changing __call__ on demand

[F. Petitjean]

Le Sun, 13 Feb 2005 13:19:03 -0500, Hans Nowak a écrit :
> Stefan Behnel wrote:
>> Hi!
>> 
>> This somewhat puzzles me:
>> 
>> Python 2.4 (#1, Feb  3 2005, 16:47:05)
>> [GCC 3.3.4 (pre 3.3.5 20040809)] on linux2
>> Type "help", "copyright", "credits" or "license" for more information.
>> 
>> .>>> class test(object):
>> ...   def __init__(self):
>> ...     self.__call__ = self.__call1   # self.__call__ is bound
>> ...   def __call1(self):
>> ...     print 1
>> ...   def __call__(self):   # self.__call__ is rebound
>> ...     print 2
>> ...
>> .>>> t = test()
>> .>>> t()
>> 2
2 because the last defined __call__ wins. Try dir(t)
>> 
>> If I take out the __call__ method completely and only set it in 
>> __init__, I get a TypeError saying that test is not callable.
This seems logical.
> 
> Note that it works just fine if you don't use a new-style class:
> 
> >>> class Test:
> ...     def __init__(self):
> ...         self.__call__ = self.foobar
> ...     def foobar(self, *args, **kwargs):
> ...         print "Called with:", args, kwargs
> ...
> >>> t = Test()
> >>> t()
> Called with: () {}
> >>> t(3, 4)
> Called with: (3, 4) {}
> >>> t(42, x=0)
> Called with: (42,) {'x': 0}
Are you sure that if you add a __call__() method, it will still work
fine ?


Regards
>