Question

如何检查VB.NET中数字的小数位数？

例如：在一个循环中我有一个if语句，在该语句中我想检查一个数字是否有四个小数位（8.9659）。

Answer 1

Dim numberAsString As String = myNumber.ToString()
Dim indexOfDecimalPoint As Integer = numberAsString.IndexOf(".")
Dim numberOfDecimals As Integer = _
    numberAsString.Substring(indexOfDecimalPoint + 1).Length

Answer 2

考虑整数值的类似方法。

Public Function NumberOfDecimalPlaces(ByVal number As Double) As Integer
    Dim numberAsString As String = number.ToString()
    Dim indexOfDecimalPoint As Integer = numberAsString.IndexOf(".")

    If indexOfDecimalPoint = -1 Then ' No decimal point in number
        Return 0
    Else
        Return numberAsString.Substring(indexOfDecimalPoint + 1).Length
    End If

End Function

Answer 3

  Public Shared Function IsInSignificantDigits(val As Double, sigDigits As Integer)
    Dim intVal As Double = val * 10 ^ sigDigits
    Return intVal = Int(intVal)
   End Function

Answer 4

对于全球化......

Public Function NumberOfDecimalPlaces(ByVal number As Double) As Integer
    Dim numberAsString As String = number.ToString(System.Globalization.CultureInfo.InvariantCulture)
    Dim indexOfDecimalPoint As Integer = numberAsString.IndexOf(".")

    If (indexOfDecimalPoint = -1) Then ' No decimal point in number
        Return 0
    Else
        Return numberAsString.Substring(indexOfDecimalPoint + 1).Length
    End If

End Function

Answer 5

此问题附带的其他一些答案建议将数字转换为字符串，然后使用“点”的字符位置作为小数位数的指示符。但这并不是一种可行的方法。如果数字有多个小数位，则会导致非常不准确的答案，并且它转换为包含指数表示法的字符串。

例如，对于公式1/11111111111111111（一个除以17个），字符串转换为“9E-17”，这意味着当它应为17时得到的答案是5.当然可以提取正确的答案。当“E-”存在时，从字符串末尾回答，但为什么在数学上可以完成呢？

这是我刚刚做的一个功能。这不是一个完美的解决方案，我没有彻底测试过，但似乎有效。

Public Function CountOfDecimalPlaces(ByVal inputNumber As Variant) As Integer
'
' This function returns the count of deciml places in a number using simple math and a loop. The
' input variable is of the Variant data type, so this function is versatile enougfh to work with
' any type of input number.
'
CountOfDecimalPlaces = 0                            'assign a default value of zero
inputNumber = VBA.CDec(inputNumber)                 'convert to Decimal for more working space
inputNumber = inputNumber - VBA.Fix(inputNumber)    'discard the digits left of the decimal
Do While inputNumber <> VBA.Int(inputNumber)        'when input = Int(input), it's done
    CountOfDecimalPlaces = CountOfDecimalPlaces + 1 'do the counting
    inputNumber = inputNumber * 10                  'move the decimal one place to the right
Loop                                                'repeat until no decimal places left
End Function

Answer 6

简单...其中n是位数

Dim n as integer = 2
Dim d as decimal = 100.123456
d = Math.Round(d, n);
MessageBox.Show(d.ToString())

响应：100.12

VB.NET中的小数位数

6 个答案: