Python 2.0 and Stackless

[Gordon McMillan]

Toby Dickenson wrote:

>Consider this "naive" function
>
>def m(y):
>   x()
>   try:
>      y()
>   finally:
>      z()
>
>Today I can be sure that a call to z is always preceeded by a single
>call to x. Stacklessness makes that assumption false.

No, I'm fairly sure that no matter what insane manipulations you might do 
with the continuation extension module, z will always be preceeded by a 
call to x. IOW, however you got to z, your path back up the graph will lead 
you to x.

- Gordon