Math.Round 方法

定义

将值舍入到最接近的整数或指定的小数位数。Rounds a value to the nearest integer or to the specified number of fractional digits.

重载

Round(Decimal, Int32, MidpointRounding)

将小数值舍入到指定数量的小数位,并为中点值使用指定的舍入规则。Rounds a decimal value to a specified number of fractional digits, and uses the specified rounding convention for midpoint values.

Round(Double, Int32, MidpointRounding)

将双精度浮点值舍入到指定数量的小数位,并为中点值使用指定的舍入规则。Rounds a double-precision floating-point value to a specified number of fractional digits, and uses the specified rounding convention for midpoint values.

Round(Double, MidpointRounding)

将双精度浮点值舍入到最接近的整数,并为中点值使用指定的舍入约定。Rounds a double-precision floating-point value to the nearest integer, and uses the specified rounding convention for midpoint values.

Round(Double, Int32)

将双精度浮点值舍入到指定数量的小数位,并将中点值舍入到最接近的偶数。Rounds a double-precision floating-point value to a specified number of fractional digits, and rounds midpoint values to the nearest even number.

Round(Double)

将双精度浮点值舍入到最接近的整数值,并将中点值舍入到最接近的偶数。Rounds a double-precision floating-point value to the nearest integral value, and rounds midpoint values to the nearest even number.

Round(Decimal, Int32)

将小数值舍入到指定数量的小数位,并将中点值舍入到最接近的偶数。Rounds a decimal value to a specified number of fractional digits, and rounds midpoint values to the nearest even number.

Round(Decimal)

将小数值舍入到最接近的整数值,并将中点值舍入到最接近的偶数。Rounds a decimal value to the nearest integral value, and rounds midpoint values to the nearest even number.

Round(Decimal, MidpointRounding)

将小数值舍入到最接近的整数,并为中点值使用指定的舍入规则。Rounds a decimal value to the nearest integer, and uses the specified rounding convention for midpoint values.

示例

除了 "备注" 部分中的示例之外,本文还包含一些示例,演示了Math.Round方法的以下重载:In addition to the examples in the Remarks section, this article includes examples that illustrate the following overloads of the Math.Round method:

Math.Round(Decimal)Math.Round(Decimal)
Math.Round(Double)Math.Round(Double)
Math.Round(Decimal, Int32)Math.Round(Decimal, Int32)
Math.Round(Decimal, MidpointRounding)Math.Round(Decimal, MidpointRounding)
Math.Round(Double, Int32)Math.Round(Double, Int32)
Math.Round(Double, MidpointRounding)Math.Round(Double, MidpointRounding)
Math.Round(Decimal, Int32, MidpointRounding)Math.Round(Decimal, Int32, MidpointRounding)
Math.Round(Double, Int32, MidpointRounding)Math.Round(Double, Int32, MidpointRounding)

备注

本文中的 C# 示例运行在 Try.NET 内联代码运行程序和演练环境中。The C# examples in this article run in the Try.NET inline code runner and playground. 选择“运行”按钮以在交互窗口中运行示例 。Select the Run button to run an example in an interactive window. 执行代码后,可通过再次选择“运行”来修改它并运行已修改的代码 。Once you execute the code, you can modify it and run the modified code by selecting Run again. 已修改的代码要么在交互窗口中运行,要么编译失败时,交互窗口将显示所有 C# 编译器错误消息。The modified code either runs in the interactive window or, if compilation fails, the interactive window displays all C# compiler error messages.

注解

本节内容:In this section:

我要调用哪种方法?Which method do I call?

您可以使用下表来选择适当的舍入方法。You can use the following table to select an appropriate rounding method. 除了Math.Round方法之外,它还包含Math.CeilingMath.FloorIn addition to the Math.Round methods, it also includes Math.Ceiling and Math.Floor.

功能To CallCall
使用舍入到最接近的约定,将数字舍入到整数。Round a number to an integer by using the rounding to nearest convention. Round(Decimal)

- 或 --or-

Round(Double)
使用指定的舍入约定将数字舍入到整数。Round a number to an integer by using a specified rounding convention. Round(Decimal, MidpointRounding)

- 或 --or-

Round(Double, MidpointRounding)
使用舍入到最接近的约定,将数字舍入到指定的小数位数。Round a number to a specified number of fractional digits by using the rounding to nearest convention. Round(Decimal, Int32)

- 或 --or-

Round(Double, Int32)
使用指定的舍入约定将数字舍入到指定的小数位数。Round a number to a specified number of fractional digits by using a specified rounding convention. Round(Decimal, Int32, MidpointRounding)

-or-

Round(Double, Int32, MidpointRounding)
使用指定的舍入约定将值舍入到指定的小数位数,并将精度减至最小。SingleRound a Single value to a specified number of fractional digits by using a specified rounding convention and minimizing the loss of precision. 将转换Single Round(Decimal, Int32, MidpointRounding)为,并调用。 DecimalConvert the Single to a Decimal and call Round(Decimal, Int32, MidpointRounding).
将数字舍入到指定的小数位数,同时最大程度地减少舍入点值中的精度问题。Round a number to a specified number of fractional digits while minimizing problems of precision in rounding midpoint values. 调用实现 "大于或等于" 比较的舍入方法。Call a rounding method that implements a "greater than or approximately equal to" comparison. 请参见舍入和精度See Rounding and precision.
将小数值舍入到大于小数值的整数。Round a fractional value to an integer that is greater than the fractional value. 例如,3.1 舍入为4。For example, round 3.1 to 4. Ceiling
将小数值舍入为小于小数值的整数。Round a fractional value to an integer that is less than the fractional value. 例如,3.9 舍入为3。For example, round 3.9 to 3. Floor

中点值和舍入约定Midpoint values and rounding conventions

舍入涉及到使用精度较低的值将数值转换为最接近的值。Rounding involves converting a numeric value with a specified precision to the nearest value with less precision. 例如,可以使用Round(Double)方法将3.4 的值舍入到3.0, Round(Double, Int32)并使用方法将3.579 值舍入到3.58。For example, you can use the Round(Double) method to round a value of 3.4 to 3.0, and the Round(Double, Int32) method to round a value of 3.579 to 3.58.

在中点值中,结果中的最小有效位后的值在两个数字之间精确到了一半。In a midpoint value, the value after the least significant digit in the result is precisely half way between two numbers. 例如,如果要将两个小数位舍入,3.47500 是中值,如果要将它舍入到整数,则7.500 为中点值。For example, 3.47500 is a midpoint value if it is to be rounded two decimal places, and 7.500 is a midpoint value if it is to be rounded to an integer. 在这些情况下,如果没有舍入约定,则无法轻松地识别最接近的值。In these cases, the nearest value can't be easily identified without a rounding convention.

Round方法支持两个用于处理中点值的舍入约定:The Round method supports two rounding conventions for handling midpoint values:

  • 远离零的舍入Rounding away from zero

    中值舍入到下一个数字,而不是零。Midpoint values are rounded to the next number away from zero. 例如,3.75 舍入为3.8,3.85 舍入到3.9,-3.75 舍入到-3.8,-3.85 舍入到-3.9。For example, 3.75 rounds to 3.8, 3.85 rounds to 3.9, -3.75 rounds to -3.8, and -3.85 rounds to -3.9. 这种形式的舍入由MidpointRounding.AwayFromZero枚举成员表示。This form of rounding is represented by the MidpointRounding.AwayFromZero enumeration member.

    远离零的舍入是最广泛的舍入形式。Rounding away from zero is the most widely known form of rounding.

  • 舍入为最接近或按下舍入Rounding to nearest, or banker's rounding

    中值舍入为最接近的偶数。Midpoint values are rounded to the nearest even number. 例如,3.75 和3.85 舍入为3.8,-3.75 和-3.85 舍入到-3.8。For example, both 3.75 and 3.85 round to 3.8, and both -3.75 and -3.85 round to -3.8. 这种形式的舍入由MidpointRounding.ToEven枚举成员表示。This form of rounding is represented by the MidpointRounding.ToEven enumeration member.

    舍入到最接近的是在财务和统计操作中使用的标准舍入形式。Rounding to nearest is the standard form of rounding used in financial and statistical operations. 它符合 IEEE Standard 754,第4部分。It conforms to IEEE Standard 754, section 4. 当在多个舍入操作中使用时,它可减少因按单一方向持续舍入点值而导致的舍入误差。When used in multiple rounding operations, it reduces the rounding error that is caused by consistently rounding midpoint values in a single direction. 在某些情况下,此舍入误差很重要。In some cases, this rounding error can be significant.

    下面的示例演示了一个方向上按一致舍入的点值导致的偏差。The following example illustrates the bias that can result from consistently rounding midpoint values in a single direction. 该示例计算Decimal值数组的 true 平均值,然后使用两种约定来计算数组中的值的值。The example computes the true mean of an array of Decimal values, and then computes the mean when the values in the array are rounded by using the two conventions. 在此示例中,如果舍入到最接近的是相同的,则为真正的平均值和结果。In this example, the true mean and the mean that results when rounding to nearest are the same. 但是,当从零进行舍入时得出的结果将因 .05 (或 3.6%)而异从真正的平均值。However, the mean that results when rounding away from zero differs by .05 (or by 3.6%) from the true mean.

    decimal[] values = { 1.15m, 1.25m, 1.35m, 1.45m, 1.55m, 1.65m };
    decimal sum = 0;
    
    // Calculate true mean.
    foreach (var value in values)
       sum += value;
    
    Console.WriteLine("True mean:     {0:N2}", sum/values.Length);
    
    // Calculate mean with rounding away from zero.
    sum = 0;
    foreach (var value in values)
       sum += Math.Round(value, 1, MidpointRounding.AwayFromZero);
    
    Console.WriteLine("AwayFromZero:  {0:N2}", sum/values.Length);
    
    // Calculate mean with rounding to nearest.
    sum = 0;
    foreach (var value in values)
       sum += Math.Round(value, 1, MidpointRounding.ToEven);
    
    Console.WriteLine("ToEven:        {0:N2}", sum/values.Length);
    // The example displays the following output:
    //       True mean:     1.40
    //       AwayFromZero:  1.45
    //       ToEven:        1.40
    
    Module Example
       Public Sub Main()
          Dim values() As Decimal = { 1.15d, 1.25d, 1.35d, 1.45d, 1.55d, 1.65d }
          Dim sum As Decimal
          
          ' Calculate true mean.
          For Each value In values
             sum += value
          Next
          Console.WriteLine("True mean:     {0:N2}", sum/values.Length)
          
          ' Calculate mean with rounding away from zero.
          sum = 0
          For Each value In values
             sum += Math.Round(value, 1, MidpointRounding.AwayFromZero)
          Next
          Console.WriteLine("AwayFromZero:  {0:N2}", sum/values.Length)
          
          ' Calculate mean with rounding to nearest.
          sum = 0
          For Each value In values
             sum += Math.Round(value, 1, MidpointRounding.ToEven)
          Next
          Console.WriteLine("ToEven:        {0:N2}", sum/values.Length)
       End Sub
    End Module
    ' The example displays the following output:
    '       True mean:     1.40
    '       AwayFromZero:  1.45
    '       ToEven:        1.40
    

默认情况下, Round该方法使用舍入到最接近的约定。By default, the Round method uses the rounding to nearest convention. 下表列出了Round方法的重载以及每个重载使用的舍入约定。The following table lists the overloads of the Round method and the rounding convention that each uses.

重载Overload 舍入约定Rounding convention
Round(Decimal) ToEven
Round(Double) ToEven
Round(Decimal, Int32) ToEven
Round(Double, Int32) ToEven
Round(Decimal, MidpointRounding) mode参数确定。Determined by mode parameter.
Round(Double, MidpointRounding) mode参数确定Determined by mode parameter
Round(Decimal, Int32, MidpointRounding) mode参数确定Determined by mode parameter
Round(Double, Int32, MidpointRounding) mode参数确定Determined by mode parameter

舍入和精度Rounding and precision

若要确定舍入运算是否包含中点值, Round方法会将原始值乘以 10n,其中n是返回值中所需的小数位数,然后确定该值的剩余小数部分是否大于或等于5。In order to determine whether a rounding operation involves a midpoint value, the Round method multiplies the original value to be rounded by 10n, where n is the desired number of fractional digits in the return value, and then determines whether the remaining fractional portion of the value is greater than or equal to .5. 这对于测试相等性稍有不同,如Double参考主题的 "测试相等性" 一节中所述,与浮点值的相等性测试是否存在问题,因为浮点格式的二进制文件存在问题表示形式和精度。This is a slight variation on a test for equality, and as discussed in the "Testing for Equality" section of the Double reference topic, tests for equality with floating-point values are problematic because of the floating-point format's issues with binary representation and precision. 这意味着,不会向后舍入的数字的小数部分稍微小于0.5 (因为精度损失)。This means that any fractional portion of a number that is slightly less than .5 (because of a loss of precision) will not be rounded upward.

重要

在舍入中点值时,舍入算法执行相等测试。When rounding midpoint values, the rounding algorithm performs an equality test. 由于浮点格式存在的二进制表示和精度问题,因此方法返回的值可能异常。Because of problems of binary representation and precision in the floating-point format, the value returned by the method can be unexpected. 有关详细信息,请参阅“舍入和精度”For more information, see "Rounding and precision".

以下示例演示了该问题。The following example illustrates the problem. 它会重复将 .1 添加到11.0,并将结果舍入到最接近的整数。It repeatedly adds .1 to 11.0 and rounds the result to the nearest integer. 无论舍入约定如何,11.5 都应该舍入到12。Regardless of the rounding convention, 11.5 should round to 12. 但是,如示例的输出所示,它不会。However, as the output from the example shows, it does not. 该示例使用 "R"标准数字格式字符串来显示浮点值的完整精度,并显示在重复添加时要舍入的值已丢失精度,并且它的值实际上是11.499999999999998 的。The example uses the "R" standard numeric format string to display the floating point value's full precision, and shows that the value to be rounded has lost precision during repeated additions, and its value is actually 11.499999999999998. 由于499999999999998小于 .5,因此不将值舍入到下一个最大整数。Because .499999999999998 is less than .5, the value is not rounded to the next highest integer. 如示例中所示,如果仅将常量值11.5 赋给Double变量,则不会出现此问题。As the example also shows, this problem does not occur if we simply assign the constant value 11.5 to a Double variable.

using System;

public class Example
{
   public static void Main()
   {
      Console.WriteLine("{0,5} {1,20:R}  {2,12} {3,15}\n", 
                        "Value", "Full Precision", "ToEven",
                        "AwayFromZero");
      double value = 11.1;
      for (int ctr = 0; ctr <= 5; ctr++)    
         value = RoundValueAndAdd(value);

      Console.WriteLine();

      value = 11.5;
      RoundValueAndAdd(value);
   }
   
   private static double RoundValueAndAdd(double value)
   {
      Console.WriteLine("{0,5:N1} {0,20:R}  {1,12} {2,15}", 
                        value, Math.Round(value, MidpointRounding.ToEven),
                        Math.Round(value, MidpointRounding.AwayFromZero));
      return value + .1;
   }
}
// The example displays the following output:
//       Value       Full Precision        ToEven    AwayFromZero
//       
//        11.1                 11.1            11              11
//        11.2                 11.2            11              11
//        11.3   11.299999999999999            11              11
//        11.4   11.399999999999999            11              11
//        11.5   11.499999999999998            11              11
//        11.6   11.599999999999998            12              12
//       
//        11.5                 11.5            12              12
Module Example
   Public Sub Main()
      Dim value As Double = 11.1

      Console.WriteLine("{0,5} {1,20:R}  {2,12} {3,15}", 
                        "Value", "Full Precision", "ToEven",
                        "AwayFromZero")
      Console.WriteLine()
      For ctr As Integer = 0 To 5    
         value = RoundValueAndAdd(value)
      Next
      Console.WriteLine()

      value = 11.5
      RoundValueAndAdd(value)
   End Sub
   
   Private Function RoundValueAndAdd(value As Double) As Double
      Console.WriteLine("{0,5:N1} {0,20:R}  {1,12} {2,15}", 
                        value, Math.Round(value, MidpointRounding.ToEven),
                        Math.Round(value, MidpointRounding.AwayFromZero))
      Return value + .1
   End Function   
End Module
' The example displays the following output:
'       Value       Full Precision        ToEven    AwayFromZero
'       
'        11.1                 11.1            11              11
'        11.2                 11.2            11              11
'        11.3   11.299999999999999            11              11
'        11.4   11.399999999999999            11              11
'        11.5   11.499999999999998            11              11
'        11.6   11.599999999999998            12              12
'       
'        11.5                 11.5            12              12

在以下情况下,舍入点值中的精度问题最可能出现在其中:Problems of precision in rounding midpoint values are most likely to arise in the following conditions:

  • 当小数值不能精确表示为浮点类型的二进制格式时。When a fractional value cannot be expressed precisely in the floating-point type's binary format.

  • 要舍入的值是从一个或多个浮点运算计算得出的。When the value to be rounded is calculated from one or more floating-point operations.

  • 当要舍入的值是Single而不Double是或Decimal时。When the value to be rounded is a Single rather than a Double or Decimal. 有关详细信息,请参阅下一节舍入和单精度浮点值For more information, see the next section, Rounding and single-precision floating-point values.

如果舍入操作缺少精度,则可以执行以下操作:In cases where the lack of precision in rounding operations is problematic, you can do the following:

  • 如果舍入操作Double调用了舍入值的重载,则可以Double将更改为Decimal Decimal值并调用舍入值的重载。If the rounding operation calls an overload that rounds a Double value, you can change the Double to a Decimal value and call an overload that rounds a Decimal value instead. Decimal尽管数据类型也有问题的表示形式和损失,但这些问题远远不常见。Although the Decimal data type also has problems of representation and loss of precision, these issues are far less common.

  • 定义一种自定义舍入算法,该算法执行 "将近相等" 测试来确定要舍入的值是否接近于中点值可接受。Define a custom rounding algorithm that performs a "nearly equal" test to determine whether the value to be rounded is acceptably close to a midpoint value. 下面的示例定义一个RoundApproximate方法,该方法检查小数值是否足以接近于中点舍入的中点值。The following example defines a RoundApproximate method that examines whether a fractional value is sufficiently near to a midpoint value to be subject to midpoint rounding. 如示例的输出所示,它会更正上一个示例中显示的舍入问题。As the output from the example shows, it corrects the rounding problem shown in the previous example.

    using System;
    
    public class Example
    {
       public static void Main()
       {
          Console.WriteLine("{0,5} {1,20:R}  {2,12} {3,15}\n", 
                            "Value", "Full Precision", "ToEven",
                            "AwayFromZero");
          double value = 11.1;
          for (int ctr = 0; ctr <= 5; ctr++)    
             value = RoundValueAndAdd(value);
    
          Console.WriteLine();
    
          value = 11.5;
          RoundValueAndAdd(value);
       }
       
       private static double RoundValueAndAdd(double value)
       {
          const double tolerance = 8e-14;
    
          Console.WriteLine("{0,5:N1} {0,20:R}  {1,12} {2,15}", 
                            value, 
                            RoundApproximate(value, 0, tolerance, MidpointRounding.ToEven),
                            RoundApproximate(value, 0, tolerance, MidpointRounding.AwayFromZero));
          return value + .1;
       }
    
       private static double RoundApproximate(double dbl, int digits, double margin, 
                                         MidpointRounding mode)
       {                                      
          double fraction = dbl * Math.Pow(10, digits);
          double value = Math.Truncate(fraction); 
          fraction = fraction - value;   
          if (fraction == 0)
             return dbl;
          
          double tolerance = margin * dbl;
          // Determine whether this is a midpoint value.
          if ((fraction >= .5 - tolerance) & (fraction <= .5 + tolerance)) {
             if (mode == MidpointRounding.AwayFromZero)
                return (value + 1)/Math.Pow(10, digits);
             else
                if (value % 2 != 0)
                   return (value + 1)/Math.Pow(10, digits);
                else
                   return value/Math.Pow(10, digits);
          }
          // Any remaining fractional value greater than .5 is not a midpoint value.
          if (fraction > .5)
             return (value + 1)/Math.Pow(10, digits);
          else
             return value/Math.Pow(10, digits);
       }
    }
    // The example displays the following output:
    //       Value       Full Precision        ToEven    AwayFromZero
    //       
    //        11.1                 11.1            11              11
    //        11.2                 11.2            11              11
    //        11.3   11.299999999999999            11              11
    //        11.4   11.399999999999999            11              11
    //        11.5   11.499999999999998            12              12
    //        11.6   11.599999999999998            12              12
    //       
    //        11.5                 11.5            12              12
    
    Module Example
       Public Sub Main()
          Dim value As Double = 11.1
    
          Console.WriteLine("{0,5} {1,20:R}  {2,12} {3,15}\n", 
                            "Value", "Full Precision", "ToEven",
                            "AwayFromZero")
          For ctr As Integer = 0 To 5    
             value = RoundValueAndAdd(value)
          Next
          Console.WriteLine()
    
          value = 11.5
          RoundValueAndAdd(value)
       End Sub
       
       Private Function RoundValueAndAdd(value As Double) As Double
          Const tolerance As Double = 8e-14
          Console.WriteLine("{0,5:N1} {0,20:R}  {1,12} {2,15}", 
                            value, 
                            RoundApproximate(value, 0, tolerance, MidpointRounding.ToEven),
                            RoundApproximate(value, 0, tolerance, MidpointRounding.AwayFromZero))
          Return value + .1
       End Function   
    
       Private Function RoundApproximate(dbl As Double, digits As Integer, margin As Double, 
                                         mode As MidpointRounding) As Double
          Dim fraction As Double = dbl * Math.Pow(10, digits)
          Dim value As Double = Math.Truncate(fraction) 
          fraction = fraction - value   
          If fraction = 0 Then Return dbl
          
          Dim tolerance As Double = margin * dbl
          ' Determine whether this is a midpoint value.
          If (fraction >= .5 - tolerance) And (fraction <= .5 + tolerance) Then
             If mode = MidpointRounding.AwayFromZero Then
                Return (value + 1)/Math.Pow(10, digits)
             Else
                If value Mod 2 <> 0 Then
                   Return (value + 1)/Math.Pow(10, digits)
                Else
                   Return value/Math.Pow(10, digits)
                End If
             End If
          End If
          ' Any remaining fractional value greater than .5 is not a midpoint value.
          If fraction > .5 Then
             Return (value + 1)/Math.Pow(10, digits)
          Else
             return value/Math.Pow(10, digits)
          End If      
       End Function
    End Module
    ' The example displays the following output:
    '       Value       Full Precision        ToEven    AwayFromZero
    '       
    '        11.1                 11.1            11              11
    '        11.2                 11.2            11              11
    '        11.3   11.299999999999999            11              11
    '        11.4   11.399999999999999            11              11
    '        11.5   11.499999999999998            12              12
    '        11.6   11.599999999999998            12              12
    '       
    '        11.5                 11.5            12              12
    

舍入和单精度浮点值Rounding and single-precision floating-point values

方法包含接受类型Decimal为和Double的参数的重载。 RoundThe Round method includes overloads that accept arguments of type Decimal and Double. 没有用于舍入类型Single的值的方法。There are no methods that round values of type Single. Single如果将值传递DoubleRound方法的重载之一,则会将其强制转换(在中C#)或转换(在 Visual Basic 中)到,并调用具有Round Double参数的对应重载。If you pass a Single value to one of the overloads of the Round method, it is cast (in C#) or converted (in Visual Basic) to a Double, and the corresponding Round overload with a Double parameter is called. 尽管这是一个扩大转换,但它通常会损失精度,如下面的示例所示。Although this is a widening conversion, it often involves a loss of precision, as the following example illustrates. 如果将16.325 的Round 值传递给方法,并使用舍入到最接近的约定舍入为两位小数位数,则结果为16.33,而不是预期的16.32结果。SingleWhen a Single value of 16.325 is passed to the Round method and rounded to two decimal places using the rounding to nearest convention, the result is 16.33 and not the expected result of 16.32.

using System;

public class Example
{
   public static void Main()
   {
      Single value = 16.325f;
      Console.WriteLine("Widening Conversion of {0:R} (type {1}) to {2:R} (type {3}): ", 
                        value, value.GetType().Name, (double) value, 
                        ((double) (value)).GetType().Name);
      Console.WriteLine(Math.Round(value, 2));
      Console.WriteLine(Math.Round(value, 2, MidpointRounding.AwayFromZero));
      Console.WriteLine();
      
      Decimal decValue = (decimal) value;
      Console.WriteLine("Cast of {0:R} (type {1}) to {2} (type {3}): ", 
                        value, value.GetType().Name, decValue, 
                        decValue.GetType().Name);
      Console.WriteLine(Math.Round(decValue, 2));
      Console.WriteLine(Math.Round(decValue, 2, MidpointRounding.AwayFromZero));
   }
}
// The example displays the following output:
//    Widening Conversion of 16.325 (type Single) to 16.325000762939453 (type Double):
//    16.33
//    16.33
//    
//    Cast of 16.325 (type Single) to 16.325 (type Decimal):
//    16.32
//    16.33
Module Example
   Public Sub Main()
      Dim value As Single = 16.325
      Console.WriteLine("Widening Conversion of {0:R} (type {1}) to {2:R} (type {3}): ", 
                        value, value.GetType().Name, CDbl(value), 
                        CDbl(value).GetType().Name)
      Console.WriteLine(Math.Round(value, 2))
      Console.WriteLine(Math.Round(value, 2, MidpointRounding.AwayFromZero))
      Console.WriteLine()
      
      Dim decValue As Decimal = CDec(value)
      Console.WriteLine("Cast of {0:R} (type {1}) to {2} (type {3}): ", 
                        value, value.GetType().Name, decValue, 
                        decValue.GetType().Name)
      Console.WriteLine(Math.Round(decValue, 2))
      Console.WriteLine(Math.Round(decValue, 2, MidpointRounding.AwayFromZero))
      Console.WriteLine()
   End Sub
End Module
' The example displays the following output:
'    Widening Conversion of 16.325 (type Single) to 16.325000762939453 (type Double):
'    16.33
'    16.33
'    
'    Cast of 16.325 (type Single) to 16.325 (type Decimal):
'    16.32
'    16.33

这种意外的结果是因为在将Single值转换Double为时精度丢失。This unexpected result is due to a loss of precision in the conversion of the Single value to a Double. 由于16.325000762939453 的Double结果值不是中点值并且大于16.325,因此始终向上舍入。Because the resulting Double value of 16.325000762939453 is not a midpoint value and is greater than 16.325, it is always rounded upward.

在许多情况下,如示例中所示,通过将Single值强制转换或转换Decimal为,可以最大程度地减少或消除精度损失。In many cases, as the example illustrates, the loss of precision can be minimized or eliminated by casting or converting the Single value to a Decimal. 请注意,因为这是收缩转换,所以需要使用强制转换运算符或调用转换方法。Note that, because this is a narrowing conversion, it requires using a cast operator or calling a conversion method.

Round(Decimal, Int32, MidpointRounding)

将小数值舍入到指定数量的小数位,并为中点值使用指定的舍入规则。Rounds a decimal value to a specified number of fractional digits, and uses the specified rounding convention for midpoint values.

public:
 static System::Decimal Round(System::Decimal d, int decimals, MidpointRounding mode);
public static decimal Round (decimal d, int decimals, MidpointRounding mode);
static member Round : decimal * int * MidpointRounding -> decimal
Public Shared Function Round (d As Decimal, decimals As Integer, mode As MidpointRounding) As Decimal

参数

d
Decimal

要舍入的小数。A decimal number to be rounded.

decimals
Int32

返回值中的小数位数。The number of decimal places in the return value.

mode
MidpointRounding

在两个数字之间时如何舍入 d 的规范。Specification for how to round d if it is midway between two other numbers.

返回

最接近 ddecimals 位小数的数字。The number nearest to d that contains a number of fractional digits equal to decimals. 如果 ddecimals 少部分数字,d 原样返回。If d has fewer fractional digits than decimals, d is returned unchanged.

异常

decimals 小于 0 或大于 28。decimals is less than 0 or greater than 28.

mode 不是 MidpointRounding 的一个有效值。mode is not a valid value of MidpointRounding.

结果超出了 Decimal 的范围。The result is outside the range of a Decimal.

注解

请参阅中点值和舍入约定,了解有关带中点值的舍入的信息。See Midpoint values and rounding conventions for information about rounding numbers with midpoint values.

重要

在舍入中点值时,舍入算法执行相等测试。When rounding midpoint values, the rounding algorithm performs an equality test. 由于浮点格式存在的二进制表示和精度问题,因此方法返回的值可能异常。Because of problems of binary representation and precision in the floating-point format, the value returned by the method can be unexpected. 有关详细信息,请参阅“舍入和精度”For more information, see "Rounding and precision".

decimals参数的值的范围为0到28。The value of the decimals argument can range from 0 to 28.

示例Example

下面的示例演示如何将Round方法MidpointRounding与枚举一起使用。The following example demonstrates how to use the Round method with the MidpointRounding enumeration.

// This example demonstrates the Math.Round() method in conjunction 
// with the MidpointRounding enumeration.
using namespace System;

void main()
{
    Decimal result = (Decimal) 0.0;
    Decimal posValue = (Decimal) 3.45;
    Decimal negValue = (Decimal) -3.45;

    // By default, round a positive and a negative value to the nearest
    // even number. The precision of the result is 1 decimal place.
    result = Math::Round(posValue, 1);
    Console::WriteLine("{0,4} = Math.Round({1,5}, 1)", result, posValue);
    result = Math::Round(negValue, 1);
    Console::WriteLine("{0,4} = Math.Round({1,5}, 1)", result, negValue);
    Console::WriteLine();

    // Round a positive value to the nearest even number, then to the
    // nearest number away from zero. The precision of the result is 1
    // decimal place.
    result = Math::Round(posValue, 1, MidpointRounding::ToEven);
    Console::WriteLine(
        "{0,4} = Math.Round({1,5}, 1, MidpointRounding.ToEven)",
        result, posValue);
    result = Math::Round(posValue, 1, MidpointRounding::AwayFromZero);
    Console::WriteLine(
        "{0,4} = Math.Round({1,5}, 1, MidpointRounding.AwayFromZero)",
        result, posValue);
    Console::WriteLine();

    // Round a negative value to the nearest even number, then to the
    // nearest number away from zero. The precision of the result is 1
    // decimal place.
    result = Math::Round(negValue, 1, MidpointRounding::ToEven);
    Console::WriteLine(
        "{0,4} = Math.Round({1,5}, 1, MidpointRounding.ToEven)",
        result, negValue);
    result = Math::Round(negValue, 1, MidpointRounding::AwayFromZero);
    Console::WriteLine(
        "{0,4} = Math.Round({1,5}, 1, MidpointRounding.AwayFromZero)",
        result, negValue);
    Console::WriteLine();
}

/*
This code example produces the following results:

3.4 = Math.Round( 3.45, 1)
-3.4 = Math.Round(-3.45, 1)

3.4 = Math.Round( 3.45, 1, MidpointRounding.ToEven)
3.5 = Math.Round( 3.45, 1, MidpointRounding.AwayFromZero)

-3.4 = Math.Round(-3.45, 1, MidpointRounding.ToEven)
-3.5 = Math.Round(-3.45, 1, MidpointRounding.AwayFromZero)

*/
decimal result = 0.0m;
decimal posValue =  3.45m;
decimal negValue = -3.45m;

// By default, round a positive and a negative value to the nearest even number. 
// The precision of the result is 1 decimal place.

result = Math.Round(posValue, 1);
Console.WriteLine("{0,4} = Math.Round({1,5}, 1)", result, posValue);
result = Math.Round(negValue, 1);
Console.WriteLine("{0,4} = Math.Round({1,5}, 1)", result, negValue);
Console.WriteLine();

// Round a positive value to the nearest even number, then to the nearest number away from zero. 
// The precision of the result is 1 decimal place.

result = Math.Round(posValue, 1, MidpointRounding.ToEven);
Console.WriteLine("{0,4} = Math.Round({1,5}, 1, MidpointRounding.ToEven)", result, posValue);
result = Math.Round(posValue, 1, MidpointRounding.AwayFromZero);
Console.WriteLine("{0,4} = Math.Round({1,5}, 1, MidpointRounding.AwayFromZero)", result, posValue);
Console.WriteLine();

// Round a negative value to the nearest even number, then to the nearest number away from zero. 
// The precision of the result is 1 decimal place.

result = Math.Round(negValue, 1, MidpointRounding.ToEven);
Console.WriteLine("{0,4} = Math.Round({1,5}, 1, MidpointRounding.ToEven)", result, negValue);
result = Math.Round(negValue, 1, MidpointRounding.AwayFromZero);
Console.WriteLine("{0,4} = Math.Round({1,5}, 1, MidpointRounding.AwayFromZero)", result, negValue);
Console.WriteLine();
/*
This code example produces the following results:

3.4 = Math.Round( 3.45, 1)
-3.4 = Math.Round(-3.45, 1)

3.4 = Math.Round( 3.45, 1, MidpointRounding.ToEven)
3.5 = Math.Round( 3.45, 1, MidpointRounding.AwayFromZero)

-3.4 = Math.Round(-3.45, 1, MidpointRounding.ToEven)
-3.5 = Math.Round(-3.45, 1, MidpointRounding.AwayFromZero)

*/
' This example demonstrates the Math.Round() method in conjunction 
' with the MidpointRounding enumeration.
Class Sample
    Public Shared Sub Main() 
        Dim result As Decimal = 0D
        Dim posValue As Decimal = 3.45D
        Dim negValue As Decimal = -3.45D
        
        ' By default, round a positive and a negative value to the nearest even number. 
        ' The precision of the result is 1 decimal place.
        result = Math.Round(posValue, 1)
        Console.WriteLine("{0,4} = Math.Round({1,5}, 1)", result, posValue)
        result = Math.Round(negValue, 1)
        Console.WriteLine("{0,4} = Math.Round({1,5}, 1)", result, negValue)
        Console.WriteLine()
        
        ' Round a positive value to the nearest even number, then to the nearest number 
        ' away from zero. The precision of the result is 1 decimal place.
        result = Math.Round(posValue, 1, MidpointRounding.ToEven)
        Console.WriteLine("{0,4} = Math.Round({1,5}, 1, MidpointRounding.ToEven)", _
                           result, posValue)
        result = Math.Round(posValue, 1, MidpointRounding.AwayFromZero)
        Console.WriteLine("{0,4} = Math.Round({1,5}, 1, MidpointRounding.AwayFromZero)", _
                           result, posValue)
        Console.WriteLine()
        
        ' Round a negative value to the nearest even number, then to the nearest number 
        ' away from zero. The precision of the result is 1 decimal place.
        result = Math.Round(negValue, 1, MidpointRounding.ToEven)
        Console.WriteLine("{0,4} = Math.Round({1,5}, 1, MidpointRounding.ToEven)", _
                            result, negValue)
        result = Math.Round(negValue, 1, MidpointRounding.AwayFromZero)
        Console.WriteLine("{0,4} = Math.Round({1,5}, 1, MidpointRounding.AwayFromZero)", _
                           result, negValue)
        Console.WriteLine()
    
    End Sub
End Class
'
'This code example produces the following results:
'
' 3.4 = Math.Round( 3.45, 1)
'-3.4 = Math.Round(-3.45, 1)
'
' 3.4 = Math.Round( 3.45, 1, MidpointRounding.ToEven)
' 3.5 = Math.Round( 3.45, 1, MidpointRounding.AwayFromZero)
'
'-3.4 = Math.Round(-3.45, 1, MidpointRounding.ToEven)
'-3.5 = Math.Round(-3.45, 1, MidpointRounding.AwayFromZero)
'

另请参阅

Round(Double, Int32, MidpointRounding)

将双精度浮点值舍入到指定数量的小数位,并为中点值使用指定的舍入规则。Rounds a double-precision floating-point value to a specified number of fractional digits, and uses the specified rounding convention for midpoint values.

public:
 static double Round(double value, int digits, MidpointRounding mode);
public static double Round (double value, int digits, MidpointRounding mode);
static member Round : double * int * MidpointRounding -> double
Public Shared Function Round (value As Double, digits As Integer, mode As MidpointRounding) As Double

参数

value
Double

要舍入的双精度浮点数。A double-precision floating-point number to be rounded.

digits
Int32

返回值中的小数数字。The number of fractional digits in the return value.

mode
MidpointRounding

在两个数字之间时如何舍入 value 的规范。Specification for how to round value if it is midway between two other numbers.

返回

最接近 valuedigits 位小数的数字。The number nearest to value that has a number of fractional digits equal to digits. 如果 valuedigits 少部分数字,value 原样返回。If value has fewer fractional digits than digits, value is returned unchanged.

异常

digits 小于 0 或大于 15。digits is less than 0 or greater than 15.

mode 不是 MidpointRounding 的一个有效值。mode is not a valid value of MidpointRounding.

注解

digits参数的值的范围为0到15。The value of the digits argument can range from 0 to 15. 请注意,15是该Double类型支持的整数和小数位数的最大值。Note that 15 is the maximum number of integral and fractional digits supported by the Double type.

请参阅中点值和舍入约定,了解有关带中点值的舍入的信息。See Midpoint values and rounding conventions for information about rounding numbers with midpoint values.

重要

在舍入中点值时,舍入算法执行相等测试。When rounding midpoint values, the rounding algorithm performs an equality test. 由于浮点格式存在的二进制表示和精度问题,因此方法返回的值可能异常。Because of problems of binary representation and precision in the floating-point format, the value returned by the method can be unexpected. 有关详细信息,请参阅“舍入和精度”For more information, see "Rounding and precision".

如果value参数的值为Double.NaN,则该方法返回Double.NaNIf the value of the value argument is Double.NaN, the method returns Double.NaN. 如果valueDouble.PositiveInfinityDouble.PositiveInfinity Double.NegativeInfinity,则该方法将分别返回或。 Double.NegativeInfinityIf value is Double.PositiveInfinity or Double.NegativeInfinity, the method returns Double.PositiveInfinity or Double.NegativeInfinity, respectively.

示例Example

下面的示例演示如何将Round(Double, Int32, MidpointRounding)方法MidpointRounding与枚举一起使用。The following example demonstrates how to use the Round(Double, Int32, MidpointRounding) method with the MidpointRounding enumeration.

 double posValue =  3.45;
 double negValue = -3.45;

 // Round a positive and a negative value using the default. 
 double result = Math.Round(posValue, 1);
 Console.WriteLine("{0,4} = Math.Round({1,5}, 1)", result, posValue);
 result = Math.Round(negValue, 1);
 Console.WriteLine("{0,4} = Math.Round({1,5}, 1)\n", result, negValue);

 // Round a positive value using a MidpointRounding value. 
 result = Math.Round(posValue, 1, MidpointRounding.ToEven);
 Console.WriteLine("{0,4} = Math.Round({1,5}, 1, MidpointRounding.ToEven)", 
                   result, posValue);
 result = Math.Round(posValue, 1, MidpointRounding.AwayFromZero);
 Console.WriteLine("{0,4} = Math.Round({1,5}, 1, MidpointRounding.AwayFromZero)\n", 
                   result, posValue);

 // Round a negative value using a MidpointRounding value. 
 result = Math.Round(negValue, 1, MidpointRounding.ToEven);
 Console.WriteLine("{0,4} = Math.Round({1,5}, 1, MidpointRounding.ToEven)", 
                   result, negValue);
 result = Math.Round(negValue, 1, MidpointRounding.AwayFromZero);
 Console.WriteLine("{0,4} = Math.Round({1,5}, 1, MidpointRounding.AwayFromZero)\n", 
                   result, negValue);
// The example displays the following output:
//        3.4 = Math.Round( 3.45, 1)
//       -3.4 = Math.Round(-3.45, 1)
//       
//        3.4 = Math.Round( 3.45, 1, MidpointRounding.ToEven)
//        3.5 = Math.Round( 3.45, 1, MidpointRounding.AwayFromZero)
//       
//       -3.4 = Math.Round(-3.45, 1, MidpointRounding.ToEven)
//       -3.5 = Math.Round(-3.45, 1, MidpointRounding.AwayFromZero)
Module Example
    Public Sub Main() 
        Dim posValue As Double = 3.45
        Dim negValue As Double = -3.45
        
        ' Round a positive and a negative value using the default.  
        Dim result As Double = Math.Round(posValue, 1)
        Console.WriteLine("{0,4} = Math.Round({1,5}, 1)", result, posValue)
        result = Math.Round(negValue, 1)
        Console.WriteLine("{0,4} = Math.Round({1,5}, 1)", result, negValue)
        Console.WriteLine()
        
        ' Round a positive value using a MidpointRounding value. 
        result = Math.Round(posValue, 1, MidpointRounding.ToEven)
        Console.WriteLine("{0,4} = Math.Round({1,5}, 1, MidpointRounding.ToEven)", 
                           result, posValue)
        result = Math.Round(posValue, 1, MidpointRounding.AwayFromZero)
        Console.WriteLine("{0,4} = Math.Round({1,5}, 1, MidpointRounding.AwayFromZero)", 
                           result, posValue)
        Console.WriteLine()
        
        ' Round a positive value using a MidpointRounding value. 
        result = Math.Round(negValue, 1, MidpointRounding.ToEven)
        Console.WriteLine("{0,4} = Math.Round({1,5}, 1, MidpointRounding.ToEven)", 
                            result, negValue)
        result = Math.Round(negValue, 1, MidpointRounding.AwayFromZero)
        Console.WriteLine("{0,4} = Math.Round({1,5}, 1, MidpointRounding.AwayFromZero)", 
                           result, negValue)
        Console.WriteLine()
    End Sub
End Module
' The example displays the following output:
'       3.4 = Math.Round( 3.45, 1)
'       -3.4 = Math.Round(-3.45, 1)
'       
'       3.4 = Math.Round( 3.45, 1, MidpointRounding.ToEven)
'       3.5 = Math.Round( 3.45, 1, MidpointRounding.AwayFromZero)
'       
'       -3.4 = Math.Round(-3.45, 1, MidpointRounding.ToEven)
'       -3.5 = Math.Round(-3.45, 1, MidpointRounding.AwayFromZero)

调用方说明

由于将十进制值表示为浮点数或对浮点值执行算术运算可能导致精度损失,因此,在某些情况下,该Round(Double, Int32, MidpointRounding)方法可能不会显示为按指定舍入的中值mode按参数。Because of the loss of precision that can result from representing decimal values as floating-point numbers or performing arithmetic operations on floating-point values, in some cases the Round(Double, Int32, MidpointRounding) method may not appear to round midpoint values as specified by the mode parameter. 下面的示例对此进行了说明,其中,2.135 舍入为2.13 而不是2.14。This is illustrated in the following example, where 2.135 is rounded to 2.13 instead of 2.14. 出现这种情况的原因是value ,此方法在内部乘以 10位数,而乘法运算在这种情况下会损失精度。This occurs because internally the method multiplies value by 10digits, and the multiplication operation in this case suffers from a loss of precision.

[!code-csharpSystem.Math.Round#3] [!code-vbSystem.Math.Round#3][!code-csharpSystem.Math.Round#3] [!code-vbSystem.Math.Round#3]

另请参阅

Round(Double, MidpointRounding)

将双精度浮点值舍入到最接近的整数,并为中点值使用指定的舍入约定。Rounds a double-precision floating-point value to the nearest integer, and uses the specified rounding convention for midpoint values.

public:
 static double Round(double value, MidpointRounding mode);
public static double Round (double value, MidpointRounding mode);
static member Round : double * MidpointRounding -> double
Public Shared Function Round (value As Double, mode As MidpointRounding) As Double

参数

value
Double

要舍入的双精度浮点数。A double-precision floating-point number to be rounded.

mode
MidpointRounding

在两个数字之间时如何舍入 value 的规范。Specification for how to round value if it is midway between two other numbers.

返回

最接近 value 的整数。The integer nearest value. 如果 value 是两个整数的中值,这两个整数一个为偶数,另一个为奇数,则 mode 确定返回两个整数中的哪一个。If value is halfway between two integers, one of which is even and the other odd, then mode determines which of the two is returned. 请注意,此方法返回 Double,而不是整数类型。Note that this method returns a Double instead of an integral type.

异常

mode 不是 MidpointRounding 的一个有效值。mode is not a valid value of MidpointRounding.

注解

请参阅中点值和舍入约定,了解有关带中点值的舍入的信息。See Midpoint values and rounding conventions for information about rounding numbers with midpoint values.

重要

在舍入中点值时,舍入算法执行相等测试。When rounding midpoint values, the rounding algorithm performs an equality test. 由于浮点格式存在的二进制表示和精度问题,因此方法返回的值可能异常。Because of problems of binary representation and precision in the floating-point format, the value returned by the method can be unexpected. 有关详细信息,请参阅“舍入和精度”For more information, see "Rounding and precision".

如果value参数的值为Double.NaN,则该方法返回Double.NaNIf the value of the value argument is Double.NaN, the method returns Double.NaN. 如果valueDouble.PositiveInfinityDouble.PositiveInfinity Double.NegativeInfinity,则该方法将分别返回或。 Double.NegativeInfinityIf value is Double.PositiveInfinity or Double.NegativeInfinity, the method returns Double.PositiveInfinity or Double.NegativeInfinity, respectively.

示例Example

Round(Double)下面的示例显示方法返回的值Round(Double, MidpointRounding) 、包含mode Round(Double, MidpointRounding) ToEven AwayFromZero参数的方法以及具有参数的方法。modeThe following example displays values returned by the Round(Double) method, the Round(Double, MidpointRounding) method with a mode argument of ToEven, and the Round(Double, MidpointRounding) method with a mode argument of AwayFromZero.

using System;

public class Example
{
   public static void Main()
   {
      Double[] values = { 12.0, 12.1, 12.2, 12.3, 12.4, 12.5, 12.6, 
                          12.7, 12.8, 12.9, 13.0 };
      Console.WriteLine("{0,-10} {1,-10} {2,-10} {3,-15}", "Value", "Default", 
                        "ToEven", "AwayFromZero");
      foreach (var value in values)
         Console.WriteLine("{0,-10:R} {1,-10} {2,-10} {3,-15}",
                           value, Math.Round(value), 
                           Math.Round(value, MidpointRounding.ToEven),
                           Math.Round(value, MidpointRounding.AwayFromZero));
   }
}
// The example displays the following output:
//       Value      Default    ToEven     AwayFromZero
//       12         12         12         12
//       12.1       12         12         12
//       12.2       12         12         12
//       12.3       12         12         12
//       12.4       12         12         12
//       12.5       12         12         13
//       12.6       13         13         13
//       12.7       13         13         13
//       12.8       13         13         13
//       12.9       13         13         13
//       13.0       13         13         13
Module Example
   Public Sub Main()
      Dim values() As Double = { 12.0, 12.1, 12.2, 12.3, 12.4, 12.5, 12.6, 
                                 12.7, 12.8, 12.9, 13.0 }
      Console.WriteLine("{0,-10} {1,-10} {2,-10} {3,-15}", "Value", "Default", 
                        "ToEven", "AwayFromZero")
      For Each value In values
         Console.WriteLine("{0,-10} {1,-10} {2,-10} {3,-15}",
                           value, Math.Round(value), 
                           Math.Round(value, MidpointRounding.ToEven),
                           Math.Round(value, MidpointRounding.AwayFromZero))
      Next
   End Sub
End Module
' The example displays the following output:
'       Value      Default    ToEven     AwayFromZero
'       12         12         12         12
'       12.1       12         12         12
'       12.2       12         12         12
'       12.3       12         12         12
'       12.4       12         12         12
'       12.5       12         12         13
'       12.6       13         13         13
'       12.7       13         13         13
'       12.8       13         13         13
'       12.9       13         13         13
'       13.0       13         13         13

调用方说明

由于将十进制值表示为浮点数或对浮点值执行算术运算可能导致精度损失,因此,在某些情况下,该Round(Double, MidpointRounding)方法可能不会显示为舍入点值最接近的整数。Because of the loss of precision that can result from representing decimal values as floating-point numbers or performing arithmetic operations on floating-point values, in some cases the Round(Double, MidpointRounding) method may not appear to round midpoint values to the nearest even integer. 在下面的示例中,由于浮点值 .1 没有有限的二进制表示形式,因此对值为11.5 的Round(Double)方法的第一次调用返回11而不是12。In the following example, because the floating-point value .1 has no finite binary representation, the first call to the Round(Double) method with a value of 11.5 returns 11 instead of 12.

[!code-csharpSystem.Math.Round#4] [!code-vbSystem.Math.Round#4][!code-csharpSystem.Math.Round#4] [!code-vbSystem.Math.Round#4]

另请参阅

Round(Double, Int32)

将双精度浮点值舍入到指定数量的小数位,并将中点值舍入到最接近的偶数。Rounds a double-precision floating-point value to a specified number of fractional digits, and rounds midpoint values to the nearest even number.

public:
 static double Round(double value, int digits);
public static double Round (double value, int digits);
static member Round : double * int -> double
Public Shared Function Round (value As Double, digits As Integer) As Double

参数

value
Double

要舍入的双精度浮点数。A double-precision floating-point number to be rounded.

digits
Int32

返回值中的小数数字。The number of fractional digits in the return value.

返回

最接近 valuedigits 位小数的数字。The number nearest to value that contains a number of fractional digits equal to digits.

异常

digits 小于 0 或大于 15。digits is less than 0 or greater than 15.

注解

digits参数的值的范围为0到15。The value of the digits argument can range from 0 to 15. 请注意,15是该Double类型支持的整数和小数位数的最大值。Note that 15 is the maximum number of integral and fractional digits supported by the Double type.

此方法使用的默认舍入约定MidpointRounding.ToEvenThis method uses the default rounding convention of MidpointRounding.ToEven. 请参阅中点值和舍入约定,了解有关带中点值的舍入的信息。See Midpoint values and rounding conventions for information about rounding numbers with midpoint values.

重要

在舍入中点值时,舍入算法执行相等测试。When rounding midpoint values, the rounding algorithm performs an equality test. 由于浮点格式存在的二进制表示和精度问题,因此方法返回的值可能异常。Because of problems of binary representation and precision in the floating-point format, the value returned by the method can be unexpected. 有关详细信息,请参阅“舍入和精度”For more information, see "Rounding and precision".

如果value参数的值为Double.NaN,则该方法返回Double.NaNIf the value of the value argument is Double.NaN, the method returns Double.NaN. 如果valueDouble.PositiveInfinityDouble.PositiveInfinity Double.NegativeInfinity,则该方法将分别返回或。 Double.NegativeInfinityIf value is Double.PositiveInfinity or Double.NegativeInfinity, the method returns Double.PositiveInfinity or Double.NegativeInfinity, respectively.

示例Example

下面的示例将具有两个小数位的双精度值舍入为具有一个小数位的双精度值。The following example rounds double values with two fractional digits to doubles that have a single fractional digit.

Math::Round(3.44, 1); //Returns 3.4.
Math::Round(3.45, 1); //Returns 3.4.
Math::Round(3.46, 1); //Returns 3.5.

Math::Round(4.34, 1); // Returns 4.3
Math::Round(4.35, 1); // Returns 4.4
Math::Round(4.36, 1); // Returns 4.4
Math.Round(3.44, 1); //Returns 3.4.
Math.Round(3.45, 1); //Returns 3.4.
Math.Round(3.46, 1); //Returns 3.5.

Math.Round(4.34, 1); // Returns 4.3
Math.Round(4.35, 1); // Returns 4.4
Math.Round(4.36, 1); // Returns 4.4
Math.Round(3.44, 1) 'Returns 3.4.
Math.Round(3.45, 1) 'Returns 3.4.
Math.Round(3.46, 1) 'Returns 3.5.

Math.Round(4.34, 1) ' Returns 4.3
Math.Round(4.35, 1) ' Returns 4.4
Math.Round(4.36, 1) ' Returns 4.4

调用方说明

由于将十进制值表示为浮点数或对浮点值执行算术运算可能导致精度损失,因此,在某些情况下,该Round(Double, Int32)方法可能不会显示为舍入点值digits小数点位置最近的偶数值。Because of the loss of precision that can result from representing decimal values as floating-point numbers or performing arithmetic operations on floating-point values, in some cases the Round(Double, Int32) method may not appear to round midpoint values to the nearest even value in the digits decimal position. 下面的示例对此进行了说明,其中,2.135 舍入为2.13 而不是2.14。This is illustrated in the following example, where 2.135 is rounded to 2.13 instead of 2.14. 出现这种情况的原因是value ,此方法在内部乘以 10位数,而乘法运算在这种情况下会损失精度。This occurs because internally the method multiplies value by 10digits, and the multiplication operation in this case suffers from a loss of precision.

[!code-csharpSystem.Math.Round#2] [!code-vbSystem.Math.Round#2][!code-csharpSystem.Math.Round#2] [!code-vbSystem.Math.Round#2]

另请参阅

Round(Double)

将双精度浮点值舍入到最接近的整数值,并将中点值舍入到最接近的偶数。Rounds a double-precision floating-point value to the nearest integral value, and rounds midpoint values to the nearest even number.

public:
 static double Round(double a);
public static double Round (double a);
static member Round : double -> double
Public Shared Function Round (a As Double) As Double

参数

a
Double

要舍入的双精度浮点数。A double-precision floating-point number to be rounded.

返回

最接近 a 的整数。The integer nearest a. 如果 a 的小数部分正好处于两个整数中间,其中一个整数为偶数,另一个整数为奇数,则返回偶数。If the fractional component of a is halfway between two integers, one of which is even and the other odd, then the even number is returned. 请注意,此方法返回 Double,而不是整数类型。Note that this method returns a Double instead of an integral type.

注解

此方法使用的默认舍入约定MidpointRounding.ToEvenThis method uses the default rounding convention of MidpointRounding.ToEven. 请参阅中点值和舍入约定,了解有关带中点值的舍入的信息。See Midpoint values and rounding conventions for information about rounding numbers with midpoint values.

重要

在舍入中点值时,舍入算法执行相等测试。When rounding midpoint values, the rounding algorithm performs an equality test. 由于浮点格式存在的二进制表示和精度问题,因此方法返回的值可能异常。Because of problems of binary representation and precision in the floating-point format, the value returned by the method can be unexpected. 有关详细信息,请参阅“舍入和精度”For more information, see "Rounding and precision".

如果a参数的值为Double.NaN,则该方法返回Double.NaNIf the value of the a argument is Double.NaN, the method returns Double.NaN. 如果aDouble.PositiveInfinityDouble.PositiveInfinity Double.NegativeInfinity,则该方法将分别返回或。 Double.NegativeInfinityIf a is Double.PositiveInfinity or Double.NegativeInfinity, the method returns Double.PositiveInfinity or Double.NegativeInfinity, respectively.

从 Visual Basic 15.8 开始,如果将由Round方法返回的值传递到任何整型转换函数,或返回Round的双精度值,则将优化双精度转换的性能。将自动转换为一个整数,并将Option Strict设置为 Off。Starting with Visual Basic 15.8, the performance of Double-to-integer conversion is optimized if you pass the value returned by the Round method to the any of the integral conversion functions, or if the Double value returned by Round is automatically converted to an integer with Option Strict set to Off. 此优化使代码可以更快运行 -- 对于进行大量到整型类型的转换的代码,可快达两倍。This optimization allows code to run faster -- up to twice as fast for code that does a large number of conversions to integer types. 下面的示例演示了此类优化转换:The following example illustrates such optimized conversions:

Dim d1 As Double = 1043.75133
Dim i1 As Integer = CInt(Math.Ceiling(d1))        ' Result: 1044

Dim d2 As Double = 7968.4136
Dim i2 As Integer = CInt(Math.Ceiling(d2))        ' Result: 7968

示例Example

下面的示例演示如何舍入为最接近的整数值。The following example demonstrates rounding to the nearest integer value.

using namespace System;

void main()
{
    Console::WriteLine("Classic Math.Round in CPP");
    Console::WriteLine(Math::Round(4.4));     // 4
    Console::WriteLine(Math::Round(4.5));     // 4
    Console::WriteLine(Math::Round(4.6));     // 5
    Console::WriteLine(Math::Round(5.5));     // 6
}
Console.WriteLine("Classic Math.Round in CSharp");
Console.WriteLine(Math.Round(4.4)); // 4
Console.WriteLine(Math.Round(4.5)); // 4
Console.WriteLine(Math.Round(4.6)); // 5
Console.WriteLine(Math.Round(5.5)); // 6
Module Module1

    Sub Main()
    Console.WriteLine("Classic Math.Round in Visual Basic")
    Console.WriteLine(Math.Round(4.4)) ' 4
    Console.WriteLine(Math.Round(4.5)) ' 4
    Console.WriteLine(Math.Round(4.6)) ' 5
    Console.WriteLine(Math.Round(5.5)) ' 6
    End Sub

End Module

调用方说明

由于将十进制值表示为浮点数或对浮点值执行算术运算可能导致精度损失,因此,在某些情况下,该Round(Double)方法可能不会显示为舍入点值最接近的整数。Because of the loss of precision that can result from representing decimal values as floating-point numbers or performing arithmetic operations on floating-point values, in some cases the Round(Double) method may not appear to round midpoint values to the nearest even integer. 在下面的示例中,由于浮点值 .1 没有有限的二进制表示形式,因此对值为11.5 的Round(Double)方法的第一次调用返回11而不是12。In the following example, because the floating-point value .1 has no finite binary representation, the first call to the Round(Double) method with a value of 11.5 returns 11 instead of 12.

[!code-csharpSystem.Math.Round#1] [!code-vbSystem.Math.Round#1][!code-csharpSystem.Math.Round#1] [!code-vbSystem.Math.Round#1]

另请参阅

Round(Decimal, Int32)

将小数值舍入到指定数量的小数位,并将中点值舍入到最接近的偶数。Rounds a decimal value to a specified number of fractional digits, and rounds midpoint values to the nearest even number.

public:
 static System::Decimal Round(System::Decimal d, int decimals);
public static decimal Round (decimal d, int decimals);
static member Round : decimal * int -> decimal
Public Shared Function Round (d As Decimal, decimals As Integer) As Decimal

参数

d
Decimal

要舍入的小数。A decimal number to be rounded.

decimals
Int32

返回值中的小数位数。The number of decimal places in the return value.

返回

最接近 ddecimals 位小数的数字。The number nearest to d that contains a number of fractional digits equal to decimals.

异常

decimals 小于 0 或大于 28。decimals is less than 0 or greater than 28.

结果超出了 Decimal 的范围。The result is outside the range of a Decimal.

注解

decimals参数的值的范围为0到28。The value of the decimals argument can range from 0 to 28.

此方法使用的默认舍入约定MidpointRounding.ToEvenThis method uses the default rounding convention of MidpointRounding.ToEven. 请参阅中点值和舍入约定,了解有关带中点值的舍入的信息。See Midpoint values and rounding conventions for information about rounding numbers with midpoint values.

重要

在舍入中点值时,舍入算法执行相等测试。When rounding midpoint values, the rounding algorithm performs an equality test. 由于浮点格式存在的二进制表示和精度问题,因此方法返回的值可能异常。Because of problems of binary representation and precision in the floating-point format, the value returned by the method can be unexpected. 有关详细信息,请参阅“舍入和精度”For more information, see "Rounding and precision".

示例Example

下面的示例将具有两个小数位的十进制值舍入为具有单个小数数字的值。The following example rounds decimal values with two fractional digits to values that have a single fractional digit.

Console.WriteLine(Math.Round(3.44m, 1)); 
Console.WriteLine(Math.Round(3.45m, 1)); 
Console.WriteLine(Math.Round(3.46m, 1)); 
Console.WriteLine();

Console.WriteLine(Math.Round(4.34m, 1)); 
Console.WriteLine(Math.Round(4.35m, 1)); 
Console.WriteLine(Math.Round(4.36m, 1)); 
// The example displays the following output:
//       3.4
//       3.4
//       3.5
//       
//       4.3
//       4.4
//       4.4
Public Module Example
   Sub Main()
      Console.WriteLine(Math.Round(3.44, 1)) 
      Console.WriteLine(Math.Round(3.45, 1)) 
      Console.WriteLine(Math.Round(3.46, 1)) 
      Console.WriteLine()
      
      Console.WriteLine(Math.Round(4.34, 1)) 
      Console.WriteLine(Math.Round(4.35, 1)) 
      Console.WriteLine(Math.Round(4.36, 1)) 
   End Sub  
End Module
' The example displays the following output:
'       3.4
'       3.4
'       3.5
'       
'       4.3
'       4.4
'       4.4

另请参阅

Round(Decimal)

将小数值舍入到最接近的整数值,并将中点值舍入到最接近的偶数。Rounds a decimal value to the nearest integral value, and rounds midpoint values to the nearest even number.

public:
 static System::Decimal Round(System::Decimal d);
public static decimal Round (decimal d);
static member Round : decimal -> decimal
Public Shared Function Round (d As Decimal) As Decimal

参数

d
Decimal

要舍入的小数。A decimal number to be rounded.

返回

最接近 d 参数的整数。The integer nearest the d parameter. 如果 d 的小数部分正好处于两个整数中间,其中一个整数为偶数,另一个整数为奇数,则返回偶数。If the fractional component of d is halfway between two integers, one of which is even and the other odd, the even number is returned. 请注意,此方法返回 Decimal,而不是整数类型。Note that this method returns a Decimal instead of an integral type.

异常

结果超出了 Decimal 的范围。The result is outside the range of a Decimal.

注解

此方法使用的默认舍入约定MidpointRounding.ToEvenThis method uses the default rounding convention of MidpointRounding.ToEven. 请参阅中点值和舍入约定,了解有关带中点值的舍入的信息。See Midpoint values and rounding conventions for information about rounding numbers with midpoint values.

重要

在舍入中点值时,舍入算法执行相等测试。When rounding midpoint values, the rounding algorithm performs an equality test. 由于浮点格式存在的二进制表示和精度问题,因此方法返回的值可能异常。Because of problems of binary representation and precision in the floating-point format, the value returned by the method can be unexpected. 有关详细信息,请参阅“舍入和精度”For more information, see "Rounding and precision".

示例Example

下面的示例演示Round(Decimal)方法。The following example demonstrates the Round(Decimal) method. 请注意, Decimal 4.5 的值舍入为4而不是5,因为此重载使用ToEven默认约定。Note that the Decimal value of 4.5 rounds to 4 rather than 5, because this overload uses the default ToEven convention.

for (decimal value = 4.2m; value <= 4.8m; value+=.1m )
   Console.WriteLine("{0} --> {1}", value, Math.Round(value));
// The example displays the following output:
//       4.2 --> 4
//       4.3 --> 4
//       4.4 --> 4
//       4.5 --> 4
//       4.6 --> 5
//       4.7 --> 5
//       4.8 --> 5
Module Example
   Public Sub Main()
      For value As Decimal = 4.2d To 4.8d Step .1d
         Console.WriteLine("{0} --> {1}", value, Math.Round(value))
      Next   
   End Sub                                                                 
End Module
' The example displays the following output:
'       4.2 --> 4
'       4.3 --> 4
'       4.4 --> 4
'       4.5 --> 4
'       4.6 --> 5
'       4.7 --> 5
'       4.8 --> 5

另请参阅

Round(Decimal, MidpointRounding)

将小数值舍入到最接近的整数,并为中点值使用指定的舍入规则。Rounds a decimal value to the nearest integer, and uses the specified rounding convention for midpoint values.

public:
 static System::Decimal Round(System::Decimal d, MidpointRounding mode);
public static decimal Round (decimal d, MidpointRounding mode);
static member Round : decimal * MidpointRounding -> decimal
Public Shared Function Round (d As Decimal, mode As MidpointRounding) As Decimal

参数

d
Decimal

要舍入的小数。A decimal number to be rounded.

mode
MidpointRounding

在两个数字之间时如何舍入 d 的规范。Specification for how to round d if it is midway between two other numbers.

返回

最接近 d 的整数。The integer nearest d. 如果 d 是两个数字的中值,这两个数字一个为偶数,另一个为奇数,则 mode 确定返回两个数字中的哪一个。If d is halfway between two numbers, one of which is even and the other odd, then mode determines which of the two is returned. 请注意,此方法返回 Decimal,而不是整数类型。Note that this method returns a Decimal instead of an integral type.

异常

mode 不是 MidpointRounding 的一个有效值。mode is not a valid value of MidpointRounding.

结果超出了 Decimal 的范围。The result is outside the range of a Decimal.

注解

请参阅中点值和舍入约定,了解有关带中点值的舍入的信息。See Midpoint values and rounding conventions for information about rounding numbers with midpoint values.

重要

在舍入中点值时,舍入算法执行相等测试。When rounding midpoint values, the rounding algorithm performs an equality test. 由于浮点格式存在的二进制表示和精度问题,因此方法返回的值可能异常。Because of problems of binary representation and precision in the floating-point format, the value returned by the method can be unexpected. 有关详细信息,请参阅“舍入和精度”For more information, see "Rounding and precision".

示例Example

Round(Decimal)下面的示例显示方法返回的值Round(Decimal, MidpointRounding) 、包含mode Round(Decimal, MidpointRounding) ToEven AwayFromZero参数的方法以及具有参数的方法。modeThe following example displays values returned by the Round(Decimal) method, the Round(Decimal, MidpointRounding) method with a mode argument of ToEven, and the Round(Decimal, MidpointRounding) method with a mode argument of AwayFromZero.

Console.WriteLine("{0,-10} {1,-10} {2,-10} {3,-15}", "Value", "Default", 
                  "ToEven", "AwayFromZero");
for (decimal value = 12.0m; value <= 13.0m; value += 0.1m)
   Console.WriteLine("{0,-10} {1,-10} {2,-10} {3,-15}",
                     value, Math.Round(value), 
                     Math.Round(value, MidpointRounding.ToEven),
                     Math.Round(value, MidpointRounding.AwayFromZero));
// The example displays the following output:
//       Value      Default    ToEven     AwayFromZero
//       12         12         12         12
//       12.1       12         12         12
//       12.2       12         12         12
//       12.3       12         12         12
//       12.4       12         12         12
//       12.5       12         12         13
//       12.6       13         13         13
//       12.7       13         13         13
//       12.8       13         13         13
//       12.9       13         13         13
//       13.0       13         13         13
Module Example
   Public Sub Main()
      Console.WriteLine("{0,-10} {1,-10} {2,-10} {3,-15}", "Value", "Default", 
                        "ToEven", "AwayFromZero")
      For value As Decimal = 12.0d To 13.0d Step .1d
         Console.WriteLine("{0,-10} {1,-10} {2,-10} {3,-15}",
                           value, Math.Round(value), 
                           Math.Round(value, MidpointRounding.ToEven),
                           Math.Round(value, MidpointRounding.AwayFromZero))
      Next
   End Sub
End Module
' The example displays the following output:
'       Value      Default    ToEven     AwayFromZero
'       12         12         12         12
'       12.1       12         12         12
'       12.2       12         12         12
'       12.3       12         12         12
'       12.4       12         12         12
'       12.5       12         12         13
'       12.6       13         13         13
'       12.7       13         13         13
'       12.8       13         13         13
'       12.9       13         13         13
'       13.0       13         13         13

另请参阅

适用于