Am 17.10.2010 19:51, schrieb TomF:
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.
Thanks Tom and Dennis.
This will teach me (hopefully) to pay attention to details next time and
I think I learned something too. I always thought the rules about
changing "global" objects where inconsistent because it works for
mutables... Turns out it's all fine since assignment doesn't work for
mutables too and assignment just happens to be the only way to "change"
immutables ;)
cheers
Paul
-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
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