Op 2005-11-04, Christopher Subich schreef <[EMAIL PROTECTED]>: > Antoon Pardon wrote: >> Op 2005-11-03, Stefan Arentz schreef <[EMAIL PROTECTED]>: >>>The model makes sense in my opinion. If you don't like it then there are >>>plenty of other languages to choose from that have decided to implement >>>things differently. >> >> >> And again this argument. Like it or leave it, as if one can't in general >> like the language, without being blind for a number of shortcomings. >> >> It is this kind of recations that make me think a number of people is >> blindly devoted to the language to the point that any criticism of >> the language becomes intollerable. > > No, it's just that a goodly number of people actually -like- the > relatively simple conceputal model of Python. > > Why /shouldn't/ > > >>>a.x = foo > > correspond exactly to > > >>>setattr(a,'x',foo) #? > > Similarly, why shouldn't > > >>>foo = a.x > > correspond exactly to > > >>>foo = getattr(a,'x') #? > > With that in mind, the logical action for > > >>>a.x = f(a.x) > > is > > >>>setattr(a,'x',f(a,'x')) #, > > and since > > >>>a.x += foo > > is equal to > > >>>a.x = A.__iadd__(a.x,foo) # (at least for new-style classes > >>> # that have __iadd__ defined. Otherwise, it falls back on > >>> # __add__(self,other) to return a new object, making this > >>> # evern more clear), > > why shouldn't this translate into > > >>>setattr(a,'x',A.__iadd__(getattr(a,'x'),foo)) #?
Well maybe because as far as I understand the same kind of logic can be applied to something like lst[f()] += foo In order to decide that this should be equivallent to lst[f()] = lst[f()] + foo. But that isn't the case. So it seems applying augmented operators is not a matter of just substituting straight translations to get the right result. > Looking at it this way, it's obvious that the setattr and getattr may do > different things, if the programmer understands that "instances (can) > look up object attributes, and (always) set instance attributes." In > fact, it is always the case (so far as I can quickly check) that += ends > up setting an instance attribute. Consider this code: Looking at lists in a similar way, it would be obvious that the __setitem__ and __getitem__ can do different things and so we should expect lst[f()] += foo to behave exactly as lst[f()] = lst[f()] + foo. -- Antoon Pardon -- http://mail.python.org/mailman/listinfo/python-list
