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why does foo show up in the output of inspect.getmembers(..., predicate=inspect.isclass) ?

$ python2.5
Python 2.5.2 (r252:60911, Aug 28 2008, 13:13:37) 
[GCC 4.1.2 20071124 (Red Hat 4.1.2-42)] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import junk
>>> import inspect
>>> inspect.getmembers(junk, predicate=inspect.isclass)
[('Dummy', <class 'junk.Dummy'>), ('foo', {'two': 2, 'one': 1})]
>>> inspect.isclass(junk.foo)
I expected that inspect would only return Dummy since that is the only class definition in the module. Apparently, though, junk.foo is a class in the eyes of the inspect module. Why is that?
Prior to Python v2.7, inspect.isclass naively assumed anything with a __bases__ attribute must be a class.
Dummy's __getattr__ makes Dummy instances appear to have every attribute (with a value of None).
Therefore, to inspect.isclass, foo appears to be a class.
Note: __getattr__ should raiseAttributeError when asked for an attribute it does not know about. (This is very different than returning None.)
                In particular the lack of AttributeError which will cause truly stupendously horrible bugs.
– Katriel
                Nov 2, 2010 at 20:51
First if all great answer Jon-Eric i just wanted to add some stuff:
if you do in ipython (what a great tool):
%psource inspect.isclass
you will get:
return isinstance(object, types.ClassType) or hasattr(object, '__bases__')
which what Jon-Eric said.
but i guess that you use python < 2.6 and this bug was already fixed, this is the code of inspect.isclass() in python2.7:
return isinstance(object, (type, types.ClassType))
                Excellent point! I was looking at the v2.6 source. I edited my answer to indicate that __bases__ is no longer used in v2.7+.
– Jon-Eric
                Nov 2, 2010 at 21:14
        
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