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Math.Round Method (Double, Int32)

Updated: March 2012

Rounds a double-precision floating-point value to a specified number of fractional digits.

Namespace:  System
Assembly:  mscorlib (in mscorlib.dll)
public static double Round(
	double value,
	int digits
)

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.

Return Value

Type: System.Double
The number nearest to value that contains a number of fractional digits equal to digits.
Exception Condition
ArgumentOutOfRangeException

digits is less than 0 or greater than 15.

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

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.

This method is equivalent to calling the Round method with a mode argument of MidpointRounding.ToEven. If there is a single digit in value to the right of the digits decimal position and its value is 5, the digit in the digits position is rounded up if it is odd, or left unchanged if it is even. If value has fewer fractional digits than digits, value is returned unchanged.

The behavior of this method follows IEEE Standard 754, section 4. This kind of rounding 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. To control the type of rounding used by the Round(Double, Int32) method, call the Math.Round(Double, Int32, MidpointRounding) overload.

If the value of value is Double.NaN, the method returns Double.NaN. If the value of value is Double.PositiveInfinity or Double.NegativeInfinity, the method returns Double.PositiveInfinity or Double.NegativeInfinity, respectively.

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) method may not appear to round midpoint values to the nearest even value in the digits decimal position. 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 10digits, and the multiplication operation in this case suffers from a loss of precision.


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));

   }
}
// The example displays the following output:
//       2.125 --> 2.12
//       2.135 --> 2.13
//       2.145 --> 2.14
//       3.125 --> 3.12
//       3.135 --> 3.14
//       3.145 --> 3.14


The following example demonstrates rounding to nearest.


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


.NET Framework

Supported in: 4, 3.5, 3.0, 2.0, 1.1, 1.0

.NET Framework Client Profile

Supported in: 4, 3.5 SP1

Portable Class Library

Supported in: Portable Class Library

Windows 7, Windows Vista SP1 or later, Windows XP SP3, Windows XP SP2 x64 Edition, Windows Server 2008 (Server Core not supported), Windows Server 2008 R2 (Server Core supported with SP1 or later), Windows Server 2003 SP2

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

Date

History

Reason

March 2012

Clarified description of midpoint value.

Customer feedback.

January 2012

Noted method behavior with Double.Nan, Double.PositiveInfinity, and Double.NegativeInfinity.

Customer feedback.

May 2010

Added the Notes for Callers section.

Customer feedback.

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Community Content Add
Annotations FAQ
Watch comparing result to a float
Be careful when comparing the result to the same result cast as a (float) (in C#)

example:

double x = 173.2121;
Console.WriteLine(Math.Round(x, 2));
Console.WriteLine((float)Math.Round(x, 2));
Console.WriteLine((Math.Round(x, 2)) == (float)Math.Round(x, 2));

output is:
173.21
173.21
false

It comes from comparing a double to a float, but it can be tricky to catch if you're not ready for it.
Behavior with INF and NaN

It would be nice to be able to lookup here how Math.Round behaves with Double values like INF and NaN, even if described in the IEEE standard. Now I have to test it myself or find this answer elsewhere.

Math.Round with Double.NaN, Double.PositiveInfinity, and Double.NegativeInfinity

Thanks for your post. We've modified the documentation to note how Double.NaN, Double.PositiveInfinity, and Double.NegativeInfinity affect the method's return value. The updated version of the documentation should appear in the near future.

If the method is passed Double.NaN, it returns Double.NaN. If it is passed Double.PositiveInfinity or Double.NegativeInfinity, it returns that value.

--Ron Petrusha
Common Language Runtime User Education
Microsoft Corporation