How do you express binary literals in Python?
Solution 1:
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
Solution 2:
>>> print int('01010101111',2)
687
>>> print int('11111111',2)
255
Another way.
Solution 3:
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'
Solution 4:
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
Solution 5:
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