On Fri, Jul 26, 2019 at 10:52:46AM -0400, Calvin Spealman wrote:
> Let's say you do this or any of the variants suggested... What does this do?
>
> a = {"foo": 1}
> b = {}
> with a:
> with b:
> foo = 0
Re-writing that to my suggested version:
with namespace("a") as a:
with namespace("b") as b:
foo = 0
will create a namespace object (a subclass of ModuleType) and bind it to
"a", containing a namespace object "b", containing a variable "foo":
assert isinstance(a, NamespaceType)
assert isinstance(a.b, NamespaceType)
assert a.b.foo == 0
That would be analogous to this existing code:
class a:
class b:
foo = 0
except classes have different scoping rules to modules.
--
Steven
_______________________________________________
Python-ideas mailing list -- python-ideas@python.org
To unsubscribe send an email to python-ideas-le...@python.org
https://mail.python.org/mailman3/lists/python-ideas.python.org/
Message archived at
https://mail.python.org/archives/list/python-ideas@python.org/message/W3XEBOHQS3AWJ3Z26GA3H24GBGECM6FL/
Code of Conduct: http://python.org/psf/codeofconduct/
