Converting an IPV6 string to uint16_t array in C++

Question

I have an IPV6 string as an example below :

"0001:0002:0003:0004:0005:0006:0007:0008"

I am trying to find the cleanest way to convert this into a uint16_t array of 8 values containing 1 - 8 .

I cannot use inet_pton()

I could create my own parsing function,however i'm trying to get sscanf to work as below without success. Any help would be appreciated. Edit : The values will always be represented in Hex format in the string.

char *pString        = "0001:0002:0003:0004:0005:0006:0007:0008";
uint16_t* uintValues = new uint16_t[8]{};
sscanf(pString,"%s%04x:", uintValues );

Answer 1

istringstream in("0001:0002:0003:0004:0005:0006:0007:0008");
uint16_t out[8];
char colon;
in >> hex >> out[0] >> colon >> out[1] >> colon >> out[2] >> colon >> out[3] >> colon >> out[4] >> colon >> out[5] >> colon >> out[6] >> colon >> out[7];

No error checking at all which may or may not be a problem.

Answer 2

Your requirements said you cannot use inet_pton, but you could copy a (BSD-licensed) implementation from FreeBSD:

/* int
 * inet_pton6(src, dst)
 *  convert presentation level address to network order binary form.
 * return:
 *  1 if `src' is a valid [RFC1884 2.2] address, else 0.
 * notice:
 *  (1) does not touch `dst' unless it's returning 1.
 *  (2) :: in a full address is silently ignored.
 * credit:
 *  inspired by Mark Andrews.
 * author:
 *  Paul Vixie, 1996.
 */
static int
inet_pton6(const char *src, u_char *dst)
{
    static const char xdigits_l[] = "0123456789abcdef",
              xdigits_u[] = "0123456789ABCDEF";
#define NS_IN6ADDRSZ    16
#define NS_INT16SZ  2
    u_char tmp[NS_IN6ADDRSZ], *tp, *endp, *colonp;
    const char *xdigits, *curtok;
    int ch, seen_xdigits;
    u_int val;

    memset((tp = tmp), '\0', NS_IN6ADDRSZ);
    endp = tp + NS_IN6ADDRSZ;
    colonp = NULL;
    /* Leading :: requires some special handling. */
    if (*src == ':')
        if (*++src != ':')
            return (0);
    curtok = src;
    seen_xdigits = 0;
    val = 0;
    while ((ch = *src++) != '\0') {
        const char *pch;

        if ((pch = strchr((xdigits = xdigits_l), ch)) == NULL)
            pch = strchr((xdigits = xdigits_u), ch);
        if (pch != NULL) {
            val <<= 4;
            val |= (pch - xdigits);
            if (++seen_xdigits > 4)
                return (0);
            continue;
        }
        if (ch == ':') {
            curtok = src;
            if (!seen_xdigits) {
                if (colonp)
                    return (0);
                colonp = tp;
                continue;
            } else if (*src == '\0') {
                return (0);
            }
            if (tp + NS_INT16SZ > endp)
                return (0);
            *tp++ = (u_char) (val >> 8) & 0xff;
            *tp++ = (u_char) val & 0xff;
            seen_xdigits = 0;
            val = 0;
            continue;
        }
        if (ch == '.' && ((tp + NS_INADDRSZ) <= endp) &&
            inet_pton4(curtok, tp) > 0) {
            tp += NS_INADDRSZ;
            seen_xdigits = 0;
            break;  /*%< '\\0' was seen by inet_pton4(). */
        }
        return (0);
    }
    if (seen_xdigits) {
        if (tp + NS_INT16SZ > endp)
            return (0);
        *tp++ = (u_char) (val >> 8) & 0xff;
        *tp++ = (u_char) val & 0xff;
    }
    if (colonp != NULL) {
        /*
         * Since some memmove()'s erroneously fail to handle
         * overlapping regions, we'll do the shift by hand.
         */
        const int n = tp - colonp;
        int i;

        if (tp == endp)
            return (0);
        for (i = 1; i <= n; i++) {
            endp[- i] = colonp[n - i];
            colonp[n - i] = 0;
        }
        tp = endp;
    }
    if (tp != endp)
        return (0);
    memcpy(dst, tmp, NS_IN6ADDRSZ);
    return (1);
}

Answer 3

I won't give you the answer but the answer to write the code of the parser yourself:

---

1/ Case insensitive

An example of an IPv6 address is:. 2001:0db8:85a3:0000:0000:8a2e:0370:7334 The hexadecimal digits are case-insensitive, but IETF recommendations suggest the use of lower case letters. The full representation of eight 4-digit groups may be simplified by several techniques, eliminating parts of the representation.

---

2/ Leading Zeroes

Leading zeroes in a group may be omitted, but each group must retain at least one hexadecimal digit.[1] Thus, the example address may be written as:

2001:db8:85a3:0:0:8a2e:370:7334

---

3/ 1 to Multiple consecutive groups of 0

One or more consecutive groups containing zeros only may be replaced with a single empty group, using two consecutive colons (::).. The substitution may only be applied once in the address, since multiple occurrences would create an ambiguous representation. Thus, the example address can be further simplified:

2001:db8:85a3::8a2e:370:7334

---

Bonus: Two Special cases.

The localhost (loopback) address, 0:0:0:0:0:0:0:1, and the IPv6 unspecified address, 0:0:0:0:0:0:0:0, are reduced to ::1 and ::, respectively.

Etc.

Enjoy!!