Python 2.0 and Stackless

[David Bolen]

Toby Dickenson <mbel44 at dial.pipex.net> writes:

> 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.

Um, of course unless multiple threads are involved, in which case your
naive function may be interrupted between its calls to x() and z() and
some other routine may call x() in between, right?

--
-- David
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