I believe this is because signal works on a thread level. Once you call signal from a thread, in this case the main thread and no other resource has a lock on your semaphore, then the signal call simply returns and puts a lock that belongs to your thread on the semaphore. The second time you call signal from the same thread, your thread already has the lock so signal does not block and you are free to continue. I think you need to test this with at least one other thread to see it in action. I haven't tested this myself in a while, but I believe what I stated is correct. I'm sure someone will chime in if I'm mistaken.

Kevin


You are describing a CriticalSection. The documentation for Semaphore is explicit that the counter will decrement even in the same thread. By inference, it's a CriticalSection that does not decrement a second time in the same thread. To wit, the documentation for Semaphore states:

    The Semaphore class is different from the CriticalSection
    and Mutex classes in this way: calling Signal in the same
    thread will cause the counter to decrement.

The main thread (controlling the GUI) is a thread, too. So, I still believe my code should block.

Jeff

You are right. I read this in the Language Reference PDF and interpreted in the wrong way. It refers to "the" lock instead of "a" lock and is somewhat confusing in light of the rest of the documentation.

Call Signal to obtain a lock on a resource. When you obtain a
lock, you can use the resource without fear that another
thread will try to use it simultaneously. If the call succeeds,
this function returns immediately and you code continues to
execute. If the lock is not available, then the thread that
called this method will wait until the lock becomes available.
When the call returns, you will have the lock, but you may
have to wait until the lock becomes available. When you are
finished using the resource, call the Release method to give it
back to the Semaphore.


The built in Language Ref does seem to be more clear and what you see would appear to be a bug.
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