Linux getting all network interface names
I need to collect all the interface names, even the ones that aren't up at the moment. Like ifconfig -a
.
getifaddrs()
is iterating through same interface name multiple times. How can I collect all the interface names just once using getifaddrs()
?
Solution 1:
You could check which entries from getifaddrs belong to the AF_PACKET family. On my system that seems to list all interfaces:
struct ifaddrs *addrs,*tmp;
getifaddrs(&addrs);
tmp = addrs;
while (tmp)
{
if (tmp->ifa_addr && tmp->ifa_addr->sa_family == AF_PACKET)
printf("%s\n", tmp->ifa_name);
tmp = tmp->ifa_next;
}
freeifaddrs(addrs);
Solution 2:
getifaddrs() will only return your interfaces addresses, not the interfaces themselves.
What if any of your interface has no address, or no address of the requested family, as suggested with the 'AF_PACKET' one ?
Here, an example where I’ve got a tunnel interface (with an OpenVPN connexion), and where I’m listing all entries from getifaddrs() for each of my network interfaces:
[0] 1: lo address family: 17 (AF_PACKET) b4:11:00:00:00:01
address family: 2 (AF_INET) address: <127.0.0.1>
address family: 10 (AF_INET6) address: <::1>
[...]
[5] 10: tun0 address family: 2 (AF_INET) address: <172.16.0.14>
[EOF]
Bam. No AF_PACKET on the "tun0" interface, but it DOES exist on the system.
You should, instead, use if_nameindex() syscall, which does exactly what you want. In other words, with no arguments, it returns a list of all interfaces on your system:
#include <net/if.h>
#include <stdio.h>
int main (void)
{
struct if_nameindex *if_nidxs, *intf;
if_nidxs = if_nameindex();
if ( if_nidxs != NULL )
{
for (intf = if_nidxs; intf->if_index != 0 || intf->if_name != NULL; intf++)
{
printf("%s\n", intf->if_name);
}
if_freenameindex(if_nidxs);
}
return 0;
}
And voilà.
Solution 3:
It seems that ifconfig -a
only lists active interfaces (at least on Fedora 19). I know I have at least one more network card that I'm not seeing. Anyway, I get the same list as:
ls -1 /sys/class/net
Which could easily be done programatically.