From: David Abrahams (dave_at_[hidden])
Date: 2006-09-20 11:33:25
"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)?
-- Dave Abrahams Boost Consulting www.boost-consulting.com
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