Double 结构

定义

表示一个双精度浮点数。Represents a double-precision floating-point number.

public value class Double : IComparable, IComparable<double>, IConvertible, IEquatable<double>, IFormattable
[System.Runtime.InteropServices.ComVisible(true)]
[System.Serializable]
public struct Double : IComparable, IComparable<double>, IConvertible, IEquatable<double>, IFormattable
type double = struct
    interface IFormattable
    interface IConvertible
Public Structure Double
Implements IComparable, IComparable(Of Double), IConvertible, IEquatable(Of Double), IFormattable
继承
Double
属性
实现

示例

下面的代码示例演示如何使用 DoubleThe following code example illustrates the use of Double:

// The Temperature class stores the temperature as a Double
// and delegates most of the functionality to the Double 
// implementation.
public ref class Temperature: public IComparable, public IFormattable
{
   // IComparable.CompareTo implementation.
public:
   virtual int CompareTo( Object^ obj )
   {
      if (obj == nullptr) return 1;
      
      if (dynamic_cast<Temperature^>(obj) )
      {
         Temperature^ temp = (Temperature^)(obj);
         return m_value.CompareTo( temp->m_value );
      }
      throw gcnew ArgumentException( "object is not a Temperature" );
   }

   // IFormattable.ToString implementation.
   virtual String^ ToString( String^ format, IFormatProvider^ provider )
   {
      if ( format != nullptr )
      {
         if ( format->Equals( "F" ) )
         {
            return String::Format( "{0}'F", this->Value.ToString() );
         }

         if ( format->Equals( "C" ) )
         {
            return String::Format( "{0}'C", this->Celsius.ToString() );
         }
      }
      return m_value.ToString( format, provider );
   }

   // Parses the temperature from a string in the form
   // [ws][sign]digits['F|'C][ws]
   static Temperature^ Parse( String^ s, NumberStyles styles, IFormatProvider^ provider )
   {
      Temperature^ temp = gcnew Temperature;

      if ( s->TrimEnd(nullptr)->EndsWith( "'F" ) )
      {
         temp->Value = Double::Parse( s->Remove( s->LastIndexOf( '\'' ), 2 ), styles, provider );
      }
      else
      if ( s->TrimEnd(nullptr)->EndsWith( "'C" ) )
      {
         temp->Celsius = Double::Parse( s->Remove( s->LastIndexOf( '\'' ), 2 ), styles, provider );
      }
      else
      {
         temp->Value = Double::Parse( s, styles, provider );
      }
      return temp;
   }

protected:
   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;
      }
   }
};
// The Temperature class stores the temperature as a Double
// and delegates most of the functionality to the Double
// implementation.
public class Temperature : IComparable, IFormattable 
{
    // IComparable.CompareTo implementation.
    public int CompareTo(object obj) {
        if (obj == null) return 1;
        
        Temperature temp = obj as Temperature;
        if (obj != null) 
            return m_value.CompareTo(temp.m_value);
        else
            throw new ArgumentException("object is not a Temperature");	
    }

    // IFormattable.ToString implementation.
    public string ToString(string format, IFormatProvider provider) {
        if( format != null ) {
            if( format.Equals("F") ) {
                return String.Format("{0}'F", this.Value.ToString());
            }
            if( format.Equals("C") ) {
                return String.Format("{0}'C", this.Celsius.ToString());
            }
        }

        return m_value.ToString(format, provider);
    }

    // Parses the temperature from a string in the form
    // [ws][sign]digits['F|'C][ws]
    public static Temperature Parse(string s, NumberStyles styles, IFormatProvider provider) {
        Temperature temp = new Temperature();

        if( s.TrimEnd(null).EndsWith("'F") ) {
            temp.Value = Double.Parse( s.Remove(s.LastIndexOf('\''), 2), styles, provider);
        }
        else if( s.TrimEnd(null).EndsWith("'C") ) {
            temp.Celsius = Double.Parse( s.Remove(s.LastIndexOf('\''), 2), styles, provider);
        }
        else {
            temp.Value = Double.Parse(s, styles, provider);
        }

        return temp;
    }

    // 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;
        }
    }
}
' Temperature class stores the value as Double
' and delegates most of the functionality 
' to the Double implementation.
Public Class Temperature
    Implements IComparable, IFormattable

    Public Overloads Function CompareTo(ByVal obj As Object) As Integer _
        Implements IComparable.CompareTo

        If TypeOf obj Is Temperature Then
            Dim temp As Temperature = CType(obj, Temperature)

            Return m_value.CompareTo(temp.m_value)
        End If

        Throw New ArgumentException("object is not a Temperature")
    End Function

    Public Overloads Function ToString(ByVal format As String, ByVal provider As IFormatProvider) As String _
        Implements IFormattable.ToString

        If Not (format Is Nothing) Then
            If format.Equals("F") Then
                Return [String].Format("{0}'F", Me.Value.ToString())
            End If
            If format.Equals("C") Then
                Return [String].Format("{0}'C", Me.Celsius.ToString())
            End If
        End If

        Return m_value.ToString(format, provider)
    End Function

    ' Parses the temperature from a string in form
    ' [ws][sign]digits['F|'C][ws]
    Public Shared Function Parse(ByVal s As String, ByVal styles As NumberStyles, ByVal provider As IFormatProvider) As Temperature
        Dim temp As New Temperature()

        If s.TrimEnd(Nothing).EndsWith("'F") Then
            temp.Value = Double.Parse(s.Remove(s.LastIndexOf("'"c), 2), styles, provider)
        Else
            If s.TrimEnd(Nothing).EndsWith("'C") Then
                temp.Celsius = Double.Parse(s.Remove(s.LastIndexOf("'"c), 2), styles, provider)
            Else
                temp.Value = Double.Parse(s, styles, provider)
            End If
        End If
        Return temp
    End Function

    ' 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

@No__t 值类型表示双精度64位数字,其值范围从负 1.79769313486232 e 308 到正 1.79769313486232 e 308,正值或负零,PositiveInfinityNegativeInfinity,而不是数字(NaN)。The Double value type represents a double-precision 64-bit number with values ranging from negative 1.79769313486232e308 to positive 1.79769313486232e308, as well as positive or negative zero, PositiveInfinity, NegativeInfinity, and not a number (NaN). 它用于表示非常大的值(例如行星或 galaxies 之间的距离)或极小的值(以千克为间隔),并且通常不精确(如从地球到另一颗太阳系的距离),Double 类型符合二元浮点算法的 IEC 60559:1989 (IEEE 754)标准。It is intended to represent values that are extremely large (such as distances between planets or galaxies) or extremely small (the molecular mass of a substance in kilograms) and that often are imprecise (such as the distance from earth to another solar system), The Double type complies with the IEC 60559:1989 (IEEE 754) standard for binary floating-point arithmetic.

本主题包括以下各节:This topic consists of the following sections:

浮点表示形式和精度Floating-Point Representation and Precision

@No__t-0 数据类型以64位二进制格式存储双精度浮点值,如下表所示:The Double data type stores double-precision floating-point values in a 64-bit binary format, as shown in the following table:

部件Part BitsBits
有效位数或尾数Significand or mantissa 0-510-51
ExponentExponent 52-6252-62
Sign (0 = 正,1 = 负值)Sign (0 = Positive, 1 = Negative) 6363

正如小数部分无法精确表示某些小数值(如1/3 或 Math.PI),二进制小数无法表示某些小数值。Just as decimal fractions are unable to precisely represent some fractional values (such as 1/3 or Math.PI), binary fractions are unable to represent some fractional values. 例如,1/10 以小数部分的形式精确表示为小数部分,以. 001100110011 表示为二进制小数,模式 "myhpccn-0011" 重复到无限大。For example, 1/10, which is represented precisely by .1 as a decimal fraction, is represented by .001100110011 as a binary fraction, with the pattern "0011" repeating to infinity. 在这种情况下,浮点值提供它所表示的数字的不精确表示形式。In this case, the floating-point value provides an imprecise representation of the number that it represents. 对原始浮点值执行其他数学运算通常会增加其精度。Performing additional mathematical operations on the original floating-point value often tends to increase its lack of precision. 例如,如果将0.1 乘以10并将0.1 添加到 1 9 次,则会看到加法,因为它涉及到8个以上的操作,因此产生的结果不太精确。For example, if we compare the result of multiplying .1 by 10 and adding .1 to .1 nine times, we see that addition, because it has involved eight more operations, has produced the less precise result. 请注意,仅当使用 "R"标准数字格式字符串显示两个 @no__t 0 个值时,此差异才明显,如有必要显示 @no__t 2 类型支持的所有17位精度。Note that this disparity is apparent only if we display the two Double values by using the "R" standard numeric format string, which if necessary displays all 17 digits of precision supported by the Double type.

using System;

public class Example
{
   public static void Main()
   {
      Double value = .1;
      Double result1 = value * 10;
      Double result2 = 0;
      for (int ctr = 1; ctr <= 10; ctr++)
         result2 += value;

      Console.WriteLine(".1 * 10:           {0:R}", result1);
      Console.WriteLine(".1 Added 10 times: {0:R}", result2);
   }
}
// The example displays the following output:
//       .1 * 10:           1
//       .1 Added 10 times: 0.99999999999999989
Module Example
   Public Sub Main()
      Dim value As Double = .1
      Dim result1 As Double = value * 10
      Dim result2 As Double
      For ctr As Integer = 1 To 10
         result2 += value
      Next
      Console.WriteLine(".1 * 10:           {0:R}", result1)
      Console.WriteLine(".1 Added 10 times: {0:R}", result2)
   End Sub
End Module
' The example displays the following output:
'       .1 * 10:           1
'       .1 Added 10 times: 0.99999999999999989

由于某些数字不能精确表示为小数部分的二进制值,因此浮点数只能近似于实数。Because some numbers cannot be represented exactly as fractional binary values, floating-point numbers can only approximate real numbers.

所有浮点数还具有有限数量的有效数字,还决定了浮点值接近于实数的准确程度。All floating-point numbers also have a limited number of significant digits, which also determines how accurately a floating-point value approximates a real number. @No__t 值最多可包含15个小数位数的精度,尽管它在内部维护最大17位数字。A Double value has up to 15 decimal digits of precision, although a maximum of 17 digits is maintained internally. 这意味着,某些浮点运算可能缺乏更改浮点值的精度。This means that some floating-point operations may lack the precision to change a floating point value. 下面的示例进行了这方面的演示。The following example provides an illustration. 它定义了一个非常大的浮点值,然后将 @no__t 的产品和一个千万亿添加到其中。It defines a very large floating-point value, and then adds the product of Double.Epsilon and one quadrillion to it. 然而,该产品太小,不能修改原始浮点值。The product, however, is too small to modify the original floating-point value. 它的最小有效位数为千分之几,而产品中最重要的位是 10-309Its least significant digit is thousandths, whereas the most significant digit in the product is 10-309.

using System;

public class Example
{
   public static void Main()
   {
      Double value = 123456789012.34567;
      Double additional = Double.Epsilon * 1e15;
      Console.WriteLine("{0} + {1} = {2}", value, additional, 
                                           value + additional);
   }
}
// The example displays the following output:
//    123456789012.346 + 4.94065645841247E-309 = 123456789012.346
Module Example
   Public Sub Main()
      Dim value As Double = 123456789012.34567
      Dim additional As Double = Double.Epsilon * 1e15
      Console.WriteLine("{0} + {1} = {2}", value, additional, 
                                           value + additional)
   End Sub
End Module
' The example displays the following output:
'   123456789012.346 + 4.94065645841247E-309 = 123456789012.346

浮点数的有限精度有几个后果:The limited precision of a floating-point number has several consequences:

  • 对于特定精度,看起来相等的两个浮点数可能不相等,因为其最小有效位不同。Two floating-point numbers that appear equal for a particular precision might not compare equal because their least significant digits are different. 在下面的示例中,将一系列数字添加在一起,并将其总计与预期的总数进行比较。In the following example, a series of numbers are added together, and their total is compared with their expected total. 尽管这两个值看起来是相同的,但对 Equals 方法的调用表示它们不是。Although the two values appear to be the same, a call to the Equals method indicates that they are not.

    using System;
    
    public class Example
    {
       public static void Main()
       {
          Double[] values = { 10.0, 2.88, 2.88, 2.88, 9.0 };
          Double result = 27.64;
          Double total = 0;
          foreach (var value in values)
             total += value;
    
          if (total.Equals(result))
             Console.WriteLine("The sum of the values equals the total.");
          else
             Console.WriteLine("The sum of the values ({0}) does not equal the total ({1}).",
                               total, result); 
       }
    }
    // The example displays the following output:
    //      The sum of the values (36.64) does not equal the total (36.64).   
    //
    // If the index items in the Console.WriteLine statement are changed to {0:R},
    // the example displays the following output:
    //       The sum of the values (27.639999999999997) does not equal the total (27.64).   
    
    Module Example
       Public Sub Main()
          Dim values() As Double = { 10.0, 2.88, 2.88, 2.88, 9.0 }
          Dim result As Double = 27.64
          Dim total As Double
          For Each value In values
             total += value
          Next
          If total.Equals(result) Then
             Console.WriteLine("The sum of the values equals the total.")
          Else
             Console.WriteLine("The sum of the values ({0}) does not equal the total ({1}).",
                               total, result) 
          End If     
       End Sub
    End Module
    ' The example displays the following output:
    '      The sum of the values (36.64) does not equal the total (36.64).   
    '
    ' If the index items in the Console.WriteLine statement are changed to {0:R},
    ' the example displays the following output:
    '       The sum of the values (27.639999999999997) does not equal the total (27.64).   
    

    如果将 Console.WriteLine(String, Object, Object) 语句中的格式项从 {0}{1} 更改为 {0:R},并使用 {1:R} 显示两个 @no__t 5 值的所有有效数字,则可以清楚地表明这两个值不相等,因为在添加运算符.If you change the format items in the Console.WriteLine(String, Object, Object) statement from {0} and {1} to {0:R} and {1:R} to display all significant digits of the two Double values, it is clear that the two values are unequal because of a loss of precision during the addition operations. 在这种情况下,可以通过调用 Math.Round(Double, Int32) 方法在执行比较之前将 @no__t 1 值舍入到所需精度来解决此问题。In this case, the issue can be resolved by calling the Math.Round(Double, Int32) method to round the Double values to the desired precision before performing the comparison.

  • 如果使用的是十进制数,则使用浮点数的算术或比较运算可能不会产生相同的结果,因为二进制浮点数可能不等于十进制数。A mathematical or comparison operation that uses a floating-point number might not yield the same result if a decimal number is used, because the binary floating-point number might not equal the decimal number. 前面的示例通过显示乘以0.1 乘以10的结果,再加上1次,阐释了这一点。A previous example illustrated this by displaying the result of multiplying .1 by 10 and adding .1 times.

    当包含小数值的数值操作的准确性非常重要时,可以使用 Decimal 而不是 @no__t 类型。When accuracy in numeric operations with fractional values is important, you can use the Decimal rather than the Double type. 当整数值超出 Int64 或 @no__t 类型范围的数值操作的准确性很重要时,请使用 BigInteger 类型。When accuracy in numeric operations with integral values beyond the range of the Int64 or UInt64 types is important, use the BigInteger type.

  • 如果涉及浮点数,值可能不会往返。A value might not round-trip if a floating-point number is involved. 如果某个操作将原始浮点数转换为另一种格式,则会将值转换为舍入,而反向运算会将转换后的窗体转换回浮点数,并且最终浮点数不等于原始的浮点数。A value is said to round-trip if an operation converts an original floating-point number to another form, an inverse operation transforms the converted form back to a floating-point number, and the final floating-point number is not equal to the original floating-point number. 往返过程可能会失败,因为在转换过程中一个或多个最小有效位会丢失或更改。The round trip might fail because one or more least significant digits are lost or changed in a conversion. 在下面的示例中,三个 @no__t 0 值将转换为字符串,并保存在一个文件中。In the following example, three Double values are converted to strings and saved in a file. 但在输出中,即使值看起来相同,还原的值也不等于原始值。As the output shows, however, even though the values appear to be identical, the restored values are not equal to the original values.

    using System;
    using System.IO;
    
    public class Example
    {
       public static void Main()
       {
          StreamWriter sw = new StreamWriter(@".\Doubles.dat");
          Double[] values = { 2.2/1.01, 1.0/3, Math.PI };
          for (int ctr = 0; ctr < values.Length; ctr++) {
             sw.Write(values[ctr].ToString());
             if (ctr != values.Length - 1)
                sw.Write("|");
          }      
          sw.Close();
          
          Double[] restoredValues = new Double[values.Length];
          StreamReader sr = new StreamReader(@".\Doubles.dat");
          string temp = sr.ReadToEnd();
          string[] tempStrings = temp.Split('|');
          for (int ctr = 0; ctr < tempStrings.Length; ctr++)
             restoredValues[ctr] = Double.Parse(tempStrings[ctr]);   
    
    
          for (int ctr = 0; ctr < values.Length; ctr++)
             Console.WriteLine("{0} {2} {1}", values[ctr], 
                               restoredValues[ctr],
                               values[ctr].Equals(restoredValues[ctr]) ? "=" : "<>");
       }
    }
    // The example displays the following output:
    //       2.17821782178218 <> 2.17821782178218
    //       0.333333333333333 <> 0.333333333333333
    //       3.14159265358979 <> 3.14159265358979
    
    Imports System.IO
    
    Module Example
       Public Sub Main()
          Dim sw As New StreamWriter(".\Doubles.dat")
          Dim values() As Double = { 2.2/1.01, 1.0/3, Math.PI }
          For ctr As Integer = 0 To values.Length - 1
             sw.Write(values(ctr).ToString())
             If ctr <> values.Length - 1 Then sw.Write("|")
          Next      
          sw.Close()
          
          Dim restoredValues(values.Length - 1) As Double
          Dim sr As New StreamReader(".\Doubles.dat")
          Dim temp As String = sr.ReadToEnd()
          Dim tempStrings() As String = temp.Split("|"c)
          For ctr As Integer = 0 To tempStrings.Length - 1
             restoredValues(ctr) = Double.Parse(tempStrings(ctr))   
          Next 
    
          For ctr As Integer = 0 To values.Length - 1
             Console.WriteLine("{0} {2} {1}", values(ctr), 
                               restoredValues(ctr),
                               If(values(ctr).Equals(restoredValues(ctr)), "=", "<>"))
          Next
       End Sub
    End Module
    ' The example displays the following output:
    '       2.17821782178218 <> 2.17821782178218
    '       0.333333333333333 <> 0.333333333333333
    '       3.14159265358979 <> 3.14159265358979
    

    在这种情况下,可以使用 "G17"标准数字格式字符串将值成功舍入,以保留 @no__t 1 值的完整精度,如下面的示例所示。In this case, the values can be successfully round-tripped by using the "G17" standard numeric format string to preserve the full precision of Double values, as the following example shows.

    using System;
    using System.IO;
    
    public class Example
    {
       public static void Main()
       {
          StreamWriter sw = new StreamWriter(@".\Doubles.dat");
          Double[] values = { 2.2/1.01, 1.0/3, Math.PI };
          for (int ctr = 0; ctr < values.Length; ctr++) 
             sw.Write("{0:G17}{1}", values[ctr], ctr < values.Length - 1 ? "|" : "" );
    
          sw.Close();
          
          Double[] restoredValues = new Double[values.Length];
          StreamReader sr = new StreamReader(@".\Doubles.dat");
          string temp = sr.ReadToEnd();
          string[] tempStrings = temp.Split('|');
          for (int ctr = 0; ctr < tempStrings.Length; ctr++)
             restoredValues[ctr] = Double.Parse(tempStrings[ctr]);   
    
    
          for (int ctr = 0; ctr < values.Length; ctr++)
             Console.WriteLine("{0} {2} {1}", values[ctr], 
                               restoredValues[ctr],
                               values[ctr].Equals(restoredValues[ctr]) ? "=" : "<>");
       }
    }
    // The example displays the following output:
    //       2.17821782178218 = 2.17821782178218
    //       0.333333333333333 = 0.333333333333333
    //       3.14159265358979 = 3.14159265358979
    
    Imports System.IO
    
    Module Example
       Public Sub Main()
          Dim sw As New StreamWriter(".\Doubles.dat")
          Dim values() As Double = { 2.2/1.01, 1.0/3, Math.PI }
          For ctr As Integer = 0 To values.Length - 1
             sw.Write("{0:G17}{1}", values(ctr), 
                      If(ctr < values.Length - 1, "|", ""))
          Next      
          sw.Close()
          
          Dim restoredValues(values.Length - 1) As Double
          Dim sr As New StreamReader(".\Doubles.dat")
          Dim temp As String = sr.ReadToEnd()
          Dim tempStrings() As String = temp.Split("|"c)
          For ctr As Integer = 0 To tempStrings.Length - 1
             restoredValues(ctr) = Double.Parse(tempStrings(ctr))   
          Next 
    
          For ctr As Integer = 0 To values.Length - 1
             Console.WriteLine("{0} {2} {1}", values(ctr), 
                               restoredValues(ctr),
                               If(values(ctr).Equals(restoredValues(ctr)), "=", "<>"))
          Next
       End Sub
    End Module
    ' The example displays the following output:
    '       2.17821782178218 = 2.17821782178218
    '       0.333333333333333 = 0.333333333333333
    '       3.14159265358979 = 3.14159265358979
    

重要

Double 值一起使用时,在某些情况下,"R" 格式说明符无法成功往返原始值。When used with a Double value, the "R" format specifier in some cases fails to successfully round-trip the original value. 若要确保 Double 值已成功往返,请使用 "G17" 格式说明符。To ensure that Double values successfully round-trip, use the "G17" format specifier.

  • Single 值的精度低于 Double 值。Single values have less precision than Double values. 转换为看似等效 @no__t 的 @no__t 0 值通常不等于 @no__t 2 值,因为精度存在差异。A Single value that is converted to a seemingly equivalent Double often does not equal the Double value because of differences in precision. 在下面的示例中,将相同除法运算的结果分配给 @no__t 0 和 @no__t 1 值。In the following example, the result of identical division operations is assigned to a Double and a Single value. 将 @no__t 0 值强制转换为 Double 后,这两个值的比较表明它们不相等。After the Single value is cast to a Double, a comparison of the two values shows that they are unequal.

    using System;
    
    public class Example
    {
       public static void Main()
       {
          Double value1 = 1/3.0;
          Single sValue2 = 1/3.0f;
          Double value2 = (Double) sValue2;
          Console.WriteLine("{0:R} = {1:R}: {2}", value1, value2, 
                                              value1.Equals(value2));
       }
    }
    // The example displays the following output:
    //        0.33333333333333331 = 0.3333333432674408: False
    
    Module Example
       Public Sub Main()
          Dim value1 As Double = 1/3
          Dim sValue2 As Single = 1/3
          Dim value2 As Double = CDbl(sValue2)
          Console.WriteLine("{0} = {1}: {2}", value1, value2, value1.Equals(value2))
       End Sub
    End Module
    ' The example displays the following output:
    '       0.33333333333333331 = 0.3333333432674408: False
    

    若要避免此问题,请使用 Double 代替 @no__t 数据类型,或使用 Round 方法,使这两个值具有相同的精度。To avoid this problem, use either the Double in place of the Single data type, or use the Round method so that both values have the same precision.

此外,由于缺少 @no__t 类型的精度,使用 @no__t 值的算术和赋值运算的结果可能会略有不同。In addition, the result of arithmetic and assignment operations with Double values may differ slightly by platform because of the loss of precision of the Double type. 例如,在 .NET Framework 的32位和64位版本中,赋给文本 Double 值的结果可能不同。For example, the result of assigning a literal Double value may differ in the 32-bit and 64-bit versions of the .NET Framework. 下面的示例说明了将文本值 4.42330604244772 E-305 和值为-4.42330604244772 E-305 的变量分配给 Double 变量时的这种差异。The following example illustrates this difference when the literal value -4.42330604244772E-305 and a variable whose value is -4.42330604244772E-305 are assigned to a Double variable. 请注意,在这种情况下,Parse(String) 方法的结果不会降低精度。Note that the result of the Parse(String) method in this case does not suffer from a loss of precision.

double value = -4.42330604244772E-305;

double fromLiteral = -4.42330604244772E-305;
double fromVariable = value;
double fromParse = Double.Parse("-4.42330604244772E-305");

Console.WriteLine("Double value from literal: {0,29:R}", fromLiteral);
Console.WriteLine("Double value from variable: {0,28:R}", fromVariable);
Console.WriteLine("Double value from Parse method: {0,24:R}", fromParse);      
// On 32-bit versions of the .NET Framework, the output is:
//    Double value from literal:        -4.42330604244772E-305
//    Double value from variable:       -4.42330604244772E-305
//    Double value from Parse method:   -4.42330604244772E-305
//
// On other versions of the .NET Framework, the output is:
//    Double value from literal:      -4.4233060424477198E-305
//    Double value from variable:     -4.4233060424477198E-305
//    Double value from Parse method:   -4.42330604244772E-305      
Dim value As Double = -4.42330604244772E-305

Dim fromLiteral As Double = -4.42330604244772E-305
Dim fromVariable As Double = value
Dim fromParse As Double = Double.Parse("-4.42330604244772E-305")

Console.WriteLine("Double value from literal: {0,29:R}", fromLiteral)
Console.WriteLine("Double value from variable: {0,28:R}", fromVariable)
Console.WriteLine("Double value from Parse method: {0,24:R}", fromParse)      
' On 32-bit versions of the .NET Framework, the output is:
'    Double value from literal:        -4.42330604244772E-305
'    Double value from variable:       -4.42330604244772E-305
'    Double value from Parse method:   -4.42330604244772E-305
'
' On other versions of the .NET Framework, the output is:
'    Double value from literal:        -4.4233060424477198E-305
'    Double value from variable:       -4.4233060424477198E-305
'    Double value from Parse method:     -4.42330604244772E-305      

测试是否相等Testing for Equality

若要被视为相等,两个 @no__t 0 值必须表示相同的值。To be considered equal, two Double values must represent identical values. 不过,由于值之间的精度差别,或由于一个或两个值的精度损失,应相同的浮点值经常会出现不相等,因为其最小有效位之间存在差异。However, because of differences in precision between values, or because of a loss of precision by one or both values, floating-point values that are expected to be identical often turn out to be unequal because of differences in their least significant digits. 因此,调用 Equals 方法来确定两个值是否相等,或调用 @no__t 1 方法来确定两个 @no__t 2 值之间的关系,通常会产生意外结果。As a result, calls to the Equals method to determine whether two values are equal, or calls to the CompareTo method to determine the relationship between two Double values, often yield unexpected results. 在下面的示例中,这很明显,其中两个明显等于 Double 值为不相等,因为第一个具有15位精度,而第二个则为17。This is evident in the following example, where two apparently equal Double values turn out to be unequal because the first has 15 digits of precision, while the second has 17.

using System;

public class Example
{
   public static void Main()
   {
      double value1 = .333333333333333;
      double value2 = 1.0/3;
      Console.WriteLine("{0:R} = {1:R}: {2}", value1, value2, value1.Equals(value2));
   }
}
// The example displays the following output:
//        0.333333333333333 = 0.33333333333333331: False
Module Example
   Public Sub Main()
      Dim value1 As Double = .333333333333333
      Dim value2 As Double = 1/3
      Console.WriteLine("{0:R} = {1:R}: {2}", value1, value2, value1.Equals(value2))
   End Sub
End Module
' The example displays the following output:
'       0.333333333333333 = 0.33333333333333331: False

采用不同的代码路径并以不同的方式操作的计算值通常证明不相等。Calculated values that follow different code paths and that are manipulated in different ways often prove to be unequal. 在下面的示例中,一 Double 值为平方,然后计算平方根以还原原始值。In the following example, one Double value is squared, and then the square root is calculated to restore the original value. 在结果的平方根除以3.51 后,第二个 Double 乘以3.51 和 squared,以还原原始值。A second Double is multiplied by 3.51 and squared before the square root of the result is divided by 3.51 to restore the original value. 尽管两个值看起来是相同的,但对 Equals(Double) 方法的调用指示它们不相等。Although the two values appear to be identical, a call to the Equals(Double) method indicates that they are not equal. 使用 "R" 标准格式字符串返回显示每个 Double 值的所有有效位的结果字符串,显示第二个值小于第一个值 .0000000000001。Using the "R" standard format string to return a result string that displays all the significant digits of each Double value shows that the second value is .0000000000001 less than the first.

using System;

public class Example
{
   public static void Main()
   {
      double value1 = 100.10142;
      value1 = Math.Sqrt(Math.Pow(value1, 2));
      double value2 = Math.Pow(value1 * 3.51, 2);
      value2 = Math.Sqrt(value2) / 3.51;
      Console.WriteLine("{0} = {1}: {2}\n", 
                        value1, value2, value1.Equals(value2)); 
      Console.WriteLine("{0:R} = {1:R}", value1, value2); 
   }
}
// The example displays the following output:
//    100.10142 = 100.10142: False
//    
//    100.10142 = 100.10141999999999
Module Example
   Public Sub Main()
      Dim value1 As Double = 100.10142
      value1 = Math.Sqrt(Math.Pow(value1, 2))
      Dim value2 As Double = Math.Pow(value1 * 3.51, 2)
      value2 = Math.Sqrt(value2) / 3.51
      Console.WriteLine("{0} = {1}: {2}", 
                        value1, value2, value1.Equals(value2)) 
      Console.WriteLine()
      Console.WriteLine("{0:R} = {1:R}", value1, value2) 
   End Sub
End Module
' The example displays the following output:
'    100.10142 = 100.10142: False
'    
'    100.10142 = 100.10141999999999

如果精度损失可能会影响比较结果,则可以采用以下任意替代方法来调用 @no__t 0 或 CompareTo 方法:In cases where a loss of precision is likely to affect the result of a comparison, you can adopt any of the following alternatives to calling the Equals or CompareTo method:

  • 调用 Math.Round 方法,以确保两个值具有相同的精度。Call the Math.Round method to ensure that both values have the same precision. 下面的示例修改了上一个示例,以使用此方法,以使两个小数值相等。The following example modifies a previous example to use this approach so that two fractional values are equivalent.

    using System;
    
    public class Example
    {
       public static void Main()
       {
          double value1 = .333333333333333;
          double value2 = 1.0/3;
          int precision = 7;
          value1 = Math.Round(value1, precision);
          value2 = Math.Round(value2, precision);
          Console.WriteLine("{0:R} = {1:R}: {2}", value1, value2, value1.Equals(value2));
       }
    }
    // The example displays the following output:
    //        0.3333333 = 0.3333333: True
    
    Module Example
       Public Sub Main()
          Dim value1 As Double = .333333333333333
          Dim value2 As Double = 1/3
          Dim precision As Integer = 7
          value1 = Math.Round(value1, precision)
          value2 = Math.Round(value2, precision)
          Console.WriteLine("{0:R} = {1:R}: {2}", value1, value2, value1.Equals(value2))
       End Sub
    End Module
    ' The example displays the following output:
    '       0.3333333 = 0.3333333: True
    

    但请注意,精度问题仍适用于中点值的舍入。Note, though, that the problem of precision still applies to rounding of midpoint values. 有关更多信息,请参见 Math.Round(Double, Int32, MidpointRounding) 方法。For more information, see the Math.Round(Double, Int32, MidpointRounding) method.

  • 测试近似相等性,而不是相等。Test for approximate equality rather than equality. 这要求您定义一个绝对值,而这两个值可以不同,但仍相等,也可以定义较小值与较大值之间的差异。This requires that you define either an absolute amount by which the two values can differ but still be equal, or that you define a relative amount by which the smaller value can diverge from the larger value.

    警告

    在测试相等性时,@no__t 0 有时用作两个 @no__t 值之间的距离的绝对度量值。Double.Epsilon is sometimes used as an absolute measure of the distance between two Double values when testing for equality. Double.Epsilon 度量值为零的值为零的可添加到 @no__t 的可能的最小值。However, Double.Epsilon measures the smallest possible value that can be added to, or subtracted from, a Double whose value is zero. 对于大多数正负 Double 值,Double.Epsilon 的值太小,无法检测到。For most positive and negative Double values, the value of Double.Epsilon is too small to be detected. 因此,除了零值以外,不建议在测试中使用它是否相等。Therefore, except for values that are zero, we do not recommend its use in tests for equality.

    下面的示例使用后一种方法来定义一个 IsApproximatelyEqual 方法,该方法可测试两个值之间的相对差异。The following example uses the latter approach to define an IsApproximatelyEqual method that tests the relative difference between two values. 它还比较了对 IsApproximatelyEqual 方法和 Equals(Double) 方法的调用结果。It also contrasts the result of calls to the IsApproximatelyEqual method and the Equals(Double) method.

    using System;
    
    public class Example
    {
       public static void Main()
       {
          double one1 = .1 * 10;
          double one2 = 0;
          for (int ctr = 1; ctr <= 10; ctr++)
             one2 += .1;
    
          Console.WriteLine("{0:R} = {1:R}: {2}", one1, one2, one1.Equals(one2));
          Console.WriteLine("{0:R} is approximately equal to {1:R}: {2}", 
                            one1, one2, 
                            IsApproximatelyEqual(one1, one2, .000000001));   
       }
    
       static bool IsApproximatelyEqual(double value1, double value2, double epsilon)
       {
          // If they are equal anyway, just return True.
          if (value1.Equals(value2))
             return true;
    
          // Handle NaN, Infinity.
          if (Double.IsInfinity(value1) | Double.IsNaN(value1))
             return value1.Equals(value2);
          else if (Double.IsInfinity(value2) | Double.IsNaN(value2))
             return value1.Equals(value2);
    
          // Handle zero to avoid division by zero
          double divisor = Math.Max(value1, value2);
          if (divisor.Equals(0)) 
             divisor = Math.Min(value1, value2);
          
          return Math.Abs((value1 - value2) / divisor) <= epsilon;           
       } 
    }
    // The example displays the following output:
    //       1 = 0.99999999999999989: False
    //       1 is approximately equal to 0.99999999999999989: True
    
    Module Example
       Public Sub Main()
          Dim one1 As Double = .1 * 10
          Dim one2 As Double = 0
          For ctr As Integer = 1 To 10
             one2 += .1
          Next
          Console.WriteLine("{0:R} = {1:R}: {2}", one1, one2, one1.Equals(one2))
          Console.WriteLine("{0:R} is approximately equal to {1:R}: {2}", 
                            one1, one2, 
                            IsApproximatelyEqual(one1, one2, .000000001))   
       End Sub
    
       Function IsApproximatelyEqual(value1 As Double, value2 As Double, 
                                     epsilon As Double) As Boolean
          ' If they are equal anyway, just return True.
          If value1.Equals(value2) Then Return True
          
          ' Handle NaN, Infinity.
          If Double.IsInfinity(value1) Or Double.IsNaN(value1) Then
             Return value1.Equals(value2)
          Else If Double.IsInfinity(value2) Or Double.IsNaN(value2)
             Return value1.Equals(value2)
          End If
          
          ' Handle zero to avoid division by zero
          Dim divisor As Double = Math.Max(value1, value2)
          If divisor.Equals(0) Then
             divisor = Math.Min(value1, value2)
          End If 
          
          Return Math.Abs((value1 - value2) / divisor) <= epsilon           
       End Function
    End Module
    ' The example displays the following output:
    '       1 = 0.99999999999999989: False
    '       1 is approximately equal to 0.99999999999999989: True
    

浮点值和异常Floating-Point Values and Exceptions

与整数类型的操作不同,在溢出或非法操作(如被零除)情况下引发异常时,具有浮点值的操作不会引发异常。Unlike operations with integral types, which throw exceptions in cases of overflow or illegal operations such as division by zero, operations with floating-point values do not throw exceptions. 相反,在异常情况下,浮点运算的结果为零、正无穷、负无穷或非数字(NaN):Instead, in exceptional situations, the result of a floating-point operation is zero, positive infinity, negative infinity, or not a number (NaN):

  • 如果浮点运算的结果对于目标格式来说太小,则结果为零。If the result of a floating-point operation is too small for the destination format, the result is zero. 当两个非常小的数字相乘时,可能会出现这种情况,如下面的示例所示。This can occur when two very small numbers are multiplied, as the following example shows.

    using System;
    
    public class Example
    {
       public static void Main()
       {
          Double value1 = 1.1632875981534209e-225;
          Double value2 = 9.1642346778e-175;
          Double result = value1 * value2;
          Console.WriteLine("{0} * {1} = {2}", value1, value2, result);
          Console.WriteLine("{0} = 0: {1}", result, result.Equals(0.0));
       }
    }
    // The example displays the following output:
    //       1.16328759815342E-225 * 9.1642346778E-175 = 0
    //       0 = 0: True
    
    Module Example
       Public Sub Main()
          Dim value1 As Double = 1.1632875981534209e-225
          Dim value2 As Double = 9.1642346778e-175
          Dim result As Double = value1 * value2
          Console.WriteLine("{0} * {1} = {2}", value1, value2, result)
          Console.WriteLine("{0} = 0: {1}", result, result.Equals(0.0))
       End Sub
    End Module
    ' The example displays the following output:
    '       1.16328759815342E-225 * 9.1642346778E-175 = 0
    '       0 = 0: True
    
  • 如果浮点运算的结果量超过目标格式的范围,则操作的结果为 PositiveInfinityNegativeInfinity,适用于结果的符号。If the magnitude of the result of a floating-point operation exceeds the range of the destination format, the result of the operation is PositiveInfinity or NegativeInfinity, as appropriate for the sign of the result. 溢出 Double.MaxValue 的操作的结果为 PositiveInfinity,溢出 Double.MinValue 的操作的结果为 NegativeInfinity,如下面的示例所示。The result of an operation that overflows Double.MaxValue is PositiveInfinity, and the result of an operation that overflows Double.MinValue is NegativeInfinity, as the following example shows.

    using System;
    
    public class Example
    {
       public static void Main()
       {
          Double value1 = 4.565e153;
          Double value2 = 6.9375e172;
          Double result = value1 * value2;
          Console.WriteLine("PositiveInfinity: {0}", 
                             Double.IsPositiveInfinity(result));
          Console.WriteLine("NegativeInfinity: {0}\n", 
                            Double.IsNegativeInfinity(result));
    
          value1 = -value1;
          result = value1 * value2;
          Console.WriteLine("PositiveInfinity: {0}", 
                             Double.IsPositiveInfinity(result));
          Console.WriteLine("NegativeInfinity: {0}", 
                            Double.IsNegativeInfinity(result));
       }
    }                                                                 
    
    // The example displays the following output:
    //       PositiveInfinity: True
    //       NegativeInfinity: False
    //       
    //       PositiveInfinity: False
    //       NegativeInfinity: True
    
    Module Example
       Public Sub Main()
          Dim value1 As Double = 4.565e153
          Dim value2 As Double = 6.9375e172
          Dim result As Double = value1 * value2
          Console.WriteLine("PositiveInfinity: {0}", 
                             Double.IsPositiveInfinity(result))
          Console.WriteLine("NegativeInfinity: {0}", 
                            Double.IsNegativeInfinity(result))
          Console.WriteLine()                  
          value1 = -value1
          result = value1 * value2
          Console.WriteLine("PositiveInfinity: {0}", 
                             Double.IsPositiveInfinity(result))
          Console.WriteLine("NegativeInfinity: {0}", 
                            Double.IsNegativeInfinity(result))
       End Sub
    End Module
    ' The example displays the following output:
    '       PositiveInfinity: True
    '       NegativeInfinity: False
    '       
    '       PositiveInfinity: False
    '       NegativeInfinity: True
    

    @no__t 为0时,还会导致被零除和正被除数的结果,并在被零除的情况下 @no__t 1 结果。PositiveInfinity also results from a division by zero with a positive dividend, and NegativeInfinity results from a division by zero with a negative dividend.

  • 如果浮点操作无效,则操作的结果为 NaNIf a floating-point operation is invalid, the result of the operation is NaN. 例如,@no__t 以下操作的结果:For example, NaN results from the following operations:

    • 除数为零的除以零。Division by zero with a dividend of zero. 请注意,除数为零的其他情况会导致 PositiveInfinity 或 @no__t 为-1。Note that other cases of division by zero result in either PositiveInfinity or NegativeInfinity.
  • 具有无效输入的任何浮点运算。Any floating-point operation with an invalid input. 例如,使用负值调用 Math.Sqrt 方法将返回 NaN,因为调用的 Math.Acos 方法的值大于1或小于1。For example, calling the Math.Sqrt method with a negative value returns NaN, as does calling the Math.Acos method with a value that is greater than one or less than negative one.

  • 参数值为 @no__t 为0的任何操作。Any operation with an argument whose value is Double.NaN.

类型转换和双结构Type conversions and the Double structure

@No__t-0 结构不定义任何显式或隐式转换运算符;相反,转换是由编译器实现的。The Double structure does not define any explicit or implicit conversion operators; instead, conversions are implemented by the compiler.

任何基元数值类型的值到 Double 的转换都是扩大转换,因此不需要显式强制转换运算符或对转换方法的调用,除非编译器显式要求它。The conversion of the value of any primitive numeric type to a Double is a widening conversion and therefore does not require an explicit cast operator or call to a conversion method unless a compiler explicitly requires it. 例如, C#编译器需要一个转换运算符用于从 DecimalDouble 的转换,而 Visual Basic 编译器则不需要。For example, the C# compiler requires a casting operator for conversions from Decimal to Double, while the Visual Basic compiler does not. 下面的示例将其他基元数值类型的最小值或最大值转换为 DoubleThe following example converts the minimum or maximum value of other primitive numeric types to a Double.

using System;

public class Example
{
   public static void Main()
   {
      dynamic[] values = { Byte.MinValue, Byte.MaxValue, Decimal.MinValue,
                           Decimal.MaxValue, Int16.MinValue, Int16.MaxValue,
                           Int32.MinValue, Int32.MaxValue, Int64.MinValue,
                           Int64.MaxValue, SByte.MinValue, SByte.MaxValue,
                           Single.MinValue, Single.MaxValue, UInt16.MinValue,
                           UInt16.MaxValue, UInt32.MinValue, UInt32.MaxValue,
                           UInt64.MinValue, UInt64.MaxValue };
      double dblValue;
      foreach (var value in values) {
         if (value.GetType() == typeof(Decimal))
            dblValue = (Double) value;
         else
            dblValue = value;
         Console.WriteLine("{0} ({1}) --> {2:R} ({3})",
                           value, value.GetType().Name,
                           dblValue, dblValue.GetType().Name);
      }
   }
}
// The example displays the following output:
//    0 (Byte) --> 0 (Double)
//    255 (Byte) --> 255 (Double)
//    -79228162514264337593543950335 (Decimal) --> -7.9228162514264338E+28 (Double)
//    79228162514264337593543950335 (Decimal) --> 7.9228162514264338E+28 (Double)
//    -32768 (Int16) --> -32768 (Double)
//    32767 (Int16) --> 32767 (Double)
//    -2147483648 (Int32) --> -2147483648 (Double)
//    2147483647 (Int32) --> 2147483647 (Double)
//    -9223372036854775808 (Int64) --> -9.2233720368547758E+18 (Double)
//    9223372036854775807 (Int64) --> 9.2233720368547758E+18 (Double)
//    -128 (SByte) --> -128 (Double)
//    127 (SByte) --> 127 (Double)
//    -3.402823E+38 (Single) --> -3.4028234663852886E+38 (Double)
//    3.402823E+38 (Single) --> 3.4028234663852886E+38 (Double)
//    0 (UInt16) --> 0 (Double)
//    65535 (UInt16) --> 65535 (Double)
//    0 (UInt32) --> 0 (Double)
//    4294967295 (UInt32) --> 4294967295 (Double)
//    0 (UInt64) --> 0 (Double)
//    18446744073709551615 (UInt64) --> 1.8446744073709552E+19 (Double)
Module Example
   Public Sub Main()
      Dim values() As Object = { Byte.MinValue, Byte.MaxValue, Decimal.MinValue,
                                 Decimal.MaxValue, Int16.MinValue, Int16.MaxValue,
                                 Int32.MinValue, Int32.MaxValue, Int64.MinValue,
                                 Int64.MaxValue, SByte.MinValue, SByte.MaxValue,
                                 Single.MinValue, Single.MaxValue, UInt16.MinValue,
                                 UInt16.MaxValue, UInt32.MinValue, UInt32.MaxValue,
                                 UInt64.MinValue, UInt64.MaxValue }
      Dim dblValue As Double
      For Each value In values
         dblValue = value
         Console.WriteLine("{0} ({1}) --> {2:R} ({3})",
                           value, value.GetType().Name,
                           dblValue, dblValue.GetType().Name)
      Next
   End Sub
End Module
' The example displays the following output:
'    0 (Byte) --> 0 (Double)
'    255 (Byte) --> 255 (Double)
'    -79228162514264337593543950335 (Decimal) --> -7.9228162514264338E+28 (Double)
'    79228162514264337593543950335 (Decimal) --> 7.9228162514264338E+28 (Double)
'    -32768 (Int16) --> -32768 (Double)
'    32767 (Int16) --> 32767 (Double)
'    -2147483648 (Int32) --> -2147483648 (Double)
'    2147483647 (Int32) --> 2147483647 (Double)
'    -9223372036854775808 (Int64) --> -9.2233720368547758E+18 (Double)
'    9223372036854775807 (Int64) --> 9.2233720368547758E+18 (Double)
'    -128 (SByte) --> -128 (Double)
'    127 (SByte) --> 127 (Double)
'    -3.402823E+38 (Single) --> -3.4028234663852886E+38 (Double)
'    3.402823E+38 (Single) --> 3.4028234663852886E+38 (Double)
'    0 (UInt16) --> 0 (Double)
'    65535 (UInt16) --> 65535 (Double)
'    0 (UInt32) --> 0 (Double)
'    4294967295 (UInt32) --> 4294967295 (Double)
'    0 (UInt64) --> 0 (Double)
'    18446744073709551615 (UInt64) --> 1.8446744073709552E+19 (Double)

此外,@no__t 值为-1、Single.PositiveInfinity 和 @no__t 3 分别转换为 @no__t、@no__t 和 @no__t 的 @no__t 值。In addition, the Single values Single.NaN, Single.PositiveInfinity, and Single.NegativeInfinity covert to Double.NaN, Double.PositiveInfinity, and Double.NegativeInfinity, respectively.

请注意,某些数值类型的值转换为 @no__t 值0值可能会导致精度损失。Note that the conversion of the value of some numeric types to a Double value can involve a loss of precision. 如示例所示,将 DecimalInt64Single 和 @no__t 3 值转换为 Double 值时,可能会丢失精度。As the example illustrates, a loss of precision is possible when converting Decimal, Int64, Single, and UInt64 values to Double values.

Double 值转换为任何其他基元数值数据类型的值是收缩转换,需要转换运算符(在中C#为)、转换方法(在 Visual Basic 中)或对 @no__t 方法的调用。The conversion of a Double value to a value of any other primitive numeric data type is a narrowing conversion and requires a cast operator (in C#), a conversion method (in Visual Basic), or a call to a Convert method. 超出目标数据类型范围的值(由目标类型的 MinValue 和 @no__t 属性定义)的行为如下表中所示。Values that are outside the range of the target data type, which are defined by the target type's MinValue and MaxValue properties, behave as shown in the following table.

目标类型Target type 结果Result
任何整型Any integral type 如果在检查的上下文中发生转换,则为 OverflowException 异常。An OverflowException exception if the conversion occurs in a checked context.

如果转换发生在未检查的上下文中(默认为C#),则转换操作会成功,但值溢出。If the conversion occurs in an unchecked context (the default in C#), the conversion operation succeeds but the value overflows.
Decimal 一个 OverflowException 异常。An OverflowException exception.
Single 对于负值,@no__t 为0。Single.NegativeInfinity for negative values.

对于正值 @no__t 为0。Single.PositiveInfinity for positive values.

此外,Double.NaNDouble.PositiveInfinity 和 @no__t 为转换为已检查上下文中的整数而引发 OverflowException,但这些值在转换为未检查的上下文中的整数时溢出。In addition, Double.NaN, Double.PositiveInfinity, and Double.NegativeInfinity throw an OverflowException for conversions to integers in a checked context, but these values overflow when converted to integers in an unchecked context. 若要转换为 Decimal,它们始终引发 @no__t。For conversions to Decimal, they always throw an OverflowException. 为了转换到 Single,它们分别转换为 Single.NaNSingle.PositiveInfinity 和 @no__t。For conversions to Single, they convert to Single.NaN, Single.PositiveInfinity, and Single.NegativeInfinity, respectively.

请注意,将 Double 值转换为另一种数值类型可能会导致精度损失。Note that a loss of precision may result from converting a Double value to another numeric type. 如果转换非整数 @no__t 值(如示例的输出所示),则当 Double 值舍入(如 Visual Basic 中)或截断时(如在中C#为),小数部分将丢失。In the case of converting non-integral Double values, as the output from the example shows, the fractional component is lost when the Double value is either rounded (as in Visual Basic) or truncated (as in C#). 若要转换为 DecimalSingle 值,@no__t 的值在目标数据类型中可能没有精确的表示形式。For conversions to Decimal and Single values, the Double value may not have a precise representation in the target data type.

下面的示例将多个 Double 值转换为多个其他数值类型。The following example converts a number of Double values to several other numeric types. 转换在 Visual Basic (默认值)和中C#的已检查上下文中发生(因为有checked关键字)。The conversions occur in a checked context in Visual Basic (the default) and in C# (because of the checked keyword). 该示例的输出显示了已检查的未检查上下文中的转换结果。The output from the example shows the result for conversions in both a checked an unchecked context. 您可以通过使用 /removeintchecks+ 编译器开关进行编译,并C#通过注释掉 checked 语句,在 Visual Basic 中未检查的上下文中执行转换。You can perform conversions in an unchecked context in Visual Basic by compiling with the /removeintchecks+ compiler switch and in C# by commenting out the checked statement.

using System;

public class Example
{
   public static void Main()
   {
      Double[] values = { Double.MinValue, -67890.1234, -12345.6789,
                          12345.6789, 67890.1234, Double.MaxValue,
                          Double.NaN, Double.PositiveInfinity,
                          Double.NegativeInfinity };
      checked {
         foreach (var value in values) {
            try {
                Int64 lValue = (long) value;
                Console.WriteLine("{0} ({1}) --> {2} (0x{2:X16}) ({3})",
                                  value, value.GetType().Name,
                                  lValue, lValue.GetType().Name);
            }
            catch (OverflowException) {
               Console.WriteLine("Unable to convert {0} to Int64.", value);
            }
            try {
                UInt64 ulValue = (ulong) value;
                Console.WriteLine("{0} ({1}) --> {2} (0x{2:X16}) ({3})",
                                  value, value.GetType().Name,
                                  ulValue, ulValue.GetType().Name);
            }
            catch (OverflowException) {
               Console.WriteLine("Unable to convert {0} to UInt64.", value);
            }
            try {
                Decimal dValue = (decimal) value;
                Console.WriteLine("{0} ({1}) --> {2} ({3})",
                                  value, value.GetType().Name,
                                  dValue, dValue.GetType().Name);
            }
            catch (OverflowException) {
               Console.WriteLine("Unable to convert {0} to Decimal.", value);
            }
            try {
                Single sValue = (float) value;
                Console.WriteLine("{0} ({1}) --> {2} ({3})",
                                  value, value.GetType().Name,
                                  sValue, sValue.GetType().Name);
            }
            catch (OverflowException) {
               Console.WriteLine("Unable to convert {0} to Single.", value);
            }
            Console.WriteLine();
         }
      }
   }
}
// The example displays the following output for conversions performed
// in a checked context:
//       Unable to convert -1.79769313486232E+308 to Int64.
//       Unable to convert -1.79769313486232E+308 to UInt64.
//       Unable to convert -1.79769313486232E+308 to Decimal.
//       -1.79769313486232E+308 (Double) --> -Infinity (Single)
//
//       -67890.1234 (Double) --> -67890 (0xFFFFFFFFFFFEF6CE) (Int64)
//       Unable to convert -67890.1234 to UInt64.
//       -67890.1234 (Double) --> -67890.1234 (Decimal)
//       -67890.1234 (Double) --> -67890.13 (Single)
//
//       -12345.6789 (Double) --> -12345 (0xFFFFFFFFFFFFCFC7) (Int64)
//       Unable to convert -12345.6789 to UInt64.
//       -12345.6789 (Double) --> -12345.6789 (Decimal)
//       -12345.6789 (Double) --> -12345.68 (Single)
//
//       12345.6789 (Double) --> 12345 (0x0000000000003039) (Int64)
//       12345.6789 (Double) --> 12345 (0x0000000000003039) (UInt64)
//       12345.6789 (Double) --> 12345.6789 (Decimal)
//       12345.6789 (Double) --> 12345.68 (Single)
//
//       67890.1234 (Double) --> 67890 (0x0000000000010932) (Int64)
//       67890.1234 (Double) --> 67890 (0x0000000000010932) (UInt64)
//       67890.1234 (Double) --> 67890.1234 (Decimal)
//       67890.1234 (Double) --> 67890.13 (Single)
//
//       Unable to convert 1.79769313486232E+308 to Int64.
//       Unable to convert 1.79769313486232E+308 to UInt64.
//       Unable to convert 1.79769313486232E+308 to Decimal.
//       1.79769313486232E+308 (Double) --> Infinity (Single)
//
//       Unable to convert NaN to Int64.
//       Unable to convert NaN to UInt64.
//       Unable to convert NaN to Decimal.
//       NaN (Double) --> NaN (Single)
//
//       Unable to convert Infinity to Int64.
//       Unable to convert Infinity to UInt64.
//       Unable to convert Infinity to Decimal.
//       Infinity (Double) --> Infinity (Single)
//
//       Unable to convert -Infinity to Int64.
//       Unable to convert -Infinity to UInt64.
//       Unable to convert -Infinity to Decimal.
//       -Infinity (Double) --> -Infinity (Single)
// The example displays the following output for conversions performed
// in an unchecked context:
//       -1.79769313486232E+308 (Double) --> -9223372036854775808 (0x8000000000000000) (Int64)
//       -1.79769313486232E+308 (Double) --> 9223372036854775808 (0x8000000000000000) (UInt64)
//       Unable to convert -1.79769313486232E+308 to Decimal.
//       -1.79769313486232E+308 (Double) --> -Infinity (Single)
//
//       -67890.1234 (Double) --> -67890 (0xFFFFFFFFFFFEF6CE) (Int64)
//       -67890.1234 (Double) --> 18446744073709483726 (0xFFFFFFFFFFFEF6CE) (UInt64)
//       -67890.1234 (Double) --> -67890.1234 (Decimal)
//       -67890.1234 (Double) --> -67890.13 (Single)
//
//       -12345.6789 (Double) --> -12345 (0xFFFFFFFFFFFFCFC7) (Int64)
//       -12345.6789 (Double) --> 18446744073709539271 (0xFFFFFFFFFFFFCFC7) (UInt64)
//       -12345.6789 (Double) --> -12345.6789 (Decimal)
//       -12345.6789 (Double) --> -12345.68 (Single)
//
//       12345.6789 (Double) --> 12345 (0x0000000000003039) (Int64)
//       12345.6789 (Double) --> 12345 (0x0000000000003039) (UInt64)
//       12345.6789 (Double) --> 12345.6789 (Decimal)
//       12345.6789 (Double) --> 12345.68 (Single)
//
//       67890.1234 (Double) --> 67890 (0x0000000000010932) (Int64)
//       67890.1234 (Double) --> 67890 (0x0000000000010932) (UInt64)
//       67890.1234 (Double) --> 67890.1234 (Decimal)
//       67890.1234 (Double) --> 67890.13 (Single)
//
//       1.79769313486232E+308 (Double) --> -9223372036854775808 (0x8000000000000000) (Int64)
//       1.79769313486232E+308 (Double) --> 0 (0x0000000000000000) (UInt64)
//       Unable to convert 1.79769313486232E+308 to Decimal.
//       1.79769313486232E+308 (Double) --> Infinity (Single)
//
//       NaN (Double) --> -9223372036854775808 (0x8000000000000000) (Int64)
//       NaN (Double) --> 0 (0x0000000000000000) (UInt64)
//       Unable to convert NaN to Decimal.
//       NaN (Double) --> NaN (Single)
//
//       Infinity (Double) --> -9223372036854775808 (0x8000000000000000) (Int64)
//       Infinity (Double) --> 0 (0x0000000000000000) (UInt64)
//       Unable to convert Infinity to Decimal.
//       Infinity (Double) --> Infinity (Single)
//
//       -Infinity (Double) --> -9223372036854775808 (0x8000000000000000) (Int64)
//       -Infinity (Double) --> 9223372036854775808 (0x8000000000000000) (UInt64)
//       Unable to convert -Infinity to Decimal.
//       -Infinity (Double) --> -Infinity (Single)
Module Example
   Public Sub Main()
      Dim values() As Double = { Double.MinValue, -67890.1234, -12345.6789,
                                 12345.6789, 67890.1234, Double.MaxValue,
                                 Double.NaN, Double.PositiveInfinity,
                                 Double.NegativeInfinity }
      For Each value In values
         Try
             Dim lValue As Int64 = CLng(value)
             Console.WriteLine("{0} ({1}) --> {2} (0x{2:X16}) ({3})",
                               value, value.GetType().Name,
                               lValue, lValue.GetType().Name)
         Catch e As OverflowException
            Console.WriteLine("Unable to convert {0} to Int64.", value)
         End Try
         Try
             Dim ulValue As UInt64 = CULng(value)
             Console.WriteLine("{0} ({1}) --> {2} (0x{2:X16}) ({3})",
                               value, value.GetType().Name,
                               ulValue, ulValue.GetType().Name)
         Catch e As OverflowException
            Console.WriteLine("Unable to convert {0} to UInt64.", value)
         End Try
         Try
             Dim dValue As Decimal = CDec(value)
             Console.WriteLine("{0} ({1}) --> {2} ({3})",
                               value, value.GetType().Name,
                               dValue, dValue.GetType().Name)
         Catch e As OverflowException
            Console.WriteLine("Unable to convert {0} to Decimal.", value)
         End Try
         Try
             Dim sValue As Single = CSng(value)
             Console.WriteLine("{0} ({1}) --> {2} ({3})",
                               value, value.GetType().Name,
                               sValue, sValue.GetType().Name)
         Catch e As OverflowException
            Console.WriteLine("Unable to convert {0} to Single.", value)
         End Try
         Console.WriteLine()
      Next
   End Sub
End Module
' The example displays the following output for conversions performed
' in a checked context:
'       Unable to convert -1.79769313486232E+308 to Int64.
'       Unable to convert -1.79769313486232E+308 to UInt64.
'       Unable to convert -1.79769313486232E+308 to Decimal.
'       -1.79769313486232E+308 (Double) --> -Infinity (Single)
'
'       -67890.1234 (Double) --> -67890 (0xFFFFFFFFFFFEF6CE) (Int64)
'       Unable to convert -67890.1234 to UInt64.
'       -67890.1234 (Double) --> -67890.1234 (Decimal)
'       -67890.1234 (Double) --> -67890.13 (Single)
'
'       -12345.6789 (Double) --> -12346 (0xFFFFFFFFFFFFCFC6) (Int64)
'       Unable to convert -12345.6789 to UInt64.
'       -12345.6789 (Double) --> -12345.6789 (Decimal)
'       -12345.6789 (Double) --> -12345.68 (Single)
'
'       12345.6789 (Double) --> 12346 (0x000000000000303A) (Int64)
'       12345.6789 (Double) --> 12346 (0x000000000000303A) (UInt64)
'       12345.6789 (Double) --> 12345.6789 (Decimal)
'       12345.6789 (Double) --> 12345.68 (Single)
'
'       67890.1234 (Double) --> 67890 (0x0000000000010932) (Int64)
'       67890.1234 (Double) --> 67890 (0x0000000000010932) (UInt64)
'       67890.1234 (Double) --> 67890.1234 (Decimal)
'       67890.1234 (Double) --> 67890.13 (Single)
'
'       Unable to convert 1.79769313486232E+308 to Int64.
'       Unable to convert 1.79769313486232E+308 to UInt64.
'       Unable to convert 1.79769313486232E+308 to Decimal.
'       1.79769313486232E+308 (Double) --> Infinity (Single)
'
'       Unable to convert NaN to Int64.
'       Unable to convert NaN to UInt64.
'       Unable to convert NaN to Decimal.
'       NaN (Double) --> NaN (Single)
'
'       Unable to convert Infinity to Int64.
'       Unable to convert Infinity to UInt64.
'       Unable to convert Infinity to Decimal.
'       Infinity (Double) --> Infinity (Single)
'
'       Unable to convert -Infinity to Int64.
'       Unable to convert -Infinity to UInt64.
'       Unable to convert -Infinity to Decimal.
'       -Infinity (Double) --> -Infinity (Single)
' The example displays the following output for conversions performed
' in an unchecked context:
'       -1.79769313486232E+308 (Double) --> -9223372036854775808 (0x8000000000000000) (Int64)
'       -1.79769313486232E+308 (Double) --> 9223372036854775808 (0x8000000000000000) (UInt64)
'       Unable to convert -1.79769313486232E+308 to Decimal.
'       -1.79769313486232E+308 (Double) --> -Infinity (Single)
'
'       -67890.1234 (Double) --> -67890 (0xFFFFFFFFFFFEF6CE) (Int64)
'       -67890.1234 (Double) --> 18446744073709483726 (0xFFFFFFFFFFFEF6CE) (UInt64)
'       -67890.1234 (Double) --> -67890.1234 (Decimal)
'       -67890.1234 (Double) --> -67890.13 (Single)
'
'       -12345.6789 (Double) --> -12346 (0xFFFFFFFFFFFFCFC6) (Int64)
'       -12345.6789 (Double) --> 18446744073709539270 (0xFFFFFFFFFFFFCFC6) (UInt64)
'       -12345.6789 (Double) --> -12345.6789 (Decimal)
'       -12345.6789 (Double) --> -12345.68 (Single)
'
'       12345.6789 (Double) --> 12346 (0x000000000000303A) (Int64)
'       12345.6789 (Double) --> 12346 (0x000000000000303A) (UInt64)
'       12345.6789 (Double) --> 12345.6789 (Decimal)
'       12345.6789 (Double) --> 12345.68 (Single)
'
'       67890.1234 (Double) --> 67890 (0x0000000000010932) (Int64)
'       67890.1234 (Double) --> 67890 (0x0000000000010932) (UInt64)
'       67890.1234 (Double) --> 67890.1234 (Decimal)
'       67890.1234 (Double) --> 67890.13 (Single)
'
'       1.79769313486232E+308 (Double) --> -9223372036854775808 (0x8000000000000000) (Int64)
'       1.79769313486232E+308 (Double) --> 0 (0x0000000000000000) (UInt64)
'       Unable to convert 1.79769313486232E+308 to Decimal.
'       1.79769313486232E+308 (Double) --> Infinity (Single)
'
'       NaN (Double) --> -9223372036854775808 (0x8000000000000000) (Int64)
'       NaN (Double) --> 0 (0x0000000000000000) (UInt64)
'       Unable to convert NaN to Decimal.
'       NaN (Double) --> NaN (Single)
'
'       Infinity (Double) --> -9223372036854775808 (0x8000000000000000) (Int64)
'       Infinity (Double) --> 0 (0x0000000000000000) (UInt64)
'       Unable to convert Infinity to Decimal.
'       Infinity (Double) --> Infinity (Single)
'
'       -Infinity (Double) --> -9223372036854775808 (0x8000000000000000) (Int64)
'       -Infinity (Double) --> 9223372036854775808 (0x8000000000000000) (UInt64)
'       Unable to convert -Infinity to Decimal.
'       -Infinity (Double) --> -Infinity (Single)

有关数值类型转换的详细信息,请参阅 .NET Framework 和类型转换表中的类型转换For more information on the conversion of numeric types, see Type Conversion in the .NET Framework and Type Conversion Tables.

浮点型功能Floating-Point Functionality

@No__t-0 结构和相关类型在以下方面提供了执行操作的方法:The Double structure and related types provide methods to perform operations in the following areas:

  • 值比较Comparison of values. 可以调用 Equals 方法来确定两个 @no__t 1 值是否相等,或 @no__t 2 方法来确定两个值之间的关系。You can call the Equals method to determine whether two Double values are equal, or the CompareTo method to determine the relationship between two values.

    @No__t-0 结构还支持一组完整的比较运算符。The Double structure also supports a complete set of comparison operators. 例如,你可以测试是否相等或不相等,或确定一个值是否大于或等于另一个值。For example, you can test for equality or inequality, or determine whether one value is greater than or equal to another. 如果其中一个操作数为数值类型,而不是 Double,则在执行比较之前将其转换为 @no__t 1。If one of the operands is a numeric type other than a Double, it is converted to a Double before performing the comparison.

    警告

    由于精度之间存在差异,因此两个 @no__t 0 的值将被视为不相等,这将影响比较的结果。Because of differences in precision, two Double values that you expect to be equal may turn out to be unequal, which affects the result of the comparison. 请参阅测试相等性部分,详细了解如何比较两个 @no__t 1 值。See the Testing for Equality section for more information about comparing two Double values.

    还可以调用 IsNaNIsInfinityIsPositiveInfinity 和 @no__t 3 方法来测试这些特殊值。You can also call the IsNaN, IsInfinity, IsPositiveInfinity, and IsNegativeInfinity methods to test for these special values.

  • 数学运算Mathematical operations. 常见算术运算(例如加法、减法、乘法和除法)由语言编译器和公共中间语言(CIL)指令实现,而不是由 Double 方法实现。Common arithmetic operations, such as addition, subtraction, multiplication, and division, are implemented by language compilers and Common Intermediate Language (CIL) instructions, rather than by Double methods. 如果数学运算中的一个操作数为数值类型,而不是 Double,则在执行该操作之前,它将转换为 @no__t 1。If one of the operands in a mathematical operation is a numeric type other than a Double, it is converted to a Double before performing the operation. 操作的结果也是 @no__t 0 值。The result of the operation is also a Double value.

    其他数学运算可通过调用 System.Math 类中的 static (Visual Basic 中 Shared)方法来执行。Other mathematical operations can be performed by calling static (Shared in Visual Basic) methods in the System.Math class. 它包括常见用于算术运算的其他方法(例如 Math.AbsMath.SignMath.Sqrt)、几何(如 Math.CosMath.Sin)以及微积分(如 Math.Log)。It includes additional methods commonly used for arithmetic (such as Math.Abs, Math.Sign, and Math.Sqrt), geometry (such as Math.Cos and Math.Sin), and calculus (such as Math.Log).

    还可以操作 Double 值中的单个位。You can also manipulate the individual bits in a Double value. @No__t-0 方法保留64位整数中的 @no__t 1 值的位模式。The BitConverter.DoubleToInt64Bits method preserves a Double value's bit pattern in a 64-bit integer. @No__t-0 方法将其位模式返回到字节数组中。The BitConverter.GetBytes(Double) method returns its bit pattern in a byte array.

  • 舍入Rounding. 舍入通常用作一种方法,用于降低由于浮点表示形式和精度问题导致的值之间的差异。Rounding is often used as a technique for reducing the impact of differences between values caused by problems of floating-point representation and precision. 可以通过调用 @no__t 的方法,将 @no__t 值舍入。You can round a Double value by calling the Math.Round method.

  • 格式设置Formatting. 可以通过调用 ToString 方法或使用复合格式设置功能,将 @no__t 值转换为其字符串表示形式。You can convert a Double value to its string representation by calling the ToString method or by using the composite formatting feature. 有关格式字符串如何控制浮点值的字符串表示形式的信息,请参阅标准数字格式字符串自定义数字格式字符串主题。For information about how format strings control the string representation of floating-point values, see the Standard Numeric Format Strings and Custom Numeric Format Strings topics.

  • 分析字符串Parsing strings. 可以通过调用 ParseTryParse 方法,将浮点值的字符串表示形式转换为 @no__t 值0。You can convert the string representation of a floating-point value to a Double value by calling either the Parse or TryParse method. 如果分析操作失败,Parse 方法会引发异常,而 TryParse 方法返回 falseIf the parse operation fails, the Parse method throws an exception, whereas the TryParse method returns false.

  • 类型转换Type conversion. @No__t-0 结构为 IConvertible 接口提供显式接口实现,该接口支持任意两种标准 .NET Framework 数据类型之间的转换。The Double structure provides an explicit interface implementation for the IConvertible interface, which supports conversion between any two standard .NET Framework data types. 语言编译器还支持将所有其他标准数值类型的值隐式转换为 Double 值。Language compilers also support the implicit conversion of values of all other standard numeric types to Double values. 将任何标准数值类型的值转换为 Double 是扩大转换,不需要强制转换运算符或转换方法的用户。Conversion of a value of any standard numeric type to a Double is a widening conversion and does not require the user of a casting operator or conversion method,

    但是,转换 @no__t 值和 @no__t 1 值可能会导致精度损失。However, conversion of Int64 and Single values can involve a loss of precision. 下表列出了每种类型的精度差异:The following table lists the differences in precision for each of these types:

    类型Type 最大精度Maximum precision 内部精度Internal precision
    Double 1515 1717
    Int64 19个十进制数字19 decimal digits 19个十进制数字19 decimal digits
    Single 7个十进制数字7 decimal digits 9个十进制数字9 decimal digits

    精确的问题会影响转换为 @no__t 1 值 @no__t 值。The problem of precision most frequently affects Single values that are converted to Double values. 在下面的示例中,两个由相同除法运算生成的值是不相等的,因为其中一个值是转换为 Double 的单精度浮点值。In the following example, two values produced by identical division operations are unequal because one of the values is a single-precision floating point value converted to a Double.

    using System;
    
    public class Example
    {
       public static void Main()
       {
          Double value = .1;
          Double result1 = value * 10;
          Double result2 = 0;
          for (int ctr = 1; ctr <= 10; ctr++)
             result2 += value;
    
          Console.WriteLine(".1 * 10:           {0:R}", result1);
          Console.WriteLine(".1 Added 10 times: {0:R}", result2);
       }
    }
    // The example displays the following output:
    //       .1 * 10:           1
    //       .1 Added 10 times: 0.99999999999999989
    
    Module Example
       Public Sub Main()
          Dim value As Double = .1
          Dim result1 As Double = value * 10
          Dim result2 As Double
          For ctr As Integer = 1 To 10
             result2 += value
          Next
          Console.WriteLine(".1 * 10:           {0:R}", result1)
          Console.WriteLine(".1 Added 10 times: {0:R}", result2)
       End Sub
    End Module
    ' The example displays the following output:
    '       .1 * 10:           1
    '       .1 Added 10 times: 0.99999999999999989
    

字段

Epsilon

表示大于零的最小正 Double 值。Represents the smallest positive Double value that is greater than zero. 此字段为常数。This field is constant.

MaxValue

表示 Double 的最大可能值。Represents the largest possible value of a Double. 此字段为常数。This field is constant.

MinValue

表示 Double 的最小可能值。Represents the smallest possible value of a Double. 此字段为常数。This field is constant.

NaN

表示不是数字 (NaN) 的值。Represents a value that is not a number (NaN). 此字段为常数。This field is constant.

NegativeInfinity

表示负无穷。Represents negative infinity. 此字段为常数。This field is constant.

PositiveInfinity

表示正无穷。Represents positive infinity. 此字段为常数。This field is constant.

方法

CompareTo(Double)

将此实例与指定的双精度浮点数进行比较,并返回一个整数,该整数指示此实例的值是小于、等于还是大于指定双精度浮点数的值。Compares this instance to a specified double-precision floating-point number and returns an integer that indicates whether the value of this instance is less than, equal to, or greater than the value of the specified double-precision floating-point number.

CompareTo(Object)

将此实例与指定对象进行比较,并返回一个整数,该整数指示此实例的值是小于、等于还是大于指定对象的值。Compares this instance to a specified object and returns an integer that indicates whether the value of this instance is less than, equal to, or greater than the value of the specified object.

Equals(Double)

返回一个值,该值指示此实例和指定的 Double 对象是否表示相同的值。Returns a value indicating whether this instance and a specified Double object represent the same value.

Equals(Object)

返回一个值,该值指示此实例是否等于指定的对象。Returns a value indicating whether this instance is equal to a specified object.

GetHashCode()

返回此实例的哈希代码。Returns the hash code for this instance.

GetTypeCode()

返回值类型 TypeCodeDoubleReturns the TypeCode for value type Double.

IsFinite(Double)

确定指定值是否为有限值(零、不正常或正常)。Determines whether the specified value is finite (zero, subnormal, or normal).

IsInfinity(Double)

返回一个值,该值指示指定数字是计算为负无穷大还是正无穷大。Returns a value indicating whether the specified number evaluates to negative or positive infinity.

IsNaN(Double)

返回一个值,该值指示指定的值是否不为数字 (NaN)。Returns a value that indicates whether the specified value is not a number (NaN).

IsNegative(Double)

确定指定值是否为负值。Determines whether the specified value is negative.

IsNegativeInfinity(Double)

返回一个值,通过该值指示指定数字是否计算为负无穷大。Returns a value indicating whether the specified number evaluates to negative infinity.

IsNormal(Double)

确定指定值是否正常。Determines whether the specified value is normal.

IsPositiveInfinity(Double)

返回一个值,通过该值指示指定数字是否计算为正无穷大。Returns a value indicating whether the specified number evaluates to positive infinity.

IsSubnormal(Double)

确定指定值是否不正常。Determines whether the specified value is subnormal.

Parse(ReadOnlySpan<Char>, NumberStyles, IFormatProvider)
Parse(String)

将数字的字符串表示形式转换为它的等效双精度浮点数。Converts the string representation of a number to its double-precision floating-point number equivalent.

Parse(String, IFormatProvider)

将指定的区域性特定格式的数字的字符串表示形式转换为它的等效双精度浮点数。Converts the string representation of a number in a specified culture-specific format to its double-precision floating-point number equivalent.

Parse(String, NumberStyles)

将指定样式的数字的字符串表示形式转换为它的等效双精度浮点数。Converts the string representation of a number in a specified style to its double-precision floating-point number equivalent.

Parse(String, NumberStyles, IFormatProvider)

将指定样式和区域性特定格式的数字的字符串表示形式转换为它的等效双精度浮点数。Converts the string representation of a number in a specified style and culture-specific format to its double-precision floating-point number equivalent.

ToString()

将此实例的数值转换为其等效的字符串表示形式。Converts the numeric value of this instance to its equivalent string representation.

ToString(IFormatProvider)

使用指定的区域性特定格式信息,将此实例的数值转换为它的等效字符串表示形式。Converts the numeric value of this instance to its equivalent string representation using the specified culture-specific format information.

ToString(String)

使用指定的格式,将此实例的数值转换为它的等效字符串表示形式。Converts the numeric value of this instance to its equivalent string representation, using the specified format.

ToString(String, IFormatProvider)

使用指定的格式和区域性特定格式信息,将此实例的数值转换为它的等效字符串表示形式。Converts the numeric value of this instance to its equivalent string representation using the specified format and culture-specific format information.

TryFormat(Span<Char>, Int32, ReadOnlySpan<Char>, IFormatProvider)
TryParse(ReadOnlySpan<Char>, Double)
TryParse(ReadOnlySpan<Char>, NumberStyles, IFormatProvider, Double)
TryParse(String, Double)

将数字的字符串表示形式转换为它的等效双精度浮点数。Converts the string representation of a number to its double-precision floating-point number equivalent. 一个指示转换是否成功的返回值。A return value indicates whether the conversion succeeded or failed.

TryParse(String, NumberStyles, IFormatProvider, Double)

将指定样式和区域性特定格式的数字的字符串表示形式转换为它的等效双精度浮点数。Converts the string representation of a number in a specified style and culture-specific format to its double-precision floating-point number equivalent. 一个指示转换是否成功的返回值。A return value indicates whether the conversion succeeded or failed.

操作员

Equality(Double, Double)

返回一个值,该值指示两个指定的 Double 值是否相等。Returns a value that indicates whether two specified Double values are equal.

GreaterThan(Double, Double)

返回一个值,该值指示指定的 Double 值是否大于另一个指定的 Double 值。Returns a value that indicates whether a specified Double value is greater than another specified Double value.

GreaterThanOrEqual(Double, Double)

返回一个值,该值指示指定的 Double 值是否大于或等于另一个指定的 Double 值。Returns a value that indicates whether a specified Double value is greater than or equal to another specified Double value.

Inequality(Double, Double)

返回一个值,该值指示两个指定的 Double 值是否不相等。Returns a value that indicates whether two specified Double values are not equal.

LessThan(Double, Double)

返回一个值,该值指示指定的 Double 值是否小于另一个指定的 Double 值。Returns a value that indicates whether a specified Double value is less than another specified Double value.

LessThanOrEqual(Double, Double)

返回一个值,该值指示指定的 Double 值是否小于或等于另一个指定的 Double 值。Returns a value that indicates whether a specified Double value is less than or equal to another specified Double value.

显式界面实现

IComparable.CompareTo(Object)
IConvertible.GetTypeCode()
IConvertible.ToBoolean(IFormatProvider)

有关此成员的说明,请参见 ToBoolean(IFormatProvider)For a description of this member, see ToBoolean(IFormatProvider).

IConvertible.ToByte(IFormatProvider)

有关此成员的说明,请参见 ToByte(IFormatProvider)For a description of this member, see ToByte(IFormatProvider).

IConvertible.ToChar(IFormatProvider)

不支持此转换。This conversion is not supported. 尝试使用此方法将引发 InvalidCastExceptionAttempting to use this method throws an InvalidCastException.

IConvertible.ToDateTime(IFormatProvider)

不支持此转换。This conversion is not supported. 尝试使用此方法将引发 InvalidCastExceptionAttempting to use this method throws an InvalidCastException

IConvertible.ToDecimal(IFormatProvider)

有关此成员的说明,请参见 ToDecimal(IFormatProvider)For a description of this member, see ToDecimal(IFormatProvider).

IConvertible.ToDouble(IFormatProvider)

有关此成员的说明,请参见 ToDouble(IFormatProvider)For a description of this member, see ToDouble(IFormatProvider).

IConvertible.ToInt16(IFormatProvider)

有关此成员的说明,请参见 ToInt16(IFormatProvider)For a description of this member, see ToInt16(IFormatProvider).

IConvertible.ToInt32(IFormatProvider)

有关此成员的说明,请参见 ToInt32(IFormatProvider)For a description of this member, see ToInt32(IFormatProvider).

IConvertible.ToInt64(IFormatProvider)

有关此成员的说明,请参见 ToInt64(IFormatProvider)For a description of this member, see ToInt64(IFormatProvider).

IConvertible.ToSByte(IFormatProvider)

有关此成员的说明,请参见 ToSByte(IFormatProvider)For a description of this member, see ToSByte(IFormatProvider).

IConvertible.ToSingle(IFormatProvider)

有关此成员的说明,请参见 ToSingle(IFormatProvider)For a description of this member, see ToSingle(IFormatProvider).

IConvertible.ToType(Type, IFormatProvider)

有关此成员的说明,请参见 ToType(Type, IFormatProvider)For a description of this member, see ToType(Type, IFormatProvider).

IConvertible.ToUInt16(IFormatProvider)

有关此成员的说明,请参见 ToUInt16(IFormatProvider)For a description of this member, see ToUInt16(IFormatProvider).

IConvertible.ToUInt32(IFormatProvider)

有关此成员的说明,请参见 ToUInt32(IFormatProvider)For a description of this member, see ToUInt32(IFormatProvider).

IConvertible.ToUInt64(IFormatProvider)

有关此成员的说明,请参见 ToUInt64(IFormatProvider)For a description of this member, see ToUInt64(IFormatProvider).

适用于

线程安全性

此类型的所有成员都是线程安全的。All members of this type are thread safe. 看似修改实例状态的成员实际上返回用新值初始化的新实例。Members that appear to modify instance state actually return a new instance initialized with the new value. 与任何其他类型一样,读取和写入包含此类型的实例的共享变量时,必须通过锁保护以保证线程安全。As with any other type, reading and writing to a shared variable that contains an instance of this type must be protected by a lock to guarantee thread safety.

另请参阅