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Possible $calc Bug? #175023 16/04/07 02:41 AM
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CopperHead Offline OP
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//echo -a $calc(1 * (10^50)) returns 100000000000000010000000000000000000000000000000000
//echo -a $calc(2 * (10^50)) returns 200000000000000020000000000000000000000000000000000
//echo -a $calc(3 * (10^50)) returns 300000000000000020000000000000000000000000000000000
//echo -a $calc(4 * (10^50)) returns 400000000000000030000000000000000000000000000000000
//echo -a $calc(5 * (10^50)) returns 500000000000000000000000000000000000000000000000000
//echo -a $calc(6 * (10^50)) returns 600000000000000050000000000000000000000000000000000
//echo -a $calc(7 * (10^50)) returns 700000000000000090000000000000000000000000000000000
//echo -a $calc(8 * (10^50)) returns 800000000000000060000000000000000000000000000000000
//echo -a $calc(9 * (10^50)) returns 900000000000000030000000000000000000000000000000000
//echo -a $calc(10 * (10^50)) returns 999999999999999990000000000000000000000000000000000

I use mIRC v6.21 Khaled Mardam-Bey and Windows 98 second Edition.

Re: Possible $calc Bug? [Re: CopperHead] #175025 16/04/07 03:17 AM
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RusselB Offline
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While this is a bug, it's not with mIRC, but rather Windows, as the numbers that you're dealing with are so large they exceed the maximum calculation for a number using a 32 bit base (which is what Windows 95 - XP use.. NB: Excepting Windows XP Pro 64)

The largest number that works on 32 bit OS (which is what you're using) is equal to 2^56

This is, fairly obviously, going to be a smaller number than the 10^50 level that you used as your start point.

btw: 2^56 = 72,057,594,037,927,936
Seventy-two quadrillion fifty-seven trillion five hundred and ninety-four billion thirty-seven million nine hundred and twenty-seven thousand nine hundred and thirty-six.

For 64 bit OS's the number would be a lot larger, however, I don't have access to a system running a 64 bit OS at this time in order to figure it out.

Re: Possible $calc Bug? [Re: RusselB] #175032 16/04/07 10:40 AM
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Sat Offline
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Sorry, but it has nothing to do with your OS or even your CPU. I'm kind of curious about how you think a 56-bit number is the largest possible on a 32-bit base though. :p

mIRC in fact uses the 64-bit IEEE 754 'double' type to store numbers, which is integer-exact up to 2^53 or 9,007,199,254,740,992.


Saturn, QuakeNet staff
Re: Possible $calc Bug? [Re: CopperHead] #175041 16/04/07 02:03 PM
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NaquadaServ Offline
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If you really need math on numbers that large it's possible to parse the math yourself, processing only one number at a time step-by-step, although this would take more CPU that just using $calc... wink


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Re: Possible $calc Bug? [Re: Sat] #175046 16/04/07 03:52 PM
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RusselB Offline
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I got that number by using $calc(2^<N>), where I manually increased <N> until it got to the point where the number displayed via mIRC did not match that of manual calculations (pencil and paper).

Re: Possible $calc Bug? [Re: RusselB] #175049 16/04/07 06:47 PM
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damn, now that's devotion. I'd have just skipped to the next thread! lol

btk


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Re: Possible $calc Bug? [Re: billythekid] #175061 16/04/07 11:49 PM
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PhantasyX Offline
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Hurray for Windows Vista's $calc(2^1023) ? :p

Last edited by PhantasyX; 16/04/07 11:49 PM.

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Re: Possible $calc Bug? [Re: NaquadaServ] #175067 17/04/07 01:46 AM
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NaquadaServ Offline
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To clarify my last post, here is an example alias...
Code:
alias add {
  var %retval | var %stepsum | var %carry 0
  var %num1 $1 | var %num2 $2 

  while (%num1 != $null) || (%num2 != $null) {
    set %setsum $calc(%carry + $right(%num1, 1) + $right(%num2, 1))
    set %num1 $left(%num1, -1)
    set %num2 $left(%num2, -1)
    if (%setsum >= 10) { set %carry 1 | dec %setsum 10 }
    else { set %carry 0 }
    set %retval %setsum $+ %retval
  }
  if (%carry) { set %retval 1 $+ %retval }
  return %retval
}

Adds two integers: $add($str(2, 100), $str(2, 100)) returns 100 4's

Last edited by NaquadaServ; 17/04/07 02:05 AM.

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Re: Possible $calc Bug? [Re: NaquadaServ] #175119 18/04/07 01:15 AM
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CopperHead Offline OP
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How do you explain the bug //echo -a $calc(1/0) ?

Re: Possible $calc Bug? [Re: CopperHead] #175120 18/04/07 01:20 AM
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PhantasyX Offline
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Division by zero is not possible in mathematics. I believe Khaled changed the function of $calc to return 0 not to long ago.


Live to Dream & Dream for Life
Re: Possible $calc Bug? [Re: PhantasyX] #175121 18/04/07 01:28 AM
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NaquadaServ Offline
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Actually division by zero is more-or-less infinity, since you can constantly divide something into/by nothing and still have something... 'A' not divided still equals 'A', no matter how many times you don't divide it... smile


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Re: Possible $calc Bug? [Re: NaquadaServ] #175125 18/04/07 02:36 AM
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CopperHead Offline OP
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Also on another topic possible bug , I can't get $not to work correctly returns 4294967193.
1/0 on some os returns 0 and some returns 1. 1/0 is still undenfine or $null.


Here a puzzle for you.
You may have seen this little proof that 2=1:
a = x [true for some a's and x's]
a+a = a+x [add a to both sides]
2a = a+x [a+a = 2a]
2a-2x = a+x-2x [subtract 2x from both sides]
2(a-x) = a+x-2x [2a-2x = 2(a-x)]
2(a-x) = a-x [x-2x = -x]
2 = 1 [divide both sides by a-x]
You may doubt that 2=1. So, where is the mistake? Think about it

Re: Possible $calc Bug? [Re: CopperHead] #175126 18/04/07 02:44 AM
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RusselB Offline
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No thanks..I've got better things to do with my time.

Re: Possible $calc Bug? [Re: CopperHead] #175127 18/04/07 03:13 AM
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maroon Offline
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Since a = x, then a-x is zero.
So when you divide both sides by [a-x], you're dividing by zero, and the result is "undefined" (not necessarily infinity).

Re: Possible $calc Bug? [Re: CopperHead] #175142 18/04/07 07:38 AM
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Om3n Offline
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As maroon pointed out, the 'problem' with this puzzle is in the assumption that a-x=1, when infact if a=x then a-x will always be 0...

2(a-x) = a-x [x-2x = -x]...
2/a-x = a-x/a-x [divide both sides by a-x]...
2/0 = 0/0 [substitution given the proof that when a=x, a-x will always be 0]


"Allen is having a small problem and needs help adjusting his attitude" - Flutterby
Re: Possible $calc Bug? [Re: CopperHead] #175146 18/04/07 09:37 AM
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Originally Posted By: CopperHead
Also on another topic possible bug , I can't get $not to work correctly returns 4294967193.

mIRC's binary operations are 32-bit.


Saturn, QuakeNet staff
Re: Possible $calc Bug? [Re: CopperHead] #175189 18/04/07 06:54 PM
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NaquadaServ Offline
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Originally Posted By: CopperHead
1/0 on some os returns 0 and some returns 1. 1/0 is still undenfine or $null.


I don't like this at all!!!

Division by zero should not return 0, or $null, or ANYTHING. It should generate an error. It upsets me that Java allows this too.


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Re: Possible $calc Bug? [Re: NaquadaServ] #175195 18/04/07 07:47 PM
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Wims Offline
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don't like that too but it should return $null since it's impossible


Looking for a good help channel about mIRC? Check #mircscripting @ irc.swiftirc.net
Re: Possible $calc Bug? [Re: Wims] #175196 18/04/07 07:57 PM
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NaquadaServ Offline
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$null in $calc evaluates to zero, so $null and 0 would be equal. Saying $null in this case would not be a good indication of an error.

$calc(4 + $calc(1/0)) == $calc(4 + $null) == 4

Although this isn't much better....

$calc(4 + 1 / 0) == 0


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Re: Possible $calc Bug? [Re: NaquadaServ] #175216 19/04/07 01:08 AM
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genius_at_work Offline
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It seems that $calc returns 0 for all invalid calculations, not just for divide by zero errors.

//echo -a $calc(1 / 0)
//echo -a $calc(1 / !)
//echo -a $calc(1 / -)
//echo -a $calc(1 / banana)
//echo -a $calc(1 /x 1)
//echo -a $calc(1 /! 1)
etc..


although... some just return the first value and ignore the invalid parts.

//echo -a $calc(1 +d 1)
//echo -a $calc(2 +d 4)
//echo -a $calc(4 -d 1)

-genius_at_work

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