Question

我正在尝试使用pycryptodome库实现AES加密的OFB模式，并且在理解应该给密码输入什么方面遇到问题。我知道pycryptodome已经实现了OFB，但是我需要将纯文本分解为字节，应用OFB的正确步骤，以ECB模式的AES密码对其进行加密，然后对其进行解密。

诸如b'\x16\xa8W\xed.)\xc4\xb8x\xd6\xcf\x7f\xf3\xe3;^'之类的字节字符串如何在python中工作？

我需要使用这样的字节字符串，将其加密，然后将其分成两半，然后与8个纯文本字节进行异或。对于我来说，理解和执行此操作最简单的方法是使用AES对IV（上面的字节字符串）进行加密，然后将其和纯文本转换为二进制，然后将它们进行异或，然后再将其转换回字节字符串。我该怎么办？

from Crypto.Cipher import AES
from Crypto.Random import get_random_bytes

plainText = b"Lorem ipsum dolor sit amet, consectetur adipiscing e"
key = b"ANAAREMEREAAAAAA"
iv = get_random_bytes(32)

print("iv: ", iv)

cipher = AES.new(key, AES.MODE_ECB)
toXor = cipher.encrypt(iv)

print("toXor: ", toXor)

"""
toXorF=first half of toXor
ivS=second half of iv

cipherText=toXorF xored to first 8 bytes of plainText
iv=ivS + toXorS
"""

打印输出：

iv:  b"v'xg;\xd7h\xfc\xf2\xa4[\r\xee]J\x1eu\xa5\xe68\xa3a\xde\x02(\xc1\xe0\xc9z\x0f\xc3n"
toXor:  b'\x97\xbex\xfc\xb6\xbb\x85\xccZ\xc4\xe4\x9d\xd6M\xf2\xd7\xb7\xbf\xd0\xbe\xa5A\xd6\xee\x07U\xe0S\x7f\x88\xea\xcd'

如果您对我的程序/解决方案的架构有任何建议，请随时向我启发。

Answer 1

我应用了t.m.adam的建议，这是完美的最终代码。

from Crypto.Cipher import AES
from Crypto.Random import get_random_bytes

plainText = b"Lorem ipsum dolor sit amet, consectetur adipiscing e"
key = b"ANAAREMEREAAAAAA"


def ofbEnc(plainText, key):
    pos = 0
    cipherTextChunks = []
    iv = get_random_bytes(16)
    originalIV = iv
    cipher = AES.new(key, AES.MODE_ECB)
    if len(plainText) % 16 != 0:
        plainText += b"1"
    while len(plainText) % 16 != 0:
        plainText += b"0"
    while pos + 16 <= len(plainText):
        toXor = cipher.encrypt(iv)
        nextPos = pos + 16
        toEnc = plainText[pos:nextPos]
        cipherText = bytes([toXor[i] ^ toEnc[i] for i in range(16)])
        cipherTextChunks.append(cipherText)
        pos += 16
        iv = toXor
    return (originalIV, cipherTextChunks)


def ofbDec(cipherTextChunks, key, iv):
    plainText = b""
    cipher = AES.new(key, AES.MODE_ECB)
    for chunk in cipherTextChunks:
        toXor = cipher.encrypt(iv)
        plainText += bytes([toXor[i] ^ chunk[i] for i in range(15)])
        iv = toXor
    while plainText[-1] == 48:
        plainText = plainText[0:-1]
    if plainText[-1] == 49:
        plainText = plainText[0:-1]
    return plainText


iv, result = ofbEnc(plainText, key)
print(iv, result)

plain = ofbDec(result, key, iv)
print(plain)

带有pycryptodome的python中的AES OFB实现

1 个答案: