python 3 problem: how to convert an extension method into a class Method

Peter Otten __peter__ at web.de
Tue Feb 26 21:54:29 CET 2013


Ethan Furman wrote:

> On 02/26/2013 09:21 AM, Robin Becker wrote:
>> In python 2 I was able to improve speed of reportlab using a C extension
>> to optimize some heavily used methods.
>>
>> so I was able to do this
>>
>>
>> class A:
>>      .....
>>      def method(self,...):
>>         ....
>>
>>
>> try:
>>      from extension import c_method
>>      import new
>>      A.method = new.instancemethod(c_method,None,A)
>> except:
>>      pass
>>
>> and if the try succeeds our method is bound as a class method ie is
>> unbound and works fine when I call it.
>>
>> In python 3 this doesn't seem to work at all. In fact the new module is
>> gone. The types.MethodType stuff doesn't seem to work.
>>
>> Is there a way in Python 3.3 to make this happen? This particular method
>> is short, but is called many times so adding python wrapping layers is
>> not a good way forward.
> 
> Dumb question, but have you tried just assigning it?  In Py3 methods are
> just normal functions...
> 
> 8<----------------------
>    class A():
>        pass
> 
>    A.method = c_method
> 8<----------------------

The problem is that functions implemented in C don't support the descriptor 
protocol (they don't have a __get__() method). So

>>> from math import sqrt
>>> class A(int):
...     pass
... 
>>> A.c = sqrt
>>> A.py = lambda self: sqrt(self)
>>> a = A(42)
>>> a.py()
6.48074069840786
>>> a.c()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: sqrt() takes exactly one argument (0 given)





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