CODEWITHSUNDEEP
pythonhexxor
user1427661
user1427661

Reputation: 11774

Xor Hex/ASCII/Conversion Issue

So I have a problem where I want to xor various hex strings, convert them to regular english strings, then re-convert them to hex strings. I'm not really familiar with working with hex or xor in any meaningful way, however. Do I need to convert the hex to binary or unicode before I perform a bitwise xor operation? If so, how do I retrieve the hex values once that is done? I've been looking into using str.encode('hex') and str.decode('hex'), but I keep getting errors saying that I am using non-hexadecimal characters. In short, I'm totally lost.

Upvotes: 1

Views: 4026

Answers (2)

pragmatic
pragmatic

Reputation: 95

@user1427661: you are seeing the same output as one of the input(say input1) because -

len(input1) > len(input2)

What you possibly can do now is apply a check on the length of the two strings and strip the larger one to match the size of the smaller one (because rest of the part is anyways useless) with something like this-

if len(input1) > len(input2):

          input1 = input1[:len(b)]

Likewise an else condition.

Let me give you a more simpler answer (ofcourse in my opinion!). You may use the in-built 'operator' package and then directly use the xor method in it.

http://docs.python.org/2/library/operator.html

Hope this helps.

Upvotes: 1

schesis
schesis

Reputation: 59118

Python has an XOR operator for integers: ^. Here's how you could use it:

>>> hex(int("123abc", 16) ^ int("def456", 16))
'0xccceea'

EDIT: testing with long hex strings as per your comment:

>>> def hexor(hex1, hex2):
...     """XOR two hex strings."""
...     xor = hex(int(hex1, 16) ^ int(hex2, 16))
...     return xor[2:].rstrip("L")  # get rid of "0x" and maybe "L"
... 
>>> import random
>>> a = "".join(random.choice("0123456789abcdef") for i in range(200))
>>> b = "".join(random.choice("0123456789abcdef") for i in range(200))
>>> a
'8db12de2f49f092620f6d79d6601618daab5ec6747266c2eea29c3493278daf82919aae6a72
64d4cf3dffd70cb1b6fde72ba2a04ac354fcb871eb60e088c2167e73006e0275287de6fc6133
56e44d7b0ff8378a0830d9d87151cbf3331382b096f02fd72'
>>> b
'40afe17fa8fbc56153c78f504e50a241df0a35fd204f8190c0591eda9c63502b41611aa9ac2
27fcd1a9faea642d89a3a212885711d024d2c973115eea11ceb6a57a6fa1f478998b94aa7d3e
993c04d24a0e1ac7c10fd834de61caefb97bcb65605f06eae'
>>> hexor(a, b)
'cd1ecc9d5c64cc47733158cd2851c3cc75bfd99a6769edbe2a70dd93ae1b8ad36878b04f0b0
43281e94053d689c3f5e45392af75b13702e7102fa3e0a990ca0db096fcff60db1f672561c0d
cfd849a945f62d4dc93f01ecaf30011c8a6849d5f6af293dc'

Upvotes: 1

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