DES加密实现（c语言）
这学期开了密码学课，听老师在课堂上讲各种加密算法各种高大上，但自high过后才发现别的先不说，单是DES自己看课本完全不懂。最后是在同学推荐下看了个视频才弄懂了是怎样一步步进行处理的，然后一时手痒就编了出来。虽然功能实现了，但程序的健壮性不是很好。目前程序是实现八位密钥与八位明文进行加密处理，密钥和明文从键盘读入，多于八位会缓冲掉，但少于八位的我目前还没想到怎么处理，还有的就是实际上明文不应该有位数限制，但由于我能力有限，这些问题（说不定还有我没发现的问题）都没解决，各位路过的大大有解决方法的麻烦指导一下，不胜感激~~
好了，废话不多说了，见代码吧~（ps.命名不是很规范，大家将就下）
#include
/*各种函数定义*/
int move_num(int r);
void xor(int *r_b, int *k3, int large);
void s_box(int r_t[32], int r_b[48]);
void erzhuanshiliu(int *k, char *key, int large);
void shizhuaner(int *k, int n, int i, int large);
void round(int r, int k2[56], int k3[48], int r_s[32], int r_b[48], int l[32], int k_c_two[48], int m_c_two[32], int e[48], char c_r[8], char c_l[8]);
void exchange(int *k1, int *k2, int *rule, int n);
void key_move(int k2[56], int n);
/*十进制转换成二进制*/
void shizhuaner(int *k, int n, int i,int large)
/* k是存放二进制的数组，n是需要进行转换的十进制数，i是第i个十进制数，large是数组大小*/
{
int j; //计数的j，没什么大的含义
large = large / 8;
j = large - 1;
do
{
k[i * large + j] = n % 2;
n = n / 2;
j--;
} while (n != 0);
while (j >= 0)
{
k[i * large + j] = 0;
j--;
}
return ;
}
/*二进制转换成十六进制*/
void erzhuanshiliu(int *k,char *key, int large)
/* k是二进制数组，key是存放十六进制的数组，large是k数组的大小*/
{
int i,j; //计数的i、j，没什么大的含义
large = large / 4;
for (i = 0; i < large; i++)
{
j = k[i * 4] * 8 + k[i * 4 + 1] * 4 + k[i * 4 + 2] * 2 + k[i * 4 + 3];
if (j >= 10)
{
switch (j)
{
case 10:key[i] = 'a'; break;
case 11:key[i] = 'b'; break;
case 12:key[i] = 'c'; break;
case 13:key[i] = 'd'; break;
case 14:key[i] = 'e'; break;
case 15:key[i] = 'f'; break;
}
}
else
{
j = j + 48;
key[i] = (char)j;
}
}
return;
}
/*循环函数*/
void round(int r, int k2[56], int k3[48], int r_s[32], int r_b[48], int l[32], int k_c_two[48], int m_c_two[32], int e[48], char c_r[8], char c_l[8])
/* r是轮数*/
/* k2是密钥经过置换1置换完的数组，k3是密钥经过置换2置换完的数组*/
/* r_s是明文处理中R的32大小时的数组，r_b是明文处理中R的48大小时的数组，l是明文处理中L的数组*/
/* k_c_two是密钥处理中置换2的置换规则数组，m_c_two是明文处理中置换2的置换规则数组，e是明文处理中e盒的置换规则数组*/
/* c_r是各轮密文的r部分,c_l是各轮密文的l部分*/
{
int i, n, r_t[32],r_c[32]; //计数的i，没什么大的含义；n存储该轮左移的数值；r_t,r_c存放中间处理数值的32大小的数组
printf("这是第%d轮密文：\n", r);
n = move_num(r); //获取第r轮密钥左移的位数
key_move(k2, n); //调用密钥左移函数
exchange(k2, k3, k_c_two, 48); //密钥进行置换2运算
exchange(r_s, r_b,e,48); //明文进行e盒扩展置换运算
xor(r_b, k3,48); //经过密钥置换2运算后的子密钥与从e盒出来的子明文异或
s_box(r_t,r_b); //调用明文处理中的s盒运算函数
exchange(r_t, r_c, m_c_two, 32); //明文进行置换2运算
for (i = 0; i < 32; i++)
r_t[i] = r_c[i];
xor(r_t, l, 32); //经过处理后的R与L异或
for (i = 0; i < 32; i++)
l[i] = r_s[i]; //准备下次循环的L
for (i = 0; i < 32; i++)
r_s[i] = r_t[i]; //准备下次循环的R
erzhuanshiliu(r_s, c_r, 32);
for (i = 0; i < 8; i++)
printf("%c ", c_r[i]); //第r轮R部分的密文
erzhuanshiliu(l, c_l, 32);
for (i = 0; i < 8; i++)
printf("%c ", c_l[i]); //第r轮L部分的密文
printf("\n");
printf("\n");
return;
}
/* 异或函数 */
void xor(int *r_b, int *k3,int large)
{
int i; //计数的i，没什么大的含义
for (i = 0; i < large; i++)
{
if (r_b[i] != k3[i]) //两数相同结果取0，两数不同取1
r_b[i] = 1;
else
r_b[i] = 0;
}
}
/* S盒运算函数 */
void s_box(int r_t[32],int r_b[48])
{
int i,hang,lie,jie; //计数的i，没什么大的含义；hang是S盒的行下标；lie是S盒的列下标；jie是根据行、列下标查询出的结果
/* S1~S8盒 */
int s1[4][16] = { { 14, 4, 13, 1, 2, 15, 11, 8, 3, 10, 6, 12, 5, 9, 0, 7 },
{ 0, 15, 7, 4, 14, 2, 13, 1, 10, 6, 12, 11, 9, 5, 3, 8 },
{ 4, 1, 14, 8, 13, 6, 2, 11, 15, 12, 9, 7, 3, 10, 5, 0 },
{ 15, 12, 8, 2, 4, 9, 1, 7, 5, 11, 3, 14, 10, 0, 6, 13 } },
s2[4][16] = { { 15, 1, 8, 14, 6, 11, 3, 4, 9, 7, 2, 13, 12, 0, 5, 10 },
{ 3, 13, 4, 7, 15, 2, 8, 14, 12, 0, 1, 10, 6, 9, 11, 5 },
{ 0, 14, 7, 11, 10, 4, 13, 1, 5, 8, 12, 6, 9, 3, 2, 15 },
{ 13, 8, 10, 1, 3, 15, 4, 2, 11, 6, 7, 12, 0, 5, 14, 9 } },
s3[4][16] = { { 10, 0, 9, 14, 6, 3, 15, 5, 1, 13, 12, 7, 11, 4, 2, 8 },
{ 13, 7, 0, 9, 3, 4, 6, 10, 2, 8, 5, 14, 12, 11, 15, 1 },
{ 13, 6, 4, 9, 8, 15, 3, 0, 11, 1, 2, 12, 5, 10, 14, 7 },
{ 1, 10, 13, 0, 6, 9, 8, 7, 4, 15, 14, 3, 11, 5, 2, 12 } },
s4[4][16] = { { 7, 13, 14, 3, 0, 6, 9, 10, 1, 2, 8, 5, 11, 12, 4, 15 },
{ 13, 8, 11, 5, 6, 15, 0, 3, 4, 7, 2, 12, 1, 10, 14, 9 },
{ 10, 6, 9, 0, 12, 11, 7, 13, 15, 1, 3, 14, 5, 2, 8, 3 },
{ 3, 15, 0, 6, 10, 1, 13, 8, 9, 4, 5, 11, 12, 7, 2, 14 } },
s5[4][16] = { { 2, 12, 4, 1, 7, 10, 11, 6, 8, 5, 3, 15, 13, 0, 14, 9 },
{ 14, 11, 2, 12, 4, 7, 13, 1, 5, 0, 15, 10, 3, 9, 8, 6 },
{ 4, 2, 1, 11, 10, 13, 7, 8, 15, 9, 12, 5, 6, 3, 0, 14 },
{ 11, 8, 12, 7, 1, 14, 2, 13, 6, 15, 0, 9, 10, 4, 5, 3 } },
s6[4][16] = { { 12, 1, 10, 15, 9, 2, 6, 8, 0, 13, 3, 4, 14, 7, 5, 11 },
{ 10, 15, 4, 2, 7, 12, 9, 5, 6, 1, 13, 14, 0, 11, 3, 8 },
{ 9, 14, 15, 5, 2, 8, 12, 3, 7, 0, 4, 10, 1, 13, 11, 6 },
{ 4, 3, 2, 12, 9, 5, 15, 10, 11, 14, 1, 7, 6, 0, 8, 13 } },
s7[4][16] = { { 4, 11, 2, 14, 15, 0, 8, 13, 3, 12, 9, 7, 5, 10, 6, 1 },
{ 13, 0, 11, 7, 4, 9, 1, 10, 14, 3, 5, 12, 2, 15, 8, 6 },
{ 1, 4, 11, 13, 12, 3, 7, 14, 10, 15, 6, 8, 0, 5, 9, 2 },
{ 6, 11, 13, 8, 1, 4, 10, 7, 9, 5, 0, 15, 14, 2, 3, 12 } },
s8[4][16] = { { 13, 2, 8, 4, 6, 15, 11, 1, 10, 9, 3, 14, 5, 0, 12, 7 },
{ 1, 15, 13, 8, 10, 3, 7, 4, 12, 5, 6, 11, 0, 14, 9, 2 },
{ 7, 11, 4, 1, 9, 12, 14, 2, 0, 6, 10, 13, 15, 3, 5, 8 },
{ 2, 1, 14, 7, 4, 10, 8, 13, 15, 12, 9, 0, 3, 5, 6, 11 } };
for (i = 0; i < 8; i++)
{
hang = r_b[i * 6] * 2 + r_b[i * 6 + 5]; //第0位与第5位决定行下标
lie = r_b[i * 6 + 1] * 8 + r_b[i * 6 + 2] * 4 + r_b[i * 6 + 3] * 2 + r_b[i * 6 + 4]; //第1、2、3、4位决定列下标
switch (i) //第i个数查询Si盒
{
case 0:jie = s1[hang][lie]; break;
case 1:jie = s2[hang][lie]; break;
case 2:jie = s3[hang][lie]; break;
case 3:jie = s4[hang][lie]; break;
case 4:jie = s5[hang][lie]; break;
case 5:jie = s6[hang][lie]; break;
case 6:jie = s7[hang][lie]; break;
case 7:jie = s8[hang][lie]; break;
}
shizhuaner(r_t, jie, i,32); //S盒为十六进制，须转换回二进制进行存放
}
}
/*左移函数*/
void key_move(int k2[56], int n)
{
int i, one, two, twentyeight, twentynight; //计数的i，没什么大的含义；one、two、twentyeight、twentynight用于存放第1、2、28、29个数
one = k2[0]; //先记录第1、2、28、29个数
two = k2[1];
twentyeight = k2[28];
twentynight = k2[29];
for (i = 0; i < (28 - n); i++) //循环赋值
{
k2[i] = k2[i + n];
k2[i + 28] = k2[i + n + 28];
}
if (n == 1) //左移1位时，特殊端点值（第27、55）赋值
{
k2[27] = one;
k2[55] = twentyeight;
}
else //左移2位时，特殊端点值（第27、28、54、55）赋值
{
k2[26] = one;
k2[27] = two;
k2[54] = twentyeight;
k2[55] = twentynight;
}
}
/* 获取第r轮左移位数 */
int move_num(int r)
{
int i; //i存储移动的位数
switch (r)
{
case 1:i = 1; break;
case 2:i = 1; break;
case 3:i = 2; break;
case 4:i = 2; break;
case 5:i = 2; break;
case 6:i = 2; break;
case 7:i = 2; break;
case 8:i = 2; break;
case 9:i = 1; break;
case 10:i = 2; break;
case 11:i = 2; break;
case 12:i = 2; break;
case 13:i = 2; break;
case 14:i = 2; break;
case 15:i = 2; break;
case 16:i = 1; break;
}
return i; //返回移动的位数
}
/* 置换函数 */
void exchange(int *k1, int *k2, int *rule,int n)
{
int i, j; //计数的i，j
for (i = 0; i < n; i++)
{
j = rule[i]; //根据rule数组进行置换，也就是赋值
k2[i] = k1[j-1];
}
return;
}
/*主函数*/
int main()
{
/* c是最终密文，c_r是密文的R部分，c_l是密文的L的部分，g是用于缓冲掉回车符，k接收从键盘输入密钥，m接收从键盘输入的明文*/
char c[16],c_r[8],c_l[8], g, k[8], m[8];
/* 计数的i，j；r是轮数；n接收类型转换后的结果；k1，k2，k3是密钥处理中不同大小的数组；m1，m2是明文处理中的数组，r_s是R 32大小时的数组，r_b是R 48大小时的数组，l是L的数组*/
int i, j, r, n, k1[64], k2[56], k3[48], m1[64], m2[64],r_s[32],r_b[48],l[32];
/* k_c_one是密钥处理的置换1，k_c_two是密钥处理的置换2，m_c_one是明文处理的置换1，e是明文处理的e盒，m_c_two是明文处理的置换2 */
int k_c_one[56] = { 57, 49, 41, 33, 25, 17, 9,
1, 58, 50, 42, 34, 26, 18,
10, 2, 59, 51, 43, 35, 27,
19, 11, 3, 60, 52, 44, 36,
63, 55, 47, 39, 31, 23, 15,
7, 62, 54, 46, 38, 30, 22,
14, 6, 61, 53, 45, 37, 29,
21, 13, 5, 28, 20, 12, 4 },
k_c_two[48] = { 14, 17, 11, 24, 1, 5,
3, 28, 15, 6, 21, 10,
23, 19, 12, 4, 26, 8,
16, 7, 27, 20, 13, 2,
41, 52, 31, 37, 47, 55,
30, 40, 51, 45, 33, 48,
44, 49, 39, 56, 34, 53,
46, 42, 50, 36, 29, 32 },
m_c_one[64] = { 58, 50, 42, 34, 26, 18, 10, 2,
60, 52, 44, 36, 28, 20, 12, 4,
62, 54, 46, 38, 30, 22, 14, 6,
64, 56, 48, 40, 32, 24, 16, 8,
57, 49, 41, 33, 25, 17, 9, 1,
59, 51, 43, 35, 27, 19, 11, 3,
61, 53, 45, 37, 29, 21, 13, 5,
63, 55, 47, 39, 31, 23, 15, 7 },
e[48] = { 32, 1, 2, 3, 4, 5,
4, 5, 6, 7, 8, 9,
8, 9, 10, 11, 12, 13,
12, 13, 14, 15, 16, 17,
16, 17, 18, 19, 20, 21,
20, 21, 22, 23, 24, 25,
24, 25, 26, 27, 28, 29,
28, 29, 30, 31, 32, 1 },
m_c_two[32] = { 16, 7, 20, 21,
29, 12, 28, 17,
1, 15, 23, 26,
5, 18, 31, 10,
2, 8, 24, 14,
32, 27, 3, 9,
19, 13, 30, 6,
22, 11, 4, 25 },
m_c_three[64] = { 40, 8, 48, 16, 56, 24, 64, 32,
39, 7, 47, 15, 55, 23, 63, 31,
38, 6, 46, 14, 54, 22, 62, 30,
37, 5, 45, 13, 53, 21, 61, 29,
36, 4, 44, 12, 52, 20, 60, 28,
35, 3, 43, 11, 51, 19, 59, 27,
34, 2, 42, 10, 50, 18, 58, 26,
33, 1, 41, 9, 49, 17, 57, 25 };
printf("**DES加密**\n");
printf("\n");
printf("请输入加密的密钥:\n");
for (i = 0; i < 8; i++) //从键盘里获取输入的密钥
{
scanf_s("%c", &k[i]);
n = (int)k[i];
shizhuaner(k1, n, i,64); //进行密钥进制转换
}
g = getchar(); //缓冲掉多于8个字符以外字符和回车符
while (g != '\n')
{
g = getchar();
}
printf("请输入需要加密的明文:\n");
for (i = 0; i < 8; i++) //从键盘里获取输入的明文
{
scanf_s("%c", &m[i]);
n = (int)m[i];
shizhuaner(m1, n, i,64); //进行明文进制转换
}
g = getchar(); //缓冲掉多于8个字符以外字符和回车符
while (g != '\n')
{
g = getchar();
}
printf("\n");
exchange(k1, k2, k_c_one, 56); //密钥进行置换1处理
exchange(m1, m2, m_c_one, 64); //明文进行密钥1处理
for (i = 0; i < 32; i++)
l[i] = m2[i]; //明文的L部分
j = 0;
for (i = 32; i < 64; i++)
{
r_s[j] = m2[i]; //明文的R部分
j++;
}
for (r = 1; r <= 16; r++) //进行16轮的循环处理
round(r,k2, k3, r_s, r_b, l, k_c_two, m_c_two, e, c_r, c_l);
for (i = 0; i < 64; i++) //将密文R的部分与密文L的部分合一起
{
m1[i] = r_s[i];
if (i >= 32)
m1[i] = l[i - 32];
}
exchange(m1, m2, m_c_three, 64); //进行明文处理的置换3
erzhuanshiliu(m2, c, 64); //二进制转换成十六进制，得到最终密文
printf("这是最终密文:\n");
for (i = 0; i < 16; i++)
printf("%c ", c[i]);
printf("\n");
return 0;
} 以下是测试代码和运行情况的截图：
这是我第一次写博客（好紧张好刺激啊~~），请各位大大多多指教撒~~