/api/java/util/concurrent/locks/StampedLock.html

class Point {
   private double x, y;
   private final StampedLock sl = new StampedLock();

   void move(double deltaX, double deltaY) { // an exclusively locked method
     long stamp = sl.writeLock();
     try {
       x += deltaX;
       y += deltaY;
     } finally {
       sl.unlockWrite(stamp);
     }
   }

   double distanceFromOrigin() { // A read-only method
     long stamp = sl.tryOptimisticRead();
     double currentX = x, currentY = y;
     if (!sl.validate(stamp)) {
        stamp = sl.readLock();
        try {
          currentX = x;
          currentY = y;
        } finally {
           sl.unlockRead(stamp);
        }
     }
     return Math.sqrt(currentX * currentX + currentY * currentY);
   }

   void moveIfAtOrigin(double newX, double newY) { // upgrade
     // Could instead start with optimistic, not read mode
     long stamp = sl.readLock();
     try {
       while (x == 0.0 && y == 0.0) {
         long ws = sl.tryConvertToWriteLock(stamp);
         if (ws != 0L) {
           stamp = ws;
           x = newX;
           y = newY;
           break;
         }
         else {
           sl.unlockRead(stamp);
           stamp = sl.writeLock();
         }
       }
     } finally {
       sl.unlock(stamp);
     }
   }
 }

并且假设可以从不同的线程调用类Point的所有方法:

为什么我们不需要将字段x和y声明为volatile?

Point#moveIfAtOrigin 方法在获取后始终会看到x和y字段的最新变化 StampedLock#readLock

当我们打电话时,有没有建立起某种记忆障碍 StampedLock#writeLock , 冲压锁#读锁

有人能指出文件中关于这一点的引用吗?