[Python-Dev] Re: __metaclass__ and __author__ are already decorators

[Paul Morrow]

Bob Ippolito wrote:

> 
> On Aug 21, 2004, at 6:15 PM, Paul Morrow wrote:
> 
>> Phillip J. Eby wrote:
>>
>>> At 05:34 PM 8/21/04 -0400, 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.
>>>
>>> Huh?  @decoration(1,1,param=True) is evaluated at the place where it 
>>> appears.  The *return value* of that expression is then called, 
>>> passing in the foo function.  In other words, the above is equivalent 
>>> to today's:
>>>     def foo(whatever):
>>>         pass
>>>     foo = decoration(1,1,param=True)(foo)
>>> except that the first assignment to 'foo' doesn't happen, only the 
>>> second one.  If the 'foo' function is a single expression, of course, 
>>> today you can do the straightforward:
>>>     foo = decoration(1,1,param=True)(lambda whatever: something)
>>> So, "@x func" is effectively a macro for "func = x(func)", where 
>>> 'func' may be a function, or another decorator.  That is:
>>>     @x
>>>     @y
>>>     @z
>>>     def foo():
>>>         ...
>>> is shorthand for 'foo = x(y(z(foo)))',
>>
>>
>> Wouldn't that actually be shorthand for foo = x()(y()(z()(foo))) ?
> 
> 
> No.
> 

Ok, I see, nevermind.  Thanks.