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Convert Dec to Binary - but Ones and Zeros

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Hi All,

Trying to convert a number to binary zeros and ones but I'm getting a result I don't understand and looks more like hex than binary.

Here's my basic code:

$myNum = 11
$myNumBin = Binary($myNum)
MsgBox(0, "Binary result", $myNumBin)

What I want is "1011", what I get is 0x0B000000.


Edited by WoodGrain
Wrong syntax
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Try this:

MsgBox(0, "", DecToBase(11, 2))

Func DecToBase($iInt, $iBase) ; for bases 2 to 9
    Local $iRem, $sRet = ''
    While $iInt > $iBase -1
        $iRem = Mod($iInt, $iBase)
        $sRet = $iRem & $sRet
        $iInt = Int(($iInt - $iRem) /$iBase)
    Return $iInt & $sRet
EndFunc ;==> DecToBase

To reverse the process you can use this:

MsgBox(0, "", BaseToDec('1011', 2))

Func BaseToDec($sInt, $iBase) ; bases 2 to 9 only
    Local $iRet = 0, $iLen = StringLen($sInt)
    For $i = 1 To $iLen
        $iRet += StringMid($sInt, $i, 1) * $iBase ^ ($iLen - $i)
    Return $iRet
EndFunc ;==> BaseToDec


Edited by czardas
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It's not so complicated, but unless you need bases 11 to 15, you might as well use Hex() and Dec(). I don't have much time this morning. You could try this old code of mine:

The code I posted in this thread can also be modified to convert larger numbers (up to Int64) by replacing the operators with operator64 functions (see my signature). Currently, these conversion functions only work for doubles ~ up to 15 decimal digits.

Edited by czardas
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18 hours ago, czardas said:

$iInt = Int(($iInt - $iRem) /$iBase)

This line could be written like this with no change in output.

$iInt = Int($iInt / $iBase)


If I posted any code, assume that code was written using the latest release version unless stated otherwise. Also, if it doesn't work on XP I can't help with that because I don't have access to XP, and I'm not going to.
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@BrewManNH Well spotted! By all means change it.

Instead of that, if I implement Operator64 functions, Int() could be removed. It should not really be needed here neither, but because of rounding inaccuracies, I added it to be on the safe side. That's also why I decided on division by a real factor (subtracting the remainder first): I don't trust double precision to always return an accurate result.

It's a while since I wrote this and I'm trying to remember everything I was thinking at the time. I must have written it that way because of the aforementioned reasons.

Edited by czardas
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