相关原理
占坑(大概不填了)
测试代码
使用aes测试（理论上所有使用CBC的分组加密都有这个问题）
readme
- 适用范围：CBC分组加密密码，PKCS#5或PKCS#7 padding方式
try_dec输入bytes的待解密文本，输出True为可通过unpad，输出False为不通过unpad- 如果给iv可以解密所有密文，否则不能解密第一组
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from Crypto.Cipher import AES
from Crypto.Util import number
from Crypto.Util.strxor import strxor
import os
def pad(plain):
'''
using pkcs#7 standard
'''
to_pad = 16 - len(plain) % 16
return plain + number.long_to_bytes(to_pad)*to_pad
def unpad(cipher):
if cipher[-1] > 16 or cipher[-1] == 0:
raise Exception
cipher_len = len(cipher)
for i in range(cipher_len - 1, cipher_len - cipher[-1] - 1, -1):
if cipher[i] != cipher[-1]:
raise Exception
return cipher[:-cipher[-1]]
iv = os.urandom(16)
key = os.urandom(16)
aes = AES.new(key, AES.MODE_CBC, iv)
plain_text = b"VT-x or AMD-v virtualization must be enabled"
padded = pad(plain_text)
cipher = aes.encrypt(padded)
def try_dec(cipher):
try:
x = aes.decrypt(cipher)
plain = unpad(x)
return True
except:
return False
cipher = iv+cipher
block_len = 16
cipher_len = len(cipher)
groups = [ cipher[i:i + block_len] for i in range(0, cipher_len, block_len) ]
groups_len = len(groups)
res = b''
for group_num in range(groups_len-2, -1, -1):
cipher_array = bytearray(cipher)
intermediary = bytearray(block_len)
flag = 0
for i in range(256):
cipher_array[block_len * group_num + block_len - 1] = i
test_cipehr = bytes(cipher_array)
if try_dec(test_cipehr):
if i == cipher[block_len * group_num + block_len - 1]: continue
intermediary[block_len - 1] = 1 ^ i
flag = 1
if not flag:
intermediary[block_len - 1] = 1 ^ cipher[block_len * group_num + block_len - 1]
for i in range(block_len - 2, -1, -1):
fill_value = block_len - i
for j in range(i+1, block_len):
cipher_array[block_len * group_num + j] = intermediary[j] ^ fill_value
for j in range(256):
cipher_array[block_len * group_num + i] = j
test_cipehr = bytes(cipher_array)
if try_dec(test_cipehr):
intermediary[i] = fill_value ^ j
res = strxor(groups[group_num], bytes(intermediary)) + res
cipher = cipher[:-block_len]
print(res)
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