byte operations (XOR) in python

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孤城傲影
孤城傲影 2021-02-04 07:00
    #!/usr/bin/env python3

import binascii


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

print(binascii.b2a_qp(var))
print(binasc         


        
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  • 2021-02-04 07:51

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

    Simple tests:

    assert encrypt1(b'hello', b'supersecretkey') == b'\x1b\x10\x1c\t\x1d'
    assert encrypt2(b'hello', b'supersecretkey') == b'\x1b\x10\x1c\t\x1d'
    

    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.

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  • 2021-02-04 08:06

    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"hello\u2764"
    >>> 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'hello\xe2\x9d\xa4'
    
    >>> bytes(decrypted).decode()
    'hello❤'
    
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