Comments on: Decimal To Binary

By: Chris Green

Chris Green — Tue, 02 Aug 2011 07:53:00 +0000

There appears to be a tradition of multiple posts here, so here goes, for 24 bits:

=CONCATENATE(
IF(MOD(E2,2^24)>=2^23,"1","0"),
IF(MOD(E2,2^23)>=2^22,"1","0"),
IF(MOD(E2,2^22)>=2^21,"1","0"),
IF(MOD(E2,2^21)>=2^20,"1","0"),
IF(MOD(E2,2^20)>=2^19,"1","0"),
IF(MOD(E2,2^19)>=2^18,"1","0"),
IF(MOD(E2,2^18)>=2^17,"1","0"),
IF(MOD(E2,2^17)>=2^16,"1","0"),
IF(MOD(E2,2^16)>=2^15,"1","0"),
IF(MOD(E2,2^15)>=2^14,"1","0"),
IF(MOD(E2,2^14)>=2^13,"1","0"),
IF(MOD(E2,2^13)>=2^12,"1","0"),
IF(MOD(E2,2^12)>=2^11,"1","0"),
IF(MOD(E2,2^11)>=2^10,"1","0"),
IF(MOD(E2,2^10)>=2^9,"1","0"),
IF(MOD(E2,2^9)>=2^8,"1","0"),
IF(MOD(E2,2^8)>=2^7,"1","0"),
IF(MOD(E2,2^7)>=2^6,"1","0"),
IF(MOD(E2,2^6)>=2^5,"1","0"),
IF(MOD(E2,2^5)>=2^4,"1","0"),
IF(MOD(E2,2^4)>=2^3,"1","0"),
IF(MOD(E2,2^3)>=2^2,"1","0"),
IF(MOD(E2,2^2)>=2^1,"1","0"),
IF(MOD(E2,2^1)>=2^0,"1","0"))

By: Chris Green

Chris Green — Fri, 29 Jul 2011 16:08:00 +0000

Was looking for a formula (not VBA), found this page, kept on looking, gave up looking, came up with the following instead.
Gives you 12-bit binary, no negative numbers, craps out on larger numbers.
Only have German version of xl on hand ... VERKETTEN > CONCAT, WENN > IF, REST > MOD
Example for cell E2:

=VERKETTEN(
WENN(REST(E2;4096)>=2048;"1";"0");
WENN(REST(E2;2048)>=1024;"1";"0");
WENN(REST(E2;1024)>=512;"1";"0");
WENN(REST(E2;512)>=256;"1";"0");
WENN(REST(E2;256)>=128;"1";"0");
WENN(REST(E2;128)>=64;"1";"0");
WENN(REST(E2;64)>=32;"1";"0");
WENN(REST(E2;32)>=16;"1";"0");
WENN(REST(E2;16)>=8;"1";"0");
WENN(REST(E2;8)>=4;"1";"0");
WENN(REST(E2;4)>=2;"1";"0");
WENN(REST(E2;2)>=1;"1";"0"))

By: John Davies

John Davies — Thu, 09 Jun 2011 22:06:00 +0000

This won't handle negative numbers. It will handle anything up to VBA's maximum string length, but it takes its input number and returns the transformation as strings. The "From" and "To" bases are decimal numbers. Either base can be anything from 2 to 36. I'm cutting & pasting from something else, so I hope I get all that right! Error checking is done elsewhere in my code.

Function ChangeBase(ByVal sFrom As String, _
sTo As String, _
ByVal nFrom As Long, _
ByVal nTo As Long) As Boolean

If Len(sFrom) = 0 Then
sTo = "0"
Else
Dim bNonZero As Boolean
Dim i As Long
For i = 1 To Len(sFrom)
If Mid$(sFrom, i, 1) <> "0" Then
bNonZero = True
Exit For
End If 'Mid$(sFrom, i, 1) <> "0"
Next i
If bNonZero Then
sTo = ""
Do 'Until Len(sFrom) = 0
If Left$(sFrom, 1) = "0" Then
sFrom = Right$(sFrom, Len(sFrom) - 1)
Else
Dim nCarry As Long
nCarry = 0
For i = 1 To Len(sFrom)
Dim nNum As Long
If CharToNum(Mid$(sFrom, i, 1), nNum) Then Err.Raise knCall, , ksCall
Dim n As Long
n = nFrom * nCarry + nNum
nCarry = n Mod nTo
Dim sNum As String
If NumToChar(n nTo, sNum) Then Err.Raise knCall, , ksCall
Mid$(sFrom, i, 1) = sNum
Next i
If NumToChar(nCarry, sNum) Then Err.Raise knCall, , ksCall
sTo = sNum & sTo
End If 'Left$(sFrom, 1) = "0"
Loop Until Len(sFrom) = 0
Else
sTo = sFrom 'All zeros
End If 'bNonZero
End If 'Len(sFrom) = 0

End Function

Function CharToNum(s As String, n As Long) As Boolean

If IsNumeric(s) Then
n = CLng(s)
Else
n = Asc(s) - 55
End If

End Function

Function NumToChar(n As Long, s As String) As Boolean

If n < 10 Then
s = CStr(n)
Else
s = Chr$(55 + n)
End If

End Function

By: fatbrainslim

fatbrainslim — Fri, 08 Apr 2011 14:35:00 +0000

sorry for my missed post to the wrong place, my below formular is for convert subnet mask to number of mask bit such as 255.255.255.0 to /24

=FIND("0",CONCATENATE(DEC2BIN(LEFT(c3,FIND(".",c3)-1)),DEC2BIN(MID(c3,FIND(".",c3)+1,FIND(".",c3,FIND(".",c3)+1)-FIND(".",c3)-1)),DEC2BIN(MID(c3,FIND(".",c3,FIND(".",c3)+1)+1,FIND(".",c3,FIND(".",c3,FIND(".",c3)+1)+1)-FIND(".",c3,FIND(".",c3)+1)-1)),DEC2BIN(RIGHT(c3,LEN(c3)-FIND(".",c3,FIND(".",c3,FIND(".",c3)+1)+1))),"0"))-1

By: fatbrainslim

fatbrainslim — Fri, 08 Apr 2011 14:33:00 +0000

try this....

By: Rick Rothstein (MVP - Excel)

Rick Rothstein (MVP - Excel) — Wed, 27 Oct 2010 21:50:00 +0000

Sorry, I forgot the "code tags"... I just noticed some of my lines have been corrupted. Here is my code again, but this time using the "code tags"...

Function DecToBin(ByVal DecimalIn As Variant, Optional NumberOfBits As Variant = 0) As Variant
If DecimalIn < 0 Then
' Negative values cannot be converted
DecToBin = CVErr(xlErrValue)
Exit Function
Else
DecimalIn = Int(CDec(DecimalIn))
Do While DecimalIn <> 0
DecToBin = Trim$(Str$(DecimalIn - 2 * Int(DecimalIn / 2))) & DecToBin
DecimalIn = Int(DecimalIn / 2)
Loop
If NumberOfBits > 0 Then
If Len(DecToBin) > NumberOfBits Then
' Calculated value exceeds specified bit size
DecToBin = CVErr(xlErrValue)
Else
DecToBin = Right$(String$(NumberOfBits, "0") & DecToBin, NumberOfBits)
End If
End If
End If
End Function

By: Rick Rothstein (MVP - Excel)

Rick Rothstein (MVP - Excel) — Wed, 27 Oct 2010 21:39:00 +0000

While this function (which works fine as a UDF) only handles positive values, it will return the Binary string for values up to 9999999999999999999999999999. I'm thinking that should be a large enough range to handle whatever you may want to throw at it. If the value you want to process is larger than Excel can properly handle as whole number, simply pass the value in as text. So, for the maximum value the function can handle, you would call it like this from VB...

MsgBox DecToBin("9999999999999999999999999999?)

or like this from the worksheet...

=DecToBin("9999999999999999999999999999?)

or, if A1 contains '9999999999999999999999999999 (note the leading apostrophe) or 9999999999999999999999999999 if formatted as Text, then like this...

=DecToBin(A1)

Okay, here is the function (note that the Analysis ToolPak has a similarly named function which uses a "2? to link the words "Dec" and "Bin" instead of the word "To" that my function does)...

Function DecToBin(ByVal DecimalIn As Variant, Optional NumberOfBits As Variant = 0) As Variant
If DecimalIn < 0 Then
' Negative values cannot be converted
DecToBin = CVErr(xlErrValue)
Exit Function
Else
DecimalIn = Int(CDec(DecimalIn))
Do While DecimalIn 0
DecToBin = Trim$(Str$(DecimalIn - 2 * Int(DecimalIn / 2))) & DecToBin
DecimalIn = Int(DecimalIn / 2)
Loop
If NumberOfBits > 0 Then
If Len(DecToBin) > NumberOfBits Then
' Calculated value exceeds specified bit size
DecToBin = CVErr(xlErrValue)
Else
DecToBin = Right$(String$(NumberOfBits, "0?) & DecToBin, NumberOfBits)
End If
End If
End If
End Function

By: User

User — Mon, 25 Oct 2010 16:47:00 +0000

Excellent code. Works great! Thanks.

By: xyz

xyz — Wed, 07 Jan 2009 10:33:00 +0000

Use this formula to convert the Binary(10100111000000) number in cell G19 to decimal number.

=SUMPRODUCT(2^(ROW(INDIRECT("1:"&LEN(G7)))-1),VALUE(MID(G7,LEN(G7)-ROW(INDIRECT("1:"&LEN(G7)))+1,1)))

By: GuM

GuM — Mon, 29 Sep 2008 09:20:00 +0000

another try…

Function Long2Bin(lNumber As Long, Optional lPlaces As Long = 8) As String

' converts a decimal number to a binary string

Static vDigits As Variant ' digits used in the result string

Dim lVal As Long ' temporary value

Dim sTemp As String ' to build up the result string

' init digits on first call of function

If IsEmpty(vDigits) Then vDigits = Array("0", "1")

' corrections for negative numbers

lVal = Abs(lNumber) ' always work on positive numbers

If lNumber 0) Or (lPlaces > 0)

sTemp = vDigits(lVal Mod 2) & sTemp ' add char in front of string

lPlaces = lPlaces - 1

lVal = lVal 2 ' integer division by 2

Loop

' one's complement for negative numbers

If lNumber