Math.Round Method (Double, Int32, MidpointRounding)
Updated: January 2012
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.
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.
- mode
- Type: System.MidpointRounding
Specification for how to round value if it is midway between two other numbers.
Return Value
Type: System.DoubleThe 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 digits decimal position is 5 -- that is, if it is halfway between the digit in the digits position and the next highest digit. mode 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.
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, 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 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, 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 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.
// 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 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.
As it’s said in document that
Math.Round(2.135, 2,MidpointRounding.AwayFromZero) would show 2.13 instead of 2.14
And
Math.Round(0.145, 2,MidpointRounding.AwayFromZero) would show 0.14 instead of 0.15
So what's the solution, I have found two ways to solve it.
Solution 1:
Convert double to decimal
Math.Round((decimal(2.135), 2,MidpointRounding.AwayFromZero) it would show 2.14 and
Math.Round((decimal)0.145, 2,MidpointRounding.AwayFromZero) it would show 0.15
Solution 2:
System.Math.Ceiling(2.135 * 100) / 100 it would show 2.14
And
System.Math.Ceiling(0.145 * 100) / 100 it would show 0.15
- 10/23/2011
- hafizmsuleman
double result = Math::Round( 0.145, 2, MidpointRounding::AwayFromZero);
result = 0.14, should be 0.15
The Imprecision of Floating Point Values
This "error" is due to the imprecision of floating point values. From the Notes to Callers section above: "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."
--Ron Petrusha
CLR Developer Content Team
Microsoft Corporation
- 9/2/2010
- andrelo-1
- 9/7/2010
- R Petrusha - MSFT
