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Let's say process A has an exclusive lock on some shared resource. Is there a way to ensure that after it completes execution, Process B will definitely get the lock (as compared to say, process C being able to get it)?

I am coding in a language that is relatively less used, but didn't want to limit my question to a language-specific tag (in case there are any language-agnostic ways of ensuring the above).

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You can't do this with a single mutex, but you could use some semaphores in addition to (or instead of) the mutex. Basically, when A is finished, it would signal B's semaphore. When B or C is finished, it signals an 'everyone else' semaphore. A and C wait on the 'everyone else' semaphore before using the shared resource.

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  • What do you mean by "B's semaphore" or "Everyone else semaphore"? I was under the impression that a semaphore is a property of the lock-able object and not a property of each thread. – rahs Sep 5 '19 at 22:27
  • B's semaphore is the semaphore B waits on and A posts to. The other semaphore is waited on by every other thread and posted by every thread except A. A semaphore is a standalone thing. It neither belongs to a thread nor an object. – Karl Bielefeldt Sep 6 '19 at 1:25
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No. Unless your language or library makes such a guarantee, you shall not assume that another process will get the lock. You should not either take a gamble on which other process could acquire the lock.

Starvation could for example occur, if process A is too fast to re-acquire the lock while process B is still waiting, or if the process scheduler is not well designed, which might result in B not getting the opportunity to acquire the lock.

Note: I assumed that there would be no deadlock since you said that A could finish its processing (and hence not be stuck waiting for another lock)

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