java - 不稳定StampedLock.unlock(长时间)行为？

我面临有关StampedLock的奇怪行为。以下是主要的有问题的代码行:

StampedLock lock = new StampedLock();
long stamp1 = lock.readLock();
System.out.printf("Read lock count: %d%n", lock.getReadLockCount());
lock.unlock(stamp1 + 2);
System.out.printf("Read lock count: %d%n", lock.getReadLockCount());

奇怪的行为是关于如何解锁“容忍”错误的读取标记。您觉得对吗？

供参考，下面是完整的代码:

public class StampedLockExample {
  static StampedLock lock = new StampedLock();

  static void println(String message, Object... args) {
    System.out.printf(message, args);
    System.out.println();
  }

  static void printReadLockCount() {
    println("Lock count=%d", lock.getReadLockCount());
  }

  static long tryReadLock() {
    long stamp = lock.tryReadLock();
    println("Gets read lock (%d)", stamp);
    printReadLockCount();
    return stamp;
  }

  static long tryWriteLock() {
    long stamp = lock.tryWriteLock();
    println("Gets write lock (%d)", stamp);
    return stamp;
  }

  static long tryConvertToReadLock(long stamp) {
    long newOne = lock.tryConvertToReadLock(stamp);
    println("Gets read lock (%d -> %d)", stamp, newOne);
    printReadLockCount();
    return newOne;
  }

  static void tryUnlock(long stamp) {
    try {
      lock.unlock(stamp);
      println("Unlock (%d) successfully", stamp);
    } catch (IllegalMonitorStateException e) {
      println("Unlock (%d) failed", stamp);
    }
    printReadLockCount();
  }

  public static void main(String[] args) {
    println("%n--- Gets two read locks ---");
    long stamp1 = tryReadLock();
    long stamp2 = tryReadLock();
    long min = Math.min(stamp1, stamp2);
    long max = Math.max(stamp1, stamp2);

    println("%n--- Tries unlock (-1 / +2 / +4) ---");
    tryUnlock(min - 1);
    tryUnlock(max + 2);
    tryUnlock(max + 4);

    println("%n--- Gets write lock ---");
    long stamp3 = tryWriteLock();

    println("%n--- Tries unlock (-1 / +1) ---");
    tryUnlock(stamp3 - 1);
    tryUnlock(stamp3 + 1);

    println("%n--- Tries write > read conversion ---");
    long stamp4 = tryConvertToReadLock(stamp3);

    println("%n--- Tries unlock last write stamp (-1 / 0 / +1) ---");
    tryUnlock(stamp3 - 1);
    tryUnlock(stamp3);
    tryUnlock(stamp3 + 1);

    println("%n--- Tries unlock (-1 / +1) ---");
    tryUnlock(stamp4 - 1);
    tryUnlock(stamp4 + 1);
  }
}

输出:

--- Gets two read locks ---
Gets read lock (257)
Lock count=1
Gets read lock (258)
Lock count=2

--- Tries unlock (-1 / +2 / +4) ---
Unlock (256) failed
Lock count=2
Unlock (260) successfully
Lock count=1
Unlock (262) successfully
Lock count=0

--- Gets write lock ---
Gets write lock (384)

--- Tries unlock (-1 / +1) ---
Unlock (383) failed
Lock count=0
Unlock (385) failed
Lock count=0

--- Tries write > read conversion ---
Gets read lock (384 -> 513)
Lock count=1

--- Tries unlock last write stamp (-1 / 0 / +1) ---
Unlock (383) failed
Lock count=1
Unlock (384) failed
Lock count=1
Unlock (385) failed
Lock count=1

--- Tries unlock (-1 / +1) ---
Unlock (512) failed
Lock count=1
Unlock (514) successfully
Lock count=0

最佳答案

简短答案:

在戳记上加上两个即表示修改了它的一部分，不需要在读模式锁中进行验证。

长答案:

该图章包含两个信息:状态序列号和有多少个读者。状态编号存储在标记的前57位中，而读取器计数存储在后7位中。因此，当您在图章上添加2时，您会将阅读器的数量从1更改为3，并保持状态编号不变。由于StampedLock仅在读取模式下获取，因此仅验证状态号，而忽略读取器计数。这是有道理的，因为读取锁应该能够以任何顺序解锁。

例如:从现有的StampedLock获取读取标记，其状态编号为4，读取器计数为1。从同一StampedLock获取第二读取标记，其状态编号为4，读取器计数为2。请注意，图章的状态编号是相同的，因为在获取图章之间，StampedLock的状态没有改变。第一个读取的标记用于解锁。第一个图章(4)的状态号与StampedLock(4)的状态号相匹配，这样就可以了。第一个图章(1)的读取器计数与StampedLock(2)的读取器计数不匹配，但是没关系，因为读取锁应该能够以任何顺序解锁。因此解锁成功。

请注意，StampedLocks were designed to be high-performing read/write locks用于内部实用程序，不能承受恶意代码，因此它在预期的范围内运行。我确实认为unlock()的Javadoc具有误导性。