use a 32 byte initialisation vector in pycrypto

Asked Jun 08 '18 at 12:59

Active Jun 08 '18 at 14:14

Viewed 457 times

I am trying to decrypt a text using python with pycrypto. But I get: ValueError: IV must be 16 bytes long

However my iv is 32 bytes long and I cant change that. My key is also 32 bytes in size Is there a way to do this with pycrypto? I could find anything in the docs.

import Crypto.Cipher.AES as AES
a=0x0
i='68fe3a4cf921122eda3df6c7d5458497'
print(i)
plain="encrypted text"
print(plain)
length = 16 - (len(plain) % 16)
plain += chr(length)*length
print(plain)
plaintext=""

for a in range(0, 0xffffff):
    s="6af6f9fae1ab621c2aba5f80ea4"+hex(a)[2:].zfill(5)
    print(s)
    cipher = AES.new(key=s, mode=AES.MODE_CBC, IV=i)
    text=cipher.decrypt(plain)
    if isinstance(text, unicode):    plaintext=text

print(plaintext)

edited Jun 08 '18 at 14:14

Alex K.

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asked Jun 08 '18 at 12:59

Jacknife

What key/block size/mode are you using that permits a 256 bit IV? – Alex K. Jun 08 '18 at 13:06
Nowhere near enough information. What algorithm are you using? Is the key in hex? – Luke Joshua Park Jun 08 '18 at 13:09
Please add your code – Spangen Jun 08 '18 at 13:14
https://pastebin.com/NrtFkYBf – Jacknife Jun 08 '18 at 13:49
2

`i` is a hex string & is likely supposed to be decoded to bytes `i.decode('hex')` which will give you the correct 16 byte length. – Alex K. Jun 08 '18 at 14:15

use a 32 byte initialisation vector in pycrypto

0 Answers0