[Python-Dev] Missing operator.call

[Brett Cannon]

On Wed, Feb 4, 2009 at 10:43, Steven Bethard <steven.bethard at gmail.com> wrote:
> On Wed, Feb 4, 2009 at 10:25 AM, Brett Cannon <brett at python.org> wrote:
>> On Wed, Feb 4, 2009 at 05:35, Hrvoje Niksic <hrvoje.niksic at avl.com> wrote:
>>> Andrew Bennetts wrote:
>>>>
>>>> A patch to add operator.caller(*args, **kwargs) may be a good idea.  Your
>>>> example would then be:
>>>>
>>>>    map(operator.caller(), lst)
>>>
>>> Regarding the name, note that I proposed operator.call (and
>>> operator.__call__) because it corresponds to the __call__ special method,
>>> which is analogous to how operator.neg corresponds to __neg__, operator.add
>>> to __add__, etc.  The term "caller" implies creation of a new object that
>>> carries additional state, such as method name in operator.methodcaller, item
>>> in operator.itemgetter, or attr in operator.attrgetter.
>>
>> Part of the problem is the term 'call' is an overloaded term. Do you
>> really mean only objects that define __call__? What about objects that
>> define __init__ and thus can be called as well? If you mean the former
>> than you have to make sure the docs are very clear about this; there
>> is a reason we got rid of callable(). If you mean the latter then
>> there is little benefit to the function since ``[x() for x in lst]``
>> gets you the same result as your map call.
>
> Not sure I follow you here. It's not the __init__ that allows you to
> do ``x()``, it's the fact that the class declares a __call__, right?
>
>>>> class C(object):
> ...     pass
> ...
>>>> C.__call__()
> <__main__.C object at 0x01A3C370>
>>>> C()
> <__main__.C object at 0x02622EB0>
>>>> str.__call__()
> ''
>>>> str()
> ''
>

I don't think so::

>>> Foo.__call__
<method-wrapper '__call__' of type object at 0x81cee0c>
>>> Foo.__call__ = lambda: None
>>> Foo.__call__
<unbound method Foo.<lambda>>
>>> Foo()
<__main__.Foo object at 0xf7f90e8c>

-Brett