How does semaphore work?

Question

Can the semaphore be lower than 0? I mean, say I have a semaphore with N=3 and I call \"down\" 4 times, then N will remain 0 but one process will be blocked?

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Answer 1

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).

Hope this way of thinking about it helps.