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C语言实现TEA系列加解密算法

时间:2019-01-06 08:46:45

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C语言实现TEA系列加解密算法

C语言实现TEA系列加解密算法

TEA加解密XTEA加解密XXTEA加解密

TEA加解密

#include <stdio.h>#include <stdint.h>//加密函数void encrypt(uint32_t* v, uint32_t* k) {uint32_t v0 = v[0], v1 = v[1], sum = 0, i;uint32_t delta = 0x9e3779b9;uint32_t k0 = k[0], k1 = k[1], k2 = k[2], k3 = k[3];for (i = 0; i < 32; i++) {sum += delta;v0 += ((v1 << 4) + k0) ^ (v1 + sum) ^ ((v1 >> 5) + k1);v1 += ((v0 << 4) + k2) ^ (v0 + sum) ^ ((v0 >> 5) + k3);}v[0] = v0; v[1] = v1;}//解密函数void decrypt(uint32_t* v, uint32_t* k) {uint32_t v0 = v[0], v1 = v[1], sum = 0xC6EF3720, i;uint32_t delta = 0x9e3779b9;uint32_t k0 = k[0], k1 = k[1], k2 = k[2], k3 = k[3];for (i = 0; i<32; i++) {v1 -= ((v0 << 4) + k2) ^ (v0 + sum) ^ ((v0 >> 5) + k3);v0 -= ((v1 << 4) + k0) ^ (v1 + sum) ^ ((v1 >> 5) + k1);sum -= delta;}v[0] = v0; v[1] = v1;}int main(){// v为要加解密的数据,两个32位无符号整数uint32_t v[2] = {1,2 };// k为加解密密钥,4个32位无符号整数,密钥长度为128位uint32_t k[4] = {1,2,3,4 };int n = sizeof(v) / sizeof(uint32_t);printf("加密前原始数据:0x%x 0x%x\n", v[0], v[1]);encrypt(v, k);printf("加密后的数据:0x%x 0x%x\n", v[0], v[1]);decrypt(v, k);printf("解密后的数据:0x%x 0x%x\n", v[0], v[1]);for (int i = 0; i < n; i++){for (int j = 0; j < sizeof(uint32_t)/sizeof(uint8_t); j++){printf("%c", (v[i] >> (j * 8)) & 0xFF);}}printf("\n");return 0;}

XTEA加解密

#include <stdio.h>#include <stdint.h>//加密函数void encrypt(unsigned int num_rounds, uint32_t v[2], uint32_t const key[4]) {unsigned int i;uint32_t v0 = v[0], v1 = v[1], sum = 0, delta = 0x9E3779B9;for (i = 0; i < num_rounds; i++) {v0 += (((v1 << 4) ^ (v1 >> 5)) + v1) ^ (sum + key[sum & 3]);sum += delta;v1 += (((v0 << 4) ^ (v0 >> 5)) + v0) ^ (sum + key[(sum >> 11) & 3]);}v[0] = v0; v[1] = v1;}//解密函数void decrypt(unsigned int num_rounds, uint32_t v[2], uint32_t const key[4]) {unsigned int i;uint32_t v0 = v[0], v1 = v[1], delta = 0x9E3779B9, sum = delta*num_rounds;for (i = 0; i < num_rounds; i++) {v1 -= (((v0 << 4) ^ (v0 >> 5)) + v0) ^ (sum + key[(sum >> 11) & 3]);sum -= delta;v0 -= (((v1 << 4) ^ (v1 >> 5)) + v1) ^ (sum + key[sum & 3]);}v[0] = v0; v[1] = v1;}int main(){// v为要加解密的数据,两个32位无符号整数uint32_t v[2] = {1,2 };// k为加解密密钥,4个32位无符号整数,密钥长度为128位uint32_t k[4] = {1,2,3,4 };int n = sizeof(v) / sizeof(uint32_t);// num_rounds,建议取值为32unsigned int r = 32;printf("加密前原始数据:0x%x 0x%x\n", v[0], v[1]);encrypt(r,v, k);printf("加密后的数据:0x%x 0x%x\n", v[0], v[1]);decrypt(r,v, k);printf("解密后的数据:0x%x 0x%x\n", v[0], v[1]);for (int i = 0; i < n; i++){for (int j = 0; j < sizeof(uint32_t) / sizeof(uint8_t); j++){printf("%c", (v[i] >> (j * 8)) & 0xFF);}}printf("\n");return 0;}

XXTEA加解密

#include <stdio.h>#include <stdint.h>#define DELTA 0x9e3779b9#define MX (((z>>5^y<<2) + (y>>3^z<<4)) ^ ((sum^y) + (key[(p&3)^e] ^ z)))void btea(uint32_t *v, int n, uint32_t const key[4]){uint32_t y, z, sum;unsigned p, rounds, e;//加密if (n > 1){rounds = 6 + 52 / n;sum = 0;z = v[n - 1];do{sum += DELTA;e = (sum >> 2) & 3;for (p = 0; p<n - 1; p++){y = v[p + 1];z = v[p] += MX;}y = v[0];z = v[n - 1] += MX;} while (--rounds);}//解密else if (n < -1){n = -n;rounds = 6 + 52 / n;sum = rounds*DELTA;y = v[0];do{e = (sum >> 2) & 3;for (p = n - 1; p>0; p--){z = v[p - 1];y = v[p] -= MX;}z = v[n - 1];y = v[0] -= MX;sum -= DELTA;} while (--rounds);}}int main(){/*原数据为:传进去的参数为:0xbc 0xa5 0xce 0x40->0x40cea5bc0xf4 0xb2 0xb2 0xe7->0xe7b2b2f40xa9 0x12 0x9d 0x12->0x129d12a90xae 0x10 0xc8 0x5b->0x5bc810ae0x3d 0xd7 0x06 0x1d->0x1d06d73d0xdc 0x70 0xf8 0xdc->0xdcf870dc*/uint32_t v[6] = {(unsigned int)0x40cea5bc, (unsigned int)0xe7b2b2f4,(unsigned int)0x129d12a9,(unsigned int)0x5bc810ae,(unsigned int)0x1d06d73d,(unsigned int)0xdcf870dc };/*密钥为字符串"flag"十六进制表示为 0x66 0x6c 0x61 0x67于是传进去的参数要转换成 0x67616c66由于密钥长度为128位,其余需填充0*/uint32_t const k[4] = {(unsigned int)0x67616c66, (unsigned int)0x0,(unsigned int)0X0, (unsigned int)0x0 };//n的绝对值表示v的长度,取正表示加密,取负表示解密int n = sizeof(v) / sizeof(uint32_t);printf("加密前原始数据:0x%x 0x%x\n", v[0], v[1]);btea(v, n, k);printf("加密后的数据:0x%x 0x%x\n", v[0], v[1]);btea(v, -n, k);printf("解密后的数据:0x%x 0x%x\n", v[0], v[1]);for (int i = 0; i < n; i++){for (int j = 0; j < sizeof(uint32_t) / sizeof(uint8_t); j++){printf("%c", (v[i] >> (j * 8)) & 0xFF);}}printf("\n");return 0;}

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