On Sun, 2014-03-09 at 18:53 -0400, David Miller wrote: > From: Ben Hutchings <ben@xxxxxxxxxxxxxxx> > Date: Sun, 09 Mar 2014 19:09:20 +0000 > > > On Thu, 2014-03-06 at 16:06 -0500, David Miller wrote: > >> From: Marc Kleine-Budde <mkl@xxxxxxxxxxxxxx> > >> Date: Wed, 5 Mar 2014 00:49:47 +0100 > >> > >> > @@ -839,7 +839,7 @@ void dev_deactivate_many(struct list_head *head) > >> > /* Wait for outstanding qdisc_run calls. */ > >> > list_for_each_entry(dev, head, unreg_list) > >> > while (some_qdisc_is_busy(dev)) > >> > - yield(); > >> > + msleep(1) > >> > } > >> > >> I don't understand this. > >> > >> yield() should really _mean_ yield. > >> > >> The intent of a yield() call, like this one here, is unambiguously > >> that the current thread cannot do anything until some other thread > >> gets onto the cpu and makes forward progress. > >> > >> Therefore it should allow lower priority threads to run, not just > >> equal or higher priority ones. > > > > Until when? > > > > yield() is not a sensible operation in a preemptive multitasking system, > > regardless of RT. > > To me it means "I've got nothing to do if other tasks want to run right > now" Yes, I even see it having this meaning when an RT task executes > it. > > How else can you interpret the intent above? The problem is that 'I've got nothing to do ... now' is information about a *point* in time, which gives the scheduler no clue as to when this task might be ready again. So far as task *state* goes, it never ceases to be ready. > If you change it to msleep(1), you're assigning an extra completely > arbitrary time limit to the yield. The code doesn't want to sleep > for 1ms, that's not what it's asking for. [...] I think you want to give up a 'time slice' to any task that's available. But that's not a meaningful concept for all schedulers. If your task is highest priority and is ready, it must run. You could drop priority temporarily, but then you need it to somehow be bumped up again at some time in the future. Well, sleeping effectively does that. I do understand that unconditionally sleeping also isn't ideal - the task that unblocks this one might already be running on another CPU and able to finish sooner than the sleep timeout. Maybe the answer is a yield_for(time) which sleeps for the given time or until there's an idle CPU, whichever is sooner. But I don't know how hard that would be to implement or how widely useful it would be. Ben. -- Ben Hutchings I say we take off; nuke the site from orbit. It's the only way to be sure.
Attachment:
signature.asc
Description: This is a digitally signed message part
