python not accepting 09 but accepts 07

Question

Does anyone why python handles the below this way.

>>> a = 099
  File "<stdin>", line 1
    a = 099
          ^
SyntaxError: invalid token
>>> a = 088
  File "<stdin>", line 1
    a = 088
          ^
SyntaxError: invalid token
>>> a = 0559
  File "<stdin>", line 1
    a = 0559
           ^
SyntaxError: invalid token
>>> a = 077
>>>

It does not seem to accept numbers starting with 0 and preceding with 8 or 9. If it is some other number, it is not throwing any error. Why is that?

`099` is interpreted as octal number, but 9 is not valid octal digit — myaut, Apr 05 '17 at 11:47
also see: http://stackoverflow.com/questions/43071916/why-does-000-evaluate-to-0-in-python-3 — Jean-François Fabre, Apr 05 '17 at 11:48

RemcoGerlich · Answer 1 · 2017-04-05T11:51:10.997

2

In Python 2, like in C, an integer literal that starts with a 0 is in octal. Digits 8 and 9 do not exist in octal (they are written 010 and 011 respectively) so that is a syntax error.

>>> 010
8
>>> 08
  File "<stdin>", line 1
    008
      ^
SyntaxError: invalid token

In Python 3, this feature not many people know about is gone. There, nonzero literals that start with a 0 are syntax errors.

edited Apr 05 '17 at 11:51

answered Apr 05 '17 at 11:47

RemcoGerlich

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not _all_ literals: http://stackoverflow.com/questions/43071916/why-does-000-evaluate-to-0-in-python-3 – Jean-François Fabre Apr 05 '17 at 11:49

python not accepting 09 but accepts 07

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