How to get netmask from bash?

Question

I want to get my netmask on Linux. It is outputted with ifconfig but I want to extract the string.

Solution 1

I should clarify that the code here works for Linux, (note comments and post about other Unices). OP asked for a Linux solution, but it would be good to change the question to "how to get netmask" in general, and have the answer combine the best way for more Unix flavors.

#!/bin/sh
ifconfig "$1" | sed -rn '2s/ .*:(.*)$/\1/p'

./script eth0

About ifconfig in Linux

Along with other traditional networking commands like netstat, arp, rarp and route, ifconfig is part of the net-tools package. Net-tools hasn't been actively developed from a really long time and there are efforts to deprecate it in favor of the newer iproute2 package. For the sake of brevity, if you want more details on this fundamental transition, here are some relevant links:

Case example: Debian Linux

Case example: Arch Linux

Man page for net-tools ifconfig (see BUGS section)

Analysis and comparison of the frameworks from a users perspective

Solution 2

/sbin/ifconfig eth0 | awk '/Mask:/{ print $4;} '

?

Solution 3

I wonder if you really only want the mask. Maybe you really want the network (or IP) with mask, to use in config files, with nmap, or whatever.

In that case, on Linux, you can also parse the output of the ip ... command.

To list all interfaces with their address and mask in CIDR notation:

ip -o -f inet addr show | awk '/scope global/ {print $2, $4}'

Or to restrict it to the default interface:

default_if=$(ip route list | awk '/^default/ {print $5}')
ip -o -f inet addr show $default_if | awk '{print $4}'

which gives me "192.168.1.61/24" on my machine.

Solution 4

if you're using an OS that outputs the mask in hex, you can do something like:

#!/usr/bin/bash

# read the mask, and strip leading 0x if it's there
hexmask=$( echo $1 | sed -e 's/^0x//' )

# loop through $hexmask in pairs
#
for (( i=0; i<${#hexmask}; i+=2 )); do
        if (( $i > 1 )); then
                # use a . to separate octets
                # but don't print a leading .
                printf "%s" "."
        fi
        printf "%d" "0x${1:$i:2}"
done

printf "\n"

Solution 5

/sbin/ifconfig eth0 | grep Mask | cut -d":" -f4

View more solutions

Comments

[sr_]

This question is hard, actually, if you don't specify your OS, e.g. FreeBSD's ifconfigs netmask line is: inet 192.168.144.120 netmask 0xffffff00 broadcast 192.168.144.255, so while the answers can be easily adjusted, there's probably no one-size-fits-all (using Bash/ifconfig only).

[michelemarcon]

You are missing | cut -c6- in order to also remove the "Mask:"

[Shane]

Specifying the awk field separator also works to remove "Mask:": /sbin/ifconfig eth0 | awk -F: '/Mask:/{print $4}'

[Oly Dungey]

If the interface in question has no netmask this solution returns '1'. The regex needs some serious rework to make it reliable/safe including the need for a proper anchor on the very important word 'Mask'.

[Oly Dungey]

This is the best answer thus far as the rest rely on a fixed number of columns using awk which isn't the case in real life with ifconfig. I would suggest a few changes though: ifconfig eth0 | grep -o 'Mask:[^\s]*' | cut -d':' -f2. The -o so that grep only returns part of the line rather than the whole lot (more efficient). Match non-whitespace characters ([^\s]*) to find end of the netmask token. Single quotes around the field delimiter for safety.

[Shane]

In Debian Stretch/net-tools 2.10-alpha I had to change to this: /sbin/ifconfig eth0 | awk '/netmask/{print $4}'

[bryn]

...it's not really necessary to match on 'scope global' or 'default' either as you can do ip -o -f inet addr show scope global and ip route list default.

[roaima]

grep 'pattern' | awk '{ action }' can usually be reduced to awk '/pattern/ { action }'

[Rk_thenewprogrammer]

on macOS I get /sbin/ifconfig | grep ask | sed -e 's/.*netmask //g' | sed -e 's/ .*//g' to output the mask in hex.