On Sun, 9 Mar 2014, Ben Hutchings wrote:

On Sun, 2014-03-09 at 18:53 -0400, David Miller wrote:

quoted

From: Ben Hutchings <redacted>
Date: Sun, 09 Mar 2014 19:09:20 +0000

quoted

On Thu, 2014-03-06 at 16:06 -0500, David Miller wrote:

quoted

From: Marc Kleine-Budde <mkl@pengutronix.de>
Date: Wed,  5 Mar 2014 00:49:47 +0100

quoted

@@ -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.

quoted

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.

what is msleep(0) defined to do? would it be any better?

David Lang