Overflow error VBA - vba

Question

I have this code below, and I'm getting an overflow error at the line:
s = s + (x Mod 10) [first line in the Do Loop]
Why? I declared x and s to be of type Double. Adding two doubles, why is this not working?
Thanks for your help.
Public Sub bidon1()
Dim i As Double, x As Double, s As Double, k As Byte, h As Byte
Dim y(1 To 6) As Double
For i = 1 To 1000000
x = i ^ 3
Do
s = s + (x Mod 10)
x = x \ 10
Loop Until x = 0
If s = x Then
k = k + 1
y(k) = x
If y(6) > 0 Then
For h = 1 To 6
Debug.Print y(h)
Next
Exit Sub
End If
End If
Next
End Sub

Answer 1

The problem is that the VBA mod operator coerces its arguments to be integers (if they are not already so). It is this implicit coercion which is causing the overflow. See this question: Mod with Doubles
On Edit:
Based on your comments, you want to be able to add together the digits in a largish integer. The following function might help:
Function DigitSum(num As Variant) As Long
'Takes a variant which represents an integer type
'such as Integer, Long or Decimal
'and returns the sum of its digits
Dim sum As Long, i As Long, s As String
s = CStr(num)
For i = 1 To Len(s)
sum = sum + Val(Mid(s, i, 1))
Next i
DigitSum = sum
End Function
The following test sub shows how it can be used to correctly get the sum of the digits in 999999^3:
Sub test()
Dim x As Variant, y As Variant
Debug.Print "Naive approach: " & DigitSum(999999 ^ 3)
y = CDec(999999)
x = y * y * y
Debug.Print "CDec approach: " & DigitSum(x)
End Sub
Output:
Naive approach: 63
CDec approach: 108
Since 999999^3 = 999997000002999999, only the second result is accurate. The first result is only the sum of the digits in the string representation of the double 999999^3 = 9.99997000003E+17