How do you express binary literals in Python?

For reference—future Python possibilities: Starting with Python 2.6 you can express binary literals using the prefix 0b or 0B:

>>> 0b101111
47

You can also use the new bin function to get the binary representation of a number:

>>> bin(173)
'0b10101101'

Development version of the documentation: What's New in Python 2.6

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>>> print int('01010101111',2)
687
>>> print int('11111111',2)
255

Another way.

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How do you express binary literals in Python?

They're not "binary" literals, but rather, "integer literals". You can express integer literals with a binary format with a 0 followed by a B or b followed by a series of zeros and ones, for example:

>>> 0b0010101010
170
>>> 0B010101
21

From the Python 3 docs, these are the ways of providing integer literals in Python:

Integer literals are described by the following lexical definitions: integer ::= decinteger | bininteger | octinteger | hexinteger decinteger ::= nonzerodigit (["_"] digit)* | "0"+ (["_"] "0")* bininteger ::= "0" ("b" | "B") (["_"] bindigit)+ octinteger ::= "0" ("o" | "O") (["_"] octdigit)+ hexinteger ::= "0" ("x" | "X") (["_"] hexdigit)+ nonzerodigit ::= "1"..."9" digit ::= "0"..."9" bindigit ::= "0" | "1" octdigit ::= "0"..."7" hexdigit ::= digit | "a"..."f" | "A"..."F" There is no limit for the length of integer literals apart from what can be stored in available memory. Note that leading zeros in a non-zero decimal number are not allowed. This is for disambiguation with C-style octal literals, which Python used before version 3.0. Some examples of integer literals: 7 2147483647 0o177 0b100110111 3 79228162514264337593543950336 0o377 0xdeadbeef 100_000_000_000 0b_1110_0101 Changed in version 3.6: Underscores are now allowed for grouping purposes in literals.

Other ways of expressing binary:

You can have the zeros and ones in a string object which can be manipulated (although you should probably just do bitwise operations on the integer in most cases) - just pass int the string of zeros and ones and the base you are converting from (2):

>>> int('010101', 2)
21

You can optionally have the 0b or 0B prefix:

>>> int('0b0010101010', 2)
170

If you pass it 0 as the base, it will assume base 10 if the string doesn't specify with a prefix:

>>> int('10101', 0)
10101
>>> int('0b10101', 0)
21

Converting from int back to human readable binary:

You can pass an integer to bin to see the string representation of a binary literal:

>>> bin(21)
'0b10101'

And you can combine bin and int to go back and forth:

>>> bin(int('010101', 2))
'0b10101'

You can use a format specification as well, if you want to have minimum width with preceding zeros:

>>> format(int('010101', 2), '{fill}{width}b'.format(width=10, fill=0))
'0000010101'
>>> format(int('010101', 2), '010b')
'0000010101'

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I've tried this in Python 3.6.9

Convert Binary to Decimal

>>> 0b101111
47

>>> int('101111',2)
47

Convert Decimal to binary

>>> bin(47)
'0b101111'

Place a 0 as the second parameter python assumes it as decimal.

>>> int('101111',0)
101111

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0 in the start here specifies that the base is 8 (not 10), which is pretty easy to see:

>>> int('010101', 0)
4161

If you don't start with a 0, then python assumes the number is base 10.

>>> int('10101', 0)
10101

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Another good method to get an integer representation from binary is to use eval()

Like so:

def getInt(binNum = 0):
    return eval(eval('0b' + str(n)))

I guess this is a way to do it too. I hope this is a satisfactory answer :D

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As far as I can tell Python, up through 2.5, only supports hexadecimal & octal literals. I did find some discussions about adding binary to future versions but nothing definite.

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I am pretty sure this is one of the things due to change in Python 3.0 with perhaps bin() to go with hex() and oct().

EDIT: lbrandy's answer is correct in all cases.

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