Truncating Double with VBA in excel

问题:

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?

回答1:

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

回答2:

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".

回答3:

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

回答4:

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