python实现SM4加解密 源码分析

 第一步：密钥拓展。

 第二步：加密

 第三步：解密（机密的过程与加密完全相同，只不过加密的时候密钥是从rk0开始的，解密是从rk31开始的）

# -*- coding: utf-8 -*-
# @Author  : 一根排骨
# @file: sm4.py
# @time: 2019/4/25 11.30
# @Software: VSCode

class sm:

    # 固定参数
    CK = [0x00070F15,0x1c232a31,0x383f464d,0x545b6269,
          0x70777e85,0x8c939aa1,0xa8afb6bd,0xc4cbd2d9,
          0xe0e7eef5,0xfc030a11,0x181f262d,0x343b4249,
          0x50575e65,0x6c737a81,0x888f969d,0xa4abb2b9,
          0xc0c7ced5,0xdce3eaf1,0xf8ff060d,0x141b2229,
          0x30373e45,0x4c535a61,0x686f767d,0x848b9299,
          0xa0a7aeb5,0xbcc3cad1,0xd8dfe6ed,0xf4fb0209,
          0x10171e25,0x2c333a41,0x484f565d,0x646b7279]
    # 常数 
    FK = [0xa3b1bac6,0x56aa3350,0x677d9197,0xb27022dc]
    # S盒
    SboxTable = [
    0xd6, 0x90, 0xe9, 0xfe, 0xcc, 0xe1, 0x3d, 0xb7, 0x16, 0xb6, 0x14, 0xc2, 0x28, 0xfb, 0x2c, 0x05,
    0x2b, 0x67, 0x9a, 0x76, 0x2a, 0xbe, 0x04, 0xc3, 0xaa, 0x44, 0x13, 0x26, 0x49, 0x86, 0x06, 0x99,
    0x9c, 0x42, 0x50, 0xf4, 0x91, 0xef, 0x98, 0x7a, 0x33, 0x54, 0x0b, 0x43, 0xed, 0xcf, 0xac, 0x62,
    0xe4, 0xb3, 0x1c, 0xa9, 0xc9, 0x08, 0xe8, 0x95, 0x80, 0xdf, 0x94, 0xfa, 0x75, 0x8f, 0x3f, 0xa6,
    0x47, 0x07, 0xa7, 0xfc, 0xf3, 0x73, 0x17, 0xba, 0x83, 0x59, 0x3c, 0x19, 0xe6, 0x85, 0x4f, 0xa8,
    0x68, 0x6b, 0x81, 0xb2, 0x71, 0x64, 0xda, 0x8b, 0xf8, 0xeb, 0x0f, 0x4b, 0x70, 0x56, 0x9d, 0x35,
    0x1e, 0x24, 0x0e, 0x5e, 0x63, 0x58, 0xd1, 0xa2, 0x25, 0x22, 0x7c, 0x3b, 0x01, 0x21, 0x78, 0x87,
    0xd4, 0x00, 0x46, 0x57, 0x9f, 0xd3, 0x27, 0x52, 0x4c, 0x36, 0x02, 0xe7, 0xa0, 0xc4, 0xc8, 0x9e,
    0xea, 0xbf, 0x8a, 0xd2, 0x40, 0xc7, 0x38, 0xb5, 0xa3, 0xf7, 0xf2, 0xce, 0xf9, 0x61, 0x15, 0xa1,
    0xe0, 0xae, 0x5d, 0xa4, 0x9b, 0x34, 0x1a, 0x55, 0xad, 0x93, 0x32, 0x30, 0xf5, 0x8c, 0xb1, 0xe3,
    0x1d, 0xf6, 0xe2, 0x2e, 0x82, 0x66, 0xca, 0x60, 0xc0, 0x29, 0x23, 0xab, 0x0d, 0x53, 0x4e, 0x6f,
    0xd5, 0xdb, 0x37, 0x45, 0xde, 0xfd, 0x8e, 0x2f, 0x03, 0xff, 0x6a, 0x72, 0x6d, 0x6c, 0x5b, 0x51,
    0x8d, 0x1b, 0xaf, 0x92, 0xbb, 0xdd, 0xbc, 0x7f, 0x11, 0xd9, 0x5c, 0x41, 0x1f, 0x10, 0x5a, 0xd8,
    0x0a, 0xc1, 0x31, 0x88, 0xa5, 0xcd, 0x7b, 0xbd, 0x2d, 0x74, 0xd0, 0x12, 0xb8, 0xe5, 0xb4, 0xb0,
    0x89, 0x69, 0x97, 0x4a, 0x0c, 0x96, 0x77, 0x7e, 0x65, 0xb9, 0xf1, 0x09, 0xc5, 0x6e, 0xc6, 0x84,
    0x18, 0xf0, 0x7d, 0xec, 0x3a, 0xdc, 0x4d, 0x20, 0x79, 0xee, 0x5f, 0x3e, 0xd7, 0xcb, 0x39, 0x48]
    # 拓展密钥
    K = []
    MK = []
    # 密钥
    key = 0
    X = []
    Y = []
    # 数据流
    txtstream = []
    # 文件缺失长度
    lenth = 0
    
    def __init__(self,key):
        self.key = key
        pass
    
    #第一步
    #密钥生成
    #经过拓展，rk0=K4，rk1=K5，以此类推
    def __key__expand__(self):
        for i in range(0,13,4):
            self.MK.append(self.__union__hex__(self.key[i:i+4]))
        for i in range(4):
            self.K.append(self.MK[i] ^ self.FK[i])
        for i in range(32):
            a = self.K[i+1] ^ self.K[i+2] ^ self.K[i+3] ^ self.getCK(i)
            b = self.__apart__hex__(a)
            c = [self.getSbox(i) for i in b]
            d = self.__union__hex__(c)
            e = d ^ (d <<13) ^ (d << 23)
            self.K.append(self.K[i] ^ e)
        

    #将在4个32位数据合并一个128位数据
    def __union__hex__(self,data):
        return int((data[0] << 24) | (data[1] << 16) | (data[2] << 8) | (data[3]))

    #将一个128位数据拆开位4个32位数据
    def __apart__hex__(self,data):
        return [int((data >> 24) & 0xff), int((data >> 16) & 0xff), int((data >> 8) & 0xff), int((data) & 0xff)]

    #获取S盒的元素
    def getSbox(self,i):
        return self.SboxTable[i]

    #获取固定参数
    def getCK(self,i):
        return self.CK[i]

    #由此代码可看出加密与解密只是密钥的顺序不同
    #加密解密
    def deal(self,modle):
        # 模式判断
        if modle == 'encrypt':
            name = 'good.txt'
        else:
            name = 'en.txt'
        self.X = []
        self.Y = []
        self.__file__open(modle,name)
        for p in range(0, len(self.txtstream), 16):
            #密文分组
            plaintxt = self.txtstream[p:p+16]
            self.X = []
            for i in range(0,13,4):
                self.X.append(self.__union__hex__(plaintxt[i:i+4]))
            if modle == 'encrypt':
                j = 4
            else:
                j = 35
            #求明文/密文
            for i in range(0,32):
                a = self.X[i+1] ^ self.X[i+2] ^ self.X[i+3] ^ self.K[j]
                if modle == 'encrypt':
                    j = j + 1
                else:
                    j = j - 1
                b = self.__apart__hex__(a)
                c = [self.getSbox(i) for i in b]
                d = self.__union__hex__(c)
                e = d ^ (d << 2) ^ (d << 10) ^ (d << 18) ^ (d << 24)
                self.X.append(self.X[i] ^ e)
            # 明文/密文逆序
            t = self.X[35]
            self.X[35] = self.X[32]
            self.X[32] = t
            t =self.X[34]
            self.X[34] = self.X[33]
            self.X[33] = t
            # 明文/密文储存
            for i in range(32,36):
                self.Y.append(self.X[i])
            if modle == 'encrypt':
                name = 'en.txt'
            else:
                name = 'de.txt'
        self.__file__save__(modle,name)
        pass

    #打开文件
    def __file__open(self,modle,name):
        self.txtstream = []
        with open(name,'rb') as f:
            a = f.read()
        for i in a:
            self.txtstream.append(i)
        if modle == 'encrypt':
            self.lenth = 16 - len(self.txtstream) % 16
            # 补全明文
            for i in range(self.lenth):
                self.txtstream.append(0)
        
    
    #保存文件
    def __file__save__(self,model,name):
        byte = []
        for i in self.Y:
            a = self.__apart__hex__(i)
            for j in a:
                byte.append(j)
        if model == 'decryption':
             byte = byte[:len(byte)-self.lenth]
        with open(name,'wb') as f:
            f.write(bytes(byte))

# 密钥
key = [0x5a] * 16
hi = sm(key)
hi.__key__expand__()
hi.deal('encrypt')
hi.deal('decryption')

本代码是用SM4算法加密一个txt文档，并解密的过程。

 

看不懂代码欢迎提问~~