DES是以Festil为基础的。我理解不深。不过这个算法最核心的估计就是f函数那里和S盒设计了,所以我也不会说。只是说des简单的实现。本人c语言基础薄弱,里面有一些语法是很臃肿的,我也不再一一修改,只是作一参考,有兴趣的可以自己来修改一下。并且这个算法我也没有实现,因为卡在了S盒数组越界问题(其实可以把越界素组再定义空间大点,但是我没这么做,因为总觉得不需要用这么大空间还分配这么多,不符合我的理念,是的,我好像过于吝啬了),问题的出现地方我也有标明,所以有兴趣的可以自己改一下,改好了也可以告诉我,或者哪位大佛觉得一眼就知道问题的也可以教我一下,我是知道越界了,但是不修改长度的前提下把bug改好,我没找到方法,不知道为什么越界了,其实内容长度也没到越界的长度。 不过我有开信息安全这门课,所以简单的过程,还是有实现的,比如进制的转换(需要用的可以改进一下,我c语言基础也不好),置换函数等,基于我之前的FEStil算法,省去了我一段时间写分组那些重复代码。好了,步骤基础都包含了~~,只剩下f函数中,s盒置换那里有越界问题,~~ 有兴趣的可以看看怎么改好,然后加解密就容易了。
add:数组越界问题已经解除,s盒下标有错误,已修改。不过还有问题,可以自行修改。
#include
#include
#include
#include
//IP表
static const int IP[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
};
//IP-1表
static const int IPNG[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
};
//E表
static const int 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
};
//P表
static const int P[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
};
//PC1表
static const int PC1[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
};
//pc2表
static const int PC2[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
};
//左移位数表
static const char CS[16] = {
1,1,2,2,2,2,2,2,1,2,2,2,2,2,2,1
};
//s盒
static const int S_Box[8][4][16] = {
// S1
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
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
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
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,
10,6,9,0,12,11,7,13,15,1,3,14,5,2,8,4,
3,15,0,6,10,1,13,8,9,4,5,11,12,7,2,14,
//S5
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
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
4,11,2,14,15,0,8,13,3,12,9,7,5,10,6,1,
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
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
};
int main()
{
int i=0,j=0;
char key[10]; //密钥
char plaintext[100]; //明文
char pGroup[10][10]; //明文字符串分组
char pGroupTest[10][10];
char cipher[10][10]; //密文字符串分组
char keyGroup[16][10]; //密钥字符串分组
int binpGroup[10][64]; //二级制明文分组,最多10个分组
int bincGroup[10][64]; //二进制密文分组,最多10组
int binkGroup[16][48]; // //二级制密钥分组,最多10个分组
void generateKey(char key[10],int binkGroup[16][48]);
void makeGroup(char plaintext[100],char pGroup[10][10]);
void PByte_Bin(char pGroup[10][10],int binpGroup[10][64],int flag);
void KByte_Bin(char key[10],int keyBin[64],int flag);
void ByteToBin(int dest[64],char c[10],int flag);
void RepalceMent(int * dest,int * src,int size,int flag);
void encryption(int binpGroup[10][64],int binkGroup[16][48],char pGroup[10][10]); //加密函数
void decryption(int bincGroup[10][64],int binkGroup[16][48],char cipher[10][10]);
//初始化
memset(plaintext,0,sizeof(plaintext)/sizeof(char));
memset(key,0,sizeof(key)/sizeof(char));
memset(pGroup,0,sizeof(pGroup) / sizeof(char) );
memset(pGroupTest,0,sizeof(pGroupTest) / sizeof(char) );
memset(cipher,0,sizeof(cipher) / sizeof(char) );
memset(keyGroup,0,sizeof(keyGroup) / sizeof(char) );
memset(binpGroup,0,sizeof(binpGroup) / sizeof(int) );
memset(bincGroup,0,sizeof(bincGroup) / sizeof(int) );
printf("请输入明文\n");
scanf("%s",plaintext);
printf("请输入密码\n");
scanf(" %s",key);
generateKey(key,binkGroup); //获取密钥二进制分组(binkGroup)
makeGroup(plaintext,pGroup); //明文
PByte_Bin(pGroup,binpGroup,0); //明文(密文)转二级制
encryption(binpGroup,binkGroup,cipher); //加密函数,输入明文二进制分组,密钥二进制分组,以及需要求得密文数组地址
//PByte_Bin(cipher,bincGroup,0); //密文字符转二进制
// decryption(binpGroup,binkGroup,pGroupTest);
return 0;
}
void RepalceMent(int * dest,int * src,int size,int flag) //置换函数,,置换的数组dest和src不能相同,不然会出现数值覆盖错误
{
int i=0;
const int * rpArray;
switch(flag)
{
case 1:rpArray = PC1;break; //1为pc1
case 2:rpArray = PC2;break; //2为pc2
case 3:rpArray = IP;break; //ip置换
case 4:rpArray = IPNG;break; //ip逆运算
case 5:rpArray = E;break; //E置换
case 6:rpArray = P;break; //p置换
}
while(size) //
{
* dest = * (src - i + (* rpArray) - 1); //按rpArray的值进行置换
src++;
i++;
size--;
rpArray++;
dest++;
}
}
//字符串与二进制的转换,flag标志0,1, 这个步骤可以改成用指针,而不是数组指针?这样就可把密码和明文的进制转换两个同样性质的函数合并了。
void ByteToBin(int dest[64],char c[10],int flag)
{
int i=0,j=0;
if(flag == 0) //flag为0表示字符转换二进制(包括字符转十进制,十进制转二进制)
{
for(i=0;i<8;i++)
{
int a = c[i] - '\0';
for(j=7;j>-1;j--)
{
while(a)
{
dest[i*8+j] = a & 1; //取得二进制数
j--;
a >>= 1; //右移一位
}
dest[i*8+j] = 0;
}
}
}
else if(flag == 1) //flag为1表示二进制转换字符
{
for(i=0;i<8;i++)
{
int cNum = 0,z=0;
for(j=0;j<8;j++)
{
z = 7-j; //注意二进制数的存储方式
int num = dest[i*8+j]; //
if(num == 1)
{
while(z)
{
num *= 2;
z--;
}
}
else
{
num = 0;
}
cNum += num;
}
c[i] = cNum + '\0';
}
c[8] = '\0';
}
}
int OctToBin(int oct,int * bin,int size,int flag) //二级制与十进制转换
{
int i=0,j=0;
int cNum = 0;
if(flag == 0) //十进制转二进制
{
int a = oct;
for(i=size-1;i>-1;i--)
{
while(a)
{
*(bin+i) = a & 1; //取得二进制数
i--;
a >>= 1; //右移一位
}
*(bin+i) = 0;
}
}
else if(flag == 1) //flag为1表示二进制转换十进制
{
int z = 0;
int num = 0;
int length = size-1;
for(i=0;i-1;j--)
{
f(IPplain[i],binkGroup[j],j,1); //flag为1表示解密
}
RepalceMent(bincGroup[i],IPplain[i],64,4); //IP-1置换
ByteToBin(bincGroup[i],cipher[i],1); //二级制转换字符串
}
}