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 <jul...@whistle.com> 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:  d...@nlsystems.com
> Nonlinear Systems Ltd.                        Phone: +44 181 442 9037
> 
> 
> 



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