Forums Scripts & Popups Mass deop (from a book) questions

Hello everyone.
I last used mIRC in 1999. I'm reading a book that has taken me on a trip down memory lane: "Malicious Mobile Code: Virus Protection for Windows" O'Reilly (2001). Nowadays, i.e. 25 years later, some concepts explained in the book sounds funny, but let's cut to the chase. From page 243 (I'm copy-pasting):

Mass deop attack

Once a hacker has taken control of a channel, he has to make sure he is the only operator. So, he has to have a script tool that will automatically "deop" any remaining operators, while making sure not to deop his own account. The script shown in Example 7-3 automates that process.

Example 7-3. Example deop attack malicious script

Alias:

Code

madeop {
  %deopv = 0
  %deopn = ""
  :xnext
  inc %deopv 1
  if ($opnick(%deopv,$$1) == $me) { goto xnext }
  if ($opnick(%deopv,$$1) == $null)
  {
    if ($len(%deopn) > 0) { mode # -oooo %deopn } goto xdone
  }
  %deopn = %deopn $opnick(%deopv,$$1)
  if (4 // %deopv) { mode # -oooo %deopn | %deopn = "" }
  goto xnext
  :xdone
}

To use it, the hacker would have to have operator status and then type /madeop {channel}.

Please note this is exactly what the book says; I didn't change anything.
I'm not familiar with mIRC's scripting language. Also, I'm on a Mac, so I can't run the script on mIRC.

That being said, is that script correct?

I see 4 // %deopv , but shouldn't it be %deopv // 4 ?
And why is he deopping four users at a time? Wouldn't three be a better choice?
Also, I suppose $opnick(%deopv,$$1) is the %deopv-nth op's name, but why not write $opnick(%deopv,#) or $opnick(#,%deopv) ?

Yes the script is almost correct, but it will only work up to mIRC 5.0, after which the order of parameter of $opnick() changed, which is why the channel name is the second parameter here.
Using $chan instead of $$1 or $1 is not correct as $chan refer to the current channel, while $1 is the first parameter passed to the alias/function, so allowing it to work on any channel.
4 // %deop is correct, checks if %deop is a multiple of 4.
Why would 3 be better than 4 deop at the same time?

The only thing that would make it not correct is the { bracket of the if statement not being on the same line as the if statement, as far as I know mIRC has never supported this.
And the usage of $chan in /mode instead of using $1 or $$1.
But the script would only break when testing with a number of opped nickname that is not a multiple of 4 or when the current channel is not the channel you want to deop on.

Here is a rewrite for current mIRC:


It would deop $modespl nicks at a time, which is the maximum number of modes you can do at the same time for any server.

The best choice would be the maximum number of mode changes the IRC network allows per line. The $modespl identifier is used for that. There is no standard so max number of mode changes in a single command can be anywhere between 1 and infinity.

If you have 100 users and deop three at a time the script would need to do "/mode #channel -ooo xxx xxx xxx" 33.333333 (actually 34 since you have one user left over) times. (100 / 3 = 33.333333) .

If you have 100 users and deop four at a time the script would need to do "/mode #channel -oooo xxx xxx xxx xxx" 25 times. (100 / 4 = 25).

From RFC 1459:
"When using the 'o' and 'b' options, a restriction on a total of three per mode command has been imposed."

so that statement is no longer valid?

The RFC's are more like guidelines than actual rules. The majority of the IRCd's that exists ignores the MODES=3 and use a higher number.

One other issue: since %deopv is being used to both iterate through $opnick() and in "/if (4 // %deopv)", after the "== $me" check skips your own nickname, it looks like %deopv and %deopn will be out of sync.