简体   繁体   中英

How to XOR two strings that contain hex numbers in python?

I have looked for an answer online, but none of them seem to solve my problem in my way (I know, I'm picky :D).

Here's the deal: I am using the string type to store two hex numbers, because the default integer type in python is not long enough for my purposes. For example like this:

S1 = "315c4eeaa8b5f8aaf9174145bf43e1784b8fa00dc71d885a804e5ee9fa40b16349c146fb778cdf2d3aff021dfff5b403b510d0d0455468aeb98622b137dae857553ccd8883a7bc37520e06e515d22c954eba5025b8cc57ee59418ce7dc6bc41556bdb36bbca3e8774301fbcaa3b83b220809560987815f65286764703de0f3d524400a19b159610b11ef3e"
S2 = "234c02ecbbfbafa3ed18510abd11fa724fcda2018a1a8342cf064bbde548b12b07df44ba7191d9606ef4081ffde5ad46a5069d9f7f543bedb9c861bf29c7e205132eda9382b0bc2c5c4b45f919cf3a9f1cb74151f6d551f4480c82b2cb24cc5b028aa76eb7b4ab24171ab3cdadb8356f"

The point is, that these are supposed to be NUMBERS, but they are stored in a string. What I want to do, is to treat these two strings as numbers, perform a bitwise XOR on the two and then get an output in a similar form - a hex number stored in a string.

I am rather new to programming and even newer to python, so I couldn't figure out a way to do this. I am not just looking for a script, I would also like to understand how it works, so please be thorough with your explanation, as I am quite the noob :D.

Python can hold your values as numbers.

See this for proof,

>>> hex(int(S1, 16))[2:-1] == S1
True

I'm simply adjusting the string, removing '0x' from the beginning and L from the end.

For your answer all you need to do is,

hex(int(S1, 16) ^ int(S2, 16))
  1. Reverse both strings to s1_r and s2_r
  2. Do a char by char xor (char1 (from s1_r) xor char2 (from s2_r))
  3. Append result to xor_str
  4. Reverse xor_str

To perform the XOR operation assuming they are long integers, you can use the long type in Python:

# Convert the hex string S1 and S2 to longs
l1=long(S1,16) 
l2=long(S2,16)
result=hex(l1 ^ l2) # Convert the XOR of the strings
# Output will be:
# '0x1013abb8b0ead34450ee04e8d507fa16552e5aa9f2cc9551acc9d71b646f8a9b4f3548f2068172b201bf0daf75bdddd0dedd861b9ccbcf7c9ce53e39ecafa9c86880fba0c600778fc7bc6e3bd60c8b0df469f5a7f1da4339f9202bdb43b97b22db69642ce5402b8ce44f86d990dbf5a2L'

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM