I need to truncate the amount of decimal places of my double value for display in a textbox. How would one achieve this with vba?
If you want to round the value, then you can use the Round function (but be aware that VBA's Round function uses Banker's rounding, also known as round-to-even, where it will round a 5 up or down; to round using traditional rounding, use Format).
If you want to truncate the value without rounding, then there's no need to use strings as in the accepted answer - just use math:
Dim lDecimalPlaces As Long: lDecimalPlaces = 2
Dim dblValue As Double: dblValue = 2.345
Dim lScale = 10 ^ lDecimalPlaces
Dim dblTruncated As Double: dblTruncated = Fix(dblValue * lScale) / lScale
This yields "2.34".
You can use Int() function. Debug.print Int(1.99543)
Or Better:
Public Function Trunc(ByVal value As Double, ByVal num As Integer) As Double
Trunc = Int(value * (10 ^ num)) / (10 ^ num)
End Function
So you can use Trunc(1.99543, 4)
==> result: 1.9954
You can either use ROUND
for FORMAT
in VBA
For example to show 2 decimal places
Dval = 1.56789
Debug.Print Round(dVal,2)
Debug.Print Format(dVal,"0.00")
Note: The above will give you 1.57
. So if you are looking for 1.56
then you can store the Dval in a string and then do this
Dim strVal As String
dVal = 1.56789
strVal = dVal
If InStr(1, strVal, ".") Then
Debug.Print Split(strVal, ".")(0) & "." & Left(Split(strVal, ".")(1), 2)
Else
Debug.Print dVal
End If
So fun story. I was messing around with a quick VB conversion function. I just want to truncate a double into an integer.
value = Int(83.768)
value == 83
Awesome, something in VB actually worked.
Oops, I forgot it doesn't work with negative numbers
value = Int(-83.768)
value == -84
... yeah that just happened. VB uses Banker rounding.
Public Function Trunc(ByVal value As Double) As Integer
' Truncate by calling Int on the Absolute value then multiply by the sign of the value.
' Int cannot truncate doubles that are negative
Trunc = (Abs(value) / value) * Int(Abs(value))
End Function
If you want specific decimal places do what Makah did only with Abs around the value so Int can truncate properly.
Public Function Trunc2(ByVal value As Double, Optional ByVal num As Integer = 1) As Double
' Truncate by calling Int on the Absolute value then multiply by the sign of the value.
' Int cannot truncate doubles that are negative
Dim sign As Integer
sign = Abs(value) / value
Trunc2 = sign * (Int(Abs(value) * (10 ^ num)) / (10 ^ num))
End Function