Does it though? So why doesn't a += b change the object id id(a) ?

As OP noted, given the existence of type(a).__iadd__[0] `a += b` essentially desugars to: `a = a.__iadd__(b)`.

list.__iadd__ is (sadly) defined as something like

    def __iadd__(self, other):
        self.extend(other)
        return self

So it's possible to have __iadd__ itself succeed modifying the list in place but then the "no-op" reassignment fail.

[0] like many data model operations there's really a bunch of fallbacks depending on what is and is not implemented

I'd almost consider this a subtle bug? None of the other += operators returns a value.

> I'd almost consider this a subtle bug?

It's not exactly a bug but it is a somewhat unexpected behaviour and IIRC Guido regretted that list.__iadd__ was overridden this way.

> None of the other += operators returns a value.

It's not the operator which returns a value, it's the data model hook. The data model requires that it return a value: https://docs.python.org/3/reference/datamodel.html?#object._...

> 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).

Really the issue is that the "in-place" hooks are simply weird.

Because if `a` and `b` are lists, `a.__iadd__(b)` mutates `a` in place, then returns it.

Right, got it.