原创 N位压缩BCD加法(ARM小端算法)---雁舞白沙出题

/*------------------------------------------------------------------------
          N位压缩BCD加法(ARM小端算法)---雁舞白沙出题
此题目的讨论见 http://bbs.21ic.com/club/bbs/ShowAnnounce.asp?id=2464366
HotPower@126.com    2007.3.13 于西安大雁塔菜地
------------------------------------------------------------------------*/
static unsigned char databuf[12];//32位BCD加法缓冲区
unsigned char DataAdd(unsigned char *,unsigned char *,unsigned int);

int main()
{
//32位BCD加法(12345678 + 23454543 = 35800221)
  (unsigned int)databuf[4] = 0x12345678;
  (unsigned int)databuf[8] = 0x23454543;
  DataAdd(&databuf[4], &databuf[8], 4);//结果在val和databuf[0]~databuf[3]中
  while(1);
}

unsigned char DataAdd(unsigned char *DataOneDptr,unsigned char *DataTwoDptr,unsigned int DataLend)
{
unsigned char One, Two, Tmp;//中间变量
unsigned int i, Sum = 0;//考虑CY位故取整型
  for (i = 0; i < DataLend; i ++) {
    One = DataOneDptr;//取出正确的被加数
 Two = DataTwoDptr;//取出正确的加数
 Sum = One + Two + Sum;//二进制求和(注意上次低位向高位的进位)
 Tmp = (One & 0xf0) + (Two & 0xf0);//为半进位做准备
 if ((Tmp != (Sum & 0xf0)) || ((Sum & 0x0f) > 9)) {//BCD码低4位调整
   Sum += 6;//凡加减10或9的都是"混球"~~~
 }
 if (Sum >= 0xa0) {//BCD码高4位调整
   Sum += 0x60;//注意"混球"~~~
 }
 databuf = Sum;//只存入低8位
 Sum >>= 8;//保留高8位做为下次低位向高位的进位
  }
  return Sum;//返回溢出标志(最高1位)
}