metaperl wrote: > On p.282 of "Python Cookbook" and in the Python docs on calling super: > http://www.python.org/download/releases/2.2.3/descrintro/#cooperation > > it is clear that the first argument to super is a class and not a type. > However, for the code below, I am getting an error when attempting to > provide a class as my first argument and don't understand why. Also, > since this is my first attempt at writing anything of any seriousness > in Python, any other feedback is welcome.
"super" only works for new-style classes. You could make "ftputilx" new-style by inheriting from object and FTPHost, but FTPHost is still old-style, and I don't know if that's okay with super. So in your case it would be advisable to use FTPHost.__init__(self) instead of super(ftputilx, self).__init__() Georg -- http://mail.python.org/mailman/listinfo/python-list
