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。
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
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