Mark Dickinson <dicki...@gmail.com> added the comment: This is correct behaviour. Note that .seconds is only one of the three attributes that contribute to the value of a timedelta: the others are .microseconds and .days. A timedelta is normalised in such a way that td.seconds and td.microseconds are always nonnegative; this means that td.days will be strictly negative for negative timedeltas. In your case the timedelta will have a .days attribute of -1, and the total time represented (e.g., in seconds, 86400.0 * td.days + td.seconds + 0.000001 * td.microseconds) will be small and negative.
Here's what I get on my system (in the py3k branch): Python 3.2a0 (py3k:80840:80842, May 7 2010, 12:29:35) [GCC 4.2.1 (Apple Inc. build 5659)] on darwin Type "help", "copyright", "credits" or "license" for more information. >>> import datetime >>> td = datetime.datetime.now() - datetime.datetime.now() >>> td.microseconds, td.seconds, td.days (998977, 86399, -1) >>> print(td) -1 day, 23:59:59.998977 >>> 1e-6 * td.microseconds + td.seconds + 86400 * td.days -0.0010230000043520704 >>> td.total_seconds() # new in Python 2.7, 3.2 -0.0010230000043520704 ---------- nosy: +mark.dickinson resolution: -> invalid status: open -> closed _______________________________________ Python tracker <rep...@bugs.python.org> <http://bugs.python.org/issue8643> _______________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: http://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com
