byte operations (XOR) in python

Question

#!/usr/bin/env python3

import binascii


var=binascii.a2b_qp("hello")
key=binascii.a2b_qp("supersecretkey")[:len(var)]

print(binascii.b2a_qp(var))
print(binascii.b2a_qp(key))


# here I want to do an XOR operation on the bytes in var and key and place them in 'encryption': encryption=var XOR key

print(binascii.b2a_qp(encrypted))

If someone could enlighten me on how I could accomplish this I would be very happy. Very new to the whole data-type conversions so yeah… reading through the python wiki is not as clear as I would like.

Asked By: Jcov

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Source

Answer 1

It looks like what you need to do is XOR each of the characters in the message with the corresponding character in the key. However, to do that you need a bit of interconversion using ord and chr, because you can only xor numbers, not strings:

>>> encrypted = [ chr(ord(a) ^ ord(b)) for (a,b) in zip(var, key) ] 
>>> encrypted
['x1b', 'x10', 'x1c', 't', 'x1d']

>>> decrypted = [ chr(ord(a) ^ ord(b)) for (a,b) in zip(encrypted, key) ]
>>> decrypted
['h', 'e', 'l', 'l', 'o']

>>> "".join(decrypted)
'hello'

Note that binascii.a2b_qp("hello") just converts a string to another string (though possibly with different encoding).

Your approach, and my code above, will only work if the key is at least as long as the message. However, you can easily repeat the key if required using itertools.cycle:

>>> from itertools import cycle
>>> var="hello"
>>> key="xy"

>>> encrypted = [ chr(ord(a) ^ ord(b)) for (a,b) in zip(var, cycle(key)) ]
>>> encrypted
['x10', 'x1c', 'x14', 'x15', 'x17']

>>> decrypted = [ chr(ord(a) ^ ord(b)) for (a,b) in zip(encrypted, cycle(key)) ]
>>> "".join(decrypted)
'hello'

To address the issue of unicode/multi-byte characters (raised in the comments below), one can convert the string (and key) to bytes, zip these together, then perform the XOR, something like:

>>> var=u"hellou2764"
>>> var
'hello❤'

>>> encrypted = [ a ^ b for (a,b) in zip(bytes(var, 'utf-8'),cycle(bytes(key, 'utf-8'))) ]
>>> encrypted
[27, 16, 28, 9, 29, 145, 248, 199]

>>> decrypted = [ a ^ b for (a,b) in zip(bytes(encrypted), cycle(bytes(key, 'utf-8'))) ]
>>> decrypted
[104, 101, 108, 108, 111, 226, 157, 164]

>>> bytes(decrypted)
b'helloxe2x9dxa4'

>>> bytes(decrypted).decode()
'hello❤'

Answered By: DNA

Answer 2

Comparison of two python3 solutions

The first one is based on zip:

def encrypt1(var, key):
    return bytes(a ^ b for a, b in zip(var, key))

The second one uses int.from_bytes and int.to_bytes:

def encrypt2(var, key, byteorder=sys.byteorder):
    key, var = key[:len(var)], var[:len(key)]
    int_var = int.from_bytes(var, byteorder)
    int_key = int.from_bytes(key, byteorder)
    int_enc = int_var ^ int_key
    return int_enc.to_bytes(len(var), byteorder)

Simple tests:

assert encrypt1(b'hello', b'supersecretkey') == b'x1bx10x1ctx1d'
assert encrypt2(b'hello', b'supersecretkey') == b'x1bx10x1ctx1d'

Performance tests with var and key being 1000 bytes long:

$ python3 -m timeit 
  -s "import test_xor;a=b'abcdefghij'*100;b=b'0123456789'*100" 
  "test_xor.encrypt1(a, b)"
10000 loops, best of 3: 100 usec per loop

$ python3 -m timeit 
  -s "import test_xor;a=b'abcdefghij'*100;b=b'0123456789'*100" 
  "test_xor.encrypt2(a, b)"
100000 loops, best of 3: 5.1 usec per loop

The integer approach seems to be significantly faster.

Answered By: Vincent

Answer 3

You can use Numpy to perform faster

import numpy as np
def encrypt(var, key):
    a = np.frombuffer(var, dtype = np.uint8)
    b = np.frombuffer(key, dtype = np.uint8)
    return (a^b).tobytes()

Answered By: Latze

byte operations (XOR) in python

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Comparison of two python3 solutions