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Scientific Number Conversions

Posted on 1998-07-27
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Last Modified: 2010-08-05
I need to convert a hex, octal, or binary number to DECIMAL.  How can I do this?  I know about the Hex and Oct functions, and can figure out decimal to binary, I just don't know how to get the number (string) back to decimal.  Also, if anyone has sample code for creating a scientific calculator (specifically the number conversions), I'd REALLY like to see it (and I'd pay extra points).
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Question by:regent
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Expert Comment

by:abrooke
ID: 1466873
This is the source I use to convert a hex stored as a string to a decimal long. How it's done: I iterate through the HexNumber from left to right each place has a value of 16^(total length minus current position in strength) I multiply this value times the actual value of the place 0-15. Add up the values and return the sum.

                                       Lonnie Brooke

Public Function HexToDecimal(HexNumber As String) As Long
   Dim tmpAr() As Byte
   Dim i&, j&, k&, t&, r&
   Dim l$, m$
   
   i = Len(HexNumber)
   ReDim tmpAr(i) As Byte
   j = 1
   k = i - 1
   While j < i + 1
      t = Asc(Mid(HexNumber, j, 1))
      Select Case t
         Case 65 To 70
            t = t - 55
         Case 48 To 57
            t = t - 48
      End Select
      r = r + ((16 ^ k) * t)
      k = k - 1
      j = j + 1
   Wend
   
   HexToDecimal = r
   
End Function

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by:
alamo earned 100 total points
ID: 1466874
The following function will convert a string representing a number in any base (2 through 36) to decimal. It handles illegal characters like Val(), by stopping the conversion.

Public Function ConvertToDecimal(sNumber As String, iBase As Integer) As Long
Dim lResult As Long
Dim i, iChar As Integer

    For i = 1 To Len(sNumber)
        iChar = Asc(UCase$(Mid$(sNumber, i, 1))) - 48
        If iChar > 9 Then iChar = iChar - 7
        If iChar < 0 Or iChar >= iBase Then Exit For 'invalid character
        lResult = lResult * iBase + iChar
    Next
    ConvertToDecimal = lResult
End Function

Hope this helps!
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