CODEWITHSUNDEEP
pythonencryptionaes
KamilKamil5481
KamilKamil5481

Reputation: 46

Python 2.7 crypto AES

I got a problem with aes in python 2.7

import pyelliptic
iv = pyelliptic.Cipher.gen_IV('aes-256-cfb')
ctx = pyelliptic.Cipher("secretkey", iv, 1, ciphername='aes-256-cfb')

ciphertext = ctx.update('test1')
ciphertext += ctx.final()

ctx2 = pyelliptic.Cipher("secretkey", iv, 0, ciphername='aes-256-cfb')

Now I don't know how to send this msg to server, and decrypt it on server, because I don't know the IV and my server can't decrypt it. The server has the secret key.

Upvotes: 0

Views: 1633

Answers (1)

Artjom B.
Artjom B.

Reputation: 61892

The IV does not need to be kept secret, but it needs to unique (random) for every encrypt operation with the same key.

Many implementations just add the IV bytes to the front of the ciphertext. You have to know how long the IV is for your implementation so that you can slice it off before decrypting.

# encrypt
ciphertext = iv + ciphertext

# decrypt
blocksize = pyelliptic.Cipher.get_blocksize('aes-256-cfb')
iv = ciphertext[0:blocksize]
ciphertext = ciphertext[blocksize:]

From the code it is apparent that the IV is generated in the same size as the cipher blocksize, so it is safe to slice a block from the ciphertext to get the IV.

Upvotes: 4

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