使用Python 实现国产SM3加密算法的方法？针对这个问题，这篇文章详细介绍了相对应的分析和解答，希望可以帮助更多想解决这个问题的小伙伴找到更简单易行的方法。

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SM3是中华人民共和国政府采用的一种密码散列函数标准，由国家密码管理局于2010年12月17日发布。主要用于报告文件数字签名及验证。

Python3代码如下：

from math import ceil

##############################################################################
#
#              国产SM3加密算法
#
##############################################################################

IV = "7380166f 4914b2b9 172442d7 da8a0600 a96f30bc 163138aa e38dee4d b0fb0e4e"
IV = int(IV.replace(" ", ""), 16)
a = []
for i in range(0, 8):
  a.append(0)
  a[i] = (IV >> ((7 - i) * 32)) & 0xFFFFFFFF
IV = a


def out_hex(list1):
  for i in list1:
    print("%08x" % i)
  print("\n")


def rotate_left(a, k):
  k = k % 32
  return ((a << k) & 0xFFFFFFFF) | ((a & 0xFFFFFFFF) >> (32 - k))


T_j = []
for i in range(0, 16):
  T_j.append(0)
  T_j[i] = 0x79cc4519
for i in range(16, 64):
  T_j.append(0)
  T_j[i] = 0x7a879d8a


def FF_j(X, Y, Z, j):
  if 0 <= j and j < 16:
    ret = X ^ Y ^ Z
  elif 16 <= j and j < 64:
    ret = (X & Y) | (X & Z) | (Y & Z)
  return ret


def GG_j(X, Y, Z, j):
  if 0 <= j and j < 16:
    ret = X ^ Y ^ Z
  elif 16 <= j and j < 64:
    # ret = (X | Y) & ((2 ** 32 - 1 - X) | Z)
    ret = (X & Y) | ((~ X) & Z)
  return ret


def P_0(X):
  return X ^ (rotate_left(X, 9)) ^ (rotate_left(X, 17))


def P_1(X):
  return X ^ (rotate_left(X, 15)) ^ (rotate_left(X, 23))


def CF(V_i, B_i):
  W = []
  for i in range(16):
    weight = 0x1000000
    data = 0
    for k in range(i * 4, (i + 1) * 4):
      data = data + B_i[k] * weight
      weight = int(weight / 0x100)
    W.append(data)

  for j in range(16, 68):
    W.append(0)
    W[j] = P_1(W[j - 16] ^ W[j - 9] ^ (rotate_left(W[j - 3], 15))) ^ (rotate_left(W[j - 13], 7)) ^ W[j - 6]
    str1 = "%08x" % W[j]
  W_1 = []
  for j in range(0, 64):
    W_1.append(0)
    W_1[j] = W[j] ^ W[j + 4]
    str1 = "%08x" % W_1[j]

  A, B, C, D, E, F, G, H = V_i
  """
  print "00",
  out_hex([A, B, C, D, E, F, G, H])
  """
  for j in range(0, 64):
    SS1 = rotate_left(((rotate_left(A, 12)) + E + (rotate_left(T_j[j], j))) & 0xFFFFFFFF, 7)
    SS2 = SS1 ^ (rotate_left(A, 12))
    TT1 = (FF_j(A, B, C, j) + D + SS2 + W_1[j]) & 0xFFFFFFFF
    TT2 = (GG_j(E, F, G, j) + H + SS1 + W[j]) & 0xFFFFFFFF
    D = C
    C = rotate_left(B, 9)
    B = A
    A = TT1
    H = G
    G = rotate_left(F, 19)
    F = E
    E = P_0(TT2)

    A = A & 0xFFFFFFFF
    B = B & 0xFFFFFFFF
    C = C & 0xFFFFFFFF
    D = D & 0xFFFFFFFF
    E = E & 0xFFFFFFFF
    F = F & 0xFFFFFFFF
    G = G & 0xFFFFFFFF
    H = H & 0xFFFFFFFF

  V_i_1 = []
  V_i_1.append(A ^ V_i[0])
  V_i_1.append(B ^ V_i[1])
  V_i_1.append(C ^ V_i[2])
  V_i_1.append(D ^ V_i[3])
  V_i_1.append(E ^ V_i[4])
  V_i_1.append(F ^ V_i[5])
  V_i_1.append(G ^ V_i[6])
  V_i_1.append(H ^ V_i[7])
  return V_i_1


def hash_msg(msg):
  # print(msg)
  len1 = len(msg)
  reserve1 = len1 % 64
  msg.append(0x80)
  reserve1 = reserve1 + 1
  # 56-64, add 64 byte
  range_end = 56
  if reserve1 > range_end:
    range_end = range_end + 64

  for i in range(reserve1, range_end):
    msg.append(0x00)

  bit_length = (len1) * 8
  bit_length_str = [bit_length % 0x100]
  for i in range(7):
    bit_length = int(bit_length / 0x100)
    bit_length_str.append(bit_length % 0x100)
  for i in range(8):
    msg.append(bit_length_str[7 - i])

  # print(msg)

  group_count = round(len(msg) / 64)

  B = []
  for i in range(0, group_count):
    B.append(msg[i * 64:(i + 1) * 64])

  V = []
  V.append(IV)
  for i in range(0, group_count):
    V.append(CF(V[i], B[i]))

  y = V[i + 1]
  result = ""
  for i in y:
    result = '%s%08x' % (result, i)
  return result


def str2byte(msg): # 字符串转换成byte数组
  ml = len(msg)
  msg_byte = []
  msg_bytearray = msg # 如果加密对象是字符串，则在此对msg做encode()编码即可，否则不编码
  for i in range(ml):
    msg_byte.append(msg_bytearray[i])
  return msg_byte


def byte2str(msg): # byte数组转字符串
  ml = len(msg)
  str1 = b""
  for i in range(ml):
    str1 += b'%c' % msg[i]
  return str1.decode('utf-8')


def hex2byte(msg): # 16进制字符串转换成byte数组
  ml = len(msg)
  if ml % 2 != 0:
    msg = '0' + msg
  ml = int(len(msg) / 2)
  msg_byte = []
  for i in range(ml):
    msg_byte.append(int(msg[i * 2:i * 2 + 2], 16))
  return msg_byte


def byte2hex(msg): # byte数组转换成16进制字符串
  ml = len(msg)
  hexstr = ""
  for i in range(ml):
    hexstr = hexstr + ('%02x' % msg[i])
  return hexstr


def KDF(Z, klen): # Z为16进制表示的比特串（str），klen为密钥长度（单位byte）
  klen = int(klen)
  ct = 0x00000001
  rcnt = ceil(klen / 32)
  Zin = hex2byte(Z)
  Ha = ""
  for i in range(int(rcnt)):
    msg = Zin + hex2byte('%08x' % ct)
    # print(msg)
    Ha = Ha + hash_msg(msg)
    # print(Ha)
    ct += 1
  return Ha[0: klen * 2]


def sm3_hash(msg, Hexstr=0):
  """
  封装方法，外部调用
  :param msg: 二进制流（如若需要传入字符串，则把str2byte方法里msg做encode()编码一下，否则不编码）
  :param Hexstr: 0
  :return: 64位SM3加密结果
  """
  if (Hexstr):
    msg_byte = hex2byte(msg)
  else:
    msg_byte = str2byte(msg)
  return hash_msg(msg_byte)


if __name__ == '__main__':
  print(sm3_hash(b'SM3Test'))# 打印结果：901053b4681483b737dd2dd9f9a7f56805aa1b03337f8c1abb763a96776b8905

本文名称：使用Python实现国产SM3加密算法的方法-创新互联
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