On 11/14/19 12:57 PM, R.Wieser wrote: > Michael, > >> nonlocal does not share or use its *caller's* variables. Rather it >> reaches into the scope of the outer function where it was defined. >> That's a very different concept than what you're proposing. > Oh blimy! You're right. Its an at compile-time thing, not a runtime > one. > > Thanks for the heads-up. > >> I know of no sane way that a function could work with the scope of >> any arbitrary caller. > The trick seems to be to emulate a "by reference" call, by using a mutable > object as the argument and stuff the value inside of it (IIRC a tuple with a > single element).
tuples are immutable, use a list or a dictionary. >> What would happen if the caller's scope didn't have any >> names that the function was looking for? > Handle it the same as any other mistake, and throw an error ? > > Regards, > Rudy Wieser > > -- Richard Damon -- https://mail.python.org/mailman/listinfo/python-list
