Is Python incorrectly handling this “arbitrary precision integer”?

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Actually in python3 whenever you divide ints you get float as a result. There is a // operator that does integer division: 

 >>> 999999999999999999999999/3
 3.333333333333333e+23
 >>> 999999999999999999999999//3
 333333333333333333333333

 >>> type(999999999999999999999999/3)
 <class 'float'>
 >>> type(999999999999999999999999//3)
 <class 'int'>

This does give the correct arbitrary precision output: 

 python -c 'print("%d" % (999999999999999999999999//3))' 
 333333333333333333333333

How to write code compatible with both python 2.2+ and 3.3

This is actually simple, just add: 

 >>> from __future__ import division

this will enable 3.X division in 2.2+ code. 

>>> from sys import version 
>>> version
'2.7.6 (default, Dec 30 2013, 14:37:40) \n[GCC 4.8.2]'
>>> from __future__ import division 
>>> type(999999999999999999999999//3)
<type 'long'>
>>> type(999999999999999999999999/3)
<type 'float'>

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