python实现SM4加解密源码分析

我不是女神ヾ 2022-04-24 13:08 138阅读 0赞

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~~