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. As far as I know, most contemporary mutexes do the same, they give priority to thread A not only because it can take the fast path, but also because (a) its cache is hot and (b) it's better to minimize context switches.