Determine if a string is a valid IPv4 address in C

Question

What would be a good way to determine if a string contains an IPv4 address? Should I use isdigit()?

Answer 1

This is a routine I wrote a while ago for an embedded system which generated various suspect patterns on a network. As such, it uses absolutely no fancy stuff like network libraries or even the standard C libraries, preferring to steer clear of all that modern stuff like string tokenizing and (shudder) regular expression libraries :-) To that end, it's suited to just about any environment you could find yourself in, and it was blindingly fast.

Although, if you're in an environment that has something like checkIp4Addess(), I'd suggest you use that instead. It's an indication of the stuff you sometimes have to put up with when doing embedded stuff (although it is a real solution).

int isValidIp4 (char *str) {
    int segs = 0;   /* Segment count. */
    int chcnt = 0;  /* Character count within segment. */
    int accum = 0;  /* Accumulator for segment. */

    /* Catch NULL pointer. */

    if (str == NULL)
        return 0;

    /* Process every character in string. */

    while (*str != '\0') {
        /* Segment changeover. */

        if (*str == '.') {
            /* Must have some digits in segment. */

            if (chcnt == 0)
                return 0;

            /* Limit number of segments. */

            if (++segs == 4)
                return 0;

            /* Reset segment values and restart loop. */

            chcnt = accum = 0;
            str++;
            continue;
        }
        /* Check numeric. */

        if ((*str < '0') || (*str > '9'))
            return 0;

        /* Accumulate and check segment. */

        if ((accum = accum * 10 + *str - '0') > 255)
            return 0;

        /* Advance other segment specific stuff and continue loop. */

        chcnt++;
        str++;
    }

    /* Check enough segments and enough characters in last segment. */

    if (segs != 3)
        return 0;

    if (chcnt == 0)
        return 0;

    /* Address okay. */

    return 1;
}