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$biton $bitoff $log #40247 07/08/03 07:56 AM
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Clubfoot Offline OP
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could anyone explain to me in any amount of detail how these identifiers work?

Re: $biton $bitoff $log #40248 07/08/03 08:57 AM
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Raccoon Offline
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$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 right-most 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 $log #40249 07/08/03 10:04 AM
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Hammer Offline
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$biton() and $bitoff() require a bit of knowledge of binary (but submitted in decimal).
Code:

  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)
* 11


Notice 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)
* 11


Starting with 1111 and turning off the 3rd bit gives 1011, 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. laugh


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Re: $biton $bitoff $log #40250 07/08/03 10:10 AM
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SladeKraven Offline
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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
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Hammer Offline
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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 non-zero 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
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SladeKraven Offline
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Are you quite done now? LOL grin
Nice long post(s).

Re: $biton $bitoff $log #40253 07/08/03 01:57 PM
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Raccoon Offline
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Quote:
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
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Hammer Offline
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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 base-2 logarithm. If you needed a base-2 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
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root Offline
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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
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Raccoon Offline
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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
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Clubfoot Offline OP
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excellent information, thanks very much guys

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

Quote:

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
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Raccoon Offline
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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
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Clubfoot Offline OP
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oh interesting, $chr($bitoff($asc(x),6)) turns a letter into a capital

Re: $biton $bitoff $log #40261 07/08/03 08:03 PM
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KingTomato Offline
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too bad its not failsave on any other character >:\


-KingTomato
Re: $biton $bitoff $log #40262 07/08/03 08:13 PM
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codemastr Offline
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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
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Hammer Offline
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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 first-hand 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. laugh


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Re: $biton $bitoff $log #40264 08/08/03 12:41 AM
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codemastr Offline
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Heh, never used logarithms? Certainly coulda fooled me! smile