Math.Round Method (Double, MidpointRounding)

Note
For examples and comprehensive usage information about this and other overloads of the Round method, see the Round reference page.

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. 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.

using System;

public class Example
{
   public static void Main()
   {
      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} --> {1}", value, Math.Round(value, 
                        MidpointRounding.AwayFromZero));
      return value + .1;
   }
}
// The example displays the following output: 
//       11.1 --> 11 
//       11.2 --> 11 
//       11.3 --> 11 
//       11.4 --> 11 
//       11.5 --> 11 
//       11.6 --> 12 
//        
//       11.5 --> 12

.NET Framework

.NET Framework Client Profile

Portable Class Library

.NET for Windows Store apps

Windows 8, Windows Server 2012, Windows 7, Windows Vista SP2, Windows Server 2008 (Server Core Role not supported), Windows Server 2008 R2 (Server Core Role supported with SP1 or later; Itanium not supported)

The .NET Framework does not support all versions of every platform. For a list of the supported versions, see .NET Framework System Requirements.

Reference