SHA 256 pseuedocode？

Question

我一直在努力研究SHA-256的工作原理.我一直在为其他算法做的一件事是我已经为算法制定了一种逐步的伪代码函数.

我试过为SHA256做同样的事情,但到目前为止我遇到了很多麻烦.

我试图弄清楚维基百科图如何工作,但除了解释函数的文本部分,我不确定我是否正确.

这是我到目前为止所拥有的:

Input is an array 8 items long where each item is 32 bits.
Output is an array 8 items long where each item is 32 bits.
Calculate all the function boxes and store those values. 
|I'll refer to them by function name
Store input, right shifted by 32 bits, into output. 
| At this point, in the out array, E is the wrong value and A is empty
Store the function boxes.
| now we need to calculate out E and out A.
| note: I've replaced the modulo commands with a bitwise AND 2^(32-1) 
| I can't figure out how the modulus adding lines up, but I think it is like this
Store (Input H + Ch + ( (Wt+Kt) AND 2^31 ) ) AND 2^31 As mod1
Store (sum1 + mod1) AND 2^31 as mod2
Store (d + mod2) AND 2^31 into output E 
|now output E is correct and all we need is output A
Store (MA + mod2) AND 2^31 as mod3
Store (sum0 + mod3) AND 2^31 into output A
|output now contains the correct hash of input.
|Do we return now or does this need to be run repeatedly?

我是否正确地获得了所有这些附加模数？还有什么是Wt和Kt？此操作也会运行一次并且您已完成或需要运行一定次数,并将输出重新用作输入.

这是顺便说一句的链接. http://en.wikipedia.org/wiki/SHA-2#Hash_function

非常感谢,Brian

Answer 1

看一下描述算法的官方标准,变量如下所述:http://csrc.nist.gov/publications/fips/fips180-4/fips-180-4.pdf

(哦,现在我看到我的答案差不多一年了啊,没关系......)

Answer 2

W_t是从正在处理的当前块导出的,而K_t是由迭代次数确定的固定常数.对于SHA256中的每个块,压缩功能重复64次.对于每次迭代0 <= t <= 63,存在特定常数K_t和导出值W_t.

我使用Python 3.6提供了自己的SHA256实现.元组K包含64个常数K_t值.该SHA256功能展示了如何W_T的值列表中的计算w ^.实现侧重于代码清晰度而非高性能.

W = 32          #Number of bits in word
M = 1 << W
FF = M - 1      #0xFFFFFFFF (for performing addition mod 2**32)

#Constants from SHA256 definition
K = (0x428a2f98, 0x71374491, 0xb5c0fbcf, 0xe9b5dba5,
     0x3956c25b, 0x59f111f1, 0x923f82a4, 0xab1c5ed5,
     0xd807aa98, 0x12835b01, 0x243185be, 0x550c7dc3,
     0x72be5d74, 0x80deb1fe, 0x9bdc06a7, 0xc19bf174,
     0xe49b69c1, 0xefbe4786, 0x0fc19dc6, 0x240ca1cc,
     0x2de92c6f, 0x4a7484aa, 0x5cb0a9dc, 0x76f988da,
     0x983e5152, 0xa831c66d, 0xb00327c8, 0xbf597fc7,
     0xc6e00bf3, 0xd5a79147, 0x06ca6351, 0x14292967,
     0x27b70a85, 0x2e1b2138, 0x4d2c6dfc, 0x53380d13,
     0x650a7354, 0x766a0abb, 0x81c2c92e, 0x92722c85,
     0xa2bfe8a1, 0xa81a664b, 0xc24b8b70, 0xc76c51a3,
     0xd192e819, 0xd6990624, 0xf40e3585, 0x106aa070,
     0x19a4c116, 0x1e376c08, 0x2748774c, 0x34b0bcb5,
     0x391c0cb3, 0x4ed8aa4a, 0x5b9cca4f, 0x682e6ff3,
     0x748f82ee, 0x78a5636f, 0x84c87814, 0x8cc70208,
     0x90befffa, 0xa4506ceb, 0xbef9a3f7, 0xc67178f2)

#Initial values for compression function
I = (0x6a09e667, 0xbb67ae85, 0x3c6ef372, 0xa54ff53a,
     0x510e527f, 0x9b05688c, 0x1f83d9ab, 0x5be0cd19)

def RR(x, b):
    '''
    32-bit bitwise rotate right
    '''
    return ((x >> b) | (x << (W - b))) & FF

def Pad(W):
    '''
    Pads a message and converts to byte array
    '''
    mdi = len(W) % 64           
    L = (len(W) << 3).to_bytes(8, 'big')        #Binary of len(W) in bits
    npad = 55 - mdi if mdi < 56 else 119 - mdi  #Pad so 64 | len; add 1 block if needed
    return bytes(W, 'ascii') + b'\x80' + (b'\x00' * npad) + L   #64 | 1 + npad + 8 + len(W)

def Sha256CF(Wt, Kt, A, B, C, D, E, F, G, H):
    '''
    SHA256 Compression Function
    '''
    Ch = (E & F) ^ (~E & G)
    Ma = (A & B) ^ (A & C) ^ (B & C)        #Major
    S0 = RR(A, 2) ^ RR(A, 13) ^ RR(A, 22)   #Sigma_0
    S1 = RR(E, 6) ^ RR(E, 11) ^ RR(E, 25)   #Sigma_1
    T1 = H + S1 + Ch + Wt + Kt
    return (T1 + S0 + Ma) & FF, A, B, C, (D + T1) & FF, E, F, G

def Sha256(M):
    '''
    Performs SHA256 on an input string 
    M: The string to process
    return: A 32 byte array of the binary digest
    '''
    M = Pad(M)          #Pad message so that length is divisible by 64
    DG = list(I)        #Digest as 8 32-bit words (A-H)
    for j in range(0, len(M), 64):  #Iterate over message in chunks of 64
        S = M[j:j + 64]             #Current chunk
        W = [0] * 64
        W[0:16] = [int.from_bytes(S[i:i + 4], 'big') for i in range(0, 64, 4)]  
        for i in range(16, 64):
            s0 = RR(W[i - 15], 7) ^ RR(W[i - 15], 18) ^ (W[i - 15] >> 3)
            s1 = RR(W[i - 2], 17) ^ RR(W[i - 2], 19) ^ (W[i - 2] >> 10)
            W[i] = (W[i - 16] + s0 + W[i-7] + s1) & FF
        A, B, C, D, E, F, G, H = DG #State of the compression function
        for i in range(64):
            A, B, C, D, E, F, G, H = Sha256CF(W[i], K[i], A, B, C, D, E, F, G, H)
        DG = [(X + Y) & FF for X, Y in zip(DG, (A, B, C, D, E, F, G, H))]
    return b''.join(Di.to_bytes(4, 'big') for Di in DG)  #Convert to byte array

if __name__ == "__main__":
    bd = Sha256('Hello World')
    print(''.join('{:02x}'.format(i) for i in bd))