将单值转换为双值时出现问题。
[BitStream提供的单值是2到6位简单的十进制数字,在许多情况下,简单到0.4、0.94、0.6等。(我应该注意,我收到的文档指出它们是Floats(在Java中),据我了解,它与.NET中的Single相同。
我最终需要将这些值加倍,因为它们将用作Point3D对象(X,Y和Z)的坐标,并最终在需要Double的其他应用程序的API中使用。
但是,当我使用CDbl(valueAsSingle)或Ctype(valueAsSingle,Double)执行转换时,该数字在Double中添加了额外的小数位,排在第9位和更高的小数位。这会导致最终需要使用这些值的应用程序出现问题。
首先,我很好奇为什么会这样?其次,如果我只是将Single转换为String,然后执行Double.TryParse(valueAsString)
,可能会出现问题供参考,这是一个非常简单的示例。
Sub Main()
Dim SingleX As Single = 0.4
Dim SingleY As Single = 0.94
Dim SingleZ As Single = 0.6
Console.WriteLine(String.Concat("SX: ", SingleX, ControlChars.NewLine, "SY: ", SingleY, ControlChars.NewLine, "SZ: ", SingleZ, ControlChars.NewLine))
Dim DoubleX As Double = CDbl(SingleX)
Dim DoubleY As Double = CDbl(SingleY)
Dim DoubleZ As Double = CDbl(SingleZ)
Console.WriteLine(String.Concat("DX: ", DoubleX, ControlChars.NewLine, "DY: ", DoubleY, ControlChars.NewLine, "DZ: ", DoubleZ))
Console.ReadLine()
End Sub
其结果是
SX: 0.4
SY: 0.94
SZ: 0.6
DX: 0.400000005960464
DY: 0.939999997615814
DZ: 0.600000023841858
[好,所以有一位同事的指点,我发现这篇Wikipedia文章讨论了单精度的准确性问题。阅读时我必须承认我的眼睛有光泽,但也许您会过得更好。
我无法与您讨论具体情况,但是ToStringing / Conversioning应该不会有太大问题。另外,您也可以按照Imrans的答案进行四舍五入。
使用以下内容
Dim DoubleX As Double = Math.Round(Convert.ToDouble(SingleX),2)
Dim DoubleY As Double = Math.Round(Convert.ToDouble(SingleY),2)
Dim DoubleZ As Double = Math.Round(Convert.ToDouble(SingleZ),2)
2是表示要多少分数的整数
so
以上代码将返回:
DX: 0.4
DY: 0.94
DZ: 0.6
我假设您正在使用.net 4.0
单独保留值(抵抗Math.Round()
诱惑)并处理输出。经过多年的尝试,我以此结束(通过C#
从VB.NET
转换为http://www.developerfusion.com/tools/convert/csharp-to-vb):
<System.Runtime.CompilerServices.Extension> _
Public Shared Function Nice(x As Double, significant_digits As Integer) As String
'Check for special numbers and non-numbers
If Double.IsInfinity(x) OrElse Double.IsNaN(x) Then
Return x.ToString()
End If
' extract sign so we deal with positive numbers only
Dim sign As Integer = Math.Sign(x)
x = Math.Abs(x)
Dim fmt As String
x = Math.Round(x, 15)
If x = 0 Then
fmt = New String("#"C, significant_digits - 1)
Return String.Format("{0:0." & fmt & "}", x)
End If
' get scientific exponent, 10^3, 10^6, ...
Dim sci As Integer = CInt(Math.Floor(Math.Log(x, 10) / 3)) * 3
' biases decimal when exponent is negative
' example 0.123 shows as 0.123 instead of 123e-3
If sci<0 Then
sci += 3
End If
' scale number to exponent found
x = x * Math.Pow(10, -sci)
' find number of digits to the left of the decimal
Dim dg As Integer = CInt(Math.Floor(Math.Log(x, 10))) + 1
' adjust decimals to display
Dim decimals As Integer = Math.Min(significant_digits - dg, 15)
' format for the decimals
fmt = New String("#"C, decimals)
Dim num = Math.Round(x, decimals)
If sci = 0 Then
'no exponent
Return String.Format("{0}{1:0." & fmt & "}", If(sign < 0, "-", String.Empty), num)
End If
Return String.Format("{0}{1:0." & fmt & "}e{2}", If(sign < 0, "-", String.Empty), num, sci)
End Function
以下是一些示例:
x Nice(x,4)
0.9f 0.9
0.96666666f 0.9667
96666f 96.67e3
9666666f 9.667e6
0.939999997615814e-5f 0.0094e-3
0.939999997615814f 0.94
0.939999997615814e-5f 0.94e3
尝试一下:将“转换为字符串”从单转换为双。
Dim SingleX as single = 0.4
Dim SingleX as single = 0.94
Dim SingleX as single = 0.8
Dim DoubleX, DoubleY, DoubleZ as double
Double.TryParse(SingleX.tostring, DoubleX)
Double.TryParse(SingleY.tostring, DoubleY)
Double.TryParse(SingleZ.tostring, DoubleZ)
DoubleX = 0.4
DoubleY = 0.94
DoubleZ = 0.6