诚征数字转换为字符串的经典算法

Question

型如unsigned char ddd[]= {0x3f,0x2a,0xfe,0xe1,0x42,0x32,0xa2,0xd2,0xbb};
转换为unsigned char sss[] = "3f2afee14232a2d2bb";
因为ddd可能比较长，用itoa一个一个转感觉效率比较低，有没有好的做法？

Answer 1

http://home.nuc.edu.cn/~titilima/readarticle.php?id=24

Answer 2

//lpBuf--  {0x3f,0x2a,0xfe,0xe1,0x42,0x32,0xa2,0xd2,0xbb};
//szString --  "3f2afee14232a2d2bb";
//nLen  --lpBuf的长度 
void  tostring(const unsigned  char* lpBuf, unsigned  char* szString, int nLen)
{ unsigned char dH, dL;
int i;
szString[ nLen * 2 ] = 0;
for(i = 0; i < nLen; i++ )
{  
                  dH = 0x30 + ((lpBuf[i]>>4) & 0x0f);
dL = 0x30 + (lpBuf[i] & 0x0f);
if( dH > 57 )  
dH = dH + 7;
if( dL > 57 )  
dL = dL + 7;
szString[ i * 2 ] = dH;
szString[ i * 2 + 1 ] = dL;
}
}

Answer 3

//调用setHexToStr(）sss = "3f2afee14232a2d2bb";
char sss[100];
unsigned char ddd[]= {0x3f,0x2a,0xfe,0xe1,0x42,0x32,0xa2,0xd2,0xbb};char SetAsciiToStr(char c)
{
   return c>9?c-0xa+'a':c+'0';
}void setHexToStr(char*sss,char*ddd)
{
    int i,j,len=sizeof(ddd);
    for(i=j=0;i<len;i++){
      sss[j++]=SetAsciiToStr(ddd[i]>>4);
      sss[j++]=SetAsciiToStr(ddd[i]&0xF);
    }
    sss[j]=0;
}

Answer 4

void CWingoDlg::OnButtonConvert() 
{
unsigned char ddd[]= {0x3f,0x2a,0xfe,0xe1,0x42,0x32,0xa2,0xd2,0x1b};
AfxMessageBox(ConvertToStr(ddd,9));
}CString CWingoDlg::ConvertToStr(unsigned char *c,int n)
{
CString str,tmp;
int i=0;
while (i<n) {
tmp.Format("%x",*(c+i++));
str += tmp;
}
return str;
}
看一下 我这个，但是不知道执行,谁的效率更好

Answer 5

http://home.nuc.edu.cn/~titilima/readarticle.php?id=24

Answer 6

3jaja的算法好像结果不对，len=sizeof(ddd);是取指针的长度，结果为4。即使传入正确的长度，结果也不正确。