python生成shadow中密码(SHA512)

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))