Python 2.5 nested auth functions in publisher.

• Key: MODPYTHON-198
• Type: Bug
• Status: Closed
• Resolution: Fixed
• Priority: Major
• Assignee: Graham Dumpleton
• Reporter: Graham Dumpleton
• Component/s: publisher
• Affects Version/s: 3.2.10
• Fix Version/s: 3.3.1

Description

Jim Gallacher wrote:

With python 2.5 I get 2 failures:

    test_publisher_auth_nested
    test_publisher_auth_method_nested

It looks like something has changed in python 2.5 introspection that is
messing up publisher.

Test script testme.py
---------------------

def testfunc():
     print 'stuff'
     def __auth__():
         print '__auth__ called'
     def __access__():
         print '__access__ called'

def main():
     func_obj = testfunc

     func_code = func_obj.func_code
     print func_code.co_names

if __name__ == '__main__':
     main()

Results
-------

$ python2.3 testme.py
('__auth__', '__access__')
$ python2.4 testme.py
('__auth__', '__access__')
$ python2.5 testme.py
()


Dan Eloff points out that information is now in co_varnames.

>>> fc.co_names
()
>>> fc.co_varnames
('__auth__', '__access__')

>>> def foo(a,b):
d = 5
def bar(c):
return c

>>> fc.co_names
()
>>> fc.co_varnames
('a', 'b', 'd', 'bar')

To get just args, try:

>>> fc.co_varnames[:fc.co_argcount]
('a', 'b')

And for just local vars:

>>> fc.co_varnames[fc.co_argcount:]
('d', 'bar')


Still need to work out if actual code objects for the functions are available in co_consts or not. Ie., need to replace:

        if "__auth__" in func_code.co_names:
            i = list(func_code.co_names).index("__auth__")
            __auth__ = func_code.co_consts[i+1]
            if hasattr(__auth__, "co_name"):
                __auth__ = new.function(__auth__, func_globals)
            found_auth = 1