Is Python incorrectly handling this "arbitrary precision integer"?
arbitrary-precision, python
Solution
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'>
Problem
Python is supposed to have "arbitrary precision integers," according to the answer in Python integer ranges. But this result is plainly not arbitrary precision: ``` $ python -c 'print("%d" % (999999999999999999999999/3))' 333333333333333327740928 ``` According to PEP 237, `bignum` is arbitrarily large (not just the size of C's `long` type). And Wikipedia says Python's `bignum` is arbitrary precision. So why the incorrect result from the above line of code?