Overflow error checking in Python

Question

import math
    if n < 0:
        raise ValueError("n must be >= 0")
    if math.floor(n) != n:
        raise ValueError("n must be exact integer")
    if n+1 == n:  # catch a value like 1e300
        raise OverflowError("n too large")
    result =1;
    factor =2;

    while factor <=n:
        result *= factor
        factor += 1

    return result

This is a factorial function. I don't understand "if n+1 == n:", then it indicates a 'too large integer' exception. Why is it that?

The `float` type has a precision problem. Python `int` does not. `1e16==1e16+1` is `True` due to the loss of precision, but `10000000000000000==10000000000000001` is True, even though they visually are the same. Python integers are only limited by available memory. Better to check that the user entered an integer and not a float. — Mark Tolonen, Dec 14 '16 at 02:02

score 0 · Accepted Answer · edited May 23 '17 at 12:08

0

Python has a limit to the size of a number that it can use. Conceptually, if Python's max number is 10,000 and you enter in the number 10,001, Python will truncate that number to 10,000, thus num == num+1.

Here's a better answer for more details: What is the maximum float in Python?.

edited May 23 '17 at 12:08

Community

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answered Dec 14 '16 at 01:31

David542

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That truncation is a smart way to indicate a too large number. Java doesn't work this way. – user697911 Dec 14 '16 at 01:41

score 0 · Answer 2 · answered Dec 14 '16 at 01:51

First off, I don't know why this program exists, because python has no limit to the highest integer (https://docs.python.org/3/library/stdtypes.html#typesnumeric). However, as the previous person says, this guards against truncation, with values like "1e300" where the difference between 10^300 and 10^300 + 1 is so small relatively that it ignores small values.

Overflow error checking in Python

2 Answers2