[issue1289118] timedelta multiply and divide by floating point

[Alexander Belopolsky]

Alexander Belopolsky <belopolsky at users.sourceforge.net> added the comment:

Indeed.  Here is what I intended:

"""
>>> from datetime import timedelta as d
>>> [d(microseconds=i + .5)//d.resolution for i in range(-10,10)]
[-10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Should this be considered a bug?

For comparison,

>>> [d.resolution*(i+0.5)//d.resolution for i in range(-10,10)]
[-10, -8, -8, -6, -6, -4, -4, -2, -2, 0, 0, 2, 2, 4, 4, 6, 6, 8, 8, 10]

and
>>> [round(i+0.5) for i in range(-10,10)]
[-10, -8, -8, -6, -6, -4, -4, -2, -2, 0, 0, 2, 2, 4, 4, 6, 6, 8, 8, 10]
"""

I checked the documentation and while it says: "If any argument is a float and there are fractional microseconds, the fractional microseconds left over from all arguments are combined and their sum is rounded to the nearest microsecond." it does not specify how half-integers should be handled.

While it may not be a bug in strict sense, it looks like the code in question can be improved.  I'll open a separate issue for this.

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<http://bugs.python.org/issue1289118>
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