> "the problem"? > Perhaps if you explain what you really want to do, someone can think the > way to do that, most likely *not* using id()
Thanks, now I know I cannot use id() for my problem. Here is my problem: I want to add a docstring translator into the Python interpreter. If the user input: >>> a = [1,2,3] >>> a.append( this translator will show the docstring of append in my native language. Because __doc__ is read only, I added a dict to the interpreter as follows: DOC[list.append.__doc__] = """ translated version of the __doc__ """ When it is the time to show docstring, the program get the translated version from DOC. This works, but I think the key of DOC is too long, so I want to use the id of list.append.__doc__ as the key; or use the id of list.append: DOC[id(list.append.__doc__)] = "..." DOC[id(list.append)] = "..." So, I asked how to get list.append from a.append, and why id(list.append.__doc__) changes. -- http://mail.python.org/mailman/listinfo/python-list
