# Math.Round Method (Double, Int32)

**Visual Studio 2008**

Updated: March 2012

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

**Namespace:**System

**Assembly:**mscorlib (in mscorlib.dll)

#### 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.DoubleThe 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.

**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

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, Windows CE, Windows Mobile for Smartphone, Windows Mobile for Pocket PC, Xbox 360, Zune

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.