2023年7月18日发(作者：)

MFC简单实现DES算法前⾔徐旭东⽼师说过学者就应该对知识抱有敬畏之⼼，所以我的博客的标题总喜欢加上“简单”⼆字，就是为了提醒⾃⼰，⾃⼰所学知识只是⽪⽑，离真理还远矣。DES 算法DES算法是密码体制中的对称密码体制，明⽂按64位进⾏分组，密钥长64位(其中有8位是奇偶校验位，不参与 DES 运算)，参数有三个：key、data、mode。即要加密或者解密的数据、密钥、加密还是解密。考虑到算法注重的是性能，且不涉及⾯向对象的思维，所以⼀开始选择 C 语⾔开发，但是为了良好的交互界⾯，所以最终选择了 MFC。基础知识DES 算法是对⼆进制数进⾏各种操作从⽽达到加密解密的作⽤，所以了解编码、字符和⼆进制数的概念⾮常重要。这也算是这个作业的意外收获吧~⾸先⼀个字节对应⼀个8位的⼆进制数。编码ASCIIGB2312GBKBig5GB18030UTF-16(标准的 Unicode)UTF-8(Unicode 的⼀种，变长编码)英⽂字符(单位:字节)1(有时候7位⼆进制数)111121中⽂字符(单位:字节)⽆222222-4(⼤部分占3个字节)不能满⾜其他国家需求为了显⽰中⽂制定(属于 ANSI 编码，其他国家皆有⾃⼰的标准)⽀持⼈名、古汉语等⽅⾯出现的罕⽤字，如：溙⽀持繁体字⽀持藏⽂、蒙⽂、维吾尔⽂等主要的少数民族⽂字为了适应全球化，超出范围的使⽤两个UTF-16即4字节保留 ASCII 字符的编码，存英⽂时节省空间备注加密流程1、输⼊64位的明⽂，按照以下 IP 置换表对明⽂进⾏置换：int IP_Table[64] = { //64位明⽂初始置换 表IP 57, 49, 41, 33, 25, 17, 9, 1, 59, 51, 43, 35, 27, 19, 11, 3, 61, 53, 45, 37, 29, 21, 13, 5, 63, 55, 47, 39, 31, 23, 15, 7, 56, 48, 40, 32, 24, 16, 8, 0, 58, 50, 42, 34, 26, 18, 10, 2, 60, 52, 44, 36, 28, 20, 12, 4, 62, 54, 46, 38, 30, 22, 14, 6};//IP置换int IP_Substitute(Type data[64]){ Type temp[64]; for (int i = 0; i < 64; i++){ temp[i] = data[IP_Table[i]]; } memcpy(data, temp, 64); return 0;}得到置换后的64位数据分成两份各32位，即 L0 和 R0( R0 也是下⼀轮的L1)。2、32位的 R0 通过以下 E 位选择表扩充置换成48位：int E_Table[48] = { //32位R扩充置换成48位 表E 31, 0, 1, 2, 3, 4, 3, 4, 5, 6, 7, 8, 7, 8, 9, 10, 11, 12, 11, 12, 13, 14, 15, 16, 15, 16, 17, 18, 19, 20, 19, 20, 21, 22, 23, 24, 23, 24, 25, 26, 27, 28, 27, 28, 29, 30, 31, 0};//32位R扩展置换成48位Int E_Substitute(Type data[48]){ Type temp[48]; for (int i = 0; i < 48; i++){ temp[i] = data[E_Table[i]]; } memcpy(data, temp, 48); return 0;}R0 由32位扩充置换成48位，等待和⼦密钥进⾏运算。3、输⼊64位的密钥，通过以下 PC_1 置换选择表去除第8,16,24,32,48,56,64位，即去除奇偶校验位后置换：int PC_1_Table[56] = { //64位密钥置换压缩成56位 表1 56, 48, 40, 32, 24, 16, 8, 0, 57, 49, 41, 33, 25, 17, 9, 1, 58, 50, 42, 34, 26, 18, 10, 2, 59, 51, 43, 35, 62, 54, 46, 38, 30, 22, 14, 6, 61, 53, 45, 37, 29, 21, 13, 5, 60, 52, 44, 36, 28, 20, 12, 4, 27, 19, 11, 3};//64位密钥置换压缩成56位int PC1_Compress(Type key[64], Type temp[56]){ for (int i = 0; i < 56; i++){ temp[i] = key[PC_1_Table[i]]; } return 0;}得到56位的密钥分成两份，各28位，即 C0 和 D0。4、C0 和 D0 分别按照循环左移表进⾏左移：//循环左移次数int MOVE[16] = { 1, 1, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 1 };//循环左移int LEFT_Shift(Type data[56], int time){ Type temp[56]; //保存将要循环移动到右边的位 memcpy(temp, data, time); memcpy(temp + time, data + 28, time); //前28位移动 memcpy(data, data + time, 28 - time); memcpy(data + 28 - time, temp, time); //后28位移动 memcpy(data + 28, data + 28 + time, 28 - time); memcpy(data + 56 - time, temp + time, time); return 0;}即左移⼀次得到 C1 和 D1，然后 C1 和 D1 左移1次得到 C2 和 D2，然后 C2 和 D2 左移2次得到 C3 和 D3，依次类推最后得到16个56位的密钥。这16个密钥再按照以下 PC_2 表进⾏置换压缩：int PC_2_Table[48] = { //56位⼦密钥置换压缩成48位 表2 13, 16, 10, 23, 0, 4, 2, 27, 14, 5, 20, 9, 22, 18, 11, 3, 25, 7, 15, 6, 26, 19, 12, 1, 40, 51, 30, 36, 46, 54, 29, 39, 50, 44, 32, 46, 43, 48, 38, 55, 33, 52, 45, 41, 49, 35, 28, 31};//56位的⼦密钥置换压缩成48位int PC2_Compress(Type key[56], Type temp[48]){ for (int i = 0; i < 48; i++){ temp[i] = key[PC_2_Table[i]]; } return 0;}//⽣成16个⼦密钥int MAKE_SubKeys(Type key[64], Type subKeys[16][48]){ Type temp[56]; PC1_Compress(key, temp);//PC1置换 for (int i = 0; i < 16; i++){//16轮跌代，产⽣16个⼦密钥 LEFT_Shift(temp, MOVE[i]);//循环左移 PC2_Compress(temp, subKeys[i]);//PC2置换，产⽣⼦密钥 } return 0;}

最终得到16个48位的⼦密钥：K1~K16。5、R0 和 K0 进⾏异或运算得到48位的新的 R0 (8⾏6列)，然后通过 S1~S8表进⾏压缩(每⾏分别对应⼀个 S 表)，最终压缩成8⾏4列，即32位，例如第⼀⾏：011010，第1和第6列组成00(⼗进制中的0)，则对应 S1表中第⼀⾏；中间4列组成1101(⼗进制中的13)，则对应第14列，也就是9，最终011010被替换成1001，所以很明显 S 表中的数最⼤不会超过15。//count=48:R与⼦密钥异或运算、count=32:R与L异或运算int XOR(Type R[48], Type L[48], int count){ for (int i = 0; i < count; i++){ R[i] ^= L[i]; } return 0;}int S[8][4][16] ={ //8⾏6列R压缩成8⾏4列 S盒 //S1 { { 14, 4, 13, 1, 2, 15, 11, 8, 3, 10, 6, 12, 5, 9, 0, 7 }, { 0, 15, 7, 4, 14, 2, 13, 1, 10, 6, 12, 11, 9, 5, 3, 8 }, { 4, 1, 14, 8, 13, 6, 2, 11, 15, 12, 9, 7, 3, 10, 5, 0 }, { 15, 12, 8, 2, 4, 9, 1, 7, 5, 11, 3, 14, 10, 0, 6, 13 } }, //S2 { { 15, 1, 8, 14, 6, 11, 3, 4, 9, 7, 2, 13, 12, 0, 5, 10 }, { 3, 13, 4, 7, 15, 2, 8, 14, 12, 0, 1, 10, 6, 9, 11, 5 }, { 0, 14, 7, 11, 10, 4, 13, 1, 5, 8, 12, 6, 9, 3, 2, 15 }, { 13, 8, 10, 1, 3, 15, 4, 2, 11, 6, 7, 12, 0, 5, 14, 9 } }, //S3 { { 10, 0, 9, 14, 6, 3, 15, 5, 1, 13, 12, 7, 11, 4, 2, 8 }, { 13, 7, 0, 9, 3, 4, 6, 10, 2, 8, 5, 14, 12, 11, 15, 1 }, { 13, 6, 4, 9, 8, 15, 3, 0, 11, 1, 2, 12, 5, 10, 14, 7 }, { 1, 10, 13, 0, 6, 9, 8, 7, 4, 15, 14, 3, 11, 5, 2, 12 } }, //S4 { { 7, 13, 14, 3, 0, 6, 9, 10, 1, 2, 8, 5, 11, 12, 4, 15 }, { 13, 8, 11, 5, 6, 15, 0, 3, 4, 7, 2, 12, 1, 10, 14, 9 }, { 10, 6, 9, 0, 12, 11, 7, 13, 15, 1, 3, 14, 5, 2, 8, 4 }, { 3, 15, 0, 6, 10, 1, 13, 8, 9, 4, 5, 11, 12, 7, 2, 14 } }, //S5 { { 2, 12, 4, 1, 7, 10, 11, 6, 8, 5, 3, 15, 13, 0, 14, 9 }, { 14, 11, 2, 12, 4, 7, 13, 1, 5, 0, 15, 10, 3, 9, 8, 6 }, { 4, 2, 1, 11, 10, 13, 7, 8, 15, 9, 12, 5, 6, 3, 0, 14 }, { 11, 8, 12, 7, 1, 14, 2, 13, 6, 15, 0, 9, 10, 4, 5, 3 } }, //S6 { { 12, 1, 10, 15, 9, 2, 6, 8, 0, 13, 3, 4, 14, 7, 5, 11 }, { 10, 15, 4, 2, 7, 12, 9, 5, 6, 1, 13, 14, 0, 11, 3, 8 }, { 9, 14, 15, 5, 2, 8, 12, 3, 7, 0, 4, 10, 1, 13, 11, 6 }, { 4, 3, 2, 12, 9, 5, 15, 10, 11, 14, 1, 7, 6, 0, 8, 13 } }, //S7 { { 4, 11, 2, 14, 15, 0, 8, 13, 3, 12, 9, 7, 5, 10, 6, 1 }, { 13, 0, 11, 7, 4, 9, 1, 10, 14, 3, 5, 12, 2, 15, 8, 6 }, { 1, 4, 11, 13, 12, 3, 7, 14, 10, 15, 6, 8, 0, 5, 9, 2 }, { 6, 11, 13, 8, 1, 4, 10, 7, 9, 5, 0, 15, 14, 2, 3, 12 } }, //S8 { { 13, 2, 8, 4, 6, 15, 11, 1, 10, 9, 3, 14, 5, 0, 12, 7 }, { 1, 15, 13, 8, 10, 3, 7, 4, 12, 5, 6, 11, 0, 14, 9, 2 }, { 7, 11, 4, 1, 9, 12, 14, 2, 0, 6, 10, 13, 15, 3, 5, 8 }, { 2, 1, 14, 7, 4, 10, 8, 13, 15, 12, 9, 0, 3, 5, 6, 11 } }};压缩后得到新的32位的 R0。//S盒置换int SBOX(Type data[48]){ int line, row, output; int cur1, cur2; for (int i = 0; i < 8; i++){ cur1 = i * 6; cur2 = i << 2; //计算在S盒中的⾏与列 line = (data[cur1] << 1) + data[cur1 + 5]; row = (data[cur1 + 1] << 3) + (data[cur1 + 2] << 2) + (data[cur1 + 3] << 1) + data[cur1 + 4]; output = S[i][line][row]; //化为2进制

data[cur2] = (output & 0X08) >> 3; data[cur2 + 1] = (output & 0X04) >> 2; data[cur2 + 2] = (output & 0X02) >> 1; data[cur2 + 3] = output & 0x01; } return 0;}6、R0 通过32位的置换表 P 置换：int P_Table[32] = { //32位R置换函数 表P 15, 6, 19, 20, 28, 11, 27, 16, 0, 14, 22, 25, 4, 17, 30, 9, 1, 7, 23, 13, 31, 26, 2, 8, 18, 12, 29, 5, 21, 10, 3, 24};//32位的R进⾏P置换int P_Substitute(Type data[32]){ Type temp[32]; for (int i = 0; i < 32; i++){ temp[i] = data[P_Table[i]]; } memcpy(data, temp, 32); return 0;}得到新的 R0 和 L0 进⾏异或运算(⽬前为⽌前⾯除了K2到K16，其他值都⽤上了)得到 R1(也就是下⼀轮的 L2 )；将⼀开始的 R0 赋值给 L1。于是得到了 L1 和 R1 ，使⽤ K1 循环上⾯的运算，知道把⼦密钥全部使⽤完。最终的得到 L16 和 R16，合并得到64位的密⽂。//count=48:R与⼦密钥异或运算、count=32:R与L异或运算int XOR(Type R[48], Type L[48], int count){ for (int i = 0; i < count; i++){ R[i] ^= L[i]; } return 0;}//交换int Swap(Type left[32], Type right[32]){ Type temp[32]; memcpy(temp, left, 32); memcpy(left, right, 32); memcpy(right, temp, 32); return 0;}7、得到的密⽂通过 IP_1 逆置换表置换：int IP_1_Table[64] = { //64位密⽂逆初始置换 表IP^-1 39, 7, 47, 15, 55, 23, 63, 31, 38, 6, 46, 14, 54, 22, 62, 30, 37, 5, 45, 13, 53, 21, 61, 29, 36, 4, 44, 12, 52, 20, 60, 28, 35, 3, 43, 11, 51, 19, 59, 27, 34, 2, 42, 10, 50, 18, 58, 26, 33, 1, 41, 9, 49, 17, 57, 25, 32, 0, 40, 8, 48, 16, 56, 24};//IP逆置换int IP_1_Substitute(Type data[64]){ Type temp[64]; for (int i = 0; i < 64; i++){ temp[i] = data[IP_1_Table[i]]; } memcpy(data, temp, 64); return 0;}置换后得到最终的加密结果，64位的密⽂。解密流程：由于加密的算法过程⼏乎是对称的，所以解密过程和加密过程⼏乎⼀样，只是在 R 和16个⼦密钥进⾏与或运算时，应该是从K16到K1。图形界⾯：Visual Studio->⽂件->新建->项⽬->MFC 应⽤程序：基于对话框：布局(其中两个 radio 成组)：给 Button 添加响应函数：void CDESDlg::OnBnClickedOk(){ // TODO: 在此添加控件通知处理程序代码 CString Spath, Dpath, Kpath; GetDlgItemText(Source_path, Spath); GetDlgItemText(Destination_path, Dpath); GetDlgItemText(Key_path, Kpath); char *Cspath = NULL, *Cdpath = NULL, *Ckpath = NULL; bool correct = true; if (!gth() || !gth() || !gth()){ MessageBox(_T("请选择⽂件！")); correct = false; } else{ Cspath = (LPSTR)(LPCTSTR)Spath; Cdpath = (LPSTR)(LPCTSTR)Dpath; Ckpath = (LPSTR)(LPCTSTR)Kpath; } if (correct){ int state; state = IsDlgButtonChecked(Encrypt); clock_t a, b; CString time; if (state) { a = clock(); DES_Encrypt(Cspath, Ckpath, Cdpath); b = clock(); (_T("加密成功，耗时%lu毫秒。"), b - a); MessageBox(time); } else { a = clock(); DES_Decrypt(Cspath, Ckpath, Cdpath); b = clock(); (_T("解密成功，耗时%lu毫秒。"), b - a); MessageBox(time); } }

}运⾏结果：新建4个⽂本⽂件：明⽂、密钥、加密结果、解密结果。明⽂和密钥中输⼊以下内容：选择⽂件，进⾏加密解密：加密和解密结果：修改解密密钥重新解密：