syntax - How do you express binary literals in Python?


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How do you express an integer as a binary number with Python literals?

I was easily able to find the answer for hex:

>>> 0x12AF
4783
>>> 0x100
256

and octal:

>>> 01267
695
>>> 0100
64

How do you use literals to express binary in Python?


Summary of Answers

  • Python 2.5 and earlier: can express binary using int('01010101111',2) but not with a literal.
  • Python 2.5 and earlier: there is no way to express binary literals.
  • Python 2.6 beta: You can do like so: 0b1100111 or 0B1100111.
  • Python 2.6 beta: will also allow 0o27 or 0O27 (second character is the letter O) to represent an octal.
  • Python 3.0 beta: Same as 2.6, but will no longer allow the older 027 syntax for octals.

所有的回答
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    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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    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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    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.