C語言實現RSA算法

本文轉載自查看原文 2020-07-05 18:21 1478 算法

引用自：https://github.com/pantaloons/RSA

　　使用C語言實現RSA算法，網上找了一圈，不是這出問題，就是那里出問題，有些根本都運行不了，錯誤太多，白白浪費時間；
　　絕望之際，在GitHub上看到了這位大佬，真的是大佬，究極膜拜；
　　話不多說，送上代碼（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，計算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          必須使用長乘法，否則這將溢出*/
 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)，計算Jacobi符號（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 ,減去2的倍數*/
 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，翻轉系數 */
 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，檢查a是否為n的歐拉見證 */
 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的精度檢查n是否可能是素數 */
 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）之間找一個隨機素數, 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)，計算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之間找到隨機指數x，使得gcd(x,phi)=1
 99  * this distribution is similarly nowhere near uniform,這種分布同樣不接近制服
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，用擴展歐幾里得法計算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.將文件fd讀入准備加密的字節數組
130  * The array will be padded with zeros until it divides the number of數組將填充零，知道它划分每個塊加密的字節數
131  * bytes encrypted per block. Returns the number of bytes read.返回讀取的字節數
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，將最后一個帶有零的塊插入密碼的信號端。 如果沒有房間，則增加一個額外的街區 */
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使用公共指數和模量對消息m進行編碼，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，用私有指數和公共模量解碼密碼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  * 使用公鑰（指數、模數)對給定長度的消息進行編碼）
175  得到的數組將是大小為len/字節，每個索引是由m=(m1m2*128m3*128^2.)給出的“字節”連續字符的加密，編碼=m^指數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  * 使用私鑰（指數、模數)解碼給定長度的密碼）
196  每個加密的數據包應該按照編碼消息表示“字節”字符。
197  返回的消息大小為len*字節。
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  * 系統降級的主要方法。 設置素數p，q，並開始編碼和
218  *解碼“text.txt”中給出的消息；
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("生成第一個隨機素數, p = %d ... ", p);
229         getchar();
230  
231         q = randPrime(SINGLE_MAX);
232         printf("生成第二個隨機素數, q = %d ... ", q);
233         getchar();
234  
235         n = p * q;
236         printf("計算p和q的乘積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("計算歐拉函數的值phi, phi = %d ... ", phi);
250     getchar();
251  
252     e = randExponent(phi, EXPONENT_MAX);
253     printf("選取一個隨機素數e, e = %d...\n獲得公鑰 (%d, %d) ... ", e, e, n);
254     getchar();
255  
256     d = inverse(e, phi);
257     printf("計算模反元素d, d = %d...\n獲得密鑰 (%d, %d) ... ", d, d, n);
258     getchar();
259  
260     printf("打開文件 \"text.txt\" 用於讀取信息\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倫將是多個字節，以發送整個塊 */
267     fclose(f);
268  
269     printf("文件 \"text.txt\" 讀取成功, 讀取到%d字節. 以%d字節的字節流編碼 ... ", len, bytes);
270     getchar();
271     printf("加密得密文為：");
272     encoded = encodeMessage(len, bytes, buffer, e, n);
273     printf("\n編碼成功完成 ... ");
274     getchar();
275  
276  
277     printf("正在解碼編碼的信息 ... ");
278     getchar();
279     printf("解碼得明文為：");
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 }

　　代碼沒有半點問題，跑就是了；

　　需要注意的是，運行代碼需要新建一個.txt文檔，里面存放你的明文，目前我試過支持數字和英文，中文不支持，其他還沒試過。把這個.txt文檔和代碼放在一個路徑里，注意代碼里的文件名和你的文件名，現在是text.txt；

　　這是運行結果，可以參考下；

運行結果

以上為所有內容，🏄沖浪🏄‍♀️碼字不容易！！！

您的贊就是最好的肯定🙆‍♂️

//還是那句話，有什么問題，歡迎指正！！！🙏

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