Trying to understand += better

[Steve Howell]

On Nov 22, 8:38 pm, Roy Smith <r... at panix.com> wrote:
> In article <4b0a01a... at dnews.tpgi.com.au>, Lie Ryan <lie.1... at gmail.com>
> wrote:
>
> > The semantic of the in-place operator is something like:
> > x += y
> > becomes
> > x = x.__iadd__(y)
>
> > thus
> > foo.bar += baz
> > becomes
> > foo.bar = foo.bar.__iadd__(baz)
>
> > So the call sequence is,
> > foo.__getattr__('bar') ==> x
> > x.__iadd__(baz) ==> y
> > foo.__setattr__('bar', y)
>
> I don't get where the __setattr__() call comes from in this situation.  I
> thought the whole idea of __iadd__(self, other) is that it's supposed to
> mutate self.  So, why is there another assignment happening after the
> __iadd__() call?

Non-mutable types can also use += syntax.

x = MagicalNumber(42)
x += 5 # x gets reassigned to MagicalNumber(47)

There is nothing that says that __iadd__ has to return self, but if
you are designing a mutable type, you will generally do that.

http://docs.python.org/reference/datamodel.html

'''
These methods are called to implement the augmented arithmetic
assignments (+=, -=, *=, /=, //=, %=, **=, <<=, >>=, &=, ^=, |=).
These methods should attempt to do the operation in-place (modifying
self) and return the result (which could be, but does not have to be,
self). If a specific method is not defined, the augmented assignment
falls back to the normal methods. For instance, to execute the
statement 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.
'''