Simplify s_char_to_int and align DECODE_V macro

[headscott]

The current implementation uses a switch with ten cases:

static LTC_INLINE int s_char_to_int(unsigned char x)
{
   switch (x)  {
      case '0': return 0;
      case '1': return 1;
      case '2': return 2;
      case '3': return 3;
      case '4': return 4;
      case '5': return 5;
      case '6': return 6;
      case '7': return 7;
      case '8': return 8;
      case '9': return 9;
      default:  return 100;
   }
}

This could be simplified to a range check and subtraction:

static LTC_INLINE int s_char_to_int(unsigned char x)
{
   if(x < '0' || '9' < x)
      return 100;
   return (int) x - '0';
}

This is shorter, clearer, and should be equally efficient. If you simplify it like that, remember to do so in src/pk/asn1/der/utctime/der_decode_utctime.c but also in src/pk/asn1/der/generalizedtime/der_decode_generalizedtime.c

You could also align the definition of DECODE_V in der_decode_utctime.c with the one in der_decode_generalizedtime.c by wrapping it in do { ... } while (0):
From:

#define DECODE_V(y, max) \
   if (x + 2 > declen) return CRYPT_INVALID_PACKET;     \
   y  = s_char_to_int(buf[x])*10 + s_char_to_int(buf[x+1]); \
   if (y >= max) return CRYPT_INVALID_PACKET;           \
   x += 2;

To:

#define DECODE_V(y, max) do {\
   if (x + 2 > declen) return CRYPT_INVALID_PACKET; \
   y  = s_char_to_int(buf[x])*10 + s_char_to_int(buf[x+1]); \
   if (y >= max) return CRYPT_INVALID_PACKET;           \
   x += 2; \
} while(0)