Math.Round Method (Double, Int32)

May 11, 2014

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 value with a number of fractional digits equal to digits.

ExceptionCondition
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 the value to the right of the digits position is halfway between x and x+1, where x represents the value in the digits position, the digit in the digits position is rounded up if it is odd, or left unchanged if it is even. If value has the same number or 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 Demo(System.Windows.Controls.TextBlock outputBlock)
   {
      double[] values = { 2.125, 2.135, 2.145, 3.125, 3.135, 3.145 };
      foreach (double value in values)
         outputBlock.Text += String.Format("{0} --> {1}", value, Math.Round(value, 2)) + Environment.NewLine;

   }
}
// 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


Windows Phone OS

Supported in: 8.1, 8.0, 7.1, 7.0

Windows Phone

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