On 2010-10-17 10:21:36 -0700, Paul Kölle said:
Am 17.10.2010 13:48, schrieb Steven D'Aprano:
On Sun, 17 Oct 2010 03:58:21 -0700, Yingjie Lan wrote:
Hi,
I played with an example related to namespaces/scoping. The result is a
little confusing:
[snip example of UnboundLocalError]
Python's scoping rules are such that if you assign to a variable inside a
function, it is treated as a local. In your function, you do this:
def f():
a = a + 1
Since a is treated as a local, when you enter the function the local a is
unbound -- it does not have a value. So the right hand side fails, since
local a does not exist, and you get an UnboundLocalError. You are trying
to get the value of local "a" when it doesn't have a value.
Steven's explanation is correct. In your example below you're altering
portions of a global data structure, not reassigning a global variable.
Put another way, there is a significant difference between:
a = 7
and:
a['x'] = 7
Only the first reassigns a global variable.
-Tom
Oh really? Can you explain the following?
>>> a = {}
>>> def foo():
... a['a'] = 'lowercase a'
... print a.keys()
...
>>> foo()
['a']
>>> a
{'a': 'lowercase a'}
>>> def bar():
... a['b'] = a['a'].replace('a', 'b')
...
>>> bar()
>>> a
{'a': 'lowercase a', 'b': 'lowercbse b'}
>>>
cheers
Paul
--
http://mail.python.org/mailman/listinfo/python-list
