Bitwise XOR of hexadecimal numbers
bit-manipulation, hex, python, xor
Solution
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
Problem
How can we XOR hexadecimal numbers in Python? For example, I want to XOR `'ABCD'` and `'12EF'`, the answer should be `'B922'`. I used the code below, but it gives the wrong results. ``` # xor two strings of different lengths def strxor(a, b): 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)) ```