On Oct 17, 11:08 am, Duncan Booth <[EMAIL PROTECTED]> wrote: > [EMAIL PROTECTED] wrote: > > Curious, do you have the relevant section in the docs that describes > > this behaviour? > > Yes, but mostly by implication. In section 3.4.7 of the docs, the sentence > before the one you quoted says: > > 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). > > The 'does not have to be self' tells you that the result of __iadd__ is > used, i.e there is still an assignment going on. > > Just read all of that paragraph carefully. It says that if there is no > __iadd__ method it considers calling __add__/__radd__. Nowhere does it say > that it handles the result of calling the methods differently.
Right, the paragraph is actually pretty clear after a second reading. I find it surprising nonetheless, as it's easy to forget to return a result when you're implementing a method that does an in-place operation, like __iadd__: >>> class C: ... def __init__(self, v): ... self.v = v ... def __iadd__(self, other): ... self.v += other ... >>> c=C(1) >>> c.v 1 >>> c += 3 >>> c >>> c is None True Paul -- http://mail.python.org/mailman/listinfo/python-list
