引用自：https://github.com/pantaloons/RSA
- 使用C语言实现RSA算法，网上找了一圈，不是这出问题，就是那里出问题，有些根本都运行不了，错误太多，白白浪费时间；
- 绝望之际，在GitHub上看到了这位大佬，真的是大佬，究极膜拜；
- 话不多说，送上代码（我有做修改）；
1 #include <stdio.h> 2 #include <stdlib.h> 3 #include <time.h> 4 #include <string.h> 5 6 #define ACCURACY 5 7 #define SINGLE_MAX 10000 8 #define EXPONENT_MAX 1000 9 #define BUF_SIZE 1024 10 11 /** 12 * Computes a^b mod c，计算a^b mod c 13 */ 14 int modpow(long long a, long long b, int c) { 15 int res = 1; 16 while(b > 0) { 17 /* Need long multiplication else this will overflow... 18 必须使用长乘法，否则这将溢出*/ 19 if(b & 1) { 20 res = (res * a) % c; 21 } 22 b = b >> 1; 23 a = (a * a) % c; /* Same deal here */ 24 } 25 return res; 26 } 27 28 /** 29 * Computes the Jacobi symbol, (a, n)，计算Jacobi符号（a,n） 30 */ 31 int jacobi(int a, int n) { 32 int twos, temp; 33 int mult = 1; 34 while(a > 1 && a != n) { 35 a = a % n; 36 if(a <= 1 || a == n) break; 37 twos = 0; 38 while(a % 2 == 0 && ++twos) a /= 2; /* Factor out multiples of 2 ,减去2的倍数*/ 39 if(twos > 0 && twos % 2 == 1) mult *= (n % 8 == 1 || n % 8 == 7) * 2 - 1; 40 if(a <= 1 || a == n) break; 41 if(n % 4 != 1 && a % 4 != 1) mult *= -1; /* Coefficient for flipping，翻转系数 */ 42 temp = a; 43 a = n; 44 n = temp; 45 } 46 if(a == 0) return 0; 47 else if(a == 1) return mult; 48 else return 0; /* a == n => gcd(a, n) != 1 */ 49 } 50 51 /** 52 * Check whether a is a Euler witness for n，检查a是否为n的欧拉见证 */ 53 int solovayPrime(int a, int n) { 54 int x = jacobi(a, n); 55 if(x == -1) x = n - 1; 56 return x != 0 && modpow(a, (n - 1)/2, n) == x; 57 } 58 59 /** 60 * Test if n is probably prime, using accuracy of k (k solovay tests)，用k的精度检查n是否可能是素数 */ 61 int probablePrime(int n, int k) { 62 if(n == 2) return 1; 63 else if(n % 2 == 0 || n == 1) return 0; 64 while(k-- > 0) { 65 if(!solovayPrime(rand() % (n - 2) + 2, n)) return 0; 66 } 67 return 1; 68 } 69 70 /** 71 * Find a random (probable) prime between 3 and n - 1在3和（n-1）之间找一个随机素数, this distribution is， 72 * nowhere near uniform, see prime gaps 73 */ 74 int randPrime(int n) { 75 int prime = rand() % n; 76 n += n % 2; /* n needs to be even so modulo wrapping preserves oddness */ 77 prime += 1 - prime % 2; 78 while(1) { 79 if(probablePrime(prime, ACCURACY)) return prime; 80 prime = (prime + 2) % n; 81 } 82 } 83 84 /** 85 * Compute gcd(a, b)，计算gcd（a,b） 86 */ 87 int gcd(int a, int b) { 88 int temp; 89 while(b != 0) { 90 temp = b; 91 b = a % b; 92 a = temp; 93 } 94 return a; 95 } 96 97 /** 98 * Find a random exponent x between 3 and n - 1 such that gcd(x, phi) = 1,在3和n-1之间找到随机指数x，使得gcd(x,phi)=1 99 * this distribution is similarly nowhere near uniform,这种分布同样不接近制服 100 */ 101 int randExponent(int phi, int n) { 102 int e = rand() % n; 103 while(1) { 104 if(gcd(e, phi) == 1) return e; 105 e = (e + 1) % n; 106 if(e <= 2) e = 3; 107 } 108 } 109 110 /** 111 * Compute n^-1 mod m by extended euclidian method，用扩展欧几里得法计算n^-1 mod m 112 */ 113 int inverse(int n, int modulus) { 114 int a = n, b = modulus; 115 int x = 0, y = 1, x0 = 1, y0 = 0, q, temp; 116 while(b != 0) { 117 q = a / b; 118 temp = a % b; 119 a = b; 120 b = temp; 121 temp = x; x = x0 - q * x; x0 = temp; 122 temp = y; y = y0 - q * y; y0 = temp; 123 } 124 if(x0 < 0) x0 += modulus; 125 return x0; 126 } 127 128 /** 129 * Read the file fd into an array of bytes ready for encryption.将文件fd读入准备加密的字节数组 130 * The array will be padded with zeros until it divides the number of数组将填充零，知道它划分每个块加密的字节数 131 * bytes encrypted per block. Returns the number of bytes read.返回读取的字节数 132 */ 133 int readFile(FILE* fd, char** buffer, int bytes) { 134 int len = 0, cap = BUF_SIZE, r; 135 char buf[BUF_SIZE]; 136 *buffer = (char *)malloc(BUF_SIZE * sizeof(char)); 137 while((r = fread(buf, sizeof(char), BUF_SIZE, fd)) > 0) { 138 if(len + r >= cap) { 139 cap *= 2; 140 *buffer = (char *)realloc(*buffer, cap); 141 } 142 memcpy(&(*buffer)[len], buf, r); 143 len += r; 144 } 145 /* Pad the last block with zeros to signal end of cryptogram. An additional block is added if there is no room，将最后一个带有零的块插入密码的信号端。 如果没有房间，则增加一个额外的街区 */ 146 if(len + bytes - len % bytes > cap) *buffer = (char *)realloc(*buffer, len + bytes - len % bytes); 147 do { 148 (*buffer)[len] = '\0'; 149 len++; 150 } 151 while(len % bytes != 0); 152 return len; 153 } 154 155 /** 156 * Encode the message m using public exponent and modulus, c = m^e mod n使用公共指数和模量对消息m进行编码，c = m^e Mod n 157 */ 158 int encode(int m, int e, int n) { 159 return modpow(m, e, n); 160 } 161 162 /** 163 * Decode cryptogram c using private exponent and public modulus, m = c^d mod n，用私有指数和公共模量解码密码c，m = c^d Mod n 164 */ 165 int decode(int c, int d, int n) { 166 return modpow(c, d, n); 167 } 168 169 /** 170 * Encode the message of given length, using the public key (exponent, modulus) 171 * The resulting array will be of size len/bytes, each index being the encryption 172 * of "bytes" consecutive characters, given by m = (m1 + m2*128 + m3*128^2 + ..), 173 * encoded = m^exponent mod modulus 174 * 使用公钥（指数、模数)对给定长度的消息进行编码） 175 得到的数组将是大小为len/字节，每个索引是由m=(m1m2*128m3*128^2.)给出的“字节”连续字符的加密，编码=m^指数mod模量 176 */ 177 int* encodeMessage(int len, int bytes, char* message, int exponent, int modulus) { 178 int *encoded = (int *)malloc((len/bytes) * sizeof(int)); 179 int x, i, j; 180 for(i = 0; i < len; i += bytes) { 181 x = 0; 182 for(j = 0; j < bytes; j++) x += message[i + j] * (1 << (7 * j)); 183 encoded[i/bytes] = encode(x, exponent, modulus); 184 #ifndef MEASURE 185 printf("%d ", encoded[i/bytes]); 186 #endif 187 } 188 return encoded; 189 } 190 191 /** 192 * Decode the cryptogram of given length, using the private key (exponent, modulus) 193 * Each encrypted packet should represent "bytes" characters as per encodeMessage. 194 * The returned message will be of size len * bytes. 195 * 使用私钥（指数、模数)解码给定长度的密码） 196 每个加密的数据包应该按照编码消息表示“字节”字符。 197 返回的消息大小为len*字节。 198 */ 199 int* decodeMessage(int len, int bytes, int* cryptogram, int exponent, int modulus) { 200 int *decoded = (int *)malloc(len * bytes * sizeof(int)); 201 int x, i, j; 202 for(i = 0; i < len; i++) { 203 x = decode(cryptogram[i], exponent, modulus); 204 for(j = 0; j < bytes; j++) { 205 decoded[i*bytes + j] = (x >> (7 * j)) % 128; 206 #ifndef MEASURE 207 if(decoded[i*bytes + j] != '\0') printf("%c", decoded[i*bytes + j]); 208 #endif 209 } 210 } 211 return decoded; 212 } 213 214 /** 215 * Main method to demostrate the system. Sets up primes p, q, and proceeds to encode and 216 * decode the message given in "text.txt" 217 * 系统降级的主要方法。 设置素数p，q，并开始编码和 218 *解码“text.txt”中给出的消息； 219 */ 220 int main(void) { 221 int p, q, n, phi, e, d, bytes, len; 222 int *encoded, *decoded; 223 char *buffer; 224 FILE *f; 225 srand(time(NULL)); 226 while(1) { 227 p = randPrime(SINGLE_MAX); 228 printf("生成第一个随机素数, p = %d ... ", p); 229 getchar(); 230 231 q = randPrime(SINGLE_MAX); 232 printf("生成第二个随机素数, q = %d ... ", q); 233 getchar(); 234 235 n = p * q; 236 printf("计算p和q的乘积n, n = pq = %d ... ", n); 237 if(n < 128) { 238 printf("Modulus is less than 128, cannot encode single bytes. Trying again ... "); 239 getchar(); 240 } 241 else break; 242 } 243 if(n >> 21) bytes = 3; 244 else if(n >> 14) bytes = 2; 245 else bytes = 1; 246 getchar(); 247 248 phi = (p - 1) * (q - 1); 249 printf("计算欧拉函数的值phi, phi = %d ... ", phi); 250 getchar(); 251 252 e = randExponent(phi, EXPONENT_MAX); 253 printf("选取一个随机素数e, e = %d...\n获得公钥 (%d, %d) ... ", e, e, n); 254 getchar(); 255 256 d = inverse(e, phi); 257 printf("计算模反元素d, d = %d...\n获得密钥 (%d, %d) ... ", d, d, n); 258 getchar(); 259 260 printf("打开文件 \"text.txt\" 用于读取信息\n"); 261 f = fopen("text.txt", "r"); 262 if(f == NULL) { 263 printf("Failed to open file \"text.txt\". Does it exist?\n"); 264 return EXIT_FAILURE; 265 } 266 len = readFile(f, &buffer, bytes); /* len will be a multiple of bytes, to send whole chunks伦将是多个字节，以发送整个块 */ 267 fclose(f); 268 269 printf("文件 \"text.txt\" 读取成功, 读取到%d字节. 以%d字节的字节流编码 ... ", len, bytes); 270 getchar(); 271 printf("加密得密文为："); 272 encoded = encodeMessage(len, bytes, buffer, e, n); 273 printf("\n编码成功完成 ... "); 274 getchar(); 275 276 277 printf("正在解码编码的信息 ... "); 278 getchar(); 279 printf("解码得明文为："); 280 decoded = decodeMessage(len/bytes, bytes, encoded, d, n); 281 282 283 printf("\nRSA算法演示完成!\n"); 284 285 free(encoded); 286 free(decoded); 287 free(buffer); 288 return EXIT_SUCCESS; 289 }
代码没有半点问题，跑就是了；
需要注意的是，运行代码需要新建一个.txt文档，里面存放你的明文，目前我试过支持数字和英文，中文不支持，其他还没试过。把这个.txt文档和代码放在一个路径里，注意代码里的文件名和你的文件名，现在是text.txt；
这是运行结果，可以参考下；
以上为所有内容，🏄冲浪🏄♀️码字不容易！！！
您的赞就是最好的肯定🙆♂️
//还是那句话，有什么问题，欢迎指正！！！🙏