Double.MinValue 字段

定义

命名空间:

System

程序集:

System.Runtime.dll

Source:

Double.cs

表示 Double 的最小可能值。 此字段为常数。

public: double MinValue = -1.7976931348623157E+308;

public const double MinValue = -1.7976931348623157E+308;

val mutable MinValue : double

Public Const MinValue As Double  = -1.7976931348623157E+308

字段值

Value = -1.7976931348623157E+308

Double

示例

下面的代码示例演示如何使用 Double.MinValue：

public ref class Temperature
{
public:
   static property double MinValue 
   {
      double get()
      {
         return Double::MinValue;
      }
   }

   static property double MaxValue 
   {
      double get()
      {
         return Double::MaxValue;
      }
   }

protected:
   // The value holder
   double m_value;

public:
   property double Value 
   {
      double get()
      {
         return m_value;
      }
      void set( double value )
      {
         m_value = value;
      }
   }

   property double Celsius 
   {
      double get()
      {
         return (m_value - 32.0) / 1.8;
      }
      void set( double value )
      {
         m_value = 1.8 * value + 32.0;
      }
   }
};

public class Temperature {
    public static double MinValue {
        get {
            return Double.MinValue;
        }
    }

    public static double MaxValue {
        get {
            return Double.MaxValue;
        }
    }

    // The value holder
    protected double m_value;

    public double Value {
        get {
            return m_value;
        }
        set {
            m_value = value;
        }
    }

    public double Celsius {
        get {
            return (m_value-32.0)/1.8;
        }
        set {
            m_value = 1.8*value+32.0;
        }
    }
}

type Temperature() =
    static member MinValue =
        Double.MinValue

    static member MaxValue =
        Double.MaxValue

    member val Value = 0. with get, set

    member this.Celsius
        with get () =
            (this.Value - 32.) / 1.8
        and set (value) =
            this.Value <- 1.8 * value + 32.

Public Class Temperature

    Public Shared ReadOnly Property MinValue() As Double
        Get
            Return Double.MinValue
        End Get
    End Property

    Public Shared ReadOnly Property MaxValue() As Double
        Get
            Return Double.MaxValue
        End Get
    End Property

    ' The value holder
    Protected m_value As Double

    Public Property Value() As Double
        Get
            Return m_value
        End Get
        Set(ByVal Value As Double)
            m_value = Value
        End Set
    End Property

    Public Property Celsius() As Double
        Get
            Return (m_value - 32) / 1.8
        End Get
        Set(ByVal Value As Double)
            m_value = Value * 1.8 + 32
        End Set
    End Property
End Class

注解

此常量的值为负 1.7976931348623157E+308。

小于 Double.MinValue 的操作的结果为 Double.NegativeInfinity。 在以下示例中， Double.NegativeInfinity 当结果小于 Double.MinValue时，由减法和乘法运算的结果。

using System;

public class Example
{
   public static void Main()
   {
      double result1 = -7.997e307 + -9.985e307;
      Console.WriteLine("{0} (Negative Infinity: {1})",
                        result1, Double.IsNegativeInfinity(result1));

      double result2 = -1.5935e250 * 7.948e110;
      Console.WriteLine("{0} (Negative Infinity: {1})",
                        result2, Double.IsNegativeInfinity(result2));
   }
}
// The example displays the following output:
//    -Infinity (Negative Infinity: True)
//    -Infinity (Negative Infinity: True)

open System

let result1 = -7.997e307 + -9.985e307
printfn $"{result1} (Negative Infinity: {Double.IsNegativeInfinity result1})"

let result2 = -1.5935e250 * 7.948e110
printfn $"{result2} (Negative Infinity: {Double.IsNegativeInfinity result2})"
// The example displays the following output:
//    -Infinity (Negative Infinity: True)
//    -Infinity (Negative Infinity: True)

Module Example
   Public Sub Main()
      Dim result1 As Double = -7.997e307 + -9.985e307
      Console.WriteLine("{0} (Negative Infinity: {1})", 
                        result1, Double.IsNegativeInfinity(result1))
      
      Dim result2 As Double = -1.5935e250 * 7.948e110
      Console.WriteLine("{0} (Negative Infinity: {1})", 
                        result2, Double.IsNegativeInfinity(result2))
   End Sub
End Module
' The example displays the following output:
'    -Infinity (Negative Infinity: True)
'    -Infinity (Negative Infinity: True)