Change Default Multiple in Length for AES Encryption

Question

I created a function, that uses Crypto.Cipher:

    import os
    from Crypto.Cipher import AES

    def encrypt_pass(user_entered_plaintext):
        encrypt_setup = AES.new(os.urandom(32), AES.MODE_CBC, os.urandom(16))
        plaintext = user_entered_plaintext
        ciphertext = encrypt_setup.encrypt(plaintext)

        print(ciphertext)
        return ciphertext

if I try to run this, like:

encrypt_pass('Test')

Im getting this error:

ValueError: Input strings must be a multiple of 16 in length

if I try changing:

encrypt_setup = AES.new(os.urandom(32), AES.MODE_CBC, os.urandom(16))

to

encrypt_setup = AES.new(os.urandom(32), AES.MODE_CBC, os.urandom(8))

I get:

ValueError: IV must be 16 bytes long

How can I force this input string minimum to 8?

If someone runs across this thread, here is a good article to follow regarding this topic: http://www.davychiu.com/blog/aes-encryptiondecryption-in-python.html — CodeTalk, Dec 22 '15 at 22:31

zaph · Accepted Answer · 2015-12-22T22:37:50.117

Just specify PKCS#7 padding. You will have to add that prior to encryption and remove it after decryption because to bozos who wrote pycrypto did not add padding as an option. (And I thought PHP mcrypt was bad!) See the link by CodeTalk.

AES encryption is block based and the block size is 16-bytes so if the input data is not a multiple of the block size padding must be added.

In the second attempt you change the iv to 8-bytes but it must be the block size of 16-byte.

But you are not saving the key or iv so you will not be able to later decrypt the data. You need to use variable for them.

Change Default Multiple in Length for AES Encryption

1 Answers1