On Oct 17, 10:00 am, Duncan Booth <[EMAIL PROTECTED]> wrote: > [EMAIL PROTECTED] wrote: > > On Oct 10, 8:23 am, "Diez B. Roggisch" <[EMAIL PROTECTED]> wrote: > <snip> > >> rebinds a. Period. Which is the _essential_ thing in my post, because > >> this rebinding semantics are what confused the OP. > > > Doesn't this depend on wether "a" supports __iadd__ or not? Section > > 3.4.7 of the docs say > > > """ > > If a specific method is not defined, the augmented operation falls > > back to the normal methods. For instance, to evaluate the expression x > > +=y, where x is an instance of a class that has an __iadd__() method, > > x.__iadd__(y) is called. If x is an instance of a class that does not > > define a __iadd__() method, x.__add__(y) and y.__radd__(x) are > > considered, as with the evaluation of x+y. > > """ > > > So if a.__iadd__ exists, a += b is executed as a.__iadd__(b), in which > > case there's no reason to rebind a. > > You misunderstand the documentation, what you quoted doesn't say that > the assignment is suppressed. If a.__iadd__ exists, a += b is executed > as a = a.__iadd__(b) > > The options for a+=b are: > > a = a.__iadd__(b) > a = a.__add__(b) > a = b.__radd__(a) > > but the assignment always happens, it is only what gets executed for the > right hand side which varies.
Curious, do you have the relevant section in the docs that describes this behaviour? Paul -- http://mail.python.org/mailman/listinfo/python-list
