Luhn Algorithm in VBA (for example for use in Excel)

gistfile1.vb

' Version 1.0.0
' You may use these functions directly in Excel: "=luhnCheck(A55)"


' probably only needed internally
Function luhnSum(InVal As String) As Integer
    Dim evenSum As Integer
    Dim oddSum As Integer
         
    evenSum = 0
    oddSum = 0
     
    Dim strLen As Integer
    strLen = Len(InVal)
     
    Dim i As Integer
    For i = strLen To 1 Step -1
        Dim digit As Integer
        digit = CInt(Mid(InVal, i, 1))
         
        If ((i Mod 2) = 0) Then
            oddSum = oddSum + digit
        Else
            digit = digit * 2
             
            If (digit > 9) Then
                digit = digit - 9
            End If
             
            evenSum = evenSum + digit
        End If
    Next i
     
    luhnSum = (oddSum + evenSum)
End Function

' for the curious
Function luhnCheckSum(InVal As String)
    luhnCheckSum = luhnSum(InVal) Mod 10
End Function

' true/false check
Function luhnCheck(InVal As String)
    luhnCheck = (luhnSum(InVal) Mod 10) = 0
End Function

' returns a number which, appended to the InVal, turns the composed number into a valid luhn number
Function luhnNext(InVal As String)
    Dim luhnCheckSumRes
    luhnCheckSumRes = luhnCheckSum(InVal)
     
    If (luhnCheckSumRes = 0) Then
        luhnNext = 0
    Else
        luhnNext = ((10 - luhnCheckSumRes))
    End If
End Function

Hello,
Using Excel 2016, Getting #VALUE! with Function luhnNext(InVal As String),
after I change the code to luhnCheckSumRes = luhnCheckSum(InVal*10) (suggested by Voidfae) and result of InVal*10 has more than 15 digits

I'm new to Excel & VBA & programming but a google search shows me that's because of floating point?

After a few try, since Function luhnNext(InVal As String) is As String,
instead of change the line to luhnCheckSumRes = luhnCheckSum(InVal*10),
I'm keeping it untouched, but add a new InVal = InVal + "0" line above that, seems working now.
Is this the right way to do it?

The doubling is alternating, so why not just use a flag to alternate. This way it doesn't matter if the length of InVal is even or odd

    Dim i As Integer
    Dim ToDouble As Boolean
    For i = strLen To 1 Step -1
        Dim digit As Integer
        digit = CInt(Mid(InVal, i, 1))

        ToDouble  = Not ToDouble 
 
        If Not ToDouble Then
            oddSum = oddSum + digit
        Else
            digit = digit * 2