SHA256算法加密工具类
代码如下，请自取
/** * @description: SHA256算法加密 * @author: luolei * @Date: 2022-10-31 17:16 */ public class SHA256 { // 用long模拟32位无符号数据 // 前32位0，后32位1的数字，控制数字在32位,4294967295 final static long OVER = 0xFFFFFFFFL; // 前8个质数的平方根的小数 // long[] h = {0x6a09e667, 0xbb67ae85, 0x3c6ef372, 0xa54ff53a, 0x510e527f, 0x9b05688c, 0x1f83d9ab, 0x5be0cd19}; static long h0 = 0x6a09e667L; static long h1 = 0xbb67ae85L; static long h2 = 0x3c6ef372L; static long h3 = 0xa54ff53aL; static long h4 = 0x510e527fL; static long h5 = 0x9b05688cL; static long h6 = 0x1f83d9abL; static long h7 = 0x5be0cd19L; // 前64个质数的立方根的小数 static long[] k = { 0x428a2f98, 0x71374491, 0xb5c0fbcf, 0xe9b5dba5, 0x3956c25b, 0x59f111f1, 0x923f82a4, 0xab1c5ed5, 0xd807aa98, 0x12835b01, 0x243185be, 0x550c7dc3, 0x72be5d74, 0x80deb1fe, 0x9bdc06a7, 0xc19bf174, 0xe49b69c1, 0xefbe4786, 0x0fc19dc6, 0x240ca1cc, 0x2de92c6f, 0x4a7484aa, 0x5cb0a9dc, 0x76f988da, 0x983e5152, 0xa831c66d, 0xb00327c8, 0xbf597fc7, 0xc6e00bf3, 0xd5a79147, 0x06ca6351, 0x14292967, 0x27b70a85, 0x2e1b2138, 0x4d2c6dfc, 0x53380d13, 0x650a7354, 0x766a0abb, 0x81c2c92e, 0x92722c85, 0xa2bfe8a1, 0xa81a664b, 0xc24b8b70, 0xc76c51a3, 0xd192e819, 0xd6990624, 0xf40e3585, 0x106aa070, 0x19a4c116, 0x1e376c08, 0x2748774c, 0x34b0bcb5, 0x391c0cb3, 0x4ed8aa4a, 0x5b9cca4f, 0x682e6ff3, 0x748f82ee, 0x78a5636f, 0x84c87814, 0x8cc70208, 0x90befffa, 0xa4506ceb, 0xbef9a3f7, 0xc67178f2}; public static String getSha256(String in) { StringBuilder stringBuilder = toBinary(in); String string = addZeroTo512(stringBuilder); return cycleCalculation(string); } // 1. 将输入的字符串转换为二进制in // string的长度限制在2^32 private static StringBuilder toBinary(String in) { StringBuilder stringBuilder = new StringBuilder(); char[] inArray = in.toCharArray(); for (int i = 0; i < inArray.length; i++) { // 还有前导的0 String binary = Integer.toBinaryString(inArray[i]); int count0 = 8 - binary.length(); for (int j = 0; j < count0; j++) { stringBuilder.append("0"); } stringBuilder.append(binary); } return stringBuilder; } // 2. 将输入的二进制的长度补齐到512的倍数，原长度l + 1（默认1位1） + k（补齐的0） + 64（64为二进制表示的输入字符串的长度） private static String addZeroTo512(StringBuilder binaryIn) { int l = binaryIn.length(); int k = 959 - (l % 512); if (k > 512) { k = k - 512; } // 默认先添加1 binaryIn.append("1"); // 添加k个0 for (int i = 0; i < k; i++) { binaryIn.append("0"); } // 添加64位的源数据长度 String lengthBinary = Integer.toBinaryString(l); int k2 = 64 - lengthBinary.length(); for (int i = 0; i < k2; i++) { binaryIn.append("0"); } binaryIn.append(lengthBinary); return binaryIn.toString(); } // 核心：循环计算 /** * @param binaryIn 通过补位后的源数据 * @return */ private static String cycleCalculation(String binaryIn) { // 1. 按照512的长度分成n个消息块 int n = binaryIn.length() / 512; for (int i = 0; i < n; i++) { // 对于每一个块 // 2. 每个消息块分成16个32位的“字” String[] wString = new String[16]; for (int j = 0; j < 16; j++) { wString[j] = binaryIn.substring(32 * j, 32 + 32 * j); } // 3. 将16个字扩充为64个字，转换方法 long[] w = new long[64]; for (int j = 0; j < wString.length; j++) { // 将二进制的string转为10进制的long w[j] = Long.parseLong(wString[j], 2); } // s0 := (w[i-15] rightrotate 7) xor (w[i-15] rightrotate 18) xor(w[i-15] rightshift 3) // s1 := (w[i-2] rightrotate 17) xor (w[i-2] rightrotate 19) xor(w[i-2] rightshift 10) // w[i] := w[i-16] + s0 + w[i-7] + s1 for (int j = 16; j < 64; j++) { long s0 = ((rightRotate(w[j - 15], 7)) ^ rightRotate(w[j - 15], 18) ^ (rightShift(w[j - 15], 3))) & OVER; long s1 = ((rightRotate(w[j - 2], 17)) ^ (rightRotate(w[j - 2], 19)) ^ (rightShift(w[j - 2], 10))) & OVER; w[j] = (w[j - 16] + s0 + w[j - 7] + s1) & OVER; } // 4. hash初始化 long a = h0; long b = h1; long c = h2; long d = h3; long e = h4; long f = h5; long g = h6; long h = h7; // 5. 64次循环 for (int j = 0; j < 64; j++) { // s0 := (a rightrotate 2) xor (a rightrotate 13) xor(a rightrotate 22) // maj := (a and b) xor (a and c) xor(b and c) // t2 := s0 + maj // s1 := (e rightrotate 6) xor (e rightrotate 11) xor(e rightrotate 25) // ch := (e and f) xor ((not e) and g) // t1 := h + s1 + ch + k[i] + w[i] // h := g // g := f // f := e // e := d + t1 // d := c // c := b // b := a // a := t1 + t2 long s0 = (rightRotate(a, 2) ^ (rightRotate(a, 13)) ^ (rightRotate(a, 22))) & OVER; long maj = ((a & b) ^ (a & c) ^ (b & c)) & OVER; long t2 = (s0 + maj) & OVER; long s1 = ((rightRotate(e, 6)) ^ (rightRotate(e, 11)) ^ (rightRotate(e, 25))) & OVER; long ch = ((e & f) ^ ((~e) & g)) & OVER; long t1 = (h + s1 + ch + k[j] + w[j]) & OVER; h = g; g = f; f = e; e = (d + t1) & OVER; d = c; c = b; b = a; a = (t1 + t2) & OVER; } h0 = (h0 + a) & OVER; h1 = (h1 + b) & OVER; h2 = (h2 + c) & OVER; h3 = (h3 + d) & OVER; h4 = (h4 + e) & OVER; h5 = (h5 + f) & OVER; h6 = (h6 + g) & OVER; h7 = (h7 + h) & OVER; } // 把h0到h7拼起来就是所需要的值了。 return getHash(h0) + getHash(h1) + getHash(h2) + getHash(h3) + getHash(h4) + getHash(h5) + getHash(h6) + getHash(h7); } // 把long型的数字转换为字符串，只取后32位的数字，不足则前补0，前面多余的则舍去 private static String getHash(long h) { String hash = Long.toHexString(h); StringBuilder result = new StringBuilder(); if (hash.length() > 8) { result.append(hash.substring(hash.length() - 8, hash.length())); } else { int count0 = 8 - hash.length(); for (int i = 0; i < count0; i++) { result.append("0"); } result.append(hash); } return result.toString(); } /** * 右旋n位，循环右移n位，对于32位无符号数字而言，末尾的数字被移到开头 * 用long模拟32位无符号数字，末尾被移出的数字替换到开头，相当于左移了 * * @param x * @param n * @return */ private static long rightRotate(long x, int n) { long wei = (0 << n) - 1; // 这个操作有点疑问 x = ((wei & (x & OVER)) << (32 - n)) | (x & OVER) >> n; return x; } /** * 按位右移n位，末尾的数字舍去，前面补0，相当于>>>，无符号右移 * * @param x * @param n * @return */ private static long rightShift(long x, int n) { // return (x&0xFFFFFFFFL)>>n; return (x & OVER) >>> n; } }
