BCD conversion code

#include <stdio.h>

// This might not be very portable but works for Linux/Windows
union
{
  unsigned char byte[4];
  unsigned long v;
} reg;

// internal dump of intermediate steps so you can see what's happening
void dump(char *s)
{
  int i;
  printf("%s: ",s);
  for (i=sizeof(reg.byte)-1;i>=0;i--) printf("%x ",reg.byte[i]);
  printf("\n");
}


int main(int argc, char **argv)
{
  char buf[65];
  int n,i=0;
  // Get a number to play with
  printf("Number (decimal)?");
  gets(buf);
  reg.v=atoi(buf);
  // loop 8 times
  while (i!=8)
    {
      printf("%d\n",i++);  // debug output
      dump("Base");
      // The idea is we are going to shift the number left
      // 8 places to the "new" location
      // but.. on each shift we will look to see if a
      // BCD digit is >=5... if so the next shift shoul
      // give us 10 but will instead give us A
      // so if that's the case, add 3 so you get 8 (or 9 or 10, etc.)
      // and double that will be 0x10, 0x11, 0x12...
      // Since the max output is 255 there is no reason
      // to test the top digit
      if ((reg.byte[1]&0xF)>=5)  reg.byte[1]+=3;
      if ((reg.byte[1]>>4&0xF)>=5)  reg.byte[1]+=0x30;
      dump("Add");
      // so now that we've adjusted things, shift
      reg.v<<=1;
      dump("Shift");
    }
  // final result is in reg.byte[2] (MSD) and packed in reg.byte[1]
  printf("\nResult: %d%d%d\n",reg.byte[2]&0xF,reg.byte[1]>>4,reg.byte[1]&0xF);
}