XOR Python 文本加密/解密

Question

I know there is a built in xor operator that can be imported in Python. I'm trying to execute the xor encryption/decryption. So far I have:

 def xor_attmpt():
    message = raw_input("Enter message to be ciphered: ")
    cipher = []
    for i in message:
        cipher.append(bin(ord(i))[2::])#add the conversion of the letters/characters
#in your message from ascii to binary withoout the 0b in the front to your ciphered message list
    cipher = "".join(cipher) 
    privvyKey = raw_input("Enter the private key: ")
    keydecrypt = []
    for j in privvyKey:
        keydecrypt.append(bin(ord(j))[2::]) #same
    keydecrypt = "".join(keydecrypt )#same

    print "key is '{0}'" .format(keydecrypt) #substitute values in string
    print "encrypted text is '{0}'" .format(cipher)
    from operator import xor
    for letter in message:
        print xor(bool(cipher), bool(keydecrypt))

This:

>  for letter in message:
    print xor(bool(cipher), bool(keydecrypt))

is where my python starts to go wrong.

The ouput looks like this

    Enter message to be ciphered: hello
Enter the private key: \@154>
key is '10111001000000110001110101110100111110'
encrypted text is '11010001100101110110011011001101111'
False
False
False
False
False

What I'm messing up on is trying to get these two binary(key and encrypted) to be compared and give true(1) or false(being 0). The xor should then give me a resulting 1 and 0 binary list from comparing the two. Any input?

Answer 1

Here's a variation of the code example from XOR Cipher Wikipedia article :

def xor(data, key): 
    return bytearray(a^b for a, b in zip(*map(bytearray, [data, key]))) 

Example (Python 2):

>>> one_time_pad = 'shared secret' 
>>> plaintext = 'unencrypted' 
>>> ciphertext = xor(plaintext, one_time_pad) 
>>> ciphertext 
bytearray(b'\x06\x06\x04\x1c\x06\x16Y\x03\x11\x06\x16') 
>>> decrypted = xor(ciphertext, one_time_pad) 
>>> decrypted
bytearray(b'unencrypted')
>>> plaintext == str(decrypted)
True

Answer 2

Code below works both ways and does not need length checking as cycle is used.

from itertools import cycle, izip
cryptedMessage = ''.join(chr(ord(c)^ord(k)) for c,k in izip(message, cycle(key)))

Answer 3

somecode = 'asdfln3j34tnonfdkjnflksdfnla'
message = 'this is my message'

def str_xor(s1, s2):
 return "".join([chr(ord(c1) ^ ord(c2)) for (c1,c2) in zip(s1,s2)])

encoded = str_xor(message, somecode)
# encoded == '\x15\x1b\r\x15L\x07@J^MT\x03\n\x1d\x15\x05\x0c\x0f'
decoded = str_xor(encoded, somecode)
# decoded == 'this is my message'

This is a naive / minimalistic implementation without error checking. Here len(somecode) >= len(message) is required.

Answer 4

for Python 3

from itertools import cycle

cryptedMessage = ''.join(chr(ord(c)^ord(k)) for c,k in zip(message, cycle(key)))