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VB Significant Digits function

Does anyone have a visual basic function that will return a passed number with a requested number of significant digits?
For example:
X = 1.233412
Y = FixSigFigs(X,3) 'I want Y to have 3 significant digits.

Function FixSigFigs( NumberToFix as double, SigFigs as Integer)

0
kar8942
Asked:
kar8942
3 Solutions
 
AlexFMCommented:
Round Function

Returns a number rounded to a specified number of decimal places.
Round(expression [,numdecimalplaces])

Dim MyVar, pi
pi = 3.14159
MyVar = Round(pi, 2) ' MyVar contains 3.14.

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Mike TomlinsonMiddle School Assistant TeacherCommented:
Public Function FixSigFigs(NumberToFix As Double, SigFigs As Integer) As Double
    Dim textNumber As String
    Dim decimalPlace As Integer
    Dim digitsToRight As Integer
   
    FixSigFigs = NumberToFix ' If no change, return original number
    textNumber = CStr(NumberToFix)
    decimalPlace = InStr(textNumber, ".")
    If decimalPlace > 0 Then
        digitsToRight = Len(textNumber) - decimalPlace
        If digitsToRight > SigFigs And SigFigs >= 0 Then
            FixSigFigs = CDbl(Left(textNumber, decimalPlace + SigFigs))
        End If
    End If
End Function
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Mike TomlinsonMiddle School Assistant TeacherCommented:
You didn't specify as to whether you wanted rounding or not.  Use the Round() function, as AlexFM suggested, if you need rounding to occur.  Otherwise, use my function if you simply want to truncate the number to the requested number of digits.
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VKCommented:
hello kar8942 !

That isn't a simple task in general.
Of course you can use the built in round function of vb.
But this function isn't country-independent.

Example:

? round(1.145,2)
 1,14
? round(1.245,2)
 1,25

In germany for example round(1.145,2) should be 1.15.
Therefore i wrote a user defined function:

Public Function RoundIt(Number As Variant, ByVal NumDigitsAfterDecimal As Long) As Variant
    Select Case True
        Case Not IsNumeric(Number)
            RoundIt = Number
        Case Else
            NumDigitsAfterDecimal = 10 ^ NumDigitsAfterDecimal
            RoundIt = Int(CDec(CDec(0.5) + CDec(Number) * CDec(NumDigitsAfterDecimal))) / CDec(NumDigitsAfterDecimal)
    End Select
End Function

regards

V.K.
0
 
Mike TomlinsonMiddle School Assistant TeacherCommented:
What did you decide to use?
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AlexFMCommented:
kar8942 got two ways: rounding and truncating. He should decide what he need. Without his response I think points should be splitted.
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kar8942Author Commented:
Neither answer actually worked, but I will split the points anyway. Thanks.
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