introspection: How to find out the class defining a method

[Holger Joukl]

Hi,
introspection works different in python 2.3:

Python 2.3.3 (#12, Feb 19 2004, 11:42:09)
[GCC 2.95.2 19991024 (release)] on sunos5
Type "help", "copyright", "credits" or "license" for more information..... def do(self):
.... pass
........ pass
....
whereas:

Python 2.1 (#3, Jun 1 2001, 15:51:25)
[GCC 2.95.2 19991024 (release)] on sunos5
Type "copyright", "credits" or "license" for more information..... def do(self):
.... pass
........ pass
....
Is there an elegant way to achieve the latter result (i.e. the base class
in which
this method actually is defined) in python 2.3, too? Couldn´t find it in
the docs.

Cheers
Holger

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[Jeremy Yallop]

Python 2.1 (#3, Jun 1 2001, 15:51:25)
[GCC 2.95.2 19991024 (release)] on sunos5
Type "copyright", "credits" or "license" for more information.... def do(self):
... pass
...... pass
...
Is there an elegant way to achieve the latter result (i.e. the base
class in which this method actually is defined) in python 2.3, too?
Couldn´t find it in the docs.

I'm not sure this qualifies as "elegant". I think it gives the right
answer when it works, though.

def class_defining_method(method):
for c in inspect.getmro(method.im_class):
if c.__dict__.has_key(method.im_func.__name__):
return c

Jeremy.