Static variable vs Class variable

paul.melis at gmail.com paul.melis at gmail.com
Wed Oct 17 10:16:09 CEST 2007


On Oct 17, 10:00 am, Duncan Booth <duncan.bo... at invalid.invalid>
wrote:
> paul.me... at gmail.com wrote:
> > On Oct 10, 8:23 am, "Diez B. Roggisch" <de... at nospam.web.de> 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




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