Document aeios.AES.rekey() and refactor arg validation

[isacben]

Hi,

This PR solves: aeios.AES.rekey() is not documented #7435

a) Documented the aeios.AES.rekey() function
b) Refactored the arguments validation. Please let me know if that was the idea of "canonicalize arg checking".

Tested with adafruit_feather_rp2040 with the following code:

import aesio
import os
from binascii import hexlify, unhexlify

plaintext = unhexlify("6bc1bee22e409f96e93d7e117393172a")
key = unhexlify("2b7e151628aed2a6abf7158809cf4f3c")
iv = unhexlify("000102030405060708090a0b0c0d0e0f")

cyphertext = bytearray(len(plaintext))
output = memoryview(cyphertext)
cipher = aesio.AES(key=key, mode=aesio.MODE_CBC, IV=iv)

# Encrypt
cipher.encrypt_into(plaintext, output)

print(end='\n\n')
print('input:     ', str(hexlify(plaintext)), end='\n')
print('encrypted: ', str(hexlify(output)), end='\n')

# Decrypt
plaintext = bytearray(len(plaintext))
cipher = aesio.AES(key=key, mode=aesio.MODE_CBC, IV=iv)

output = memoryview(plaintext)
cipher.decrypt_into(cyphertext, output)
print('decrypted: ', str(hexlify(output)), end='\n')

# Rekey
plaintext = unhexlify("6bc1bee22e409f96e93d7e117393172a")
key2 = unhexlify("2b7e151628aed2a6abf7158809cAAAAA")
cyphertext = bytearray(len(plaintext))
output = memoryview(cyphertext)

cipher.rekey(key=key2, IV=iv) # update key of the existing cipher object

# Encrypt
cipher.encrypt_into(plaintext, output)

print(end='\n')
print('Rekey...', end='\n')
print('input:     ', str(hexlify(plaintext)), end='\n')
print('encrypted: ', str(hexlify(output)), end='\n')

# Decrypt
plaintext = bytearray(len(plaintext))
cipher = aesio.AES(key=key2, mode=aesio.MODE_CBC, IV=iv)

output = memoryview(plaintext)
cipher.decrypt_into(cyphertext, output)
print('decrypted: ', str(hexlify(output)), end='\n\n')

Output:

input:      b'6bc1bee22e409f96e93d7e117393172a'
encrypted:  b'7649abac8119b246cee98e9b12e9197d'
decrypted:  b'6bc1bee22e409f96e93d7e117393172a'

Rekey...
input:      b'6bc1bee22e409f96e93d7e117393172a'
encrypted:  b'ff7e2905e0af08d89121429ee15b311b'
decrypted:  b'6bc1bee22e409f96e93d7e117393172a'

Thank you!

[tannewt]

One question. Looks good otherwise. Thank you!

[tannewt]

One more spot. Thanks!

[tannewt]

Thank you!