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
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