Messages in this thread | | | Date | Fri, 14 Nov 2008 16:41:45 -0500 (EST) | From | Mikulas Patocka <> | Subject | Re: Active waiting with yield() |
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On Fri, 14 Nov 2008, Peter Zijlstra wrote:
> On Fri, 2008-11-14 at 14:34 -0500, Mikulas Patocka wrote: > > > > On Fri, 14 Nov 2008, Alan Cox wrote: > > > > > > * driver unload --- check the count of outstanding requests and call > > > > yield() repeatedly until it goes to zero, then unload. > > > > > > Use a wakeup when the request count hits zero > > > > > > > * reduced size of data structures (and reduced cache footprint for the hot > > > > path that actually processes requests) > > > > > > The CPU will predict the non-wakeup path if that is normal. You can even > > > make the wakeup use something like > > > > > > if (exiting & count == 0) > > > > > > to get the prediction righ > > > > > > > The downside of yield is slower unloading of the driver by few tens of > > > > miliseconds, but the user doesn't really care about fractions of a second > > > > when unloading drivers. > > > > > > And more power usage, plus extremely rude behaviour when virtualising. > > > > How these unlikely cases can be rude? > > > > If I have a race condition that gets triggered just for one user in the > > world when repeatedly loading & unloading a driver for an hour, and I use > > yield() to solve it, what's wrong with it? A wait queue increases cache > > footprint for every user. (even if I use preallocated hashed wait queue, > > it still eats a cacheline to access it and find out that it's empty) > > Depending on the situation, yield() might be a NOP and therefore not > wait at all and possibly lock up the machine. > > Consider the task in question to be the highest priority RT task on the > system, you doing: while (!condition) yield(); will lock up the system, > because whatever is to make condition true will never get a chance to > run (not considering SMP). > > Clearly you don't understand it, please refrain from using it. Use > regular condition variables (waitqueues).
So, use msleep(1) instead of yield() ?
Mikulas
> The rules about yield are: > > - You're likely wrong, don't use it. > - Seriously, you don't need it. > - If you still think you do, goto 1. > > In all of the kernel there is 1 valid use (and it might only be in the > -rt kernel - didn't check mainline recently). > > The _ONLY_ valid use case of yield(), is if you have two equal priority > FIFO threads that co-depend. And that situation is almost always > avoidable. >
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