Math.Round Method (Double, MidpointRounding)
[ This article is for Windows Phone 8 developers. If you’re developing for Windows 10, see the latest documentation. ]
Rounds a double-precision floating-point value to the nearest integer. A parameter specifies how to round the value if it is midway between two other numbers.
Assembly: mscorlib (in mscorlib.dll)
Parameters
- value
- Type: System.Double
A double-precision floating-point number to be rounded.
- mode
- Type: System.MidpointRounding
Specification for how to round value if it is midway between two other numbers.
Return Value
Type: System.DoubleThe integer nearest value. If value is halfway between two integers, one of which is even and the other odd, then mode determines which of the two is returned.
| Exception | Condition |
|---|---|
| ArgumentException | mode is not a valid value of System.MidpointRounding. |
The mode parameter controls how value is rounded if the fractional portion of value is halfway between the one's digit and a value one greater than the one's digit. mode can have either of two values:
MidpointRounding.ToEven. If the one's digit 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 one's digit is always rounded up to the next digit. This is the most commonly known rounding method. It is known as symmetric arithmetic rounding.
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 CallersBecause 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 { private static System.Windows.Controls.TextBlock outputBlock; public static void Demo(System.Windows.Controls.TextBlock outputBlock) { Example.outputBlock = outputBlock; double value = 11.1; for (int ctr = 0; ctr <= 5; ctr++) value = RoundValueAndAdd(value); outputBlock.Text += Environment.NewLine; value = 11.5; RoundValueAndAdd(value); } private static double RoundValueAndAdd(double value) { outputBlock.Text += String.Format("{0} --> {1}", value, Math.Round(value, MidpointRounding.AwayFromZero)) + Environment.NewLine; 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
The following code example demonstrates how to use the Round method with the MidpointRounding enumeration. Although the code example rounds decimal numbers, the Round method rounds double-precision floating-point numbers in a similar way.