Shutting off wireless auto-config in Mac OS X

As a matter of security and simply less annoyance, I prefer my computers to not connect willy-nilly to just any wireless network in range. OS X currently doesn't connect to random open networks, but it does always look for them, and by default it prompts you to connect to new ones. In OS X, all you have to do is un-check the "Ask to join new networks" on the AirPort adapter in Network preferences to stop this behavior.



Once you do that, go into the Advanced preferences and remove all those random access points that you've connected to in the past, leaving only the ones you know and trust on the list. If your access point has a default-ish name (like WLAN, Default, linksys, etc) you should probably change it so that your computer doesn't join up to the first "linksys" network it runs across.

On Linux and BSD, it's easy. You simply have to try -- and mean it -- to get on a wireless network. You don't just accidentally connect.

Also, it's a cold day in hell. Last week, I bought my wife a new computer that came loaded with Windows Vista Home Premium. I can only use it for a few minutes at a time before I have the urge to go take a shower with a steel wool pad to try to get Vista off of me, but I can't for the life of me figure out how to kill Vista's auto-join feature while letting it connect to preferred networks only. If you have any tips, drop us a line. It definitely isn't like XP. The only thing I saw told me to kill the Wireless Autoconfig service (maybe called something a little different) and all that did was completely disable wireless access on Vista.

IP Subnetting - more fun with newLISP

I decided it would be fun to try to make an IP subnet calculator with newLISP.

Thanks to Elica and Lutz on the newLISP discussion boards. I needed some help with the logic. There's probably a way to compact this code down to about 3 lines somehow, but I'm still a newLISP n00b. I stuck with the logic examples that I have a firm understanding of, but the discussion yielded some interesting results.

Here's what I came up with

#!/usr/bin/newlisp
# newLISP IP Address calculator by ax0n
# ax0n (at) h-i-r.net
(define (iptostr ip4)
# Converts an integer to an IP in dotted decimal notation
(string
  (mod (/ ip4 0x1000000) 0x100) "."
  (mod (/ ip4 0x10000) 0x100) "."
  (mod (/ ip4 0x100) 0x100) "."
  (mod ip4 0x100))
)

(define (iptonum ip4str)
# Converts an IP string to an integer
 (map set '(one two three four) (parse ip4str "."))
 (+ (* 0x1000000 (float one)) (* 0x10000 (float two))
  (* 0x100 (float three)) (float four))
)

(cond(
 (< (length (main-args)) 3)
 # Display usage if no args passed
 (println "usage: ipcalc.lsp ip-address/maskbits")
 (println "ex:    ipcalc.lsp 192.168.1.20/24")
 )
 (true
  (set 'ipstr (last(main-args)))
  (map set '(ipaddr bits) (parse ipstr "/"))
  (set 'binip (iptonum ipaddr))
  (set 'netmask (& 0xffffffff ( << 0xffffffff (- 32 (int bits)))))
  (set 'netaddr (& binip netmask))
  (set 'bcast (& 0xffffffff (| binip (~ netmask))))
  (println "host IP:    " ipaddr )
  (println "netmask:    " (iptostr netmask) )
  (println "network:    " (iptostr netaddr) )
  (println "broadcast:  " (iptostr bcast))
  (println "Host range: " (iptostr (+ netaddr 1))" - "(iptostr (- bcast 1)))
 )
)
(exit)

Running it by itself gives you a syntax help page.

-bash-3.2$ ./ipcalc.lsp
usage: ipcalc.lsp ip-address/maskbits
ex: ipcalc.lsp 192.168.1.20/24

You have to provide an IP Address and mask in CIDR Notation. It does the rest!

-bash-3.2$ ./ipcalc.lsp 192.168.0.49/24
host IP:    192.168.0.49
netmask:    255.255.255.0
network:    192.168.0.0
broadcast:  192.168.0.255
Host range: 192.168.0.1 - 192.168.0.254

You can download the script here:
http://stuff.h-i-r.net/blogstuff/ipcalc.lsp

IP subnetting

This really isn't that hard, but I had someone ask for help on one of the forums I participate in, and thought I'd share the answer here.

The person needed to determine the following given an IP/Mask of 146.141.219.47/18

• Subnet Mask
• Network Address
• First usable host address
• Last usable host address

The answer is to use binary math and boolean logic to solve the problems.

/18 means there is an 18 bit subnet mask on that IP address. This is neither a Class C (24 bit subnet mask) or Class B (16 bits). Although, "Classes" are kind of obsolete these days. It's just an 18 bit subnet mask. Straightforward.

In binary, it looks like this (decimal underneath):

11111111 11111111 11000000 00000000
    255      255      192        0

To get the network address, you take the IP Address in binary, and do a boolean AND against the subnet mask. A boolean AND will only return a 1 where both numbers have a 1, and return a 0 everywhere else.

11111111 11111111 11000000 00000000 <-- subnet mask      255.255.192.0
10010010 10001101 11011011 00101111 <-- IP Address       146.141.219.47
======== ======== ======== ========
10010010 10001101 11000000 00000000 <-- Network Address  146.141.192.0

The first usable host IP address is always the first IP after the network address, in this case, 146.141.192.1.

To get the broadcast address, you simply change all of the non-masked bits from the network address (in this case, the last 14 bits) to 1.

10010010 10001101 11111111 11111111 <-- Broadcast        146.141.255.255

You need to know the broadcast address to come up with the last usable host IP address, which is always one IP below the broadcast address. The last usable IP in the network for this example is 146.141.255.254.