Why does Python change the value of an integer when there is a 0 in front of it?

Question

I implemented a function converting an integer number to its representation as a string intToStr() (code below).

For testing I\'ve passed in some values

Answer 1

An integer literal starting with a 0 is interpreted as an octal number, base 8:

>>> 01223
659

This has been changed in Python 3, where integers with a leading 0 are considered errors:

>>> 01223
  File "<stdin>", line 1
    01223
        ^
SyntaxError: invalid token
>>> 0o1223
659

You should never specify an integer literal with leading zeros; if you meant to specify an octal number, use 0o to start it, otherwise strip those zeros.

Answer 2

Numbers that start with a 0 are interpreted as octal numbers. If it starts with 0x it's hexa decimal.

Answer 3

As others have said that's because of octal numbers. But I strongly suggest you to change your function to:

>>> from functools import partial
>>> force_decimal = partial(int, base=10)
>>> force_decimal("01")
1
>>> force_decimal("0102301")
102301

This way you will explicitly force the conversion to base 10. And int wont be inferring it for you.

Answer 4

A leading zero causes Python to interpret your number as octal (base-8).

To strip out the zeros (assuming num is a string), do:

num.lstrip("0")