block/lambda

[Jean-Paul Calderone]

On Tue, 29 Jul 2008 07:26:38 -0700 (PDT), "jiri.zahradil at gmail.com" <jiri.zahradil at gmail.com> wrote:
>
>> 2. Will it be possible in Python 3.0 to do the following:
>>
>> >>> def dotimes(n, callable):
>>
>>         for i in range(n): callable()
>>
>> >>> def block():
>>
>>         nonlocal i
>>         for j in range(i):
>>                 print j,
>>         print
>
>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():
>        for j in range(i):
>            print j,
>        print
>>>> block()
>0 1 2 3 4 5 6 7 8 9
>

Python doesn't have dynamic scoping.

  >>> def dotimes(n, callable):
  ...     for i in range(n):
  ...             callable()
  ...
  >>> def block():
  ...     for j in range(i):
  ...             print j,
  ...     print
  ...
  >>> def f():
  ...     dotimes(5, block)
  ...
  >>> f()
  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