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From: Peter Dimov (pdimov_at_[hidden])
Date: 2006-09-20 13:24:32

David Abrahams wrote:
> "Peter Dimov" <pdimov_at_[hidden]> writes:
>> Anthony Williams wrote:
>>> The case given in Alexander's email above is:
>>> A: lock [lockcount == 0]
>>> B: lock [lockcount == 1, slow path]
>>> A: unlock [lockcount == 0, slow path/post sema]
>>> A: lock - [lockcount == 1, slow path]
>>> He claims that the second A: lock takes the slow path, when it could
>>> have taken the fast path because the lock is free. I disagree. The
>>> lock is NOT free, since B is waiting on it. If A can take the fast
>>> path here, that's opening the door for B to be starved.
>> This is correct. The mutex does not guarantee fairness (for the above
>> definition of fairness) at the expense of performance. It is indeed
>> true that in heavily contended scenarios thread A can finish well
>> ahead of thread B; however, the total time taken by threads A+B can
>> be (and is) significantly less than that with the "fairer" CS design.
> Probably dumb question: does this reasoning apply when A and B need to
> run "continuously" (e.g. in a server)?

I think that it does, because A and B are roles, not permanent labels. At
the last lock above, A is the currently running thread; B is a thread that
is currently blocked on the mutex. Since the scheduler distrubutes the
"currently running" role among the threads based on its notion of fairness,
a synchronization primitive that favors the currently running thread is as
fair as the scheduler in the long run.

I don't claim to be an expert on this stuff, though. :-)

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