Raymond Hettinger added the comment: > If I understand it right, in a simple case like this: ... > calling super is equivalent to calling object.__setattr__,
It is not equivalent. Instances of Foo() would behave equivalently but it might do something different for subclasses of Foo. If you're interested in learning more about super(), have a look at: http://rhettinger.wordpress.com/2011/05/26/super-considered-super/ In the meantime, I'm closing this because we do not make that blanket advice to always use super() instead of a direct call to a parent. Sometimes you want one and sometimes you want the other depending on what you're trying to do. ---------- resolution: -> not a bug status: open -> closed _______________________________________ Python tracker <rep...@bugs.python.org> <http://bugs.python.org/issue21814> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: https://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
