this is problematic.
you cannot add a new element before the pending firing because you can't
tell how far into the present trigger you are.
On Thu, 8 Jul 1999, Doug Rabson wrote:
> On Thu, 8 Jul 1999, Seigo Tanimura wrote:
>
> > On Wed, 7 Jul 1999 19:46:38 -0700 (PDT),
> > Julian Elischer <[EMAIL PROTECTED]> said:
> >
> > julian> With your scheme the clock needs to be always running at elevated speed.
> > julian> Possibly you might have a startup routine that turns on the elevated
> > julian> frequency, (basically does an 'aquire_timer0()' ) I would say that you
> > julian> would have more success in implementing your finetimer() by using
> > julian> "aquiretimer0" than the other way around.
> >
> > I agree that acquire_timer0() would give more freedom to the ticks
> > to callout. Then I tried figuring out how to manage multiple
> > callouts using acquire_timer0(), which is something like below.
> >
> >
> > Let C the callout queue, and c_i a callout. (0 <= i < I) Next define f(c_i) as
> > the callout function of c_i, and dt_rem(c_i) the time span between c_(i-1) and
> > c_i. (dt_rem(c_-1) is defined as zero) We use the time span to avoid traversing
> > though the queue to update the time tags on the callouts.
> >
> > (footnote: I'd better write in TeX :-<)
> >
> > Queueing a new callout c' to be made in t' involves a problem to find the
> > maximum j (which is an integer, j >= 0) satisfying a constraint
> >
> > t' > \sum_(k=0)^(j) dt_rem(c_k)
> >
> > where the right hand side of the inequality is the time span after which
> > the callout c_k is made. Then c' is inserted after c_j and new dt_rem(c_(j+1))
> > and dt_rem(c_(j+2)) are determined. Now we can acquire_timer0() with dt_rem(c_0).
> >
> > In clkintr(), we dequeue c_0 from C, and make a callout to f(c_0). Then
> > acquire_timer0() is called once more with the new dt_rem(c_0). dt_rem(c_i) is
> > the difference of callout times, so they need not be updated on every clkintr().
> >
> >
> > Although the computational cost in clkintr() is generaly O(1), the queueing cost
> > is O(I). Not sure whether we can reduce it or not (will it really make a trouble?)
> >
> >
> > How does it sound?
>
> If I understand this correctly, you are suggesting that we program timer0
> so that we only take interrupts when a finetimer is due to fire? If so,
> then it sounds very good. The idea of taking 6000+ interrupts/sec made me
> uneasy, even though most would return without doing any work.
>
> --
> Doug Rabson Mail: [EMAIL PROTECTED]
> Nonlinear Systems Ltd. Phone: +44 181 442 9037
>
>
>
To Unsubscribe: send mail to [EMAIL PROTECTED]
with "unsubscribe freebsd-hackers" in the body of the message