Op 2006-02-08, Scott David Daniels schreef <[EMAIL PROTECTED]>:
> Steven D'Aprano wrote:
>> On Wed, 08 Feb 2006 13:58:13 +1100, Delaney, Timothy (Tim) wrote:
>>
>>> adam johnson wrote:
>>>
>>>> Hi All.
>>>> I was wondering why defining a __call__ attribute for a module
>>>> doesn't make it actually callable.
>>> For the same reason that the following doesn't work
>> [snip example]
>>> The __call__ attribute must be defined on the class (or type) - not on
>>> the instance. A module is an instance of <type 'module'>.
>>
>> That's not a _reason_, it is just a (re-)statement of fact. We know that
>> defining a __call__ method on a module doesn't make it callable. Why not?
>> The answer isn't "because defining a __call__ method on a module or an
>> instance doesn't make it callable", that's just avoiding the question.
>
> You missed it. Steven D'Aprano was telling you why, and all you heard
> was the no. He stated a more general principal which controls why
> modules in particular are not callable. It is not a design decision
> about modules; it is a design decision about classes and instances.
>
> class SomeClass(object):
> def __call__(self): return 'Text'
>
> class AnotherClass(object):
> def __repr__(self): return 'Missive'
>
> name = SomeClass()() # this works
> name = AnotherClass()() # this doesn't
> obj = AnotherClass() # build an instance
> def fun(*args): return 'more text' # *args so nearly any call works
> obj.__call__ = fun # tack a function onto an instance
> obj() # note this doesn't call the function.
>
> Now, if you think the last _should_ do the call, then let's step
> back to classes.
>
> class SomeClass(object):
> def __call__(self): return 'Text'
>
> Now the SomeClass object (which is a subclass of object) has an
> attribute named "__call__". Should that define how the expression
> SomeClass()
> is evaluated? Should that return the string 'Text' or create a
> new instance of SomeClass?
This make me wonder. Would it be possible to do something with
metaclasses so that after
class SomeClass(MetaClass):
...
SomeClass() will be equivalent to MetaClass.__call__(SomeClass)
Just curious.
--
Antoon Pardon
--
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