bitwise XOR of hex numbers in python

Question

how can we XOR hex numbers in python eg. I want to xor 'ABCD' to '12EF'. answer should be B922.

i used below code but it is returning garbage value

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

key ='12ef'
m1='abcd'
print  strxor(key,m1)

Answer 1

Whoa. You're really over-complicating it by a very long distance. Try:

>>> print hex(0x12ef ^ 0xabcd)
0xb922

You seem to be ignoring these handy facts, at least:

• Python has native support for hexadecimal integer literals, with the 0x prefix.
• "Hexadecimal" is just a presentation detail; the arithmetic is done in binary, and then the result is printed as hex.
• There is no connection between the format of the inputs (the hexadecimal literals) and the output, there is no such thing as a "hexadecimal number" in a Python variable.
• The hex() function can be used to convert any number into a hexadecimal string for display.

If you already have the numbers as strings, you can use the int() function to convert to numbers, by providing the expected base (16 for hexadecimal numbers):

>>> print int("12ef", 16)
4874

So you can do two conversions, perform the XOR, and then convert back to hex:

>>> print hex(int("12ef", 16) ^ int("abcd", 16))
0xb922

Answer 2

If the two hex strings are the same length and you want a hex string output then you might try this.

def hexxor(a, b):    # xor two hex strings of the same length
    return "".join(["%x" % (int(x,16) ^ int(y,16)) for (x, y) in zip(a, b)])

Answer 3

If the strings are the same length, then I would go for '%x' % () of the built-in xor ( ^ ).

Examples -

>>>a = '290b6e3a'
>>>b = 'd6f491c5'
>>>'%x' % (int(a,16)^int(b,16))
'ffffffff'
>>>c = 'abcd'
>>>d = '12ef'
>>>'%x' % (int(a,16)^int(b,16))
'b922'

If the strings are not the same length, truncate the longer string to the length of the shorter using a slice longer = longer[:len(shorter)]

Answer 4

here's a better function

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 5

For performance purpose, here's a little code to benchmark these two alternatives:

#!/bin/python

def hexxorA(a, b):
    if len(a) > len(b):
        return "".join(["%x" % (int(x,16) ^ int(y,16)) for (x, y) in zip(a[:len(b)], b)])
    else:
        return "".join(["%x" % (int(x,16) ^ int(y,16)) for (x, y) in zip(a, b[:len(a)])])

def hexxorB(a, b):
    if len(a) > len(b):
        return '%x' % (int(a[:len(b)],16)^int(b,16))
    else:
        return '%x' % (int(a,16)^int(b[:len(a)],16))

def testA():
    strstr = hexxorA("b4affa21cbb744fa9d6e055a09b562b87205fe73cd502ee5b8677fcd17ad19fce0e0bba05b1315e03575fe2a783556063f07dcd0b9d15188cee8dd99660ee751", "5450ce618aae4547cadc4e42e7ed99438b2628ff15d47b20c5e968f086087d49ec04d6a1b175701a5e3f80c8831e6c627077f290c723f585af02e4c16122b7e2")
    if not int(strstr, 16) == int("e0ff3440411901bd57b24b18ee58fbfbf923d68cd88455c57d8e173d91a564b50ce46d01ea6665fa6b4a7ee2fb2b3a644f702e407ef2a40d61ea3958072c50b3", 16):
        raise KeyError
    return strstr

def testB():
    strstr = hexxorB("b4affa21cbb744fa9d6e055a09b562b87205fe73cd502ee5b8677fcd17ad19fce0e0bba05b1315e03575fe2a783556063f07dcd0b9d15188cee8dd99660ee751", "5450ce618aae4547cadc4e42e7ed99438b2628ff15d47b20c5e968f086087d49ec04d6a1b175701a5e3f80c8831e6c627077f290c723f585af02e4c16122b7e2")
    if not int(strstr, 16) == int("e0ff3440411901bd57b24b18ee58fbfbf923d68cd88455c57d8e173d91a564b50ce46d01ea6665fa6b4a7ee2fb2b3a644f702e407ef2a40d61ea3958072c50b3", 16):
        raise KeyError
    return strstr

if __name__ == '__main__':
    import timeit
    print("Time-it 100k iterations :")
    print("\thexxorA: ", end='')
    print(timeit.timeit("testA()", setup="from __main__ import testA", number=100000), end='s\n')
    print("\thexxorB: ", end='')
    print(timeit.timeit("testB()", setup="from __main__ import testB", number=100000), end='s\n')

Here are the results :

Time-it 100k iterations :
    hexxorA: 8.139988073991844s
    hexxorB: 0.240523161992314s

Seems like '%x' % (int(a,16)^int(b,16)) is faster then the zip version.