Obtaining local IP address using getaddrinfo() C function?

Question

I'm trying to obtain my local (not the external) IP address using the getaddrinfo() function, but I saw the examples provided here, and they where too complex for my needs. Also saw other posts and most of them really wanted to get the external IP, not the local one.

Could anyone provide a link to a simple example (or a simple example) about how to obtain my own local IP address using this function ?

Just to be clear when I say local, if a router is 192.168.0.1 , my local IP address could be something like 192.168.0.x ( just an example ).

getaddrinfo was made to have an uniform interface between ipv4 and ipv6, this is why it looks a little complex. (If it get the answer I'll let you know later) — Aif, Jan 27 '10 at 10:59
Thanks! , yes, it looks complex and I don't want to use gethostbyname() because it seems to be deprecated... — Goles, Jan 27 '10 at 11:03
The last example on that page is how to do this... why is that too complex? Use AF_INET if you really want only an IPv4 address... but don't do that, make it support IPv6 as well. — Andrew McGregor, Jan 27 '10 at 11:30
If you're using C++, have you considered using the ASIO networking library? http://asio.sourceforge.net/ — , Jan 27 '10 at 12:08
Link you provided as example is broken. Please update it if possible. — Fusion, Jul 28 '19 at 09:32

caf · Accepted Answer · 2011-06-28T04:04:53.723

getaddrinfo() isn't for obtaining your local IP address - it's for looking up names and/or services to socket addresses. To obtain the local IP address(es), the function you want is getifaddrs() - here's a minimal example:

#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <sys/socket.h>
#include <netinet/in.h>
#include <arpa/inet.h>
#include <net/if.h>
#include <ifaddrs.h>
#include <errno.h>

int main(int argc, char *argv[])
{
    struct ifaddrs *myaddrs, *ifa;
    void *in_addr;
    char buf[64];

    if(getifaddrs(&myaddrs) != 0)
    {
        perror("getifaddrs");
        exit(1);
    }

    for (ifa = myaddrs; ifa != NULL; ifa = ifa->ifa_next)
    {
        if (ifa->ifa_addr == NULL)
            continue;
        if (!(ifa->ifa_flags & IFF_UP))
            continue;

        switch (ifa->ifa_addr->sa_family)
        {
            case AF_INET:
            {
                struct sockaddr_in *s4 = (struct sockaddr_in *)ifa->ifa_addr;
                in_addr = &s4->sin_addr;
                break;
            }

            case AF_INET6:
            {
                struct sockaddr_in6 *s6 = (struct sockaddr_in6 *)ifa->ifa_addr;
                in_addr = &s6->sin6_addr;
                break;
            }

            default:
                continue;
        }

        if (!inet_ntop(ifa->ifa_addr->sa_family, in_addr, buf, sizeof(buf)))
        {
            printf("%s: inet_ntop failed!\n", ifa->ifa_name);
        }
        else
        {
            printf("%s: %s\n", ifa->ifa_name, buf);
        }
    }

    freeifaddrs(myaddrs);
    return 0;
}

This is the best answer I have ever seen on this question. – xmllmx Mar 20 '13 at 15:48 — xmllmx, Mar 20 '13 at 15:48

score -2 · Answer 2 · edited Nov 19 '11 at 17:09

-2

Pass the hostname after using gethostname(), to gethostbyname()

int gethostname(char *hostname, size_t size);

edited Nov 19 '11 at 17:09

Mark Hall

53,938
9
94
111

answered Nov 19 '11 at 16:59

robert

5
1

No, the hostname may not be resolvable. It's not good to try to resolve it. It's a common programming error though, so some systems guarantee that the hostname is resolvable by returning "127.0.1.1" just so there isn't an error. Therefore, you can't even rely on the return because it might not even be a useful address. – Nicholas Wilson Aug 23 '13 at 15:51

Obtaining local IP address using getaddrinfo() C function?

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