问题
#!/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 :(.
回答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"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❤'
回答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):
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.
来源:https://stackoverflow.com/questions/29408173/byte-operations-xor-in-python