[Python-Dev] Re: __metaclass__ and __author__ are already decorators
Paul Morrow
pm_mon at yahoo.com
Sun Aug 22 00:29:56 CEST 2004
Paul Morrow wrote:
> Bob Ippolito wrote:
>
>>
>> On Aug 21, 2004, at 5:34 PM, Paul Morrow wrote:
>>
>>> Phillip J. Eby wrote:
>>>
>>>> At 05:15 PM 8/21/04 -0400, Paul Morrow wrote:
>>>>
>>>>> Christophe Cavalaria wrote:
>>>>>
>>>>>> can it be ? There's also the fact that it can't handle named
>>>>>> parameters
>>>>>> like a regular function call. You can't write that :
>>>>>> def foo():
>>>>>> __decoration__ = (1,1,param=True)
>>>>>
>>>>>
>>>>>
>>>>>
>>>>> As far as I know, we can't do that with the current decorator
>>>>> proposals either.
>>>>
>>>>
>>>> @decoration(1,1,param=True)
>>>> def foo(whatever):
>>>> pass
>>>
>>>
>>>
>>> Ok, then whatever changes you've made to the Python system to support
>>> that would allow the same syntax to be used in what I'm suggesting.
>>>
>>> def foo(whatever):
>>> __decoration__ = (1,1,param=True)
>>>
>>> def decoration(decoratedFunc, a, b, param=False):
>>> __decorator__ = True
>>> __version__ = '0.1'
>>> # body of 'decoration' decorator function goes here...
>>
>>
>>
>> Congratulations, this by far the worst suggestion yet! I'm -Inf on
>> this :)
>>
>
> Thanks.
>
>
>> @decoration(1,1,param=True) makes no changes whatsoever to the Python
>> system. Everything after the @ is really just an expression. The @
>> just implies that the result of the next class or def gets put through
>> the result of the @expression before it is thrown into the namespace.
>>
>
> It seems that writing a decorator is going to be a bizarre experience.
> In the example, I would need to write a function named 'decoration' that
> returns a function that will recieve a function (foo) to be decorated
> and then return a function. Does that sound about right?
>
> What would functions like 'decoration' typically look like? Could you
> show a short code snippet?
>
Nevermind. I got wrapped around the axle. I get it now. Thanks.
More information about the Python-Dev
mailing list