On Mon, Aug 31, 2015 at 3:33 PM, Alexander Belopolsky < alexander.belopolsky@gmail.com> wrote:

On Mon, Aug 31, 2015 at 3:20 PM, Tim Peters <tim.peters@gmail.com> wrote:

...

def __hash__(self): if self._hashcode == -1: tzoff = self.utcoffset() if tzoff is None: self._hashcode = hash(self.replace(first=True)._getstate()[0]) else: days = _ymd2ord(self.year, self.month, self.day) seconds = self.hour * 3600 + self.minute * 60 + self.second self._hashcode = hash(timedelta(days, seconds, self.microsecond) - tzoff) return self._hashcode

So it's the case that two datetimes that compare true may have different hashes, when they represent the earlier and later times in a fold. I didn't say "it's a puzzle" lightly ;-)

Yes, it looks like I have a bug there, but isn't fixing it just a matter of moving self.replace(first=True) up two lines? Is there a bigger puzzle? Certainly x == y ⇒ hash(x) == hash(y) is the implication that I intend to preserve in all cases.

I think I admitted defeat too soon. Can you present a specific case where "two datetimes that compare true have different hashes"? There may be some subtlety due to the fact that we ignore tzinfo in == if it is the same for both sides, but when we compute hash(), we don't know what's on the other side. It is hard to tell without a specific example. I thought I got it right when I wrote the code above, but it is possible I missed some case.