1.MD5是什么？
一种单向的加密方法，也就是说只有从明文到密文这一步，没有密文到明文这一步。
2.为什么用MD5呢？
作为识别ID，我们实际上并不需要知道它具体是多少，只需要保证两点，唯一的明文对着唯一密文，有高度的安全性，就行了。
3.MD5算法基本过程？
任意长度字符串输入->转化成byte数组->求余补长->分组->每组都做四轮数据变换->固定128位信息
4.第一步
将字符串转化为byte数组：
byte[] inputBytes = Encoding.Default.GetBytes (_input);利用C#的Encoding里的方法即可。
5.第二步
求余补长：
1.求余，求余的目的是为了把每64个byte分成一组，也就是64*8=512个bit，512又将依据每32个的合成16个int，作为四轮数据变换的输入端。
2.当然，由于输入的字符串是任意长的，显然不能保证整除64之类的，所以我们必须对剩下的byte做补长，把他补成64位，补长规则是这样的：对于余数不足56位的，首先将他补至56位，补足规则是：填充1个1，N个0；对于56位以上的，补充规则都是一样的，总是将最后8位替换成byte数组的长度信息。
int byteLen = inputBytes.Length;
int []groups=null;
int groupCount=0;
groupCount=byteLen/64;
int rest=byteLen%64;
byte [] tempBytes=new byte[64];
if(rest<=56)
{
for(int i=0;i<rest;i++)
tempBytes[i]=inputBytes[byteLen-rest+i];
if(rest<56)
{
tempBytes[rest]=(byte)(1<<7);
for(int i=1;i<56-rest;i++)
tempBytes[rest+i]=0;
}
long len=(long)(byteLen<<3);
for(int i=0;i<8;i++)
{
tempBytes[56+i]=(byte)(len&0xFFL);
len=len>>8;
}
groups=divGroup(tempBytes,0);
trans(groups);
}
else
{
for(int i=0;i<rest;i++)
tempBytes[i]=inputBytes[byteLen-rest+i];
long len=(long)(byteLen<<3);
for(int i=0;i<8;i++)
{
tempBytes[56+i]=(byte)(len&0xFFL);
len=len>>8;
}
groups=divGroup(tempBytes,0);
trans (groups);
}
代码中的divgroup，trans，就是后面的分组，变换操作，我们稍后再说。
当然，对于byte数组的前N组64位信息组，也都要执行这个操作：
for(int step=0;step<groupCount;step++)
{
groups=divGroup(inputBytes,step*64);
trans(groups);
}
6.第三步
分组：对于64个byte重分成16个int：
private static int[] divGroup(byte[] inputBytes,int index)
{
int [] temp=new int[16];
for(int i=0;i<16;i++)
{
temp[i]=b2iu(inputBytes[4*i+index])|
(b2iu(inputBytes[4*i+1+index]))<<8|
(b2iu(inputBytes[4*i+2+index]))<<16|
(b2iu(inputBytes[4*i+3+index]))<<24;
}
return temp;
}
public static int b2iu(byte b)
{
return b < 0 ? b & 0x7F + 128 : b;
}其实就是按着顺序，每4个作为int即可。
7.第四步：
最为复杂的一步：对输入的16个int做各种变换，并把效应叠加到最后的结果上：
首先有4种基础操作F、G、H、I：
private static long F(long x, long y, long z) {
return (x & y) | ((~x) & z);
}
private static long G(long x, long y, long z) {
return (x & z) | (y & (~z));
}
private static long H(long x, long y, long z) {
return x ^ y ^ z;
}
private static long I(long x, long y, long z) {
return y ^ (x | (~z));
}接着有在这基础上的4种复杂一点的运算FF、GG、HH、II：
rivate static long FF(long a, long b, long c, long d, int x, long s,
long ac) {
a += (F(b, c, d)&0xFFFFFFFFL) + x + ac;
a = ((a&0xFFFFFFFFL)<< (int)s) | ((int)(a&0xFFFFFFFFL) >> (int)(32 - s));
a += b;
return (a&0xFFFFFFFFL);
}
private static long GG(long a, long b, long c, long d, int x, long s,
long ac) {
a += (G(b, c, d)&0xFFFFFFFFL) + x + ac;
a = ((a&0xFFFFFFFFL) <<(int) s) | ((int)(a&0xFFFFFFFFL) >> (int)(32 - s));
a += b;
return (a&0xFFFFFFFFL);
}
private static long HH(long a, long b, long c, long d, int x, long s,
long ac) {
a += (H(b, c, d)&0xFFFFFFFFL) + x + ac;
a = ((a&0xFFFFFFFFL) << (int)s) | ((int)(a&0xFFFFFFFFL) >> (int)(32 - s));
a += b;
return (a&0xFFFFFFFFL);
}
private static long II(long a, long b, long c, long d, int x, long s,
long ac) {
a += (I(b, c, d)&0xFFFFFFFFL) + x + ac;
a = ((a&0xFFFFFFFFL) <<(int) s) | ((int)(a&0xFFFFFFFFL) >> (int)(32 - s));
a += b;
return (a&0xFFFFFFFFL);
}然而我们要对着16个int做4组16轮的变换，每一组中的一轮，对应着16个int的一个数，4组分别用FF、GG、HH、II操作的一种。
先看看操作中用到的数据：
private static long A=0x67452301L;
private static long B=0xefcdab89L;
private static long C=0x98badcfeL;
private static long D=0x10325476L;
static int S11 = 7;
static int S12 = 12;
static int S13 = 17;
static int S14 = 22;
static int S21 = 5;
static int S22 = 9;
static int S23 = 14;
static int S24 = 20;
static int S31 = 4;
static int S32 = 11;
static int S33 = 16;
static int S34 = 23;
static int S41 = 6;
static int S42 = 10;
static int S43 = 15;
static int S44 = 21;
private static long [] result={A,B,C,D};
trans函数最终形态：
private static void trans(int[] groups)
{
long a = result[0], b = result[1], c = result[2], d = result[3];
a = FF(a, b, c, d, groups[0], S11, 0xd76aa478L); /* 1 */
d = FF(d, a, b, c, groups[1], S12, 0xe8c7b756L); /* 2 */
c = FF(c, d, a, b, groups[2], S13, 0x242070dbL); /* 3 */
b = FF(b, c, d, a, groups[3], S14, 0xc1bdceeeL); /* 4 */
a = FF(a, b, c, d, groups[4], S11, 0xf57c0fafL); /* 5 */
d = FF(d, a, b, c, groups[5], S12, 0x4787c62aL); /* 6 */
c = FF(c, d, a, b, groups[6], S13, 0xa8304613L); /* 7 */
b = FF(b, c, d, a, groups[7], S14, 0xfd469501L); /* 8 */
a = FF(a, b, c, d, groups[8], S11, 0x698098d8L); /* 9 */
d = FF(d, a, b, c, groups[9], S12, 0x8b44f7afL); /* 10 */
c = FF(c, d, a, b, groups[10], S13, 0xffff5bb1L); /* 11 */
b = FF(b, c, d, a, groups[11], S14, 0x895cd7beL); /* 12 */
a = FF(a, b, c, d, groups[12], S11, 0x6b901122L); /* 13 */
d = FF(d, a, b, c, groups[13], S12, 0xfd987193L); /* 14 */
c = FF(c, d, a, b, groups[14], S13, 0xa679438eL); /* 15 */
b = FF(b, c, d, a, groups[15], S14, 0x49b40821L); /* 16 */
a = GG(a, b, c, d, groups[1], S21, 0xf61e2562L); /* 17 */
d = GG(d, a, b, c, groups[6], S22, 0xc040b340L); /* 18 */
c = GG(c, d, a, b, groups[11], S23, 0x265e5a51L); /* 19 */
b = GG(b, c, d, a, groups[0], S24, 0xe9b6c7aaL); /* 20 */
a = GG(a, b, c, d, groups[5], S21, 0xd62f105dL); /* 21 */
d = GG(d, a, b, c, groups[10], S22, 0x2441453L); /* 22 */
c = GG(c, d, a, b, groups[15], S23, 0xd8a1e681L); /* 23 */
b = GG(b, c, d, a, groups[4], S24, 0xe7d3fbc8L); /* 24 */
a = GG(a, b, c, d, groups[9], S21, 0x21e1cde6L); /* 25 */
d = GG(d, a, b, c, groups[14], S22, 0xc33707d6L); /* 26 */
c = GG(c, d, a, b, groups[3], S23, 0xf4d50d87L); /* 27 */
b = GG(b, c, d, a, groups[8], S24, 0x455a14edL); /* 28 */
a = GG(a, b, c, d, groups[13], S21, 0xa9e3e905L); /* 29 */
d = GG(d, a, b, c, groups[2], S22, 0xfcefa3f8L); /* 30 */
c = GG(c, d, a, b, groups[7], S23, 0x676f02d9L); /* 31 */
b = GG(b, c, d, a, groups[12], S24, 0x8d2a4c8aL); /* 32 */
a = HH(a, b, c, d, groups[5], S31, 0xfffa3942L); /* 33 */
d = HH(d, a, b, c, groups[8], S32, 0x8771f681L); /* 34 */
c = HH(c, d, a, b, groups[11], S33, 0x6d9d6122L); /* 35 */
b = HH(b, c, d, a, groups[14], S34, 0xfde5380cL); /* 36 */
a = HH(a, b, c, d, groups[1], S31, 0xa4beea44L); /* 37 */
d = HH(d, a, b, c, groups[4], S32, 0x4bdecfa9L); /* 38 */
c = HH(c, d, a, b, groups[7], S33, 0xf6bb4b60L); /* 39 */
b = HH(b, c, d, a, groups[10], S34, 0xbebfbc70L); /* 40 */
a = HH(a, b, c, d, groups[13], S31, 0x289b7ec6L); /* 41 */
d = HH(d, a, b, c, groups[0], S32, 0xeaa127faL); /* 42 */
c = HH(c, d, a, b, groups[3], S33, 0xd4ef3085L); /* 43 */
b = HH(b, c, d, a, groups[6], S34, 0x4881d05L); /* 44 */
a = HH(a, b, c, d, groups[9], S31, 0xd9d4d039L); /* 45 */
d = HH(d, a, b, c, groups[12], S32, 0xe6db99e5L); /* 46 */
c = HH(c, d, a, b, groups[15], S33, 0x1fa27cf8L); /* 47 */
b = HH(b, c, d, a, groups[2], S34, 0xc4ac5665L); /* 48 */
a = II(a, b, c, d, groups[0], S41, 0xf4292244L); /* 49 */
d = II(d, a, b, c, groups[7], S42, 0x432aff97L); /* 50 */
c = II(c, d, a, b, groups[14], S43, 0xab9423a7L); /* 51 */
b = II(b, c, d, a, groups[5], S44, 0xfc93a039L); /* 52 */
a = II(a, b, c, d, groups[12], S41, 0x655b59c3L); /* 53 */
d = II(d, a, b, c, groups[3], S42, 0x8f0ccc92L); /* 54 */
c = II(c, d, a, b, groups[10], S43, 0xffeff47dL); /* 55 */
b = II(b, c, d, a, groups[1], S44, 0x85845dd1L); /* 56 */
a = II(a, b, c, d, groups[8], S41, 0x6fa87e4fL); /* 57 */
d = II(d, a, b, c, groups[15], S42, 0xfe2ce6e0L); /* 58 */
c = II(c, d, a, b, groups[6], S43, 0xa3014314L); /* 59 */
b = II(b, c, d, a, groups[13], S44, 0x4e0811a1L); /* 60 */
a = II(a, b, c, d, groups[4], S41, 0xf7537e82L); /* 61 */
d = II(d, a, b, c, groups[11], S42, 0xbd3af235L); /* 62 */
c = II(c, d, a, b, groups[2], S43, 0x2ad7d2bbL); /* 63 */
b = II(b, c, d, a, groups[9], S44, 0xeb86d391L); /* 64 */
result[0] += a;
result[1] += b;
result[2] += c;
result[3] += d;
result[0]=result[0]&0xFFFFFFFFL;
result[1]=result[1]&0xFFFFFFFFL;
result[2]=result[2]&0xFFFFFFFFL;
result[3]=result[3]&0xFFFFFFFFL;
}
8.最后一步
将result中的64X4按16进制输出即可：
static string[] hexs={"0","1","2","3","4","5","6","7","8","9","A","B","C","D","E","F"};
string resStr="";
long temp=0;
for(int i=0;i<4;i++)
{
for(int j=0;j<4;j++)
{
temp=result[i]&0x0FL;
string a=hexs[(int)(temp)];
result[i]=result[i]>>4;
temp=result[i]&0x0FL;
resStr+=hexs[(int)(temp)]+a;
result[i]=result[i]>>4;
}
}