Messages in this thread | | | Date | Tue, 18 Jun 2002 13:12:44 -0400 (EDT) | From | "Richard B. Johnson" <> | Subject | Re: Question about sched_yield() |
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On Tue, 18 Jun 2002, Chris Friesen wrote:
> David Schwartz wrote: > > > > >And you seem to have a misconception about sched_yield, too. If a > > >machine has n tasks, half of which are doing CPU-intense work and the > > >other half of which are just yielding... why on Earth would the yielding > > >tasks get any noticeable amount of CPU use? > > > > Because they are not blocking. They are in an endless CPU burning loop. They > > should get CPU use for the same reason they should get CPU use if they're the > > only threads running. They are always ready-to-run. > > > > >Quite frankly, even if the supposed standard says nothing of this... I > > >do not care: calling sched_yield in a loop should not show up as a CPU > > >hog. > > > > It has to. What if the only task running is: > > > > while(1) sched_yield(); > > > > What would you expect? > > If there is only the one task, then sure it's going to be 100% cpu on that task. > > However, if there is anything else other than the idle task that wants to run, > then it should run until it exhausts its timeslice. > > One process looping on sched_yield() and another one doing calculations should > result in almost the entire system being devoted to calculations. > > Chris >
It's all in the accounting. Use usleep(0) if you want it to "look good".
Cheers, Dick Johnson
Penguin : Linux version 2.4.18 on an i686 machine (797.90 BogoMips).
Windows-2000/Professional isn't.
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