高手进，请教DES算法，帮忙看看，急啊！分不够在加！！解决后在送100
int DES(unsigned char *bufferin,unsigned char *bufferout, unsigned char *key,long mode)
{
//密钥变换为56字节(去掉校验位)
static unsigned char pc1[56] = {
56, 48, 40, 32, 24, 16, 8,
0, 57, 49, 41, 33, 25, 17,
9, 1, 58, 50, 42, 34, 26,
18, 10, 2, 59, 51, 43, 35,
62, 54, 46, 38, 30, 22, 14,
6, 61, 53, 45, 37, 29, 21,
13, 5, 60, 52, 44, 36, 28,
20, 12, 4, 27, 19, 11, 3 };
//56字节变换为48 字节(数据压缩)
static unsigned char pc2[48] = {
13, 16, 10, 23, 0, 4,
2, 27, 14, 5, 20, 9,
22, 18, 11, 3, 25, 7,
15, 6, 26, 19, 12, 1,
40, 51, 30, 36, 46, 54,
29, 39, 50, 44, 32, 47,
43, 48, 38, 55, 33, 52,
45, 41, 49, 35, 28, 31 };
//32字节变换为48字节(数据扩展)
static unsigned char exp[48] = {
31, 0, 1, 2, 3, 4,
3, 4, 5, 6, 7, 8,
7, 8, 9, 10, 11, 12,
11, 12, 13, 14, 15, 16,
15, 16, 17, 18, 19, 20,
19, 20, 21, 22, 23, 24,
23, 24, 25, 26, 27, 28,
27, 28, 29, 30, 31, 0 };
//64位数据IP(Initial Permutation)变换表
static unsigned char ip[64] = {
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,
56, 48, 40, 32, 24, 16, 8, 0,
58, 50, 42, 34, 26, 18, 10, 2,
60, 52, 44, 36, 28, 20, 12, 4,
62, 54, 46, 38, 30, 22, 14, 6 };
//数据逆置换(Final Permutation)
static unsigned char ip_1[64] = {
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,
32, 0, 40, 8, 48, 16, 56, 24 };
//Permutation P
static unsigned char pp[32] = {
15, 6, 19, 20,
28, 11, 27, 16,
0, 14, 22, 25,
4, 17, 30, 9,
1, 7, 23, 13,
31, 26, 2, 8,
18, 12, 29, 5,
21, 10, 3, 24 };
/* INITIALIZE THE TABLES */
/* Table - s1 */
static unsigned char 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 };
/* Table - s2 */
static unsigned char 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 };
/* Table - s3 */
static unsigned char 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 };
/* Table - s4 */
static unsigned char 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, 4,
3, 15, 0, 6, 10, 1, 13, 8, 9, 4, 5, 11, 12, 7, 2, 14 };
/* Table - s5 */
static unsigned char 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 };
/* Table - s6 */
static unsigned char 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 };
/* Table - s7 */
static unsigned char 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 };
/* Table - s8 */
static unsigned char 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 };
/* 密钥生成中的循环左移位的累计次数*/
static unsigned char totrot[] = {1,2,4,6,8,10,12, 14, 15, 17, 19, 21, 23, 25, 27, 28 };
/*----------------------------------------------*/
//long mode = 1; //模式，1：加密，2：解密
//unsigned char bufferin[9], bufferout[9]; //明文，密文
/*----------------------------------------------*/
long i, j, k;
long rotshift; //密钥移位次数
//long keylen, buflen; //密钥长度,明文长度
unsigned char keybuf[65]; //密钥，密钥64字节缓冲区
unsigned char keyreal[57], keys[17][49]; //实际使用56字节密钥,48字节密钥数组
unsigned char srcbuf[65], dstbuf[65]; //明文，密文64字节缓冲区
unsigned char L[17][33], R[17][33], LR[65], RL[65]; //加密时临时数据左右两部分
unsigned char E[17][49]; //R数组的扩展数据
unsigned char B[9][7], BB[33], P[33]; //E和K异或后的缓冲数组
unsigned char C[17][29], D[17][29], CD[57]; //56字节密钥的左右两部分
unsigned char temp1, temp2, m, n, x;
//1.变换密钥
//密钥不足8字节则用0补足(或自定义)
//keylen = strlen((const char*)key);
//if(keylen<8)
// memset(key+keylen, 0, (8-keylen));
//将8字节密钥转换为64字节字串
for(i=0;i<8;i++)
{
j = *(key+i);
keybuf[8*i] = (j / 128) % 2;
keybuf[8*i+1] = (j / 64) % 2;
keybuf[8*i+2] = (j / 32) % 2;
keybuf[8*i+3] = (j / 16) % 2;
keybuf[8*i+4] = (j / 8) % 2;
keybuf[8*i+5] = (j / 4) % 2;
keybuf[8*i+6] = (j / 2) % 2;
keybuf[8*i+7] = (j / 1) % 2;
}
//根据pc1进行变换成56字节，去掉奇偶校验位
for(i=0;i<56;i++)
{
keyreal[i] = keybuf[pc1[i]];
}
//将56字节密钥分为左右两部分C[0],D[0]
for(i=0;i<28;i++)
{
C[0][i] = keyreal[i];
D[0][i] = keyreal[i+28];
}
//循环16次(i从1开始)
for(i=1;i<17;i++)
{
//根据加密或解密确定密钥顺序
if(mode) //加密
rotshift = totrot[i-1];
else //解密
rotshift = totrot[16-i];
//1)左移固定位数得到C[i]和D[i];
for(j=0;j<28;j++)
{
C[i][j] = C[0][j];
D[i][j] = D[0][j];
}
for(j=0;j<rotshift;j++)
{
temp1 = C[i][0];
temp2 = D[i][0];
for(k=0;k<27;k++)
{
C[i][k] = C[i][k+1];
D[i][k] = D[i][k+1];
}
C[i][27] = temp1;
D[i][27] = temp2;
}
//2)将C[i]D[i]用pc2化简为48位k[i];
for(j=0;j<28;j++)
{
CD[j] = C[i][j];
CD[j+28] = D[i][j];
}
for(j=0;j<48;j++)
{
keys[i][j] = CD[pc2[j]];
}
}
问题点数：88、回复次数：8Top
1 楼xingfuniao（幸福鸟）回复于 2006-06-01 00:40:12 得分 0
//2.数据处理
//若明文不足8字节则补0(或自定义)
//buflen = strlen((const char*)bufferin);
//if(buflen<8)
// memset(bufferin+buflen, 0, (8-buflen));
//将8字节数据转换为64字节字串
for(i=0;i<8;i++)
{
j = *(bufferin+i);
srcbuf[i*8] = (j / 128) % 2;
srcbuf[i*8+1] = (j / 64) % 2;
srcbuf[i*8+2] = (j / 32) % 2;
srcbuf[i*8+3] = (j / 16) % 2;
srcbuf[i*8+4] = (j / 8) % 2;
srcbuf[i*8+5] = (j / 4) % 2;
srcbuf[i*8+6] = (j / 2) % 2;
srcbuf[i*8+7] = (j / 1) % 2;
}
//将srcbuf按ip进行变换
for(i=0;i<64;i++)
LR[i] = srcbuf[ip[i]];
//将64字节数据转换为两部分L[0],R[0]
for(i=0;i<32;i++)
{
L[0][i] = LR[i];
R[0][i] = LR[i+32];
}
//循环16次(i从1开始)，用密钥加密数据
for(i=1;i<17;i++)
{
//1)将32位的R[i-1]按exp扩展为48位的E[i-1];
for(j=0;j<48;j++)
{
E[i-1][j] = R[i-1][exp[j]];
}
//2)异或E[i-1]和K[i];
for(j=0;j<48;j++)
{
keys[i][j] = keys[i][j] ^ E[i-1][j];
}
//3)将异或结果分为8个6位长的部分B[8]
for(j=0;j<8;j++)
{
B[j][0] = keys[i][j*6];
B[j][1] = keys[i][j*6+1];
B[j][2] = keys[i][j*6+2];
B[j][3] = keys[i][j*6+3];
B[j][4] = keys[i][j*6+4];
B[j][5] = keys[i][j*6+5];
}
//4)循环用S表替换(j从1开始)
for(j=0;j<8;j++)
{
//a)B[j]第1位和第6位组合为M，作为S[j]的行号
m = 2 * B[j][0] + B[j][5];
//b)B[j]第2到5位组合为N，作为S[j]的列号
n = 2 * (2 * (2 * B[j][1] + B[j][2]) + B[j][3]) + B[j][4];
//c)用S[j][M][N]来取代B[j]
switch(j)
{
case 0:
x = s1[m][n];
break;
case 1:
x = s2[m][n];
break;
case 2:
x = s3[m][n];
break;
case 3:
x = s4[m][n];
break;
case 4:
x = s5[m][n];
break;
case 5:
x = s6[m][n];
break;
case 6:
x = s7[m][n];
break;
case 7:
x = s8[m][n];
break;
}
BB[j*4] = (x / 8) % 2;
BB[j*4 + 1] = (x / 4) % 2;
BB[j*4 + 2] = (x / 2) % 2;
BB[j*4 + 3] = (x / 1) % 2;
}
//5)将B[1]到B[8]按P组合得到p
for(j=0;j<32;j++)
{
P[j] = BB[pp[j]];
}
//6)R[i] = p xor L[i-1];L[i] = R[i-1];
for(j=0;j<32;j++)
{
R[i][j] = P[j] ^ L[i-1][j];
L[i][j] = R[i-1][j];
}
}
//3.组合变换后的R[16]L[16]按ip_1变换得到最后结果
for(i=0;i<32;i++)
{
RL[i] = R[16][i];
RL[i+32] = L[16][i];
}
for(i=0;i<64;i++)
{
dstbuf[i] = RL[ip_1[i]];
}
//将64字节数据转换为8字节
for (i = 0; i < 8; i++)
{
*(bufferout + i) = 0x00;
for (k = 0; k < 7; k++)
*(bufferout + i) = ((*(bufferout + i)) + dstbuf[8*i+k]) * 2;
*(bufferout + i) = *(bufferout + i) + dstbuf[8*i+7];
}
return 0;
}
这个函数我怎样去使用呢，我没看懂，对DES算法不熟悉，大狭告诉我怎样使用这个函数来进行加密？
例如我现在有个密钥'57415443484441544154696d65434f53’,我想通过这个密钥对一个16进制的随机数加密如‘47972DC678443425’，想得到一个数，该怎样使用这个函数，
或者大狭要是有关于DES的完整的应用，也提供给小弟学习一下。要纯C开发的。
分不够在加！！！！！Top
2 楼happytang（一只叫苏格拉底的猪）回复于 2006-06-01 01:29:58 得分 30
网上很多DES的资料的
DES算法的入口参数有三个：Key、Data、Mode。其中Key为8个字节共64位，是DES算法的工作密钥；Data也为8个字节64位，是要被加密或被解密的数据；Mode为DES的工作方式，有两种：加密或解密。
DES算法是这样工作的：如Mode为加密，则用Key 去把数据Data进行加密， 生成Data的密码形式（64位）作为DES的输出结果；如Mode为解密，则用Key去把密码形式的数据Data解密，还原为Data的明码形式（64位）作为DES的输出结果。在通信网络的两端，双方约定一致的Key，在通信的源点用Key对核心数据进行DES加密，然后以密码形式在公共通信网（如电话网）中传输到通信网络的终点，数据到达目的地后，用同样的Key对密码数据进行解密，便再现了明码形式的核心数据。这样，便保证了核心数据（如PIN、MAC等）在公共通信网中传输的安全性和可靠性。
D.E.S 是分块加密的，将明文分割成 64 BITS 的块, 然后它们一个个接起来 。它使用56位密钥对64位的数据块进行加密，并对64bits的数据块进行16轮编码。与每轮编码时，一个48bits的“每轮”密钥值由56bits的完整密钥得出来。
http://topic.csdn.net/t/20050510/20/3996245.htmlTop
3 楼happytang（一只叫苏格拉底的猪）回复于 2006-06-01 01:39:35 得分 0
//long mode = 1; //模式，1：加密，2：解密
//unsigned char bufferin[9], bufferout[9]; //明文，密文
/*----------------------------------------------*/
仔细看看程序，输入输出都写得很清楚
不难得。还不是3des，IdesTop
4 楼jixingzhong（瞌睡虫·星辰）回复于 2006-06-01 08:50:53 得分 0
int DES(unsigned char *bufferin,unsigned char *bufferout, unsigned char *key,long mode)
例如我现在有个密钥'57415443484441544154696d65434f53’,我想通过这个密钥对一个16进制的随机数加密如‘47972DC678443425’，想得到一个数，该怎样使用这个函数，
＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝＝
密钥'57415443484441544154696d65434f53’保存在一个 unsigned char *中，
传递给这个函数的 unsigned char *key
要加密的数据如‘47972DC678443425’，传递给unsigned char *bufferin
mode 根据实际选择，传入合适的参数
（mode＝1或其他非0数据 为加密 或者 mode ＝0 为解密）
unsigned char *bufferout 为加密后数据的保存区域Top
5 楼jixingzhong（瞌睡虫·星辰）回复于 2006-06-01 08:56:32 得分 58
main()
{
unsigned char in[100] = "47972DC678443425";
unsigned char key[] = "57415443484441544154696d65434f53";
unsigned char out[100];
DES(in, out, key, 1); //加密操作，数据保存于 out 中
puts(out);
DES(out, in, key, 0); //对 out 中的数据进行还原解密操作
puts(in);
}Top
6 楼xingfuniao（幸福鸟）回复于 2006-06-01 10:34:37 得分 0
不过我想用这个密钥和数据加密后得到如下的结果，用DES得不出来，所以应该是3DES加密后得出来的
密钥：57415443484441544154696d65434f53
数据：47972DC678443425
加密后结果：19ECE1DC4C0BD6A6
上述数据为16进制数。
请大家提供3DES算法给小弟学习一下。要纯C开发的。
分不够在加！！！！！
已经新开贴，
http://community.csdn.net/Expert/topic/4793/4793089.xml?temp=.8899805
Top
7 楼jixingzhong（瞌睡虫·星辰）回复于 2006-06-01 10:56:00 得分 0
那楼主结了这个帖子吧 ...
否则两个帖子都回复很麻烦 ...Top
8 楼jixingzhong（瞌睡虫·星辰）回复于 2006-06-01 10:59:50 得分 0
先搞清除 3DES 流程就可以了 ...
其实还是一个 DES
只不过 反复了一下而已 ...
需要两个密钥，
似乎加密时候是 A加密，B解密，A解密 ...
解密的时候反一下就可以了 ～Top
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