Math.Round Method

Definition

Namespace:

System

Assemblies:

netstandard.dll, System.Runtime.dll

Rounds a value to the nearest integer or to the specified number of fractional digits.

Overloads

Round(Double, Int32, MidpointRounding)	Rounds a double-precision floating-point value to a specified number of fractional digits using the specified rounding convention.
Round(Decimal, Int32, MidpointRounding)	Rounds a decimal value to a specified number of fractional digits using the specified rounding convention.
Round(Double, MidpointRounding)	Rounds a double-precision floating-point value to an integer using the specified rounding convention.
Round(Double, Int32)	Rounds a double-precision floating-point value to a specified number of fractional digits, and rounds midpoint values to the nearest even number.
Round(Decimal, Int32)	Rounds a decimal value to a specified number of fractional digits, and rounds midpoint values to the nearest even number.
Round(Double)	Rounds a double-precision floating-point value to the nearest integral value, and rounds midpoint values to the nearest even number.
Round(Decimal)	Rounds a decimal value to the nearest integral value, and rounds midpoint values to the nearest even number.
Round(Decimal, MidpointRounding)	Rounds a decimal value an integer using the specified rounding convention.

Examples

In addition to the examples in the Remarks section, this article includes examples that illustrate the following overloads of the Math.Round method:

Math.Round(Decimal) Math.Round(Double) Math.Round(Decimal, Int32) Math.Round(Decimal, MidpointRounding) Math.Round(Double, Int32) Math.Round(Double, MidpointRounding) Math.Round(Decimal, Int32, MidpointRounding) Math.Round(Double, Int32, MidpointRounding)

Remarks

In this section:

• Which method do I call?
• Midpoint values and rounding conventions
• Rounding and precision
• Rounding and single-precision floating point values
• Examples of individual overloads

Which method do I call?

You can use the following table to select an appropriate rounding method. In addition to the Math.Round methods, it also includes Math.Ceiling and Math.Floor.

To	Call
Round a number to an integer by using the rounding-to-nearest convention.	Round(Decimal) -or- Round(Double)
Round a number to an integer by using a specified rounding convention.	Round(Decimal, MidpointRounding) -or- Round(Double, MidpointRounding)
Round a number to a specified number of fractional digits by using the rounding to nearest convention.	Round(Decimal, Int32) -or- Round(Double, Int32)
Round a number to a specified number of fractional digits by using a specified rounding convention.	Round(Decimal, Int32, MidpointRounding) -or- Round(Double, Int32, MidpointRounding)
Round a Single value to a specified number of fractional digits by using a specified rounding convention and minimizing the loss of precision.	Convert the Single to a Decimal and call Round(Decimal, Int32, MidpointRounding).
Round a number to a specified number of fractional digits while minimizing problems of precision in rounding midpoint values.	Call a rounding method that implements a "greater than or approximately equal to" comparison. See Rounding and precision.
Round a fractional value to an integer that is greater than the fractional value. For example, round 3.1 to 4.	Ceiling
Round a fractional value to an integer that is less than the fractional value. For example, round 3.9 to 3.	Floor

Midpoint values and rounding conventions

Rounding involves converting a numeric value with a specified precision to a value with less precision. For example, you can use the Round(Double) method to round a value of 3.4 to 3.0, and the Round(Double, Int32) method to round a value of 3.579 to 3.58.

In a midpoint value, the value after the least significant digit in the result is precisely half way between two numbers. For example, 3.47500 is a midpoint value if it is to be rounded to two decimal places, and 7.500 is a midpoint value if it is to be rounded to an integer. In these cases, if the round-to-nearest strategy is used, the nearest value can't be easily identified without a rounding convention.

The Round method supports two rounding conventions for handling midpoint values:

• Rounding away from zero

  Midpoint values are rounded to the next number away from zero. For example, 3.75 rounds to 3.8, 3.85 rounds to 3.9, -3.75 rounds to -3.8, and -3.85 rounds to -3.9. This form of rounding is represented by the MidpointRounding.AwayFromZero enumeration member.

• Rounding to nearest even, or banker's rounding

  Midpoint values are rounded to the nearest even number. For example, both 3.75 and 3.85 round to 3.8, and both -3.75 and -3.85 round to -3.8. This form of rounding is represented by the MidpointRounding.ToEven enumeration member.

Note

In .NET Core 3.0 and later versions, three additional rounding strategies are available through the MidpointRounding enumeration. These strategies are used in all cases, not just for midpoint values as MidpointRounding.ToEven and MidpointRounding.AwayFromZero are.

Rounding away from zero is the most widely known form of rounding, while rounding to nearest even is the standard in financial and statistical operations. It conforms to IEEE Standard 754, section 4. When used in multiple rounding operations, rounding to nearest even reduces the rounding error that is caused by consistently rounding midpoint values in a single direction. In some cases, this rounding error can be significant.

The following example illustrates the bias that can result from consistently rounding midpoint values in a single direction. The example computes the true mean of an array of Decimal values, and then computes the mean when the values in the array are rounded by using the two conventions. In this example, the true mean and the mean that results when rounding to nearest are the same. However, the mean that results when rounding away from zero differs by .05 (or by 3.6%) from the true mean.

C#

decimal[] values = { 1.15m, 1.25m, 1.35m, 1.45m, 1.55m, 1.65m };
decimal sum = 0;

// Calculate true mean.
foreach (var value in values)
    sum += value;

Console.WriteLine("True mean:     {0:N2}", sum / values.Length);

// Calculate mean with rounding away from zero.
sum = 0;
foreach (var value in values)
    sum += Math.Round(value, 1, MidpointRounding.AwayFromZero);

Console.WriteLine("AwayFromZero:  {0:N2}", sum / values.Length);

// Calculate mean with rounding to nearest.
sum = 0;
foreach (var value in values)
    sum += Math.Round(value, 1, MidpointRounding.ToEven);

Console.WriteLine("ToEven:        {0:N2}", sum / values.Length);

// The example displays the following output:
//       True mean:     1.40
//       AwayFromZero:  1.45
//       ToEven:        1.40

By default, the Round method uses the round to nearest even convention. The following table lists the overloads of the Round method and the rounding convention that each uses.

Overload	Rounding convention
Round(Decimal)	ToEven
Round(Double)	ToEven
Round(Decimal, Int32)	ToEven
Round(Double, Int32)	ToEven
Round(Decimal, MidpointRounding)	Determined by mode parameter.
Round(Double, MidpointRounding)	Determined by mode parameter
Round(Decimal, Int32, MidpointRounding)	Determined by mode parameter
Round(Double, Int32, MidpointRounding)	Determined by mode parameter

Rounding and precision

In order to determine whether a rounding operation involves a midpoint value, the Round method multiplies the original value to be rounded by 10ⁿ, where n is the desired number of fractional digits in the return value, and then determines whether the remaining fractional portion of the value is greater than or equal to .5. This is a slight variation on a test for equality, and as discussed in the "Testing for Equality" section of the Double reference topic, tests for equality with floating-point values are problematic because of the floating-point format's issues with binary representation and precision. This means that any fractional portion of a number that is slightly less than .5 (because of a loss of precision) will not be rounded upward.

The following example illustrates the problem. It repeatedly adds .1 to 11.0 and rounds the result to the nearest integer. 11.5 should round to 12 using either of the midpoint-rounding conventions (ToEven or AwayFromZero). However, as the output from the example shows, it does not. The example uses the "R" standard numeric format string to display the floating point value's full precision, and shows that the value to be rounded has lost precision during repeated additions, and its value is actually 11.499999999999998. Because .499999999999998 is less than .5, the midpoint-rounding conventions don't come into play and the value is rounded down. As the example also shows, this problem does not occur if you assign the constant value 11.5 to a Double variable.

C#

public static void Example()
{
    Console.WriteLine("{0,5} {1,20:R}  {2,12} {3,15}\n",
                      "Value", "Full Precision", "ToEven",
                      "AwayFromZero");
    double value = 11.1;
    for (int ctr = 0; ctr <= 5; ctr++)
        value = RoundValueAndAdd(value);

    Console.WriteLine();

    value = 11.5;
    RoundValueAndAdd(value);
}

private static double RoundValueAndAdd(double value)
{
    Console.WriteLine("{0,5:N1} {0,20:R}  {1,12} {2,15}",
                      value, Math.Round(value, MidpointRounding.ToEven),
                      Math.Round(value, MidpointRounding.AwayFromZero));
    return value + .1;
}

// The example displays the following output:
//       Value       Full Precision        ToEven    AwayFromZero
//
//        11.1                 11.1            11              11
//        11.2                 11.2            11              11
//        11.3   11.299999999999999            11              11
//        11.4   11.399999999999999            11              11
//        11.5   11.499999999999998            11              11
//        11.6   11.599999999999998            12              12
//
//        11.5                 11.5            12              12

Problems of precision in rounding midpoint values are most likely to arise in the following conditions:

• When a fractional value cannot be expressed precisely in the floating-point type's binary format.

• When the value to be rounded is calculated from one or more floating-point operations.

• When the value to be rounded is a Single rather than a Double or Decimal. For more information, see the next section, Rounding and single-precision floating-point values.

In cases where the lack of precision in rounding operations is problematic, you can do the following:

• If the rounding operation calls an overload that rounds a Double value, you can change the Double to a Decimal value and call an overload that rounds a Decimal value instead. Although the Decimal data type also has problems of representation and loss of precision, these issues are far less common.

• Define a custom rounding algorithm that performs a "nearly equal" test to determine whether the value to be rounded is acceptably close to a midpoint value. The following example defines a RoundApproximate method that examines whether a fractional value is sufficiently near to a midpoint value to be subject to midpoint rounding. As the output from the example shows, it corrects the rounding problem shown in the previous example.

  C#

  public static void Example()
  {
      Console.WriteLine("{0,5} {1,20:R}  {2,12} {3,15}\n",
                        "Value", "Full Precision", "ToEven",
                        "AwayFromZero");
      double value = 11.1;
      for (int ctr = 0; ctr <= 5; ctr++)
          value = RoundValueAndAdd(value);
  
      Console.WriteLine();
  
      value = 11.5;
      RoundValueAndAdd(value);
  }
  
  private static double RoundValueAndAdd(double value)
  {
      const double tolerance = 8e-14;
  
      Console.WriteLine("{0,5:N1} {0,20:R}  {1,12} {2,15}",
                        value,
                        RoundApproximate(value, 0, tolerance, MidpointRounding.ToEven),
                        RoundApproximate(value, 0, tolerance, MidpointRounding.AwayFromZero));
      return value + .1;
  }
  
  private static double RoundApproximate(double dbl, int digits, double margin,
                                    MidpointRounding mode)
  {
      double fraction = dbl * Math.Pow(10, digits);
      double value = Math.Truncate(fraction);
      fraction = fraction - value;
      if (fraction == 0)
          return dbl;
  
      double tolerance = margin * dbl;
      // Determine whether this is a midpoint value.
      if ((fraction >= .5 - tolerance) & (fraction <= .5 + tolerance))
      {
          if (mode == MidpointRounding.AwayFromZero)
              return (value + 1) / Math.Pow(10, digits);
          else
             if (value % 2 != 0)
              return (value + 1) / Math.Pow(10, digits);
          else
              return value / Math.Pow(10, digits);
      }
      // Any remaining fractional value greater than .5 is not a midpoint value.
      if (fraction > .5)
          return (value + 1) / Math.Pow(10, digits);
      else
          return value / Math.Pow(10, digits);
  }
  
  // The example displays the following output:
  //       Value       Full Precision        ToEven    AwayFromZero
  //
  //        11.1                 11.1            11              11
  //        11.2                 11.2            11              11
  //        11.3   11.299999999999999            11              11
  //        11.4   11.399999999999999            11              11
  //        11.5   11.499999999999998            12              12
  //        11.6   11.599999999999998            12              12
  //
  //        11.5                 11.5            12              12
  

Rounding and single-precision floating-point values

The Round method includes overloads that accept arguments of type Decimal and Double. There are no methods that round values of type Single. If you pass a Single value to one of the overloads of the Round method, it is cast (in C#) or converted (in Visual Basic) to a Double, and the corresponding Round overload with a Double parameter is called. Although this is a widening conversion, it often involves a loss of precision, as the following example illustrates. When a Single value of 16.325 is passed to the Round method and rounded to two decimal places using the rounding to nearest convention, the result is 16.33 and not the expected result of 16.32.

C#

Single value = 16.325f;
Console.WriteLine("Widening Conversion of {0:R} (type {1}) to {2:R} (type {3}): ",
                  value, value.GetType().Name, (double)value,
                  ((double)(value)).GetType().Name);
Console.WriteLine(Math.Round(value, 2));
Console.WriteLine(Math.Round(value, 2, MidpointRounding.AwayFromZero));
Console.WriteLine();

Decimal decValue = (decimal)value;
Console.WriteLine("Cast of {0:R} (type {1}) to {2} (type {3}): ",
                  value, value.GetType().Name, decValue,
                  decValue.GetType().Name);
Console.WriteLine(Math.Round(decValue, 2));
Console.WriteLine(Math.Round(decValue, 2, MidpointRounding.AwayFromZero));

// The example displays the following output:
//    Widening Conversion of 16.325 (type Single) to 16.325000762939453 (type Double):
//    16.33
//    16.33
//
//    Cast of 16.325 (type Single) to 16.325 (type Decimal):
//    16.32
//    16.33

This unexpected result is due to a loss of precision in the conversion of the Single value to a Double. Because the resulting Double value of 16.325000762939453 is not a midpoint value and is greater than 16.325, it is always rounded upward.

In many cases, as the example illustrates, the loss of precision can be minimized or eliminated by casting or converting the Single value to a Decimal. Note that, because this is a narrowing conversion, it requires using a cast operator or calling a conversion method.

Round(Double, Int32, MidpointRounding)

Source:

Math.cs

Rounds a double-precision floating-point value to a specified number of fractional digits using the specified rounding convention.

C#

public static double Round(double value, int digits, MidpointRounding mode);

Parameters

value

Double

A double-precision floating-point number to be rounded.

digits

Int32

The number of fractional digits in the return value.

mode

MidpointRounding

One of the enumeration values that specifies which rounding strategy to use.

Returns

Double

The number that has digits fractional digits that value is rounded to. If value has fewer fractional digits than digits, value is returned unchanged.

Exceptions

ArgumentOutOfRangeException

digits is less than 0 or greater than 15.

ArgumentException

mode is not a valid value of MidpointRounding.

Remarks

The value of the digits argument can range from 0 to 15. The maximum number of integral and fractional digits supported by the Double type is 15.

See Midpoint values and rounding conventions for information about rounding numbers with midpoint values.

Important

When rounding midpoint values, the rounding algorithm performs an equality test. Because of problems of binary representation and precision in the floating-point format, the value returned by the method can be unexpected. For more information, see Rounding and precision.

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

Example

The following example demonstrates how to use the Round(Double, Int32, MidpointRounding) method with the MidpointRounding enumeration.

C#

// Round a positive and a negative value using the default.
double result = Math.Round(3.45, 1);
Console.WriteLine($"{result,4} = Math.Round({3.45,5}, 1)");
result = Math.Round(-3.45, 1);
Console.WriteLine($"{result,4} = Math.Round({-3.45,5}, 1)\n");

// Round a positive value using a MidpointRounding value.
result = Math.Round(3.45, 1, MidpointRounding.ToEven);
Console.WriteLine($"{result,4} = Math.Round({3.45,5}, 1, MidpointRounding.ToEven)");
result = Math.Round(3.45, 1, MidpointRounding.AwayFromZero);
Console.WriteLine($"{result,4} = Math.Round({3.45,5}, 1, MidpointRounding.AwayFromZero)");
result = Math.Round(3.47, 1, MidpointRounding.ToZero);
Console.WriteLine($"{result,4} = Math.Round({3.47,5}, 1, MidpointRounding.ToZero)\n");

// Round a negative value using a MidpointRounding value.
result = Math.Round(-3.45, 1, MidpointRounding.ToEven);
Console.WriteLine($"{result,4} = Math.Round({-3.45,5}, 1, MidpointRounding.ToEven)");
result = Math.Round(-3.45, 1, MidpointRounding.AwayFromZero);
Console.WriteLine($"{result,4} = Math.Round({-3.45,5}, 1, MidpointRounding.AwayFromZero)");
result = Math.Round(-3.47, 1, MidpointRounding.ToZero);
Console.WriteLine($"{result,4} = Math.Round({-3.47,5}, 1, MidpointRounding.ToZero)\n");

// The example displays the following output:

//         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(3.47, 1, MidpointRounding.ToZero)

//         -3.4 = Math.Round(-3.45, 1, MidpointRounding.ToEven)
//         -3.5 = Math.Round(-3.45, 1, MidpointRounding.AwayFromZero)
//         -3.4 = Math.Round(-3.47, 1, MidpointRounding.ToZero)

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 10^digits, and the multiplication operation in this case suffers from a loss of precision.

C#

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

See also

• Ceiling(Double)
• Floor(Double)
• Round(Decimal, Int32, MidpointRounding)

Round(Decimal, Int32, MidpointRounding)

Source:

Math.cs

Rounds a decimal value to a specified number of fractional digits using the specified rounding convention.

C#

public static decimal Round(decimal d, int decimals, MidpointRounding mode);

Parameters

d

Decimal

A decimal number to be rounded.

decimals

Int32

The number of decimal places in the return value.

mode

MidpointRounding

One of the enumeration values that specifies which rounding strategy to use.

Returns

Decimal

The number with decimals fractional digits that d is rounded to. If d has fewer fractional digits than decimals, d is returned unchanged.

Exceptions

ArgumentOutOfRangeException

decimals is less than 0 or greater than 28.

ArgumentException

mode is not a valid value of MidpointRounding.

OverflowException

The result is outside the range of a Decimal.

Remarks

See Midpoint values and rounding conventions for information about rounding numbers with midpoint values.

Important

When rounding midpoint values, the rounding algorithm performs an equality test. Because of problems of binary representation and precision in the floating-point format, the value returned by the method can be unexpected. For more information, see Rounding and precision.

The value of the decimals argument can range from 0 to 28.

Example

The following example demonstrates how to use the Round method with the MidpointRounding enumeration.

C#

decimal result;

// Round a positive value using different strategies.
// The precision of the result is 1 decimal place.

result = Math.Round(3.45m, 1, MidpointRounding.ToEven);
Console.WriteLine($"{result} = Math.Round({3.45m}, 1, MidpointRounding.ToEven)");
result = Math.Round(3.45m, 1, MidpointRounding.AwayFromZero);
Console.WriteLine($"{result} = Math.Round({3.45m}, 1, MidpointRounding.AwayFromZero)");
result = Math.Round(3.47m, 1, MidpointRounding.ToZero);
Console.WriteLine($"{result} = Math.Round({3.47m}, 1, MidpointRounding.ToZero)\n");

// Round a negative value using different strategies.
// The precision of the result is 1 decimal place.

result = Math.Round(-3.45m, 1, MidpointRounding.ToEven);
Console.WriteLine($"{result} = Math.Round({-3.45m}, 1, MidpointRounding.ToEven)");
result = Math.Round(-3.45m, 1, MidpointRounding.AwayFromZero);
Console.WriteLine($"{result} = Math.Round({-3.45m}, 1, MidpointRounding.AwayFromZero)");
result = Math.Round(-3.47m, 1, MidpointRounding.ToZero);
Console.WriteLine($"{result} = Math.Round({-3.47m}, 1, MidpointRounding.ToZero)\n");

/*
This code example produces the following results:

3.4 = Math.Round(3.45, 1, MidpointRounding.ToEven)
3.5 = Math.Round(3.45, 1, MidpointRounding.AwayFromZero)
3.4 = Math.Round(3.47, 1, MidpointRounding.ToZero)

-3.4 = Math.Round(-3.45, 1, MidpointRounding.ToEven)
-3.5 = Math.Round(-3.45, 1, MidpointRounding.AwayFromZero)
-3.4 = Math.Round(-3.47, 1, MidpointRounding.ToZero)
*/

See also

• Ceiling(Double)
• Floor(Double)
• Round(Decimal, Int32, MidpointRounding)

Round(Double, MidpointRounding)

Source:

Math.cs

Rounds a double-precision floating-point value to an integer using the specified rounding convention.

C#

public static double Round(double value, MidpointRounding mode);

Parameters

value

Double

A double-precision floating-point number to be rounded.

mode

MidpointRounding

One of the enumeration values that specifies which rounding strategy to use.

Returns

Double

The integer that value is rounded to. This method returns a Double instead of an integral type.

Exceptions

ArgumentException

mode is not a valid value of MidpointRounding.

Remarks

See Midpoint values and rounding conventions for information about rounding numbers with midpoint values.

Important

When rounding midpoint values, the rounding algorithm performs an equality test. Because of problems of binary representation and precision in the floating-point format, the value returned by the method can be unexpected. For more information, see Rounding and precision.

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

Example

The following example displays values returned by the Round(Double, MidpointRounding) method with different mode values.

C#

Double[] values = { 12.0, 12.1, 12.2, 12.3, 12.4, 12.5, 12.6,
                  12.7, 12.8, 12.9, 13.0 };
Console.WriteLine($"{"Value",-10} {"Default",-10} {"ToEven",-10} {"AwayFromZero",-15} {"ToZero",-15}");

foreach (var value in values)
    Console.WriteLine($"{value,-10:R} {Math.Round(value),-10} " +
        $"{Math.Round(value, MidpointRounding.ToEven),-10} " +
        $"{Math.Round(value, MidpointRounding.AwayFromZero),-15} " +
        $"{Math.Round(value, MidpointRounding.ToZero),-15}");

// The example displays the following output:
//       Value      Default    ToEven     AwayFromZero    ToZero
//       12         12         12         12              12
//       12.1       12         12         12              12
//       12.2       12         12         12              12
//       12.3       12         12         12              12
//       12.4       12         12         12              12
//       12.5       12         12         13              12
//       12.6       13         13         13              12
//       12.7       13         13         13              12
//       12.8       13         13         13              12
//       12.9       13         13         13              12
//       13         13         13         13              13

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, 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.

C#

using System;

public class Example
{
   public static void Main()
   {
      double value = 11.1;
      for (int ctr = 0; ctr <= 5; ctr++)
         value = RoundValueAndAdd(value);

      Console.WriteLine();

      value = 11.5;
      RoundValueAndAdd(value);
   }

   private static double RoundValueAndAdd(double value)
   {
      Console.WriteLine("{0} --> {1}", value, Math.Round(value,
                        MidpointRounding.AwayFromZero));
      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

See also

• Ceiling(Double)
• Floor(Double)
• Round(Decimal, MidpointRounding)

Round(Double, Int32)

Source:

Math.cs

Rounds a double-precision floating-point value to a specified number of fractional digits, and rounds midpoint values to the nearest even number.

C#

public static double Round(double value, int digits);

Parameters

value

Double

A double-precision floating-point number to be rounded.

digits

Int32

The number of fractional digits in the return value.

Returns

Double

The number nearest to value that contains a number of fractional digits equal to digits.

Exceptions

ArgumentOutOfRangeException

digits is less than 0 or greater than 15.

Remarks

The value of the digits argument can range from 0 to 15. The maximum number of integral and fractional digits supported by the Double type is 15.

This method uses the default rounding convention of MidpointRounding.ToEven. See Midpoint values and rounding conventions for information about rounding numbers with midpoint values.

Important

When rounding midpoint values, the rounding algorithm performs an equality test. Because of problems of binary representation and precision in the floating-point format, the value returned by the method can be unexpected. For more information, see Rounding and precision.

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

Example

The following example rounds double values with two fractional digits to doubles that have a single fractional digit.

C#

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

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 10^digits, and the multiplication operation in this case suffers from a loss of precision.

C#

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

See also

• Ceiling(Double)
• Floor(Double)
• Round(Decimal, Int32)

Round(Decimal, Int32)

Source:

Math.cs

Rounds a decimal value to a specified number of fractional digits, and rounds midpoint values to the nearest even number.

C#

public static decimal Round(decimal d, int decimals);

Parameters

d

Decimal

A decimal number to be rounded.

decimals

Int32

The number of decimal places in the return value.

Returns

Decimal

The number nearest to d that contains a number of fractional digits equal to decimals.

Exceptions

ArgumentOutOfRangeException

decimals is less than 0 or greater than 28.

OverflowException

The result is outside the range of a Decimal.

Remarks

The value of the decimals argument can range from 0 to 28.

This method uses the default rounding convention of MidpointRounding.ToEven. For information about rounding numbers with midpoint values, see Midpoint values and rounding conventions.

Important

When rounding midpoint values, the rounding algorithm performs an equality test. Because of problems of binary representation and precision in the floating-point format, the value returned by the method can be unexpected. For more information, see Rounding and precision.

Example

The following example rounds decimal values with two fractional digits to values that have a single fractional digit.

C#

Console.WriteLine(Math.Round(3.44m, 1));
Console.WriteLine(Math.Round(3.45m, 1));
Console.WriteLine(Math.Round(3.46m, 1));
Console.WriteLine();

Console.WriteLine(Math.Round(4.34m, 1));
Console.WriteLine(Math.Round(4.35m, 1));
Console.WriteLine(Math.Round(4.36m, 1));

// The example displays the following output:
//       3.4
//       3.4
//       3.5
//
//       4.3
//       4.4
//       4.4

See also

• Ceiling(Double)
• Floor(Double)
• Round(Decimal, Int32)

Round(Double)

Source:

Math.cs

Rounds a double-precision floating-point value to the nearest integral value, and rounds midpoint values to the nearest even number.

C#

public static double Round(double a);

Parameters

a

Double

A double-precision floating-point number to be rounded.

Returns

Double

The integer nearest a. If the fractional component of a is halfway between two integers, one of which is even and the other odd, then the even number is returned. Note that this method returns a Double instead of an integral type.

Remarks

This method uses the default rounding convention of MidpointRounding.ToEven. For information about rounding numbers with midpoint values, see Midpoint values and rounding conventions.

Important

When rounding midpoint values, the rounding algorithm performs an equality test. Because of problems of binary representation and precision in the floating-point format, the value returned by the method can be unexpected. For more information, see Rounding and precision.

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

Starting with Visual Basic 15.8, the performance of Double-to-integer conversion is optimized if you pass the value returned by the Round method to the any of the integral conversion functions, or if the Double value returned by Round is automatically converted to an integer with Option Strict set to Off. This optimization allows code to run faster -- up to twice as fast for code that does a large number of conversions to integer types. The following example illustrates such optimized conversions:

VB

Dim d1 As Double = 1043.75133
Dim i1 As Integer = CInt(Math.Ceiling(d1))        ' Result: 1044

Dim d2 As Double = 7968.4136
Dim i2 As Integer = CInt(Math.Ceiling(d2))        ' Result: 7968

Example

The following example demonstrates rounding to the nearest integer value.

C#

Console.WriteLine("Classic Math.Round in CSharp");
Console.WriteLine(Math.Round(4.4)); // 4
Console.WriteLine(Math.Round(4.5)); // 4
Console.WriteLine(Math.Round(4.6)); // 5
Console.WriteLine(Math.Round(5.5)); // 6

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) 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.

C#

using System;

public class Example
{
   public static void Main()
   {
      double value = 11.1;
      for (int ctr = 0; ctr <= 5; ctr++)
         value = RoundValueAndAdd(value);

      Console.WriteLine();

      value = 11.5;
      RoundValueAndAdd(value);
   }

   private static double RoundValueAndAdd(double value)
   {
      Console.WriteLine("{0} --> {1}", value, Math.Round(value));
      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

See also

• Ceiling(Double)
• Floor(Double)
• Round(Decimal)

Round(Decimal)

Source:

Math.cs

Rounds a decimal value to the nearest integral value, and rounds midpoint values to the nearest even number.

C#

public static decimal Round(decimal d);

Parameters

d

Decimal

A decimal number to be rounded.

Returns

Decimal

The integer nearest the d parameter. If the fractional component of d is halfway between two integers, one of which is even and the other odd, the even number is returned. Note that this method returns a Decimal instead of an integral type.

Exceptions

OverflowException

The result is outside the range of a Decimal.

Examples

The following example demonstrates the Round(Decimal) method. The Decimal value of 4.5 rounds to 4 rather than 5, because this overload uses the default ToEven convention.

C#

for (decimal value = 4.2m; value <= 4.8m; value+=.1m )
   Console.WriteLine("{0} --> {1}", value, Math.Round(value));
// The example displays the following output:
//       4.2 --> 4
//       4.3 --> 4
//       4.4 --> 4
//       4.5 --> 4
//       4.6 --> 5
//       4.7 --> 5
//       4.8 --> 5

Remarks

This method uses the default rounding convention of MidpointRounding.ToEven. For information about rounding numbers with midpoint values, see Midpoint values and rounding conventions.

Important

When rounding midpoint values, the rounding algorithm performs an equality test. Because of problems of binary representation and precision in the floating-point format, the value returned by the method can be unexpected. For more information, see Rounding and precision.

See also

• Ceiling(Double)
• Floor(Double)
• Round(Decimal)

Round(Decimal, MidpointRounding)

Source:

Math.cs

Rounds a decimal value an integer using the specified rounding convention.

C#

public static decimal Round(decimal d, MidpointRounding mode);

Parameters

d

Decimal

A decimal number to be rounded.

mode

MidpointRounding

One of the enumeration values that specifies which rounding strategy to use.

Returns

Decimal

The integer that d is rounded to. This method returns a Decimal instead of an integral type.

Exceptions

ArgumentException

mode is not a valid value of MidpointRounding.

OverflowException

The result is outside the range of a Decimal.

Remarks

For information about rounding numbers with midpoint values, see Midpoint values and rounding conventions.

Important

When rounding midpoint values, the rounding algorithm performs an equality test. Because of problems of binary representation and precision in the floating-point format, the value returned by the method can be unexpected. For more information, see Rounding and precision.

Example

The following example displays values returned by the Round(Decimal, MidpointRounding) method with different mode values.

C#

Console.WriteLine($"{"Value",-10} {"Default",-10} {"ToEven",-10} {"AwayFromZero",-15} {"ToZero",-15}");
for (decimal value = 12.0m; value <= 13.0m; value += 0.1m)
    Console.WriteLine($"{value,-10} {Math.Round(value),-10} " +
        $"{Math.Round(value, MidpointRounding.ToEven),-10} " +
        $"{Math.Round(value, MidpointRounding.AwayFromZero),-15} " +
        $"{Math.Round(value, MidpointRounding.ToZero),-15}");

// The example displays the following output:
//       Value      Default    ToEven     AwayFromZero    ToZero
//       12.0       12         12         12              12
//       12.1       12         12         12              12
//       12.2       12         12         12              12
//       12.3       12         12         12              12
//       12.4       12         12         12              12
//       12.5       12         12         13              12
//       12.6       13         13         13              12
//       12.7       13         13         13              12
//       12.8       13         13         13              12
//       12.9       13         13         13              12
//       13.0       13         13         13              13

See also

• Ceiling(Double)
• Floor(Double)
• Round(Decimal, MidpointRounding)