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

Answer 1

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.