以下（ ）封锁序列违反两阶段锁协议。 A: Slock A … Slock B … Xlock C … Unlock A … Unlock B … Unlock C B: Slock A … Slock B … Xlock C … Unlock C … Unlock B … Unlock A C: Slock A … Slock B … Xlock C … Unlock B … Unlock C … Unlock A D: Slock A … Unlock A … Slock B … Xlock C …Unlock B … Unlock C

Slock A … Slock B … Xlock C …… Unlock C … Unlock B … Unlock A违反两段锁协议。

事务T1符合两段锁封锁法的序列是（） A: LOCK S(A)...LOCKS S(B) ...UNLOCK (B) ... UNLOCK (A) B: LOCK S(A) UNLOCK (A) LOCK S(B) UNLOCK (A) C: LOCK S(B) LOCK X(A) UNLOCK (A) LOCK S(C) D: UNLOCK (A) UNLOCK (B) LOCK X(A) LOCK X(B)

已知事务T1的封锁序列为：LOCKS(A)…LOCKS(B)…LOCKX(C) …UNLOCK(B)…UNLOCK(A)…UNLOCK(C) 事务T2的封锁序列为：LOCKS(A)…UNLOCK(A)…LOCKS(B) …LOCKX(C)…UNLOCK(C)…UNLOCK(B) 则遵守两段封锁协议的事务是（　　）

解锁scott用户的正确命令是_______。 A: update user scott account unlock B: alert user scott account unlock C: update user scott unlock D: alert user scott unlock

下面()是使用Application对象时防止竞争的代码?Application. ; //锁定Application对象Application[“counter”]=(int) Application[“counter”]+1;Application. ; //解除对Application对象的锁定 A: lock unlock B: Lock UnLock C: Lock() UnLock() D: UnLock() Lock()

[A] break [C] define[B] elaborate [D] unlock A: break B: elaborate C: define D: unlock

关于Application对象的方法，下面说法正确的是（） A: 只有Lock方法 B: 只有Unlock方法 C: 没有Lock方法和Unlock方法 D: 有Lock方法和Unlock方法

He asked whether he was to lock the car or leave it (unlock) ______.

In order to unlock your own creativity, you need to _________ into the sources of inspiration.