在很久很久以前写过一篇文章讲linux中shadow文件的数据构成(https://blog.csdn.net/cracker_zhou/article/details/50817743)。

当然在文章末尾提到了使用python的crypt模块可以生成或者校验linux密码的有效性。但是crypt模块在windows上并不能使用。同时在文章中提交密码的生成依赖于算法(md5,sha512之类)和盐值(salt)，但是并没有说清楚salt是如何使用的。

研究发现，密码的生成比我想象的复杂的多，以$6为例并不是单纯的sha512(pass+salt)或者sha512(salt+passwd)，而是经历了一系列复杂的运算而成。

下面python代码参考了node版本: https://github.com/mvo5/sha512crypt-node

同时给出官方的C语言版本: https://www.akkadia.org/drepper/SHA-crypt.txt

Tip: 测试代码在python3.6版本编译通过，其他python版本需要做微调

import hashlib,math

def rstr_sha512(text: bytes) -> bytes:

sha512 = hashlib.sha512()

sha512.update(text)

return sha512.digest()

def _extend(source: bytes, size_ref: int) -> bytes :

extended = b"";

for i in range(math.floor(size_ref/64)):

extended += source;

extended += source[:size_ref % 64]

return extended;

def _sha512crypt_intermediate(password: bytes,salt: bytes) -> bytes:

#digest_a = rstr_sha512(password + salt)

digest_b = rstr_sha512(password + salt + password)

digest_b_extended = _extend(digest_b,len(password))

intermediate_input = password + salt + digest_b_extended

passwd_len = len(password)

while passwd_len!=0:

if passwd_len&1 == 1:

intermediate_input += digest_b

else:

intermediate_input += password

passwd_len >>= 1

return rstr_sha512(intermediate_input)

def _sha512crypt(password :bytes,salt :bytes,rounds :int) -> bytes:

digest_a = _sha512crypt_intermediate(password, salt)

p = _extend(rstr_sha512(password*len(password)),len(password))

s = _extend(rstr_sha512(salt*(16+digest_a[0])),len(salt))

digest = digest_a

for i in range(rounds):

c_input = b""

if i&1 :

c_input += p

else:

c_input += digest

if i % 3:

c_input += s

if i % 7:

c_input += p

if i & 1:

c_input += digest

else:

c_input += p

digest = rstr_sha512(c_input)

return digest

def sha512crypt(password :bytes,salt :bytes, rounds=5000) -> str:

salt = salt[:16] # max 16 bytes for salt

input = _sha512crypt(password, salt, rounds)

tab = "./0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"

order = [ 42, 21, 0, 1, 43, 22, 23, 2, 44, 45, 24, 3,

4, 46, 25, 26, 5, 47, 48, 27, 6, 7, 49, 28,

29, 8, 50, 51, 30, 9, 10, 52, 31, 32, 11, 53,

54, 33, 12, 13, 55, 34, 35, 14, 56, 57, 36, 15,

16, 58, 37, 38, 17, 59, 60, 39, 18, 19, 61, 40,

41, 20, 62, 63]

output = ""

for i in range(0,len(input),3):

# special case for the end of the input

if i+1 >= len(order): # i == 63

char_1 = input[order[i+0]] & 0b00111111

char_2 = (input[order[i+0]] & 0b11000000) >> 6

output += tab[char_1] + tab[char_2]

else:

char_1 = input[order[i+0]] & 0b00111111

char_2 = (((input[order[i+0]] & 0b11000000) >> 6) |

(input[order[i+1]] & 0b00001111) << 2)

char_3 = (

((input[order[i+1]] & 0b11110000) >> 4) |

(input[order[i+2]] & 0b00000011) << 4)

char_4 = (input[order[i+2]] & 0b11111100) >> 2

output += tab[char_1] + tab[char_2] + tab[char_3] + tab[char_4]

if rounds!=5000:

return "$6$rounds={}${}${}".format(rounds,salt.decode("utf-8"),output)

else:

return "$6${}${}".format(salt.decode("utf-8"),output)

if __name__ == "__main__":

# 与crypt.crypt("123456","$6$123456") 运算结果一致

print(sha512crypt(b"123456",b"123456",5000))