For reference—future Python possibilities:
Starting with Python 2.6 you can express binary literals using the prefix0b or0B:

>>> 0b10111147

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

>>> bin(173)'0b10101101'

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

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

Another way.

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 a0 followed by aB orb followed by a series of zeros and ones, for example:

>>> 0b0010101010170>>> 0B01010121

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

Integer literals are described by the following lexical definitions:

integer      ::=  decinteger | bininteger | octinteger | hexintegerdecinteger   ::=  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    0b1001101113     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 the0b or0B prefix:

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

If you pass it0 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 combinebin andint 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'

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

I've tried this in Python 3.6.9

Convert Binary to Decimal

>>> 0b10111147>>> 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

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

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.

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.

• python
• syntax
• binary
• integer
• literals