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Math.Round Method (Double, Int32, MidpointRounding)

.NET Framework 3.5

Updated: May 2010

Rounds a double-precision floating-point value to the specified number of fractional digits. A parameter specifies how to round the value if it is midway between two other numbers.

Namespace:  System
Assembly:  mscorlib (in mscorlib.dll)
C#
C++
JScript
VB
Copy 
public static double Round(
	double value,
	int digits,
	MidpointRounding mode
)

Parameters

value
Type: System.Double
A double-precision floating-point number to be rounded.
digits
Type: System.Int32
The number of fractional digits in the return value.
mode
Type: System.MidpointRounding
Specification for how to round value if it is midway between two other numbers.

Return Value

Type: System.Double
The number nearest to value that has a number of fractional digits equal to digits. If the number of fractional digits in value is less than digits, value is returned unchanged.
Exception Condition
ArgumentOutOfRangeException

digits is less than 0 or greater than 15.

ArgumentException

mode is not a valid value of System.MidpointRounding.

The digits parameter specifies the number of fractional digits in the return value and ranges from 0 to 15. If digits is zero, then an integer is returned.

The mode parameter controls how value is rounded if the first digit in value to the right of the decimal position represented by the digits parameter is 5—that is, if it is halfway between the digit in the digits position and the next highest digit. The mode parameter can have one of the following two values:

  • MidpointRounding.ToEven . If the digit in the digits position is odd, it is changed to an even digit. Otherwise, it is left unchanged. This behavior follows IEEE Standard 754, section 4. It is sometimes called rounding to nearest or banker's rounding. It minimizes rounding errors that result from consistently rounding a midpoint value in a single direction.

  • MidpointRounding.AwayFromZero . The digit in the digits position is always rounded up to the next digit. This is the most commonly known rounding method. It is known as symmetric arithmetic rounding.

The maximum total number of integral and fractional digits that can be returned is 15. If the rounded value contains more than 15 digits, the 15 most significant digits are returned. If the rounded value contains 15 or fewer digits, the integral digits and as many fractional digits as the digits parameter specifies are returned.

Notes to Callers:

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. This is illustrated in the following example, where 2.135 is rounded to 2.13 instead of 2.14. This occurs because internally the method multiplies value by 10 * digits, and the multiplication operation in this case suffers from a loss of precision.

C#
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using System;

public class Example
{
   public static void Main()
   {
      double[] values = { 2.125, 2.135, 2.145, 3.125, 3.135, 3.145 };
      foreach (double value in values)
         Console.WriteLine("{0} --> {1}", value, 
                           Math.Round(value, 2, MidpointRounding.AwayFromZero));

   }
}
// The example displays the following output:
//       2.125 --> 2.13
//       2.135 --> 2.13
//       2.145 --> 2.15
//       3.125 --> 3.13
//       3.135 --> 3.14
//       3.145 --> 3.15

The following code example demonstrates the Round method in conjunction with the MidpointRounding enumeration. Although the code example rounds decimal numbers, the Round method rounds double-precision floating-point numbers in a similar way.

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

class Sample 
{
    public static void Main() 
    {
    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)

*/

Windows 7, Windows Vista, Windows XP SP2, Windows XP Media Center Edition, Windows XP Professional x64 Edition, Windows XP Starter Edition, Windows Server 2008 R2, Windows Server 2008, Windows Server 2003, Windows Server 2000 SP4, Windows Millennium Edition, Windows 98

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

.NET Framework

Supported in: 3.5, 3.0, 2.0

Date

History

Reason

May 2010

Added the Notes for Callers section.

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Same problem

Math.Round(34.785, 2, MidpointRounding.AwayFromZero) yields 34.78 instead of 34.79.

Precision in Rounding

This is the issue discussed in the "Notes to Callers" section of the documentation.

Note how the problem of precision happens here. To round a value to a specified number of fractional digits, the Round method multiples the floating point value by 10^digits. If we display each value using the "R" standard format string, we can see all of the fractional digits used in the rounding operation:

      Console.WriteLine(Math.Round(value, 2, MidpointRounding.AwayFromZero));
      Console.WriteLine("{0:R} --> {1:R}", value, value * Math.Pow(10, 2));      
      // Displays:
      //       34.78
      //       34.785 --> 3478.4999999999995


So the problem is that the intermediate result loses precision and becomes slightly less than .5. Since it is no longer a midpoint value, it is correctly rounded down.

--Ron Petrusha
Common Language Runtime User Education
Microsoft Corporation

History
I found a bug ?
I'll try to explain :
look this example :
Module
Module1 
Sub Main()
Dim num AsDouble = 1.25
Console.WriteLine(Math.Round(num, 1, MidpointRounding.AwayFromZero))
num = 1.225
Console.WriteLine(Math.Round(num, 2, MidpointRounding.AwayFromZero))
num = 1.2225
Console.WriteLine(Math.Round(num, 3, MidpointRounding.AwayFromZero))
num = 1.22225
Console.WriteLine(Math.Round(num, 4, MidpointRounding.AwayFromZero))
num = 1.222225
Console.WriteLine(Math.Round(num, 5, MidpointRounding.AwayFromZero))
num = 1.2222225
Console.WriteLine(Math.Round(num, 6, MidpointRounding.AwayFromZero))
EndSub
End
Module


The output of this sample console application is :
1,3
1,23
1,223
1,2223
1,22222
1,222223

Is there something wrong ?
In my application I must use five decimal digits and this round method.

Regards
Andrea

Precision in Rounding

This is the issue discussed in the "Notes to Callers" section of the documentation. Also, see my response to the "Same Error" post.

--Ron Petrusha
Common Language Runtime User Education
Microsoft Corporation

History
My finding (or solution)

I have similar finding and I'd wrongly posted to Math.Round(Decimal, digit, MidpointRound).

Perhaps the problem of rounding 1.222225 to 5 digits is like this:

Math.Round(1.222225, 5, MidpointRounding.AwayFromZero)

= Math.Round(1.222225 * 100000, 0, MidpointRounding.AwayFromZero) / 100000

= Math.Round(122222.49999999998, 0, MidpointRound.AwayFromZero) / 100000

= 122222 / 100000

= 1.22222

One solution to 'correctly' round 1.222225 to 5 digits it to specify the data type as decimal instead of double, i.e. Math.Round(1.222225d, 5, MidpointRounding.AwayFromZero) = 1.22223D

Precision in Rounding

This is the issue discussed in the "Notes to Callers" section of the documentation. Also, see my response to the "Same Error" post.

It is true that specifying the numeric value as a Decimal rather than a Double produces a more precise result in a rounding operation. However, the Decimal type still can suffer from a loss of precision, so it is not necessarily the case that it will eliminate all rounding error caused by a lack of precision.

--Ron Petrusha
Common Language Runtime User Education
Microsoft Corporation

History