In exploring how my local openVPN virtual interface utun0 works, I came across the following data, and I don't know how to make sense of it. (I'm on a Mac)

$ netstat -nr
Routing tables

Destination        Gateway            Flags        Refs      Use   Netif Expire
0/1                 UGSc           61        0   utun0
default        UGSc            7        0     en0           UGSc            1        0   utun0            UHr           110       12   utun0      UGSc            2        0     en0

It looks like "0/1" is CIDR notation. Is that correct? If so, I have follow up questions. From my understanding, an interface is chosen according to which subnet(s) match the destination ip. With 0/1, only ip addresses whose first bit is 0 would match -- which means only ip address >= would match. Is that true? I could believe that except then I get this

$ ip route get via dev utun0  src

So now I'm really confused what "0/1" means and why that route trumps the default route.


0/1 would actually mean anything <, however, I still get this:

$ ip route get via dev utun0  src

So ip addresses both greater and lesser than go through the router. How? Why?

I do see a 128/1 as well:

$ netstat -nr
Routing tables

Destination        Gateway            Flags        Refs      Use   Netif Expire
0/1                 UGSc           63        0   utun0
default        UGSc            5        0     en0
128.0/1             UGSc           42        0   utun0

So @Teun Vink seems to be correct.

  • What is the router model?
    – Ron Maupin
    Commented Aug 29, 2016 at 17:33
  • How did you do "ip route get" on a Mac? Or had you switched to a Linux box for that? I happen to be looking for a similar command for the Mac. Commented Mar 3, 2019 at 20:07
  • It's very possible that I brew installed something to get that cli command. But I don't remember what if I did Commented Mar 4, 2019 at 14:44
  • Or maybe you actually typed "route get" which would work on a Mac. Commented Jul 19, 2022 at 0:49

2 Answers 2


Some VPNs push the default gateway (a /0 netmask) as two /1 networks: 0/1 and 128/1. Since a more specific route always wins, this forces traffic to be routed via the VPN instead of over the default gateway.

  • What would happen if I have two VPN clients installed? I assume what would happen is that when I turn VPN 1 on, it would be fine, but when I turn VPN 2 on (assuming it also does the 0/1 128/1 trick), it would fail because those subnets are already in use. Does that sound right? Commented Aug 29, 2016 at 19:19
  • Assuming both VPNs push the same routes: yes, then you'd have two equal routes, so that wouldn't work as expected.
    – Teun Vink
    Commented Aug 30, 2016 at 5:07

The first bit is for the netmask not for the address, this route mean all IP starting from with netmask so

Network:            0 0000000.00000000.00000000.00000000
HostMin:              0 0000000.00000000.00000000.00000001
HostMax:      0 1111111.11111111.11111111.11111110

Network:          1 0000000.00000000.00000000.00000000
HostMin:            1 0000000.00000000.00000000.00000001
HostMax:      1 1111111.11111111.11111111.11111110

the routing daemon take the biggest netmask it knows to select the best route. So whenever you have a default route, you always have another route with biggest netmask using 2 net with /1 netmask

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