1

After a few years of contemplating setting up an old laptop as a server in a home network, and the world as a whole I guess, I'd like to learn more about IP addresses.

How can I find what IP addresses are stored on my computer now? This way I can look at interesting files now. I can possibly use them later to setup IP addresses to configure programs.

Although and tags appear under this question, a GUI solution is perfectly welcome in place of . grep doesn't have to used to fulfill tag and if binary files can be interpreted that is all well and good too.

2

Use grep

You can use grep to find all files containing something that looks like an IPv4 IP address. Be aware there will be false positives. For example the file:

/usr/src/linux-headers-4.14.30-041430/include/linux/oid_registry.h

at line 48 will contain:

OID_smimeAuthenticatedAttrs,    /* 1.2.840.113549.1.9.16.2.11 */

It sort of looks like an IP address in the comments but it is not.

Initially start with a count of all the lines containing an IP address on your system:

$ sudo time grep -rnwI --exclude-dir={boot,dev,media,mnt,lib,proc,root,run,sys,/tmp,tmpfs,var} -E "([0-9]{1,3}[\.]){3}[0-9]{1,3}" / | wc
27.76user 13.17system 1:31.06elapsed 44%CPU (0avgtext+0avgdata 10416maxresident)k
12451744inputs+0outputs (2major+2098minor)pagefaults 0swaps
  17164  122083 3138285

Let's break down the commands

  • sudo prevents "Permission denied" errors
  • time tells us how long it takes to run, ie. 1 minute 31 seconds
  • grep is the command that searches for strings in files
  • -rnwI are the arguments (aka parameters) passed to grep. The r means recursive meaning sub-directories are processed. The n argument prints the line number the search string occurs in the file. The I argument tells it to ignore binary files. If binary files were include the number of files would increase from 17,164 to 22,253 on my system. You can't open binary files and make any sense of them though.
  • -exclude-dir= are the directories to exclude from the search. Without this list grep can take 53 hours to complete: `grep`ing all files for a string takes a long time
  • -E is the argument to grep that tells it the search string is about to follow.
  • "([0-9]{1,3}[\.]){3}[0-9]{1,3}" is the search string to look forward. Explained in more detail below.
  • / tells grep to start at the root directory. However the excluded directories will be skipped.
  • | is the pipe command sending all output to the wc command instead of the screen.
  • wc is the "word count" command. It counts the number lines, number of words and number of characters passed to it. In our case it is 17164 lines, 122,083 words and 3,138,285 characters. Commas added for clarity.

Breaking down "([0-9]{1,3}[\.]){3}[0-9]{1,3}"

As shown earlier, the search string passed to grep is "([0-9]{1,3}[\.]){3}[0-9]{1,3}". Here is how it works:

"([0-9]{1,3}[\.]){3}[0-9]{1,3}"
    ^    ^   ^    ^   ^    ^
    |    |   |    |   |    +---- count of digits must be 1 to 3 
    |    |   |    |   +--------- look for digits 0 through 9
    |    |   |    +------------- patterns 1 to 3 digits of 0-9 followed by . occurs 3 times
    |    |   +------------------ count of 1 to 3 digits must be followed by .
    |    +---------------------- count of digits is 1 to 3
    +--------------------------- look for digits 0 to 9

Seeing the output instead of word count

To see the actual output instead of just the word count remove | wc from the end of the command line:

$ sudo time grep -rnwI --exclude-dir={boot,dev,media,mnt,lib,proc,root,run,sys,/tmp,tmpfs,var} -E "([0-9]{1,3}[\.]){3}[0-9]{1,3}" /

    (... SNIP ...)
/usr/bin/printer-profile:176:   OUT="nc 192.168.1.12 9100 < xxx.prn"
/opt/google/chrome/default_apps/external_extensions.json:23:    "external_version": "0.0.0.6"
/opt/google/chrome/product_logo_32.xpm:330:"    [.}.}.|.1.2.3.4.5.6.7.8.9.0.a.b.8.c.d.e.f.g.h.h.i.j.k.l.    ",
28.52user 12.54system 1:31.78elapsed 44%CPU (0avgtext+0avgdata 9516maxresident)k
12793352inputs+0outputs (3major+1884minor)pagefaults 0swaps

The listing is too long to fit in this answer. Note the last file found is a false positive:

/opt/google/chrome/product_logo_32.xpm

because it doesn't contain a real IP address:

[.}.}.|.1.2.3.4.5.6.7.8.9.0.a.b.8.c.d.e.f.g.h.h.i.j.k.l.

Restrict your search to /etc directory at first

To narrow down search to more meaningful short-list use:

$ sudo time grep -rnI -E "([0-9]{1,3}[\.]){3}[0-9]{1,3}" /etc/etc/hosts:1:127.0.0.1 localhost
/etc/hosts:2:127.0.1.1  alien
/etc/cron.daily/google-earth:47:Version: GnuPG v1.4.2.2 (GNU/Linux)
    (... SNIP ...)
/etc/cups/cups-browsed.conf:77:# BrowseDeny 192.168.1.13
/etc/cups/cups-browsed.conf:78:# BrowseDeny 192.168.3.0/24
/etc/cups/cups-browsed.conf:79:# BrowseDeny 192.168.3.0/255.255.255.0
0.04user 0.03system 0:00.19elapsed 40%CPU (0avgtext+0avgdata 2800maxresident)k
22384inputs+0outputs (1major+181minor)pagefaults 0swaps
  • This is insuffient as it fails to find IPv6 addresses. Also: 0xc0.0xa8.0x01.0x6e, 0xC0A8016E and 3232235886 are all valid ways to write the IPv4 address usually written as 192.168.1.110. So it is not such an easy job to really find all IP addresses. – Sebastian Stark Apr 9 '18 at 4:04
  • @SebastianStark I did find some Stack Exchange answers on how to grep on IPv6 addresses but have not quite got it figured out yet. You can post an answer if you know how, else I will update my answer after I figure it out. – WinEunuuchs2Unix Apr 9 '18 at 4:07
  • Forgot octal notation for IPv4: 0300.0250.0001.0156 – Sebastian Stark Apr 9 '18 at 4:09
  • 2
    It is not so clear what you are trying to do, except pose a little bit esoteric text processing question. – Sebastian Stark Apr 9 '18 at 4:14
  • 2
    A substantial number of IP addresses on your computer will be only in RAM, like for instance the routing table or the ARP cache. Most of them you will probably be able to retrieve from the /proc file system or with specialised tools (arp(8), ip(8)). – Sebastian Stark Apr 9 '18 at 4:28

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