Maciej Blizi?ski wrote: > calling B::call_bar(): B::bar() > calling B::call___bar(): A::__bar()
(BTW, there is no :: operator in Python. It should be, e. g., B.bar().) > In the latter case, it calls the base class' implementation. It > probably goes along with Python's spec, but I found it surprising. __-prepended names are dynamically mangled to include the class name. Hence, B.__bar doesn't overwrite A.__bar in B. > I don't want to expose the __bar() function outside, In principle, you do. Always. Public or Private is just convention in Python. Even __-prepended names can be accessed from outside. Couldn't find it in the docs though. Quite inconcise. Regards, Björn -- BOFH excuse #120: we just switched to FDDI. -- http://mail.python.org/mailman/listinfo/python-list
