Question

I have a program where an external component passes me a string which contains an IP address. I then need to turn it into a URI. For IPv4 this is easy; I prepend http:// and append /. However, for IPv6 I need to also surround it in brackets [].

Is there a standard sockets API call to determine the address family of the address?

Answer 1

Use getaddrinfo() and set the hint flag AI_NUMERICHOST, family to AF_UNSPEC, upon successfull return from getaddrinfo, the resulting struct addrinfo .ai_family member will be either AF_INET or AF_INET6.

EDIT, small example

#include <sys/types.h>
#include <stdio.h>
#include <string.h>
#include <sys/socket.h>
#include <netdb.h>

int main(int argc, char *argv[])
{
    struct addrinfo hint, *res = NULL;
    int ret;

    memset(&hint, '\0', sizeof hint);

    hint.ai_family = PF_UNSPEC;
    hint.ai_flags = AI_NUMERICHOST;

    ret = getaddrinfo(argv[1], NULL, &hint, &res);
    if (ret) {
        puts("Invalid address");
        puts(gai_strerror(ret));
        return 1;
    }
    if(res->ai_family == AF_INET) {
        printf("%s is an ipv4 address\n",argv[1]);
    } else if (res->ai_family == AF_INET6) {
        printf("%s is an ipv6 address\n",argv[1]);
    } else {
        printf("%s is an is unknown address format %d\n",argv[1],res->ai_family);
    }

   freeaddrinfo(res);
   return 0;
}

$ ./a.out 127.0.0.1
127.0.0.1 is an ipv4 address
$ ./a.out ff01::01
ff01::01 is an ipv6 address