java - 不稳定的 StampedLock.unlock(long) 行为？

Question

我正面临关于 StampedLock 的奇怪行为.以下是主要有问题的代码行:

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());

奇怪的行为是关于解锁如何“容忍”错误的读取标记。你觉得正确吗？

---

完整代码供引用:

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

Answer 1

简答:

向戳记添加两个是修改其中不需要在读取模式锁中验证的部分。

长答案:

邮票包含两条信息:州序号和有多少读者。状态号存储在标记的前 57 位中，阅读器计数存储在后 7 位中。因此，当您将 2 添加到邮票时，您会将读者计数从 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，因此仅验证状态编号而忽略读取器计数。这是有道理的，因为读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，状态编号为 4，读取器计数为 1。第二个读取标记从同一个 StampedLock 获取，状态编号为 4，读取器计数of 2. 请注意，邮票的状态编号是相同的，因为 StampedLock 的状态在邮票的获取之间没有改变。第一个读取标记用于解锁。第一个图章 (4) 的州编号与 StampedLock (4) 的州编号相匹配，所以没问题。第一个标记 (1) 的读取器计数与 StampedLock (2) 的读取器计数不匹配，但这无关紧要，因为读锁应该能够以任何顺序解锁。至此解锁成功。

请注意 StampedLocks were designed to be high-performing read/write locks对于内部实用程序，无法承受恶意编码，因此它在其预期范围内运行。我确实认为 unlock() 的 Javadoc虽然具有误导性。