Semaphore释疑

java api中Semaphore(信号量)，用于控制有限资源的并发访问。API也非常好理解，不过有几个需要注意的地方:

1. Semaphore是纯粹的应用级控制“锁”，使用简单的volitale变量作为信号量信息，通过acquire、release、reduce等显式的可以修改此信号量数字。
2. 它并没有维护任何锁，也不是控制reentrant的，它不会维护信号和thread的关系。
3. Semaphore的初始值可以为0，甚至可以为负数。对于acquire调用（信号down），它只会比较现在信号值与0的大小关系，如果<=0那么将不能获得授权。
4. 对于release（信号up），只是简单的对信号值进行原子增加，经过多次的release，信号值可以超过初始的阀值。
5. 对于Semaphore(0/-N)的场景，有特殊的使用，这种信号控制，在可以acquire之前，必须经过约定的足够多的release之后才可以被使用。

参考:http://stackoverflow.com/questions/1221322/how-does-semaphore-work

 

Calling down when it's 0 should not work. Calling up when it's 3 does work. (I am thinking of Java).

 

Let me add some more. Many people think of locks like (binary) semaphores (ie - N = 1, so the value of the semaphore is either 0 (held) or 1 (not held)). But this is not quite right. A lock has a notion of "ownership" so it may be "reentrant". That means that a thread that holds a lock, is allowed to call lock() again (effectively moving the count from 0 to -1), because the thread already holds the lock and is allowed to "reenter" it. Locks can also be non reentrant. A lock holder is expected to call unlock() the same number of times as lock().

 

Semaphores have no notion of ownership, so they cannot be reentrant, although as many permits as are available may be acquired. That means a thread needs to block when it encounters a value of 0, until someone increments the semaphore.

 

Also, in what I have seen (which is Java), you can increment the semaphore greater than N, and that also sort of has to do with ownership: a Semaphore has no notion of ownership so anybody can give it more permits. Unlike a thread, where whenever a thread calls unlock() without holding a lock, that is an error. (In java it will throw an exception).