同时在看一本书《从零开始-自己动手写区块链》,
这书讲得易懂,我也动手实践一下。
这个算法和python3本身的实现相同,
所以,同样的字串,摘要是相同的。
import struct import binascii import hashlib # 64个常数K _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) # 初始化缓存 _H = (0x6a09e667, 0xbb67ae85, 0x3c6ef372, 0xa54ff53a, 0x510e527f, 0x9b05688c, 0x1f83d9ab, 0x5be0cd19) # 定义sha1_256类 class sha_256: # 输入参数为明文 def __init__(self, m=None): # 初始化明文 self.buffer = b'' # 输入明文长度 self.counter = 0 self.H = _H self.K = _K if m: self.update(m) # 定义循环右移方法 def rotr(self, x, y): return ((x >> y) | (x << (32 - y))) & 0xFFFFFFFF # 定义对单个分组进行操作的方法 def operate(self, c): # 定义长度为64的空列表w w = [0] * 64 # 将单个分组转换为16个32位的字,并填充w列表的前16位 w[0:16] = struct.unpack('!16L', c) # 填充w列表的后48位 for i in range(16, 64): s0 = self.rotr(w[i - 15], 7) ^ self.rotr(w[i - 15], 18) ^ (w[i - 15] >> 3) s1 = self.rotr(w[i - 2], 17) ^ self.rotr(w[i - 2], 19) ^ (w[i - 2] >> 10) w[i] = (w[i - 16] + s0 + w[i - 7] + s1) & 0xFFFFFFFF a, b, c, d, e, f, g, h = self.H # 执行64步迭代操作 for i in range(64): s0 = self.rotr(a, 2) ^ self.rotr(a, 13) ^ self.rotr(a, 22) maj = (a & b) ^ (a & c) ^ (b & c) t2 = s0 + maj s1 = self.rotr(e, 6) ^ self.rotr(e, 11) ^ self.rotr(e, 25) ch = (e & f) ^ ((~e) & g) t1 = h + s1 + ch + self.K[i] + w[i] h = g g = f f = e e = (d + t1) & 0xFFFFFFFF d = c c = b b = a a = (t1 + t2) & 0xFFFFFFFF # 更新缓存 self.H = [(x + y) & 0xFFFFFFFF for x, y in zip(self.H, [a, b, c, d, e, f, g, h])] # 定义更新N个分组缓存的方法 def update(self, m): if not m: return # 获取明文 self.buffer = m # 获取明文长度 self.counter = len(m) # 计算明文长度表示的后64位 length = struct.pack('!Q', int(self.counter * 8)) # 对前N-1个分组进行哈希过程 while len(self.buffer) >= 64: self._operate(self.buffer[:64]) self.buffer = self.buffer[64:] # 填充未处理的第N个分组到512位或1024位,并进行哈希过程 mdi = self.counter % 64 # 如果第N个分组小于56,则填充至512位 if mdi < 56: padlen = 55 - mdi self.buffer += (b'\x80' + (b'\x00' * padlen) + length) self.operate(self.buffer) # 否则,填充到1024位 else: padlen = 119 - mdi self.buffer += (b'\x80' + (b'\x00' * padlen) + length) for i in range(2): self.operate(self.buffer[i * 64:(i + 1) * 64]) # 输出明文摘要, 字节串类型 def digest(self): return struct.pack('!8L', *self.H) # 输出明文搞要, 十六进制字符串类型 def hexdigest(self): return binascii.hexlify(self.digest()).decode() if __name__ == "__main__": print(sha_256(b'chengang').digest()) print(hashlib.sha256(b'chengang').digest()) print(sha_256(b'chengang').hexdigest()) print(hashlib.sha256(b'chengang').hexdigest())
输出看看吧。
b'\x85A\xa3\xe3\xc1{\x04}\xff\x1d)5wFK\x10L=\x1f\xb4"\x88a=f\x835\xc7\xc1\x89_\t' b'\x85A\xa3\xe3\xc1{\x04}\xff\x1d)5wFK\x10L=\x1f\xb4"\x88a=f\x835\xc7\xc1\x89_\t' 8541a3e3c17b047dff1d293577464b104c3d1fb42288613d668335c7c1895f09 8541a3e3c17b047dff1d293577464b104c3d1fb42288613d668335c7c1895f09 Process finished with exit code 0