how to do bitwise exclusive or of two strings in python?

Question

I would like to perform a bitwise exclusive or of two strings in python, but xor of strings are not allowed in python. How can I do it ?

Answer 1

You can convert the characters to integers and xor those instead:

l = [ord(a) ^ ord(b) for a,b in zip(s1,s2)]

Here's an updated function in case you need a string as a result of the XOR:

def sxor(s1,s2):    
    # convert strings to a list of character pair tuples
    # go through each tuple, converting them to ASCII code (ord)
    # perform exclusive or on the ASCII code
    # then convert the result back to ASCII (chr)
    # merge the resulting array of characters as a string
    return ''.join(chr(ord(a) ^ ord(b)) for a,b in zip(s1,s2))

See it working online: ideone

Answer 2

the one liner for python3 is :

def bytes_xor(a, b) :
    return bytes(x ^ y for x, y in zip(a, b))

where a, b and the returned value are bytes() instead of str() of course

can't be easier, I love python3 :)

Answer 3

For bytearrays you can directly use XOR:

>>> b1 = bytearray("test123")
>>> b2 = bytearray("321test")
>>> b = bytearray(len(b1))
>>> for i in range(len(b1)):
...   b[i] = b1[i] ^ b2[i]

>>> b
bytearray(b'GWB\x00TAG')

Answer 4

Based on William McBrine's answer, here is a solution for fixed-length strings which is 9% faster for my use case:

import itertools
import struct
def make_strxor(size):
    def strxor(a, b, izip=itertools.izip, pack=struct.pack, unpack=struct.unpack, fmt='%dB' % size):
        return pack(fmt, *(a ^ b for a, b in izip(unpack(fmt, a), unpack(fmt, b))))
    return strxor
strxor_3 = make_strxor(3)
print repr(strxor_3('foo', 'bar'))

Answer 5

If the strings are not even of equal length, you can use this

def strxor(a, b):     # xor two strings of different lengths
    if len(a) > len(b):
        return "".join([chr(ord(x) ^ ord(y)) for (x, y) in zip(a[:len(b)], b)])
    else:
        return "".join([chr(ord(x) ^ ord(y)) for (x, y) in zip(a, b[:len(a)])])

Answer 6

I've found that the ''.join(chr(ord(a)^ord(b)) for a,b in zip(s,m)) method is pretty slow. Instead, I've been doing this:

fmt = '%dB' % len(source)
s = struct.unpack(fmt, source)
m = struct.unpack(fmt, xor_data)
final = struct.pack(fmt, *(a ^ b for a, b in izip(s, m)))