Can't define __call__ within __init__?

[Steve Holden]

Neal Becker wrote:
> Steven D'Aprano wrote:
> 
>> On Wed, 10 Mar 2010 08:12:14 -0500, Neal Becker wrote:
>>
>>> Want to switch __call__ behavior.  Why doesn't this work?  What is the
>>> correct way to write this?
>>>
>>> class X (object):
>>>     def __init__(self, i):
>>>         if i == 0:
>>>             def __call__ (self):
>>>                 return 0
>>>         else:
>>>             def __call_ (self):
>>>                 return 1
>>
>> Others have already pointed out that there are two reasons that won't
>> work:
>>
>> (1) you define __call__ as a local variable of the __init__ method which
>> then disappears as soon as the __init__ method completes; and
>>
>> (2) special methods like __call__ are only called on the class, not the
>> instance, so you can't give each instance its own method.
>>
>>
>> Perhaps the best way to solve this is to use delegation:
>>
>>
>> def zero_returner():
>>     return 0
>>
>> def one_returner():
>>     return 1
>>
>>
>> class X (object):
>>     def __init__(self, i):
>>         if i == 0:
>>             self.func = zero_returner
>>         else:
>>             self.func = one_returner
>>     def __call__(self, *args, **kwargs):
>>         return self.func(*args, **kwargs)
>>
>>
>> zero_returner and one_returner can be any callable object, not
>> necessarily a function.
>>
>> Of course, all this assumes that your solution isn't even simpler:
>>
>> class X (object):
>>     def __init__(self, i):
>>         self.i = i
>>     def __call__(self):
>>         return self.i
>>
>> but I assume if it was that simple, you would have done that already.
>>
>>
>>
> The example I showed was just a toy problem.  The real problem is
> I expect to call a function many times, and I want to avoid the overhead of 
> the 'if blah' everytime.
> 
This is a premature optimization. First, make it work. Then (if it
doesn't work fast enough) make it work faster.

regards
 Steve
-- 
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