Martin Panter added the comment:
Calling exec() with only one argument is equivalent to exec(..., globals(),
locals()). It does not create a new scope for names. So an equivalent of your
three-level example is more like
>>> i = 'global'
>>> def f():
... i = 'nonlocal'
... class_locals = dict()
... exec("print(i)\ni = 'local'\nprint(i)\n", globals(), class_locals)
...
>>> f()
global
local
If exec() worked like a function rather than a class, the first print(i) would
trigger an UnboundLocalError instead:
>>> i = 'global'
>>> def f():
... i = 'nonlocal'
... def g():
... print(i)
... i = 'local'
... g()
...
>>> f()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 6, in f
File "<stdin>", line 4, in g
UnboundLocalError: local variable 'i' referenced before assignment
In your first exec() example, i='nonlocal' is passed to exec() via the default
locals parameter, and the exec() uses that value rather than deferring to its
globals. To be a free variable, “i” has to be used but not defined. Even if you
dropped the “i = 'local' ” assignment, it is still defined via the implicit
locals parameter.
Your proposal for “Interaction with dynamic features” sounds reasonable.
----------
nosy: +martin.panter
_______________________________________
Python tracker <rep...@bugs.python.org>
<http://bugs.python.org/issue26225>
_______________________________________
_______________________________________________
Python-bugs-list mailing list
Unsubscribe:
https://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
