sha-256 python实现中的错误
sha-256 python实现中的错误
提问于 2018-05-25 08:38:27
在过去的几天里，我一直在研究像SHA-256这样的散列算法是如何实际工作的。为了更好地理解这个过程，我决定用python (3.6)实现它。通过将各种网站(如wiki和NIST页面)的概念拼接在一起，我构建了以下代码。我知道这个过程接近于它应该是什么样子，但是当我测试它时，结果与它们应该是什么样子略有不同。例如，使用SHA-256散列的'a‘应该是'ca978112ca1bbdcafac231b39a23dc4da786eff8147c4e72b9807785afee48bb’，但是我的程序返回'ca978112ca1bbdcafac231b319a23dc4da786eff81147c4e72b9807785afee48bb'.答案非常相似，但并不相同。我做错了什么。感谢您花时间研究它。
def hash(string):
return(process(pad(string)))
def pad(string):
data = ""
length = len(string) * 8
for c in string:
data += bin(ord(c))[2:].zfill(8)
data += "1"
while len(data)%512 != 448:
data += "0"
data += bin(length)[2:].zfill(64)
return data
def process(bins):
main_block=[
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]
h0 = 0x6a09e667
h1 = 0xbb67ae85
h2 = 0x3c6ef372
h3 = 0xa54ff53a
h4 = 0x510e527f
h5 = 0x9b05688c
h6 = 0x1f83d9ab
h7 = 0x5be0cd19
for c in chunks(bins, 512):
words = chunks(c, 32)
w = [0]*64
w[:15] = [int(n, 2) for n in words]
for i in range(16, len(w)):
tmp1 = rightRotate(w[i-15], 7) ^ rightRotate(w[i-15], 18) ^ rightShift(w[i-15], 3)
tmp2 = rightRotate(w[i-2], 17) ^ rightRotate(w[i-2], 19) ^ rightShift(w[i-2], 10)
w[i] = (w[i-16] + tmp1 + w[i-7] + tmp2) & 0xffffffff
a = h0
b = h1
c = h2
d = h3
e = h4
f = h5
g = h6
h = h7
for i in range(0, 64):
s1 = rightRotate(e, 6) ^ rightRotate(e, 11) ^ rightRotate(e, 25)
ch = g ^ (e & (f ^ g))
tmp1 = h + s1 + ch + main_block[i] + w[i]
s0 = rightRotate(a, 2) ^ rightRotate(a, 13) ^ rightRotate(a, 22)
maj = (a&b) ^ (a&c) ^ (b&c)
tmp2 = s0 + maj
h = g
g = f
f = e
e = d + tmp1 & 0xffffffff
d = c
c = b
b = a
a = tmp1 + tmp2 & 0xffffffff
h0 += a
h1 += b
h2 += c
h3 += d
h4 += e
h5 += f
h6 += g
h7 += h
return '%08x%08x%08x%08x%08x%08x%08x%08x' % (h0, h1, h2, h3, h4, h5, h6, h7)
def rightShift(x, n):
return (x & 0xffffffff) >> n
def rightRotate(x, y):
return (((x & 0xffffffff) >> (y & 31)) | (x << (32 - (y & 31)))) & 0xffffffff
def chunks(l, n):
return [l[i:i+n] for i in range(0, len(l), n)]
string = "a"
print(hash(string))回答 1
Stack Overflow用户
回答已采纳
发布于 2018-05-25 08:44:58
在计算后，您无法屏蔽h*，从而导致虚假的前导“1”。
h0 &= 0xffffffff
...
h7 &= 0xffffffff页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/50519967
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