Programmatically extract private IP address(es)

Anything in the private IP space will always start with one of three IP address blocks.

• 24-bit block - 10.X.X.X
• 20-bit block - 172.16.X.X - 172.31.X.X
• 16-bit block - 192.168.X.X

So just grep for the above types of IP addresses.

$ ifconfig | grep 'inet addr' | cut -d ':' -f 2 | awk '{ print $1 }' | \
      grep -E '^(192\.168|10\.|172\.1[6789]\.|172\.2[0-9]\.|172\.3[01]\.)'
192.168.1.20

Details

The grep I'm using makes use of regular expressions. In this case we're looking for the following patterns:

• 192.168
• 10.
• 172.1[6789].
• 172.2[0-9].
• 172.3[01].

Additionally we're being explicit in only matching numbers which start with one of these patterns. The anchor (^) is providing this capability to us.

More Examples

If we add the following lines to a file just to test the grep out.

$ cat afile 
192.168.0.1
10.11.15.3
1.23.3.4
172.16.2.4

We can then test it like so:

$ cat afile | grep -E '^(192\.168|10\.|172\.1[6789]\.|172\.2[0-9]\.|172\.3[01]\.)'
192.168.0.1
10.11.15.3
172.16.2.4