Re: decimal to binary conversion

[Dave Peterson]

You could break up the value into pieces:

=TEXT(DEC2BIN(INT(A1/512)),REPT("0",9))&TEXT(DEC2BIN(MOD(A1,512)),REPT("0",9))

This'll work until =512^2-1 (or 262,143).

Then you'd have to break it up more.

tam wrote:
>
> How do I convert a decimal number >511 to binary?

--

Dave Peterson

[Ron Rosenfeld]

On Thu, 29 Sep 2005 06:23:05 -0700, tam <t...@discussions.microsoft.com> wrote:

>How do I convert a decimal number >511 to binary?

You can use this little beauty from Harlan Grove:

========================

Pick your number apart into powers of 512. For numbers from 0 to -1+2^36,

=DEC2BIN(INT(x/2^27),9)&DEC2BIN(INT(MOD(x,2^27)/2^18),9)
&DEC2BIN(INT(MOD(x,2^18)/2^9),9)&DEC2BIN(MOD(x,2^9),9)

==============================

Or you can use a UDF. Here is a general function to convert any base to any
other base in the range stated in the UDF; and also handle decimal places:

================================
Function BaseConvert(Num, FromBase As Integer, _
ToBase As Integer, Optional DecPlace As Long) _
As String

'by Ron Rosenfeld
'Handles from base 2 to base 62 by differentiating small and capital letters

Dim LDI As Integer 'Leading Digit Index
Dim i As Integer, j As Integer
Dim Temp, Temp2
Dim Digits()
Dim r
Dim DecSep As String

DecSep = Application.International(xlDecimalSeparator)

On Error GoTo HANDLER

If FromBase > 62 Or ToBase > 62 _
Or FromBase < 2 Or ToBase < 2 Then
BaseConvert = "Base out of range"
Exit Function
End If

If InStr(1, Num, "E") And FromBase = 10 Then
Num = CDec(Num)
End If

'Convert to Base 10
LDI = InStr(1, Num, DecSep) - 2
If LDI = -2 Then LDI = Len(Num) - 1

j = LDI

Temp = Replace(Num, DecSep, "")
For i = 1 To Len(Temp)
Temp2 = Mid(Temp, i, 1)
Select Case Temp2
Case "A" To "Z"
Temp2 = Asc(Temp2) - 55
Case "a" To "z"
Temp2 = Asc(Temp2) - 61
End Select
If Temp2 >= FromBase Then
BaseConvert = "Invalid Digit"
Exit Function
End If
r = CDec(r + Temp2 * FromBase ^ j)
j = j - 1
Next i

If r <> 0 Then LDI = Fix(CDec(Log(r) / Log(ToBase)))
If r < 1 Then LDI = 0

ReDim Digits(LDI)

For i = UBound(Digits) To 0 Step -1
Digits(i) = Format(Fix(r / ToBase ^ i))
r = CDbl(r - Digits(i) * ToBase ^ i)
Select Case Digits(i)
Case 10 To 35
Digits(i) = Chr(Digits(i) + 55)
Case 36 To 62
Digits(i) = Chr(Digits(i) + 61)
End Select
Next i

Temp = StrReverse(Join(Digits, "")) 'Integer portion
ReDim Digits(DecPlace)

If r <> 0 Then
Digits(0) = DecSep
For i = 1 To UBound(Digits)
Digits(i) = Format(Fix(r / ToBase ^ -i))
r = CDec(r - Digits(i) * ToBase ^ -i)
Select Case Digits(i)
Case 10 To 35
Digits(i) = Chr(Digits(i) + 55)
Case 36 To 62
Digits(i) = Chr(Digits(i) + 61)
End Select
Next i
End If

BaseConvert = Temp & Join(Digits, "")

Exit Function
HANDLER: MsgBox ("Error: " & Err.Number & " " & Err.Description & vbLf & _
"Number being converted: " & Num)

End Function
======================

--ron

[karar...@gmail.com]

hello can you help fast ? tonight /
How do I convert a decimal number >23.125 to binary?

[GS]

> hello can you help fast ? tonight /
> How do I convert a decimal number >23.125 to binary?

''''''''''''''''''
'by Rick Rothstein
''''''''''''''''''
Function DecToBin(ByVal DecimalIn As Variant, Optional NumberOfBits As
Variant) As String
' The DecimalIn argument is limited to 79228162514264337593543950266
' (approximately 96-bits) - large numerical values must be entered
' as a String value to prevent conversion to scientific notation.

DecToBin = "": DecimalIn = CDec(DecimalIn)
Do While DecimalIn <> 0
DecToBin = Trim$(Str$(DecimalIn - 2 * Int(DecimalIn / 2))) &
DecToBin
DecimalIn = Int(DecimalIn / 2)
Loop
If Not IsMissing(NumberOfBits) Then
If Len(DecToBin) > NumberOfBits Then
DecToBin = "Error - Number too large for bit size"
Else
DecToBin = Right$(String$(NumberOfBits, "0") & DecToBin,
NumberOfBits)
End If
End If
End Function

--
Garry

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[Ron Rosenfeld]

On Wed, 27 Nov 2013 15:36:41 -0500, GS <g...@somewhere.net> wrote:

>> hello can you help fast ? tonight /
>> How do I convert a decimal number >23.125 to binary?
>
>''''''''''''''''''
>'by Rick Rothstein
>''''''''''''''''''
>Function DecToBin(ByVal DecimalIn As Variant, Optional NumberOfBits As
>Variant) As String

When I try to implement this, it only converts the integer part, not the decimal part.

I usually use the BAseChange function in xNumbers (do a web search for this free add-in), for stuff like that.

[GS]

> On Wed, 27 Nov 2013 15:36:41 -0500, GS <g...@somewhere.net> wrote:
>
>>> hello can you help fast ? tonight /
>>> How do I convert a decimal number >23.125 to binary?
>>
>> ''''''''''''''''''
>> 'by Rick Rothstein
>> ''''''''''''''''''
>> Function DecToBin(ByVal DecimalIn As Variant, Optional NumberOfBits
>> As Variant) As String
>
> When I try to implement this, it only converts the integer part, not
> the decimal part.

I seem to recall something about this, now that you mention it. I've
never used/tested it myself, though, to be honest!

>
> I usually use the BAseChange function in xNumbers (do a web search
> for this free add-in), for stuff like that.

[joeu2004]

<karar...@gmail.com> wrote:
> hello can you help fast ? tonight /
> How do I convert a decimal number >23.125 to binary?

What do you mean by "binary"?

Do you mean 10111.001 for 23.125?

If so, what do you want for 24.05?

Ostensibly, it is 11000.000011001100(1100)... (repeating the last 4 bits).
0.05 cannot be represented exactly in binary.

Or do you really mean that you want the 64-bit binary floating-point
representation?

If so, hex representation might be best; for example, &h40380CCC,CCCCCCCD
for 23.05.