How do I manipulate bits in Python?

Question

How do I manipulate bits in Python?

In C I could, for example, zero out bit #10 in a 32 bit unsigned value like so:

unsigned long value = 0xdeadbeef;
value &= ~(1<<10);

How do I do that in Python ?

python

asked on Stack Overflow Sep 29, 2008 by

Eden

9 Answers

Bitwise operations on Python ints work much like in C. The &, | and ^ operators in Python work just like in C. The ~ operator works as for a signed integer in C; that is, ~x computes -x-1.

You have to be somewhat careful with left shifts, since Python integers aren't fixed-width. Use bit masks to obtain the low order bits. For example, to do the equivalent of shift of a 32-bit integer do (x << 5) & 0xffffffff.

answered on Stack Overflow Sep 29, 2008 by

Fredrik Johansson • edited Apr 20, 2018 by

heemayl

value = 0xdeadbeef
value &= ~(1<<10)

answered on Stack Overflow Sep 29, 2008 by

Greg Hewgill

Some common bit operations that might serve as example:

def get_bit(value, n):
    return ((value >> n & 1) != 0)

def set_bit(value, n):
    return value | (1 << n)

def clear_bit(value, n):
    return value & ~(1 << n)

Usage e.g.

>>> get_bit(5, 2)
True
>>> get_bit(5, 1)
False
>>> set_bit(5, 1)
7
>>> clear_bit(5, 2)
1 
>>> clear_bit(7, 2)
3

answered on Stack Overflow Apr 20, 2018 by

Boern

You should also check out BitArray, which is a nice interface for dealing with sequences of bits.

answered on Stack Overflow Sep 30, 2008 by user19883

Have you tried copying and pasting your code into the Python REPL to see what will happen?

>>> value = 0xdeadbeef
>>> value &= ~(1<<10)
>>> hex (value)
'0xdeadbaef'

answered on Stack Overflow Sep 29, 2008 by

John Millikin

Omit the 'unsigned long', and the semi-colons are not needed either:

value = 0xDEADBEEF
value &= ~(1<<10)
print value
"0x%08X" % value

answered on Stack Overflow Sep 29, 2008 by

Jonathan Leffler

Python has C style bit manipulation operators, so your example is literally the same in Python except without type keywords.

value = 0xdeadbeef
value &= ~(1 << 10)

answered on Stack Overflow Sep 29, 2008 by

Martin W

If you're going to do a lot of bit manipulation ( and you care much more about readability rather than performance for your application ) then you may want to create an integer wrapper to enable slicing like in Verilog or VHDL:

 import math
 class BitVector:
     def __init__(self,val):
         self._val = val

     def __setslice__(self,highIndx,lowIndx,newVal):
         assert math.ceil(math.log(newVal)/math.log(2)) <= (highIndx-lowIndx+1)

         # clear out bit slice
         clean_mask = (2**(highIndx+1)-1)^(2**(lowIndx)-1)

         self._val = self._val ^ (self._val & clean_mask)
         # set new value
         self._val = self._val | (newVal<<lowIndx)

     def __getslice__(self,highIndx,lowIndx):
         return (self._val>>lowIndx)&(2L**(highIndx-lowIndx+1)-1)

 b = BitVector(0)
 b[3:0]   = 0xD
 b[7:4]   = 0xE
 b[11:8]  = 0xA
 b[15:12] = 0xD

 for i in xrange(0,16,4):
     print '%X'%b[i+3:i]

Outputs:

 D
 E
 A
 D

answered on Stack Overflow Sep 29, 2008 by

Ross Rogers

a = int('00001111', 2)
b = int('11110000', 2)
bin(a & b)[2:].zfill(8)
bin(a | b)[2:].zfill(8)
bin(a << 2)[2:].zfill(8)
bin(a >> 2)[2:].zfill(8)
bin(a ^ b)[2:].zfill(8)
int(bin(a | b)[2:].zfill(8), 2)

answered on Stack Overflow Aug 8, 2017 by

witold-gren

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