$biton $bitoff $log
#40247
07/08/03 07:56 AM

Joined: Aug 2003
Posts: 17
Clubfoot
OP
Pikka bird

OP
Pikka bird
Joined: Aug 2003
Posts: 17 
could anyone explain to me in any amount of detail how these identifiers work?



Re: $biton $bitoff $log
#40248
07/08/03 08:57 AM

Joined: Feb 2003
Posts: 2,800
Raccoon
Hoopy frood

Hoopy frood
Joined: Feb 2003
Posts: 2,800 
$biton(A,N) Returns the A value with the Nth bit turned on. $bitoff(A,N) Returns the A value with the Nth bit turned off. $log(N) Returns the natural logarithm of N.
$biton() and $bitoff() take a decimal value (a base 10 normal number) and returns the result if binary (base 2) bit N is turned into a 1 or a 0.
For Example: base10 42 == base2 101010 To do this conversion, use $base(42,10,2) $biton(42,1) == 43 101011 $bitoff(43,1) == 42 101010 Positions are counted backwards from right to left, so 1 is the rightmost position in red.
As for $log(N), you will simply have to take Geometry, Trig or Calculus to learn that. (i forget)
 Raccoon
Well. At least I won lunch. Good philosophy, see good in bad, I like!



Re: $biton $bitoff $log
#40249
07/08/03 10:04 AM

Joined: Dec 2002
Posts: 1,321
Hammer
Hoopy frood

Hoopy frood
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Posts: 1,321 
$biton() and $bitoff() require a bit of knowledge of binary (but submitted in decimal).
0000 = 0 0100 = 4 1000 = 8 1100 = 12
0001 = 1 0101 = 5 1001 = 9 1101 = 13
0010 = 2 0110 = 6 1010 = 10 1110 = 14
0011 = 3 0111 = 7 1011 = 11 1111 = 15 So, let's say we want to start with 3 (0011) and see what that would be if we turned ON the 4th bit. //echo a * $biton(3, 4) * 11Notice that 3 is 0011 and we turned on the 4th bit, which becomes 1011, which is 11 from the chart above. If we started with 15 (1111) and wanted to see what the value would be if we turned OFF the 3rd bit: //echo a * $bitoff(15,3) * 11Starting with 1111 and turning off the 3rd bit gives 1 011, which is 11. These are basically useless as they are given here, but become very useful when you start using variables that you don't know the value ahead of time during any given execution of $biton or $bitoff. $log() is a mathematical function, just as $sin(), $cos(), $tan(), etc. are. $log(N) returns the natural logarithm of the value you give it. Assume that X raised to the Yth power equals Z. Translated into mIRC scripting, this is %Z = $calc(%X ^ %Y). But what if we wanted to find %Y if we knew %X and %Z? That's where $log() comes in. A natural logarithm uses as its base ( %X) the constant e (which is approximately equal to 2.7182818). Exponential and logarithmic functions with base e occur in many practical applications, including those involving growth and decay, continuous compounding of interest, alternating currents and learning curves. If you have need of this function, you'll know it and mIRC provides a way to use it; if you don't, you can safely ignore it.
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Re: $biton $bitoff $log
#40250
07/08/03 10:10 AM

Joined: Dec 2002
Posts: 3,547
SladeKraven
Hoopy frood

Hoopy frood
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Posts: 3,547 
Anything else you want to add to that Hammer? Got enough information there? LOL..



Re: $biton $bitoff $log
#40251
07/08/03 11:05 AM

Joined: Dec 2002
Posts: 1,321
Hammer
Hoopy frood

Hoopy frood
Joined: Dec 2002
Posts: 1,321 
Well, I had intended to delve into using & (bitwise comparison) for use in IF conditions to check to see if a bit is turned on, as well as $and(), $or() and $xor(), but since that information wasn't requested, I didn't go into them. However, for completeness' sake, I will do so now.
For an AND result to be true, both inputs (or bits) must be true (1). Otherwise, the result is false (0). This is useful in checking if a bit is turned on (all bits 0 except the bit you want to check  if you get a nonzero answer back, then the bit was on). It's also the way you turn off a bit (all bits 1 except the bit to turn off).
1 AND 1 = 1 1 AND 0 = 0 0 AND 1 = 0 0 AND 0 = 0
For an OR result to be true, at least one input must be true. If both are 0, then the result is false. This is useful in turning on a bit (all bits 0 except the bit you want to set  if you get the same answer back, then the bit was already on).
1 OR 1 = 1 1 OR 0 = 1 0 OR 1 = 1 0 OR 0 = 0
For an XOR (eXclusive OR) result to be true, one and only one input must be true. If neither or both are true, the result is false. This is useful in toggling bit states (all bits 0 except the bit you wish to toggle).
1 XOR 1 = 0 1 XOR 0 = 1 0 XOR 1 = 1 0 XOR 0 = 0
Just to refresh your memory of the bits and their decimal values we'll be working with throughout, here they are again:
1 = 0001 2 = 0010 4 = 0100 8 = 1000
and our base number is: 5 = 0101
[color:#00007F]if (N & M)
//if (5 & 1) echo a * Bit is on  else echo a * Bit is off * Bit is on (because the 1st bit  which is worth 1 in decimal  is turned on)
//if (5 & 2) echo a * Bit is on  else echo a * Bit is off * Bit is off (because the 2nd bit  which is worth 2 in decimal  is turned off)
//if (5 & 4) echo a * Bit is on  else echo a * Bit is off * Bit is on (because the 3rd bit  which is worth 4 in decimal  is turned on)
//if (5 & 8) echo a * Bit is on  else echo a * Bit is off * Bit is off (because the 4th bit  which is worth 8 in decimal  is turned off)
$and(N,M)
//echo a * $and(5,14) * 4
//echo a * $and(5,13) * 5
//echo a * $and(5,11) * 1
//echo a * $and(5,7) * 5
$or(N,M)
//echo a * $or(5,1) * 5
//echo a * $or(5,2) * 7
//echo a * $or(5,4) * 5
//echo a * $or(5,8) * 13
$xor(N,M) Starting with 5 (0101) and toggling bits as we go:
//echo a * $xor(5,1) * 4 (0100  toggled bit 1 off)
//echo a * $xor(5,2) * 7 (0111  toggled bit 2 on)
//echo a * $xor(5,4) * 1 (0001  toggled bit 3 off)
//echo a * $xor(5,8) * 13 (1101  toggled bit 4 on)[/color]
:tongue:
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Re: $biton $bitoff $log
#40252
07/08/03 11:11 AM

Joined: Dec 2002
Posts: 3,547
SladeKraven
Hoopy frood

Hoopy frood
Joined: Dec 2002
Posts: 3,547 
Are you quite done now? LOL Nice long post(s).



Re: $biton $bitoff $log
#40253
07/08/03 01:57 PM

Joined: Feb 2003
Posts: 2,800
Raccoon
Hoopy frood

Hoopy frood
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Posts: 2,800 
Assume that X raised to the Yth power equals Z. Translated into mIRC scripting, this is %Z = $calc(%X ^ %Y). But what if we wanted to find %Y if we knew %X and %Z? That's where $log() comes in. *scratches his head* So what if I have %x ^ %y = %z, where %x = 2 and %z = 256, how do I solve for %y (8) using $log?  Raccoon
Well. At least I won lunch. Good philosophy, see good in bad, I like!



Re: $biton $bitoff $log
#40254
07/08/03 02:36 PM

Joined: Dec 2002
Posts: 1,321
Hammer
Hoopy frood

Hoopy frood
Joined: Dec 2002
Posts: 1,321 
As I said before, $log() is a natural logarithm; that means it doesn't use %x = 2, nor %x = 10, but rather %x = e...which is calculated as an infinite series, like this:
e = 1 + 1/(1!) + 1/(2!) + 1/(3!) + 1/(4!) + 1/(5!) + 1/(6!) + 1/(7!) + 1/(8!) + 1/(9!) + •••
or more simply,
e = 2.71828182845904523536028747135266249775724709369995
This is your %x (base). This number appears very, very often in calculus.
That is why $log(256) returns 5.545177 instead of 8, which would be true of a base2 logarithm. If you needed a base2 logarithm, you might be able to $com over to Excel.Application (or ADODB.Connection and use a connection string to get to an Excel worksheet!) and use Excel's =Log(256,2) to return 8.
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Re: $biton $bitoff $log
#40255
07/08/03 04:20 PM

Joined: Jul 2003
Posts: 35
root
Ameglian cow

Ameglian cow
Joined: Jul 2003
Posts: 35 
Jesus christ hammer, let me get a <expletive deleted> pen cuz i ran out of paper printing those posts. Got enough detail there? I feel like i could write a <insert word here> with that already O.o
indepth
Last edited by Hammer; 07/08/03 04:42 PM.



Re: $biton $bitoff $log
#40256
07/08/03 04:34 PM

Joined: Feb 2003
Posts: 2,800
Raccoon
Hoopy frood

Hoopy frood
Joined: Feb 2003
Posts: 2,800 
You post the word '<expletive deleted>' but censor the word 'book'?
Last edited by Hammer; 07/08/03 04:42 PM.
Well. At least I won lunch. Good philosophy, see good in bad, I like!



Re: $biton $bitoff $log
#40257
07/08/03 04:41 PM

Joined: Aug 2003
Posts: 17
Clubfoot
OP
Pikka bird

OP
Pikka bird
Joined: Aug 2003
Posts: 17 
excellent information, thanks very much guys



Re: $biton $bitoff $log
#40258
07/08/03 05:12 PM

Joined: Dec 2002
Posts: 2,809
codemastr
Hoopy frood

Hoopy frood
Joined: Dec 2002
Posts: 2,809 
Well honestly, you have a TON of great info there, which is why I was suprised to see you say use $com to use a base 2 log... There is an incredibly simple formula that allows you to take the log of any base. To take the log2(N) (log base 2 of N) you do log(N)/log(2) Or more generally logB(N) = log(N)/log(B) Thats the "change of base" formula, it's very easy to use, and you can easily make an mIRC alias for it: alias nlog { return $calc($log($1)/$log($2)) } (Note that due to mIRC's methods of rounding, it won't give 100% precise answers. But, anyway, For Racoons question, So what if I have %x ^ %y = %z, where %x = 2 and %z = 256, how do I solve for %y (8) using $log?
$calc($log(256)/$log(2)) = 8. Or, using the alias I included just $nlog(256,2) As I said, due to mIRC's rounding method it doesn't give you an exact answer, mIRC says it is 8.000001, but if you do that same expression in calc.exe, you get 8.



Re: $biton $bitoff $log
#40259
07/08/03 05:39 PM

Joined: Feb 2003
Posts: 2,800
Raccoon
Hoopy frood

Hoopy frood
Joined: Feb 2003
Posts: 2,800 
Thanks codemastr, that's actually quite handy.
As for the rounding issue, one can simply $round($calc($log($1)/$log($2)),5) or even less decimal places as desired.
However, it would be nice if Khaled added this function internally, as it does pose a problem when working with even moderately large numbers.
Well. At least I won lunch. Good philosophy, see good in bad, I like!



Re: $biton $bitoff $log
#40260
07/08/03 07:43 PM

Joined: Aug 2003
Posts: 17
Clubfoot
OP
Pikka bird

OP
Pikka bird
Joined: Aug 2003
Posts: 17 
oh interesting, $chr($bitoff($asc(x),6)) turns a letter into a capital



Re: $biton $bitoff $log
#40261
07/08/03 08:03 PM

Joined: Jan 2003
Posts: 3,012
KingTomato
Hoopy frood

Hoopy frood
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Posts: 3,012 
too bad its not failsave on any other character >:\
KingTomato



Re: $biton $bitoff $log
#40262
07/08/03 08:13 PM

Joined: Dec 2002
Posts: 2,809
codemastr
Hoopy frood

Hoopy frood
Joined: Dec 2002
Posts: 2,809 
Yeah, imho mIRC needs a way to do more accurate calculations. I wrote a little script that converts a decimal into a fraction, and it works great, so I was testing it out and I tried $dec2frac(.0001) And I got 1/9951 as a result. When I first saw it, I thought I had a bug or something, because clearly, 1/9951 is NOT .0001. But, if you go and type $calc(1/9951) you do get .0001 as a result because mIRC has such imprecise rounding. It really makes dealing with any math that requires a lot of accuracy impossible.



Re: $biton $bitoff $log
#40263
08/08/03 12:38 AM

Joined: Dec 2002
Posts: 1,321
Hammer
Hoopy frood

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Posts: 1,321 
codemastr: The reason I didn't include that formula is because I didn't know that formula. I am no mathematician; in fact, I have never used a logarithm in my life, never having ever found a need for it. The information I posted was from the research I did in order to write the post, not from any firsthand personal acquaintance with the subject itself. I am certainly glad there are folks around, such as yourself, who understand such things and can clarify any mistakes I make.
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Re: $biton $bitoff $log
#40264
08/08/03 12:41 AM

Joined: Dec 2002
Posts: 2,809
codemastr
Hoopy frood

Hoopy frood
Joined: Dec 2002
Posts: 2,809 
Heh, never used logarithms? Certainly coulda fooled me!




