[calpaterson]

    >>> from datetime import datetime, timedelta
    >>> datetime.now() - (datetime.now() + timedelta(days=1))
    datetime.timedelta(-2, 86399, 999969)

[jbg_]

I’m not entirely sure what you were trying to demonstrate, but clearly the result of datetime.now() changes between the two invocations (it has microsecond precision).

Try:

    >>> from datetime import datetime, timedelta
    >>> now = datetime.now()
    >>> now - (now + timedelta(days=1))
    datetime.timedelta(-1)

[calpaterson]

datetime.timedelta internally uses a triple of (days,seconds,microseconds) to represent it's data and exposes that to users of the class.

That a timedelta can be represented in multiple ways is alone quite surprising but some of the representations that happen can be very confusing. I think the example I gave which represents "-1 day and a few microseconds" as "-2 days + 86399 milliseconds + 999969 microseconds" is very surprising and it commonly happens in practice.

For comparison, here's what postgres does:

    postgres=> select now() - (now() + '12 milliseconds'::interval + '1 day'::interval);
           ?column?        
    -----------------------
     -1 days -00:00:00.012
    (1 row)

[sdispater]

That's a problem Pendulum is trying to solve

    d1 = datetime(2012, 1, 1, 1, 2, 3, tzinfo=pytz.UTC)
    d2 = datetime(2011, 12, 31, 22, 2, 3, tzinfo=pytz.UTC)
    delta = d2 - d1
    delta.days
    -1
    delta.seconds
    75600
    
    d1 = Pendulum(2012, 1, 1, 1, 2, 3)
    d2 = Pendulum(2011, 12, 31, 22, 2, 3)
    delta = d2 - d1
    delta.days
    0
    delta.hours
    -3