On Tue, 29 Jul 2008 07:26:38 -0700 (PDT), "jiri.zahradil@ gmail.com" <jiri.zahradil@ gmail.comwrote:
Python doesn't have dynamic scoping.
... for i in range(n):
... callable()
...
... for j in range(i):
... print j,
... print
...
... dotimes(5, block)
...
Traceback (most recent call last):
File "<stdin>", line 1, in ?
File "<stdin>", line 2, in f
File "<stdin>", line 3, in dotimes
File "<stdin>", line 2, in block
NameError: global name 'i' is not defined
The "nonlocal" keyword in Python 3 won't do this, either. It's for
referencing names in outer lexical scopes, not outer dynamic scopes.
Jean-Paul
>
>
>dotimes seems ok and what is wrong with that function "block"? You do
>not need to specify that i is "nonlocal", global i will be used.
>
for j in range(i):
print j,
print
>0 1 2 3 4 5 6 7 8 9
>
>2. Will it be possible in Python 3.0 to do the following:
>>
>>
> for i in range(n): callable()
>>
>>
> nonlocal i
> for j in range(i):
> print j,
> print
>>
>>def dotimes(n, callable):
> for i in range(n): callable()
>>
>>def block():
> nonlocal i
> for j in range(i):
> print j,
>dotimes seems ok and what is wrong with that function "block"? You do
>not need to specify that i is "nonlocal", global i will be used.
>
>>>i=10
>>>def block():
>>>def block():
print j,
>>>block()
>
>>def dotimes(n, callable):
... callable()
...
>>def block():
... print j,
...
>>def f():
...
>>f()
File "<stdin>", line 1, in ?
File "<stdin>", line 2, in f
File "<stdin>", line 3, in dotimes
File "<stdin>", line 2, in block
NameError: global name 'i' is not defined
>>>
referencing names in outer lexical scopes, not outer dynamic scopes.
Jean-Paul