[Tutor] Octal confusion, please explain why. Python 3.2

[Jordan]

Thank you for the detailed answer, now I understand and I understand
that each number format is an integer just with a different base and
cosmetic appearance.

On 06/02/2012 01:51 PM, Steven D'Aprano wrote:
> Jordan wrote:
>> Hello, first off I am using Python 3.2 on Linux Mint 12 64-bit.
>> I am confused as to why I can not successfully compare a variable that
>> was created as an octal to a variable that is converted to an octal in a
>> if statement yet print yields that they are the same octal value.
>
> Because the oct() function returns a string, not a number.
Ahh, I see. That makes sense next time I will read the documentation on
the functions that I am using.
> py> oct(63)
> '0o77'
>
> In Python, strings are never equal to numbers:
>
> py> 42 == '42'
> False
>
>
> Octal literals are just regular integers that you type differently:
>
> py> 0o77
> 63
>
> "Octals" aren't a different type or object; the difference between the
> numbers 0o77 and 63 is cosmetic only. 0o77 is just the base 8 version
> of the base 10 number 63 or the binary number 0b111111.
>
>
> Because octal notation is only used for input, there's no need to
> covert numbers to octal strings to compare them:
>
> py> 0o77 == 63  # no different to "63 == 63"
> True
>
>
> The same applies for hex and bin syntax too:
>
> py> 0x1ff == 511
> True
>
>
>