WhyCan Forum(哇酷开发者社区)

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#include <string.h>
#include <stdio.h>

#define MX (z>>5^y<<2) + (y>>3^z<<4)^(sum^y) + (k[p&3^e]^z)

//注意：delta的取值是随机的，但是为了避免不良的取值，采取的是
//黄金分割数（根号5－2）/2与2的32次方的乘积。为0x9e3779b9。  
//在解密中，sum＝delta×round，如：delta×32＝13C6EF3720
  
//v表示为运算的长整型数据的首地址
//k为长整型的密钥的首地址
//n表示以32bit为基本单位的要运算的组元个数，正表示加密，负表示解密

long xxtea(long* v, long n, long* k);

long xxtea(long* v, long n, long* k)
{
 unsigned long z=v[n-1], y=v[0], sum=0, e, DELTA=0x9e3779b9;
 long p, q;

 if(n > 1)
 {/* 加密过程 */
  q = 6 + 52/n;
  while(q-- > 0)
  {
   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;
  }
  return 0;
 }
 else
 if(n < -1)
 {/* 解密过程 */
  n = -n; q = 6 + 52/n; sum = q*DELTA;
  while(sum != 0)
  {
   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;
  }
  return 0;
 } 
 return 1;
}

void  xxtea_test(void)
{
 unsigned long buff[4];
 unsigned long test[4]={0x11223344,0x55667788,0x99AABBCC,0xDDEEFF00};
 unsigned long keys[4]={0x12345678,0x9ABCDEF0,0x12345678,0x9ABCDEF0};
 
 while(1)
 {
  printf("\n\rxxtea test: \n\r");
  memcpy(buff,test,16); printf("0x%08x,0x%08x,0x%08x,0x%08x\n\r",buff[0],buff[1],buff[2],buff[3]);
  xxtea( buff, 4,keys); printf("0x%08x,0x%08x,0x%08x,0x%08x\n\r",buff[0],buff[1],buff[2],buff[3]);
  xxtea( buff,-4,keys); printf("0x%08x,0x%08x,0x%08x,0x%08x\n\r",buff[0],buff[1],buff[2],buff[3]);
  
  while(1);
 }
}