[Tim]
>> I can see one kind of annoyance that would remain:
>>
>> dt2 = dt1 + a_timedelta
>>
>> is currently specified to force dt2.fold==0 even if dt1.fold==1. But
>> that may not make good sense.
[Alex]
> Note that dt2.fold==0 even if dt1.fold==1 *and* a_timedelta==timedelta(0).
Yup.
> This is what I call "fold-unaware" arithmetic. It is consistent with
> dt2==dt1.
Heh - setting
dt2.fold = random.randrange(2)
would also be consistent with dt2 == dt1. That is, "==" ignores both
`fold`s entirely in this case.
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