Hi Hakril, I think this question is probably off-topic for this list. This list is for development of the Python compiler, not for development with Python, and questions like this should probably go to python-l...@python.org (also mirrored as comp.lang.python if you have Usenet access). But I have a brief observation below:
On Wed, Oct 23, 2013 at 07:29:03PM +0200, hakril lse wrote: > For example: > > # A.descr is a descriptor > # B inherit from A > # C inherit from B > > c = C() > c.descr > super(C, c).descr > super(B, c).descr > > In these 3 cases the __get__ method is called with the same arguments > that are : __get__(descr, c, C). > > If this behavior is really expected: Could you explain why ? because > it means that I am missing something obvious. > Because, at first sight, the 'type' argument seems to be the perfect > place to get the type of the 'real calling class'. I'm afraid I don't understand what you mean by "real calling class", if it is not the class of the instance doing the calling. I have a feeling that perhaps you think that calls to super are equivalent to "the parent of the class", and so you expect: c.desc => desc.__get__(c, C) super(C, c).desc => desc.__get__(c, B) super(B, c).desc => desc.__get__(c, A) but that would not be correct. super is equivalent to "the next class in the MRO of the instance doing the calling", and in all cases, the "real-calling instance" is c, and the "real calling class" is the class of c, namely C. Extended discussion of this should go to python-list, but I think I have the basics right. -- Steven _______________________________________________ Python-Dev mailing list Python-Dev@python.org https://mail.python.org/mailman/listinfo/python-dev Unsubscribe: https://mail.python.org/mailman/options/python-dev/archive%40mail-archive.com
