原创 字节倒序的最快算法[原创]

2011-1-5 21:31 3327 7 8 分类: MCU/ 嵌入式
   最近碰到一个字节倒序的问题,程序需要有最快的执行速度,通过请教和集中测试,发现查表的方法是最快的,好在只有256情况,对于空间占用并不大.实现的方法中,方法四局限于51单片机,但其他方法是通用的.虽然这个问题并没有多大难度,但有时还是起到作用.程序源码经过测试是正确的,请大家放心引用.

//BY 51FLY 2008.8.22
#i nclude <reg52.h>
#define  uint8   unsigned char               
#define  int8    signed   char                 
#define  uint16  unsigned short              
#define  int16   signed   short


sbit A_0=ACC^0;//方便位操作
sbit A_1=ACC^1;
sbit A_2=ACC^2;
sbit A_3=ACC^3;
sbit A_4=ACC^4;
sbit A_5=ACC^5;
sbit A_6=ACC^6;
sbit A_7=ACC^7;

sbit B_0=B^0;//方便位操作
sbit B_1=B^1;
sbit B_2=B^2;
sbit B_3=B^3;
sbit B_4=B^4;
sbit B_5=B^5;
sbit B_6=B^6;
sbit B_7=B^7;

uint8 code zjdx[256]=
{
0x00,0x80,0x40,0xC0,0x20,0xA0,0x60,0xE0,0x10,0x90,
0x50,0xD0,0x30,0xB0,0x70,0xF0,0x08,0x88,0x48,0xC8,
0x28,0xA8,0x68,0xE8,0x18,0x98,0x58,0xD8,0x38,0xB8,
0x78,0xF8,0x04,0x84,0x44,0xC4,0x24,0xA4,0x64,0xE4,
0x14,0x94,0x54,0xD4,0x34,0xB4,0x74,0xF4,0x0C,0x8C,
0x4C,0xCC,0x2C,0xAC,0x6C,0xEC,0x1C,0x9C,0x5C,0xDC,
0x3C,0xBC,0x7C,0xFC,0x02,0x82,0x42,0xC2,0x22,0xA2,
0x62,0xE2,0x12,0x92,0x52,0xD2,0x32,0xB2,0x72,0xF2,
0x0A,0x8A,0x4A,0xCA,0x2A,0xAA,0x6A,0xEA,0x1A,0x9A,
0x5A,0xDA,0x3A,0xBA,0x7A,0xFA,0x06,0x86,0x46,0xC6,
0x26,0xA6,0x66,0xE6,0x16,0x96,0x56,0xD6,0x36,0xB6,
0x76,0xF6,0x0E,0x8E,0x4E,0xCE,0x2E,0xAE,0x6E,0xEE,
0x1E,0x9E,0x5E,0xDE,0x3E,0xBE,0x7E,0xFE,0x01,0x81,
0x41,0xC1,0x21,0xA1,0x61,0xE1,0x11,0x91,0x51,0xD1,
0x31,0xB1,0x71,0xF1,0x09,0x89,0x49,0xC9,0x29,0xA9,
0x69,0xE9,0x19,0x99,0x59,0xD9,0x39,0xB9,0x79,0xF9,
0x05,0x85,0x45,0xC5,0x25,0xA5,0x65,0xE5,0x15,0x95,
0x55,0xD5,0x35,0xB5,0x75,0xF5,0x0D,0x8D,0x4D,0xCD,
0x2D,0xAD,0x6D,0xED,0x1D,0x9D,0x5D,0xDD,0x3D,0xBD,
0x7D,0xFD,0x03,0x83,0x43,0xC3,0x23,0xA3,0x63,0xE3,
0x13,0x93,0x53,0xD3,0x33,0xB3,0x73,0xF3,0x0B,0x8B,
0x4B,0xCB,0x2B,0xAB,0x6B,0xEB,0x1B,0x9B,0x5B,0xDB,
0x3B,0xBB,0x7B,0xFB,0x07,0x87,0x47,0xC7,0x27,0xA7,
0x67,0xE7,0x17,0x97,0x57,0xD7,0x37,0xB7,0x77,0xF7,
0x0F,0x8F,0x4F,0xCF,0x2F,0xAF,0x6F,0xEF,0x1F,0x9F,
0x5F,0xDF,0x3F,0xBF,0x7F,0xFF
};
/****************************************************************************
* 功能:字节倒序
****************************************************************************/
uint8 B2B1(uint8  xx)
{
  uint8 yy=0;
  yy=((xx<<7)&0x80)|((xx<<5)&0x40)|((xx<<3)&0x20)|((xx<<1)&0x10)|((xx>>7)&0x01)|((xx>>5)&0x02)|((xx>>3)&0x04)|((xx>>1)&0x08);
  return yy;
}
/****************************************************************************
* 功能:字节倒序
****************************************************************************/
uint8 B2B2(uint8  xx)
{
  uint8 yy=0;
  uint8 j;
  for(j=0;j<7;j++)
  {
     if((xx&1)==1)
  yy=yy|1;

     xx=xx>>1;
     yy=yy<<1;
  }
  if((xx&1)==1)
  yy=yy|1;

  return yy;
}
/****************************************************************************
* 功能:字节倒序
****************************************************************************/
uint8 B2B3(uint8  xx)
{
   return  zjdx[xx];
}

/****************************************************************************
* 功能:字节倒序
****************************************************************************/
uint8 B2B4(uint8  xx)
{
        ACC=xx;
        B_0=A_7;
        B_1=A_6;
        B_2=A_5;
        B_3=A_4;
        B_4=A_3;
        B_5=A_2;
        B_6=A_1;
        B_7=A_0;
        return B;
}
//---------------------------------------------------------
void main(void)
{
uint8 ss;

 ss=B2B1(0x73);      //中间
  ss=B2B2(0x73);     //最慢
   ss=B2B3(0x73);    //最快
      ss=B2B4(0x73); //次之

 ss=B2B1(0x54);
  ss=B2B2(0x54);
   ss=B2B3(0x54);
     ss=B2B4(0x54);

 ss=B2B1(0x8e);
  ss=B2B2(0x8e);
   ss=B2B3(0x8e);
      ss=B2B4(0x8e);


 ss=B2B1(0xcd);
  ss=B2B2(0xcd);
   ss=B2B3(0xcd);
      ss=B2B4(0xcd);


while(1);
}                

PARTNER CONTENT

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用户377235 2014-6-23 17:41

正好用到,思路很值得学习!

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