How to turn a hex string into an unsigned char array?

Question

For example, I have a cstring \"E8 48 D8 FF FF 8B 0D\" (including spaces) which needs to be converted into the equivalent unsigned char array {0xE8,0x48,0

Answer 1

The old C way, do it by hand ;-) (there is many shorter ways, but I'm not golfing, I'm going for run-time).

enum { NBBYTES = 7 };
char res[NBBYTES+1];
const char * c = "E8 48 D8 FF FF 8B 0D";
const char * p = c;
int i = 0;

for (i = 0; i < NBBYTES; i++){
    switch (*p){
    case '0': case '1': case '2': case '3': case '4':
    case '5': case '6': case '7': case '8': case '9':
      res[i] = *p - '0';
    break;
    case 'A': case 'B': case 'C': case 'D': case 'E': case 'F':
      res[i] = *p - 'A' + 10;
    break;
   default:
     // parse error, throw exception
     ;
   }
   p++;
   switch (*p){
   case '0': case '1': case '2': case '3': case '4':
   case '5': case '6': case '7': case '8': case '9':
      res[i] = res[i]*16 + *p - '0';
   break;
   case 'A': case 'B': case 'C': case 'D': case 'E': case 'F':
      res[i] = res[i]*16 + *p - 'A' + 10;
   break;
   default:
      // parse error, throw exception
      ;
   }
   p++;
   if (*p == 0) { continue; }
   if (*p == ' ') { p++; continue; }
   // parse error, throw exception
}

// let's show the result, C style IO, just cout if you want C++
for (i = 0 ; i < 7; i++){
   printf("%2.2x ", 0xFF & res[i]);
}
printf("\n");

Now another one that allow for any number of digit between numbers, any number of spaces to separate them, including leading or trailing spaces (Ben's specs):

#include 
#include 

int main(){
    enum { NBBYTES = 7 };
    char res[NBBYTES];
    const char * c = "E8 48 D8 FF FF 8B 0D";
    const char * p = c;
    int i = -1;

    res[i] = 0;
    char ch = ' ';
    while (ch && i < NBBYTES){
       switch (ch){
       case '0': case '1': case '2': case '3': case '4':
       case '5': case '6': case '7': case '8': case '9':
          ch -= '0' + 10 - 'A';
       case 'A': case 'B': case 'C': case 'D': case 'E': case 'F':
          ch -= 'A' - 10;
          res[i] = res[i]*16 + ch;
          break;
       case ' ':
         if (*p != ' ') {
             if (i == NBBYTES-1){
                 printf("parse error, throw exception\n");
                 exit(-1);
            }
            res[++i] = 0;
         }
         break;
       case 0:
         break;
       default:
         printf("parse error, throw exception\n");
         exit(-1);
       }
       ch = *(p++);
    }
    if (i != NBBYTES-1){
        printf("parse error, throw exception\n");
        exit(-1);
    }

   for (i = 0 ; i < 7; i++){
      printf("%2.2x ", 0xFF & res[i]);
   }
   printf("\n");
}

No, it's not really obfuscated... but well, it looks like it is.