how to dynamically create a function object (from a code object)?

Petri Savolainen petri.savolainen at
Wed Jul 2 00:43:03 CEST 2003

After reading the manuals and googling around a bit, I thought I'd use 
the 'compile' built-in to create a code object. Then, using either 
new.function() or types.FunctionType(), create a function object out of 
the code object. The function object can then be turned into a method 
for example using types.MethodType(). Right? Well, on Windows 98, using 
python 2.2.2 (or 2.3b2):

 >>> c=compile('def a(msg): return msg','<nowhere>','exec')
 >>> f=types.FunctionType(c,globals(),'a')
 >>> f
 >>> <code object ? at 00F8A9E0, file "<nowhere>", line 1>
 >>> f('hello')

Traceback (most recent call last):
   File "<pyshell#136>", line 1, in -toplevel-
TypeError: ?() takes no arguments (1 given)

 >>> f()
 >>> a('hello')
 >>> 'hello'

This is, well, not what I would have expected.

After peeking around in the code object, I found out its 'co_const' 
instance variable also contains a code object - which, it seems, should 
really be fed to the function creation methods:

 >>> c.co_consts
(<code object a at 00F8A960, file "<nowhere>", line 1>, None)
 >>> f=types.FunctionType(c.co_consts[0],globals(),'a')
 >>> f('hello')

Which is the behaviour I would have expected in the first place!

I would really like to know what I am doing wrong here, or any 
clarification regarding what is going on above... I dare not hope having 
found a bug :-P



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