On 28/10/2006, at 10:01 PM, Dan Eloff wrote:
On 10/28/06, Graham Dumpleton <[EMAIL PROTECTED]> wrote:
Dan, the code that needs to be updated is:
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
Note how it accesses code objects for functions from co_consts. Do
they still appear
to be there in Python 2.5? Are you able to work out some code that
does the same
thing as this?
Using the test function:
def foo(a,b):
d = 5
def __auth__(req):
return True
e = d + 5
fc = foo.func_code
import new
func_globals = globals()
for i, var_name in enumerate(fc.co_varnames):
if var_name == '__auth__':
__auth__ = fc.co_consts[i-fc.co_argcount+1]
if hasattr(__auth__, 'co_name'):
__auth__ = new.function(__auth__, func_globals)
found_auth = 1
break
__auth__
<function __auth__ at 0x01159830>
I am curious as to the hasattr(__auth__, 'co_name') section. Is there
any case where this is not true? (and does it make sense to say
found_auth = 1 if it isn't?)
The co_name check is making sure it is a code object as opposed to a
dictionary or some other constant.
See:
http://www.modpython.org/live/current/doc-html/hand-pub-alg-auth.html
for what __auth__ can be.
Graham
