IPv6 parsing in C

Question

I wanted to know how I can parse an IPv6 address in C and convert it to a 128 bit value?

So a hex address like 1:22:333:aaaa:b:c:d:e needs to be converted to its 128 bit equivalent binary. The problem is the IP address could be of the type ::2 and its variant since they are valid IPv6 address.

The input is from the keyboard and hence is in ASCII format.

Answer 1

You can use POSIX inet_pton to convert a string to a struct in6_addr.

#include <arpa/inet.h>

  ...

const char *ip6str = "::2";
struct in6_addr result;

if (inet_pton(AF_INET6, ip6str, &result) == 1) // success!
{
    //successfully parsed string into "result"
}
else
{
    //failed, perhaps not a valid representation of IPv6?
}

Answer 2

getaddrinfo() can understand IPv6 addresses. Pass AF_INET6 to it in the hints, as well as AI_NUMERICHOST (to prevent a DNS lookup). Linux has it, Windows has it as of Windows XP.

Answer 3

You can use getaddrinfo() POSIX function. It is more flexible than inet_pton(), for example it automatically detects IPv4 and IPv6 address formats, it can resolve even hostnames (using DNS resolving) and port/service names (using /etc/services).

#include <sys/types.h>
#include <netdb.h>
#include <netdb.h>

....

const char *ip6str = "::2";

struct sockaddr_storage result;
socklen_t result_len;

struct addrinfo *res = NULL;
struct addrinfo hints;
memset(&hints, 0, sizeof(struct addrinfo));
hints.ai_family = PF_UNSPEC;
hints.ai_socktype = SOCK_STREAM;
hints.ai_flags = AI_DEFAULT | AI_NUMERICHOST | AI_NUMERICSERV;

rc = getaddrinfo(ip6str, NULL, &hints, &res);
if (rc != 0)
{
    fprintf(stderr, "Failure to parse host '%s': %s (%d)", ip6str, gai_strerror(rc), rc);
    return -1;
}

if (res == NULL)
{
    // Failure to resolve 'ip6str'
    fprintf(stderr, "No host found for '%s'", ip6str);
    return -1;
}

// We use the first returned entry
result_len = res->ai_addrlen;
memcpy(&result, res->ai_addr, res->ai_addrlen);

freeaddrinfo(res);

The IPv6 address is stored in the struct sockaddr_storage result variable.

if (result.ss_family == AF_INET6) // Ensure that we deal with IPv6
{
    struct sockaddr_in6 * sa6 = (struct sockaddr_in6 *) &result;
    struct in6_addr * in6 = &sa6->sin6_addr;
    in6->s6_addr[0]; // This is a first byte of the IPv6
    in6->s6_addr[15]; // This is a last byte of the IPv6
}

Answer 4

To parse IPv6 in C, you need to build yourself a utility function, which tokenized string (colon for hex blocks, and forward-slash for subnet bits).

1. Tokenize raw IPv6 string into smaller substring.
2. Convert non-empty substring into hex blocks. (ASCII to decimal conversion)
3. Expand hex block into 2-bytes by padding zero in front. (only leading zeroes get trimmed)
4. Complete IPv6 should have 8 hex blocks, calculate missing hex-block(s). (zeroes grouping can happen only once)
5. Reinsert missing hex-block. (use index of the empty substring)

Answer 5

Rosetta has samples in several languages: https://rosettacode.org/wiki/Parse_an_IP_Address

Answer 6

In Windows, you can use WSAStringToAddress, which is available since Windows 2000.

Answer 7

if you can use boost, something like this should work:

#include<boost/asio.hpp>

using boost::asio::ip;

bool parseIpv6String(std::string ipv6_string, char* dest){
    try{
        address_v6 addr = address_v6::from_string(ipv6_string);
        memcpy(dest,addr.to_bytes().data(), 16);
    }catch(...){
        return false;
    }
    return true;
}

It is a bit more portable than for example POSIX specific functions.