Python 2.5 nested auth functions in publisher.
----------------------------------------------
Key: MODPYTHON-198
URL: http://issues.apache.org/jira/browse/MODPYTHON-198
Project: mod_python
Issue Type: Bug
Components: publisher
Affects Versions: 3.2.10
Reporter: Graham Dumpleton
Assigned To: Graham Dumpleton
Fix For: 3.3
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
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