On Tue, Sep 1, 2015 at 1:23 AM, Tim Peters <[email protected]> wrote:
> 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. > Note that dt2.fold==0 even if dt1.fold==1 *and* a_timedelta==timedelta(0). This is what I call "fold-unaware" arithmetic. It is consistent with dt2==dt1.
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