5

About half an hour thinking "what am i doing wrong!?" on the 5-lines code.. because Python3 is somehow rounding big integers. Anyone know why there is a problem such:

Python2:

int(6366805760909027985741435139224001        # This is 7**40.
    / 7) == 909543680129861140820205019889143 # 7**39

Python3:

int(6366805760909027985741435139224001 
    / 7) == 909543680129861204865300750663680 # I have no idea what this is.
Eryk Sun
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inexxt
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2 Answers2

7

Python 3 is not "rounding big integers". What it does is that it will return a float after division. Hence, in Python 2:

>>> 4/2
2

while in Python 3:

>>> 4/2
2.0

The reason for this is simple. In Python 2, / being integer division when you use integers have some surprising results:

>>> 5/2
2

Ooops. In Python 3 this is fixed:

>>> 5/2
2.5

This means that in Python 3, your division returns a float:

>>> 6366805760909027985741435139224001/7
9.095436801298612e+32

This float has less accuracy than the digits you need. You then convert this to an integer with int(), and you get a number you don't expect.

You should instead use integer division (in both Python 2 and Python 3):

>>> 6366805760909027985741435139224001//7
909543680129861140820205019889143L

(The trailing L means it's a long integer, in Python 3 the long and the normal integer is merged, so there is no trailing L).

Lennart Regebro
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5

In Python 3 / is floating point division so it may not treat your arguments like integers. Use 

//

to do integer division in Python 3.

James King
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    you could also use `//` in Python 2 i.e., the same source could be used on Python 2 and 3. `(7**40 // 7) == 7**39`. `from __future__ import division` enables Python 3 behaviour for `/` on Python 2. – jfs Dec 28 '13 at 03:55