AES加密解密C语言实现 - 图文

快毕业了，最后一个课程设计，《基于Windows Socket的安全通信》，内容就是基于AES加密的SOCKET通信，貌似挺简单，不过要用VC++6.0开发，C++我确实没有任何代码经验，虽然不是强制性，但由于机房里各种纠结，只能用它了（用Java没有挑战性，封装得太好了...也算熟悉下VC++吧）

先搞定AES算法，基本变换包括SubBytes（字节替代）、ShiftRows（行移位）、MixColumns（列混淆）、AddRoundKey(轮密钥加） 其算法一般描述为

明文及密钥的组织排列方式 ByteSubstitution（字节替代） 非线性的字节替代，单独处理每个字节： 求该字节在有限域GF(28)上的乘法逆，\被映射为自身，即对于α∈GF(28)，求β∈GF(28)， 使得α·β=β·α=1mod(x8+x4+x2+x+1)。 对上一步求得的乘法逆作仿射变换 yi=xi + x(i+4)mod8 + x(i+6)mod8 + x(i+7)mod8 + ci (其中ci是6310即011000112的第i位），用矩阵表示为 本来打算把求乘法逆和仿射变换算法敲上去，最后还是放弃了...直接打置换表 1 unsigned char sBox[] = 2 { /* 0 1 2 3 4 5 6 7 8 9 a b c d e f */ 3 0x63,0x7c,0x77,0x7b,0xf2,0x6b,0x6f,0xc5,0x30,0x01,0x67,0x2b,0xfe,0xd7,0xab,0x76, /*0*/ 4 0xca,0x82,0xc9,0x7d,0xfa,0x59,0x47,0xf0,0xad,0xd4,0xa2,0xaf,0x9c,0xa4,0x72,0xc0, /*1*/ 5 0xb7,0xfd,0x93,0x26,0x36,0x3f,0xf7,0xcc,0x34,0xa5,0xe5,0xf1,0x71,0xd8,0x31,0x15, /*2*/ 6 0x04,0xc7,0x23,0xc3,0x18,0x96,0x05,0x9a,0x07,0x12,0x80,0xe2,0xeb,0x27,0xb2,0x75, /*3*/ 7 0x09,0x83,0x2c,0x1a,0x1b,0x6e,0x5a,0xa0,0x52,0x3b,0xd6,0xb3,0x29,0xe3,0x2f,0x84, /*4*/ 8 0x53,0xd1,0x00,0xed,0x20,0xfc,0xb1,0x5b,0x6a,0xcb,0xbe,0x39,0x4a,0x4c,0x58,0xcf, /*5*/ 9 0xd0,0xef,0xaa,0xfb,0x43,0x4d,0x33,0x85,0x45,0xf9,0x02,0x7f,0x50,0x3c,0x9f,0xa8, /*6*/ 10 0x51,0xa3,0x40,0x8f,0x92,0x9d,0x38,0xf5,0xbc,0xb6,0xda,0x21,0x10,0xff,0xf3,0xd2, /*7*/ 11 0xcd,0x0c,0x13,0xec,0x5f,0x97,0x44,0x17,0xc4,0xa7,0x7e,0x3d,0x64,0x5d,0x19,0x73, /*8*/ 12 0x60,0x81,0x4f,0xdc,0x22,0x2a,0x90,0x88,0x46,0xee,0xb8,0x14,0xde,0x5e,0x0b,0xdb, /*9*/ 13 0xe0,0x32,0x3a,0x0a,0x49,0x06,0x24,0x5c,0xc2,0xd3,0xac,0x62,0x91,0x95,0xe4,0x79, /*a*/ 14 0xe7,0xc8,0x37,0x6d,0x8d,0xd5,0x4e,0xa9,0x6c,0x56,0xf4,0xea,0x65,0x7a,0xae,0x08, /*b*/ 15 0xba,0x78,0x25,0x2e,0x1c,0xa6,0xb4,0xc6,0xe8,0xdd,0x74,0x1f,0x4b,0xbd,0x8b,0x8a, /*c*/ 16 0x70,0x3e,0xb5,0x66,0x48,0x03,0xf6,0x0e,0x61,0x35,0x57,0xb9,0x86,0xc1,0x1d,0x9e, /*d*/ 17 0xe1,0xf8,0x98,0x11,0x69,0xd9,0x8e,0x94,0x9b,0x1e,0x87,0xe9,0xce,0x55,0x28,0xdf, /*e*/ 18 0x8c,0xa1,0x89,0x0d,0xbf,0xe6,0x42,0x68,0x41,0x99,0x2d,0x0f,0xb0,0x54,0xbb,0x16 /*f*/ 19 }; 下面是逆置换表，解密时使用 1 unsigned char invsBox[256] = 2 { /* 0 1 2 3 4 5 6 7 8 9 a b c d e 3 0x52,0x09,0x6a,0xd5,0x30,0x36,0xa5,0x38,0xbf,0x40,0xa3,0x9e,0x81,0xf3,0xd7,0x4 0x7c,0xe3,0x39,0x82,0x9b,0x2f,0xff,0x87,0x34,0x8e,0x43,0x44,0xc4,0xde,0xe9,0x5 0x54,0x7b,0x94,0x32,0xa6,0xc2,0x23,0x3d,0xee,0x4c,0x95,0x0b,0x42,0xfa,0xc3,0x6 0x08,0x2e,0xa1,0x66,0x28,0xd9,0x24,0xb2,0x76,0x5b,0xa2,0x49,0x6d,0x8b,0xd1,0x7 0x72,0xf8,0xf6,0x64,0x86,0x68,0x98,0x16,0xd4,0xa4,0x5c,0xcc,0x5d,0x65,0xb6,0x8 0x6c,0x70,0x48,0x50,0xfd,0xed,0xb9,0xda,0x5e,0x15,0x46,0x57,0xa7,0x8d,0x9d,0x9 0x90,0xd8,0xab,0x00,0x8c,0xbc,0xd3,0x0a,0xf7,0xe4,0x58,0x05,0xb8,0xb3,0x45,0x10 0xd0,0x2c,0x1e,0x8f,0xca,0x3f,0x0f,0x02,0xc1,0xaf,0xbd,0x03,0x01,0x13,0x8a,0x11 0x3a,0x91,0x11,0x41,0x4f,0x67,0xdc,0xea,0x97,0xf2,0xcf,0xce,0xf0,0xb4,0xe6,0x12 0x96,0xac,0x74,0x22,0xe7,0xad,0x35,0x85,0xe2,0xf9,0x37,0xe8,0x1c,0x75,0xdf,0x13 0x47,0xf1,0x1a,0x71,0x1d,0x29,0xc5,0x89,0x6f,0xb7,0x62,0x0e,0xaa,0x18,0xbe,0x14 0xfc,0x56,0x3e,0x4b,0xc6,0xd2,0x79,0x20,0x9a,0xdb,0xc0,0xfe,0x78,0xcd,0x5a,0x15 0x1f,0xdd,0xa8,0x33,0x88,0x07,0xc7,0x31,0xb1,0x12,0x10,0x59,0x27,0x80,0xec,0x16 0x60,0x51,0x7f,0xa9,0x19,0xb5,0x4a,0x0d,0x2d,0xe5,0x7a,0x9f,0x93,0xc9,0x9c,0x17 0xa0,0xe0,0x3b,0x4d,0xae,0x2a,0xf5,0xb0,0xc8,0xeb,0xbb,0x3c,0x83,0x53,0x99,0x18 0x17,0x2b,0x04,0x7e,0xba,0x77,0xd6,0x26,0xe1,0x69,0x14,0x63,0x55,0x21,0x0c,0x19 };

这里遇到问题了，本来用纯c初始化数组很正常，封装成类以后发现不能初始化，不管是声明、构造函数都无法初始化，百歌谷度了一通后没有任何答案，无奈只能在构造函数中声明一个局部变量数组并初始化，然后用memcpy，（成员变量名为Sbox/InvSbox，局部变量名sBox/invsBox） 1 2 3 4 5 6 7 8 9 10 11 void AES::SubBytes(unsigned char state[][4]) { int r,c; for(r=0; r<4; r++) { for(c=0; c<4; c++) { state[r][c] = Sbox[state[r][c]]; } } } ShiftRows（行移位变换） 行移位变换完成基于行的循环位移操作，变换方法： 即行移位变换作用于行上，第0行不变，第1行循环左移1个字节，第2行循环左移2个字节，第3行循环左移3个字节。 1 2 3 4 5 6 7 8 9 10 11 void AES::ShiftRows(unsigned char state[][4]) { unsigned char t[4]; int r,c; for(r=1; r<4; r++) { for(c=0; c<4; c++) { t[c] = state[r][(c+r)%4]; } for(c=0; c<4; c++) 12 13 14 15 16 { state[r][c] = t[c]; } } } MixColumns（列混淆变换） 逐列混合，方法： b(x) = (03·x3 + 01·x2 + 01·x + 02) · a(x) mod(x4 + 1) 矩阵表示形式： 1 2 3 4 5 6 7 void AES::MixColumns(unsigned char state[][4]) { unsigned char t[4]; int r,c; for(c=0; c< 4; c++) { for(r=0; r<4; r++)