I am able to get the IPv4 address of a given interface with following code
int fd;
char ipv4[33];
char ifname[] = "eth0";
struct ifreq ifr;
fd = socket(AF_INET, SOCK_DGRAM, 0);
ifr.ifr_addr.sa_family = AF_INET;
strncpy(ifr.ifr_name, ifname, IFNAMSIZ-1);
ioctl(fd, SIOCGIFADDR, &ifr);
close(fd);
snprintf(ipv4, 33, "%s", inet_ntoa(((struct sockaddr_in *)&ifr.ifr_addr)->sin_addr));
What is the easiest way to get the IPv6 address of a given interface, using C ?
*** I AM POSTING THE CODE WHICH I COULD USE TO MAKE IT BELOW ..
int8_t find_device_ipv6(const char *ifname, char *ipv6, int8_t ipv6_size)
{
FILE *f;
int ret, scope, prefix;
unsigned char _ipv6[16];
char dname[IFNAMSIZ];
char address[INET6_ADDRSTRLEN];
char *scopestr;
f = fopen("/proc/net/if_inet6", "r");
if (f == NULL) {
return -1;
}
while (19 == fscanf(f,
" %2hhx%2hhx%2hhx%2hhx%2hhx%2hhx%2hhx%2hhx%2hhx%2hhx%2hhx%2hhx%2hhx%2hhx%2hhx%2hhx %*x %x %x %*x %s",
&_ipv6[0],
&_ipv6[1],
&_ipv6[2],
&_ipv6[3],
&_ipv6[4],
&_ipv6[5],
&_ipv6[6],
&_ipv6[7],
&_ipv6[8],
&_ipv6[9],
&_ipv6[10],
&_ipv6[11],
&_ipv6[12],
&_ipv6[13],
&_ipv6[14],
&_ipv6[15],
&prefix,
&scope,
dname)) {
if (strcmp(ifname, dname) != 0) {
continue;
}
if (inet_ntop(AF_INET6, _ipv6, address, sizeof(address)) == NULL) {
continue;
}
snprintf(ipv6, ipv6_size, "%s", address);
}
fclose(f);
return 0;
}
I assume your OS is Linux.
To find out how it is done, try strace ifconfig eth0
.
You see that it is using /proc/net/if_inet6
and some ioctl, notably SIOCGIFADDR
As Ctx commented, you probably want getifaddrs(3)