AES原始碼（java)實現過程解析
阿新 • • 發佈：2018-12-11
在CSDN上找了一下AES的原始碼(免積分的)，但是沒有註釋，有註釋的都需要積分，所以我自己理了一遍，順便把原始碼也分享給大家。
/** * AES實現過程演示 * * @author Bloomeet * @time 2018/9/23 19:55 */ public class AESclass { static int Nr = 10;//輪數 static int key[] = {0x2B, 0x7E, 0x15, 0x16, 0x28, 0xAE, 0xD2, 0xA6, 0xAB, 0xF7, 0x15, 0x88, 0x09, 0xCF, 0x4F, 0x3C};//16進位制金鑰 static int text[] = {0x32, 0x43, 0xF6, 0xA8, 0x88, 0x5A, 0x30, 0x8D, 0x31, 0x31, 0x98, 0xA2, 0xE0, 0x37, 0x07, 0x34};//16進位制明文 static int sbox[] = {0x63, 0x7C, 0x77, 0x7B, 0xF2, 0x6B, 0x6F, 0xC5, 0x30, 0x01, 0x67, 0x2B, 0xFE, 0xD7, 0xAB, 0x76, 0xCA, 0x82, 0xC9, 0x7D, 0xFA, 0x59, 0x47, 0xF0, 0xAD, 0xD4, 0xA2, 0xAF, 0x9C, 0xA4, 0x72, 0xC0, 0xB7, 0xFD, 0x93, 0x26, 0x36, 0x3F, 0xF7, 0xCC, 0x34, 0xA5, 0xE5, 0xF1, 0x71, 0xD8, 0x31, 0x15, 0x04, 0xC7, 0x23, 0xC3, 0x18, 0x96, 0x05, 0x9A, 0x07, 0x12, 0x80, 0xE2, 0xEB, 0x27, 0xB2, 0x75, 0x09, 0x83, 0x2C, 0x1A, 0x1B, 0x6E, 0x5A, 0xA0, 0x52, 0x3B, 0xD6, 0xB3, 0x29, 0xE3, 0x2F, 0x84, 0x53, 0xD1, 0x00, 0xED, 0x20, 0xFC, 0xB1, 0x5B, 0x6A, 0xCB, 0xBE, 0x39, 0x4A, 0x4C, 0x58, 0xCF, 0xD0, 0xEF, 0xAA, 0xFB, 0x43, 0x4D, 0x33, 0x85, 0x45, 0xF9, 0x02, 0x7F, 0x50, 0x3C, 0x9F, 0xA8, 0x51, 0xA3, 0x40, 0x8F, 0x92, 0x9D, 0x38, 0xF5, 0xBC, 0xB6, 0xDA, 0x21, 0x10, 0xFF, 0xF3, 0xD2, 0xCD, 0x0C, 0x13, 0xEC, 0x5F, 0x97, 0x44, 0x17, 0xC4, 0xA7, 0x7E, 0x3D, 0x64, 0x5D, 0x19, 0x73, 0x60, 0x81, 0x4F, 0xDC, 0x22, 0x2A, 0x90, 0x88, 0x46, 0xEE, 0xB8, 0x14, 0xDE, 0x5E, 0x0B, 0xDB, 0xE0, 0x32, 0x3A, 0x0A, 0x49, 0x06, 0x24, 0x5C, 0xC2, 0xD3, 0xAC, 0x62, 0x91, 0x95, 0xE4, 0x79, 0xE7, 0xC8, 0x37, 0x6D, 0x8D, 0xD5, 0x4E, 0xA9, 0x6C, 0x56, 0xF4, 0xEA, 0x65, 0x7A, 0xAE, 0x08, 0xBA, 0x78, 0x25, 0x2E, 0x1C, 0xA6, 0xB4, 0xC6, 0xE8, 0xDD, 0x74, 0x1F, 0x4B, 0xBD, 0x8B, 0x8A, 0x70, 0x3E, 0xB5, 0x66, 0x48, 0x03, 0xF6, 0x0E, 0x61, 0x35, 0x57, 0xB9, 0x86, 0xC1, 0x1D, 0x9E, 0xE1, 0xF8, 0x98, 0x11, 0x69, 0xD9, 0x8E, 0x94, 0x9B, 0x1E, 0x87, 0xE9, 0xCE, 0x55, 0x28, 0xDF, 0x8C, 0xA1, 0x89, 0x0D, 0xBF, 0xE6, 0x42, 0x68, 0x41, 0x99, 0x2D, 0x0F, 0xB0, 0x54, 0xBB, 0x16,};//S盒表 static int RCon[];//金鑰的常量陣列 static int word[][];//輪金鑰擴充套件後的陣列，大小為 word[44][4] public static void main(String[] args) { keyExpansion();//金鑰 int i, r = 0; text = addRoundKey(text, 0);//初始輪金鑰加 r++; for (; r <= Nr; r++) { // 迭代10輪 for (i = 0; i < 16; i++) { text[i] = subByte(text[i]);// 位元組替代 } text = shiftRows(text); // 行移位 if (r != 10) {// 注意：AES加密最後一輪沒有列混合 text = mixColumn(text);// 列混合 } text = addRoundKey(text, r);// 輪金鑰加 } //輸出加密結果 for (i = 0; i < 16; i++) { System.out.print(Integer.toHexString(text[i]) + " "); if (i % 4 == 3) { System.out.println(); } } System.out.println(); } /** * 金鑰陣列擴充套件 */ static void keyExpansion() { RCon = new int[10];//輪常量為固定值 RCon[0] = 0x01; RCon[1] = 0x02; RCon[2] = 0x04; RCon[3] = 0x08; RCon[4] = 0x10; RCon[5] = 0x20; RCon[6] = 0x40; RCon[7] = 0x80; RCon[8] = 0x1B; RCon[9] = 0x36; word = new int[44][4];//44組輪金鑰 int i, j; int temp[]; for (i = 0; i < 4; i++) { for (j = 0; j < 4; j++) { word[i][j] = key[i * 4 + j]; //把初始金鑰放入陣列 } } /*通過金鑰計算規則計算餘下陣列 * *1.如果i不是4的倍數，那麼第i列由如下等式確定： *W[i]=W[i-4]⨁W[i-1] *2.如果i是4的倍數，那麼第i列由如下等式確定： *W[i]=W[i-4]⨁T(W[i-1]) *其中，T是一個有點複雜的函式。函式T由3部分組成：字迴圈(每次迴圈一位)、位元組代換（s盒）和輪常量異或。 */ for (i = 4; i < 44; i++) { temp = word[i - 1];//看作W[i-1] if (i % 4 == 0) { //i為4的倍數 進入函式運算 W[i-1]=T(W[i-1]) temp = subWord(rotWord(temp)); temp[0] = temp[0] ^ RCon[i / 4 - 1]; } for (j = 0; j < 4; j++) { word[i][j] = word[i - 4][j] ^ temp[j];//相當於W[i]=W[i-4]⨁W[i-1] } } } //金鑰擴充套件中的移位 static int[] rotWord(int[] word) { int[] rot = new int[4]; int i; for (i = 0; i < 4; i++) { rot[i] = word[(i + 1) % 4]; } return rot; } //金鑰擴充套件中的4個位元組的代換（4個位元組為一組） 例如 A2 BE C4 D5 static int[] subWord(int[] word) { int sub[] = new int[4]; int i; for (i = 0; i < 4; i++) { sub[i] = subByte(word[i]); } return sub; } //S盒的單個位元組代換 例如 AE static int subByte(int w) { int x = w / 16; int y = w % 16; return sbox[x * 16 + y]; } //輪金鑰加 static int[] addRoundKey(int[] text, int round) { int[] add = new int[16]; int i, j; for (i = 0; i < 4; i++) { for (j = 0; j < 4; j++) { // System.out.print(Integer.toHexString(word[4 * round + i][j]) + ",");//第round輪的輪金鑰 add[4 * i + j] = text[4 * i + j] ^ word[4 * round + i][j]; //逐位元異或 } } return add; } //行移位（迴圈移位）規則：第0行移0位 ..... 第3行移3位 static int[] shiftRows(int[] text) { int[] shift = new int[16]; int i, j; for (i = 0; i < 4; i++) { for (j = 0; j < 4; j++) { shift[4 * i + j] = text[4 * ((i + j) % 4) + j]; } } return shift; } //列混合 static int[] mixColumn(int[] text) { int[] mix = new int[16]; int[] mass = {2, 3, 1, 1}; int i, j, u; for (i = 0; i < 16; i++) { u = 0; for (j = 0; j < 4; j++) { u = u ^ fieldMulit(text[(i / 4) * 4 + j], mass[(4 - i % 4 + j) % 4]); } mix[i] = u; } return mix; } static int fieldMulit(int x, int y) { String xString = Integer.toBinaryString(x); int i, j, mul = 0, tem = y; for (i = 0; i < xString.length(); i++) { if (xString.charAt(i) == '1') { for (j = 1; j < xString.length() - i; j++) { tem = tem << 1; if (tem > 255) { tem = tem % 256; tem = tem ^ 0x1b; } } mul = mul ^ tem; tem = y; } } return mul; } }