Making sense of Stackless

Christian Tismer tismer at tismer.com
Sun Feb 27 21:33:05 CET 2000


Darrell wrote:
> 
> Thinking about how I could use Stackless and how it's different from using
> callbacks on a object. I decided to view continuations as objects and here's
> were it lead.
> 
> If an instance can be thought of as a stack frame then a continuation can be
> thought of as an instance with a constructor and a run method.
> 
> import continuation
> 
> def construct():
>     tripFlag=1
>     c=continuation.caller()
>     tf=c.update(tripFlag)
>     if tripFlag==1:
>             # Don't understand how tripFlag stops being equal to 1
>         return c
>     return tripFlag

What do you want to achieve with the update() ?
The value just passes through. update is used to
either update a co in *your* frame to that
assignment position, or to update a different
frame's co to that frame's current true state.
Both don't seem to apply here since your caller()
has not changed its state.
I guess you can leave that out.

It is a bit hard to use just a single function for both
initialisation and the action. I don't see an easy way
to avoid the "if someinit" stuff.
Well, here the same as a single function:

def classFunc():
    print 'Construct:'
    x=1
    y=2
    c=continuation.current()
    if c != None:
        return c
    print 'Run:'
    x=x+1
    print x+y

This has the same effect as your construct() call.
current() is the same as caller(0), btw.
What happens here?

At the "c=" position, we have a snapshot. We return this
snapshot. When you call this continuation later, you
are repeating this assignment, but this time you are
implicitly passing None. A good thing, since this
auto-clears the otherwise cyclic reference!

Well, it would be a speed improvement if we could save that
if part. You could try a c.update(), but this doesn't work
since c is dead already. Maybe I should provide a method
update_release() that does the same thing, but also
frees the connection? Note that this would combine
update() and deleting co.link into one step.

cheers - chris

-- 
Christian Tismer             :^)   <mailto:tismer at appliedbiometrics.com>
Applied Biometrics GmbH      :     Have a break! Take a ride on Python's
Kaunstr. 26                  :    *Starship* http://starship.python.net
14163 Berlin                 :     PGP key -> http://wwwkeys.pgp.net
PGP Fingerprint       E182 71C7 1A9D 66E9 9D15  D3CC D4D7 93E2 1FAE F6DF
     we're tired of banana software - shipped green, ripens at home




More information about the Python-list mailing list