operating system - What happens when Binary Semaphore is Signalled twice? ie. s=1;wait(s); signal(s)signal(s); Does s becomes 0 or it remains 1? -


what happens when binary semaphore signalled twice? i.e. suppose s binary semaphore variable s=1

now following--

wait(s),signal(s),signal(s)

does s becomes 0 or remains 1?

the binary semaphore used protect resource, 1 process may access @ time.

this process wait, which, assuming semaphore 1 (resource isn't busy) make semaphore =0.

when process finished resource signal, allowing other processes access resource, making semaphore =1.

there's no reason why process signal twice, nor there reason why other process signal before doing wait (let me know if you've thought of one), 2 successive signals error. if writing os primitive implementation i'd judge better idea leave signal @ 1 put 0 , lock resource out time.

the standard texts on semaphores talk incrementing , decrementing count, that's because it's useful way of thinking counting semaphores. way actual semaphores behave in actual os when they're used in odd way depends on implementation, , not worth examining general theory.


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