本文整理汇总了VB.NET中System.Double.Epsilon字段的典型用法代码示例。如果您正苦于以下问题：VB.NET Double.Epsilon字段的具体用法？VB.NET Double.Epsilon怎么用？VB.NET Double.Epsilon使用的例子？那么恭喜您, 这里精选的字段代码示例或许可以为您提供帮助。您也可以进一步了解该字段所在类System.Double的用法示例。
在下文中一共展示了Double.Epsilon字段的3个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的VB.NET代码示例。
示例1: Example
Module Example
Public Sub Main()
Dim values() As Double = { 0, Double.Epsilon, Double.Epsilon * .5 }
For ctr As Integer = 0 To values.Length - 2
For ctr2 As Integer = ctr + 1 To values.Length - 1
Console.WriteLine("{0:r} = {1:r}: {2}", _
values(ctr), values(ctr2), _
values(ctr).Equals(values(ctr2)))
Next
Console.WriteLine()
Next
End Sub
End Module输出：
0 = 4.94065645841247E-324: False 0 = 0: True 4.94065645841247E-324 = 0: False
示例2: Example
Module Example
Public Sub Main()
Dim values() As Double = { 0.0, Double.Epsilon }
For Each value In values
Console.WriteLine(GetComponentParts(value))
Console.WriteLine()
Next
End Sub
Private Function GetComponentParts(value As Double) As String
Dim result As String = String.Format("{0:R}: ", value)
Dim indent As Integer = result.Length
' Convert the double to an 8-byte array.
Dim bytes() As Byte = BitConverter.GetBytes(value)
' Get the sign bit (byte 7, bit 7).
result += String.Format("Sign: {0}{1}",
If((bytes(7) And &H80) = &H80, "1 (-)", "0 (+)"),
vbCrLf)
' Get the exponent (byte 6 bits 4-7 to byte 7, bits 0-6)
Dim exponent As Integer = (bytes(7) And &H07F) << 4
exponent = exponent Or ((bytes(6) And &HF0) >> 4)
Dim adjustment As Integer = If(exponent <> 0, 1023, 1022)
result += String.Format("{0}Exponent: 0x{1:X4} ({1}){2}",
New String(" "c, indent), exponent - adjustment,
vbCrLf)
' Get the significand (bits 0-51)
Dim significand As Long = ((bytes(6) And &H0F) << 48)
significand = significand Or (bytes(5) << 40)
significand = significand Or (bytes(4) << 32)
significand = significand Or (bytes(3) << 24)
significand = significand Or (bytes(2) << 16)
significand = significand Or (bytes(1) << 8)
significand = significand Or bytes(0)
result += String.Format("{0}Mantissa: 0x{1:X13}{2}",
New String(" "c, indent), significand, vbCrLf)
Return result
End Function
End Module输出：
0: Sign: 0 (+) Exponent: 0xFFFFFC02 (-1022) Mantissa: 0x0000000000000 4.94065645841247E-324: Sign: 0 (+) Exponent: 0xFFFFFC02 (-1022) Mantissa: 0x0000000000001
示例3: Test
public class Test
public Shared Sub Main
Dim D As Double
D = 5
Console.WriteLine(D.Epsilon)
End Sub
End class