Le vendredi 11 Février 2005 20:11, Steven Bethard a écrit : > Is there a good way to determine if an object is a numeric type? > Generally, I avoid type-checks in favor of try/except blocks, but I'm > not sure what to do in this case: > > def f(i): > ... > if x < i: > ... > > The problem is, no error will be thrown if 'i' is, say, a string: > > py> 1 < 'a' > True > py> 10000000000 < 'a' > True > > But for my code, passing a string is bad, so I'd like to provide an > appropriate error. >
This is a very bad Python feature that might very well be fixed in version 3.0 according to your own reply to a previous thread. This problem clearly shows that this Python feature does hurt. Here's a transcript of the answer : > Yes, that rule being to compare objects of different types by their type > names (falling back to the address of the type object if the type names > are the same, I believe). Of course, this is arbitrary, and Python does > not guarantee you this ordering -- it would not raise backwards > compatibility concerns to, say, change the ordering in Python 2.5. > > > What was the goal behind this rule ? > > I believe at the time, people thought that comparison should be defined > for all Python objects. Guido has since said that he wishes the > decision hadn't been made this way, and has suggested that in Python > 3.0, objects of unequal types will not have a default comparison. > > Probably this means ripping the end off of default_3way_compare and > raising an exception. As Fredrik Lundh pointed out, they could, if they > wanted to, also rip out the code that special-cases None too. > > Steve Regards Francis Girard -- http://mail.python.org/mailman/listinfo/python-list
