# Math.Round Method

## Definition

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

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

## Examples

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

## Remarks

In this section:

### 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 the nearest 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 two decimal places, and 7.500 is a midpoint value if it is to be rounded to an integer. In these cases, 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 away from zero is the most widely known form of rounding.

• Rounding to nearest, 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.

Rounding to nearest is the standard form of rounding used in financial and statistical operations. It conforms to IEEE Standard 754, section 4. When used in multiple rounding operations, it 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.

``````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
``````
``````Module Example
Public Sub Main()
Dim values() As Decimal = { 1.15d, 1.25d, 1.35d, 1.45d, 1.55d, 1.65d }
Dim sum As Decimal

' Calculate true mean.
For Each value In values
sum += value
Next
Console.WriteLine("True mean:     {0:N2}", sum/values.Length)

' Calculate mean with rounding away from zero.
sum = 0
For Each value In values
sum += Math.Round(value, 1, MidpointRounding.AwayFromZero)
Next
Console.WriteLine("AwayFromZero:  {0:N2}", sum/values.Length)

' Calculate mean with rounding to nearest.
sum = 0
For Each value In values
sum += Math.Round(value, 1, MidpointRounding.ToEven)
Next
Console.WriteLine("ToEven:        {0:N2}", sum/values.Length)
End Sub
End Module
' The example displays the following output:
'       True mean:     1.40
'       AwayFromZero:  1.45
'       ToEven:        1.40
``````

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

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.

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 following example illustrates the problem. It repeatedly adds .1 to 11.0 and rounds the result to the nearest integer. Regardless of the rounding convention, 11.5 should round to 12. 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 value is not rounded to the next highest integer. As the example also shows, this problem does not occur if we simply assign the constant value 11.5 to a Double variable.

``````using System;

public class Example
{
public static void Main()
{
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++)

Console.WriteLine();

value = 11.5;
}

{
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
``````
``````Module Example
Public Sub Main()
Dim value As Double = 11.1

Console.WriteLine("{0,5} {1,20:R}  {2,12} {3,15}",
"Value", "Full Precision", "ToEven",
"AwayFromZero")
Console.WriteLine()
For ctr As Integer = 0 To 5
Next
Console.WriteLine()

value = 11.5
End Sub

Private Function RoundValueAndAdd(value As Double) As Double
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
End Function
End Module
' 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.

``````:::code language="csharp" source="~/samples/snippets/csharp/VS_Snippets_CLR_System/system.math.round.overload/cs/precision2.cs" interactive="try-dotnet" id="Snippet8":::
``````

#### 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.

``````using System;

public class Example
{
public static void Main()
{
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
``````
``````Module Example
Public Sub Main()
Dim value As Single = 16.325
Console.WriteLine("Widening Conversion of {0:R} (type {1}) to {2:R} (type {3}): ",
value, value.GetType().Name, CDbl(value),
CDbl(value).GetType().Name)
Console.WriteLine(Math.Round(value, 2))
Console.WriteLine(Math.Round(value, 2, MidpointRounding.AwayFromZero))
Console.WriteLine()

Dim decValue As Decimal = CDec(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))
Console.WriteLine()
End Sub
End Module
' 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(Decimal, Int32, MidpointRounding)

Rounds a decimal value to a specified number of fractional digits, and uses the specified rounding convention for midpoint values.

``````public:
static System::Decimal Round(System::Decimal d, int decimals, MidpointRounding mode);``````
``public static decimal Round (decimal d, int decimals, MidpointRounding mode);``
``static member Round : decimal * int * MidpointRounding -> decimal``
``Public Shared Function Round (d As Decimal, decimals As Integer, mode As MidpointRounding) As Decimal``

#### Parameters

d
Decimal

A decimal number to be rounded.

decimals
Int32

The number of decimal places in the return value.

mode
MidpointRounding

Specification for how to round `d` if it is midway between two other numbers.

#### Returns

Decimal

The number nearest to `d` that contains a number of fractional digits equal to `decimals`. If `d` has fewer fractional digits than `decimals`, `d` is returned unchanged.

#### Exceptions

`decimals` is less than 0 or greater than 28.

`mode` is not a valid value of MidpointRounding.

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.

``````// This example demonstrates the Math.Round() method in conjunction
// with the MidpointRounding enumeration.
using namespace System;

void main()
{
Decimal result = (Decimal) 0.0;
Decimal posValue = (Decimal) 3.45;
Decimal negValue = (Decimal) -3.45;

// 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)

*/
``````
``````decimal result = 0.0m;
decimal positiveValue = 3.45m;
decimal negativeValue = -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(positiveValue, 1);
Console.WriteLine(\$"{result} = Math.Round({positiveValue}, 1)");
result = Math.Round(negativeValue, 1);
Console.WriteLine(\$"{result} = Math.Round({negativeValue}, 1)");
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(positiveValue, 1, MidpointRounding.ToEven);
Console.WriteLine(\$"{result} = Math.Round({positiveValue}, 1, MidpointRounding.ToEven)");
result = Math.Round(positiveValue, 1, MidpointRounding.AwayFromZero);
Console.WriteLine(\$"{result} = Math.Round({positiveValue}, 1, MidpointRounding.AwayFromZero)");
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(negativeValue, 1, MidpointRounding.ToEven);
Console.WriteLine(\$"{result} = Math.Round({negativeValue}, 1, MidpointRounding.ToEven)");
result = Math.Round(negativeValue, 1, MidpointRounding.AwayFromZero);
Console.WriteLine(\$"{result} = Math.Round({negativeValue}, 1, MidpointRounding.AwayFromZero)");
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)

*/
``````
``````' This example demonstrates the Math.Round() method in conjunction
' with the MidpointRounding enumeration.
Class Sample
Public Shared Sub Main()
Dim result As Decimal = 0D
Dim posValue As Decimal = 3.45D
Dim negValue As Decimal = -3.45D

' 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()

End Sub
End Class
'
'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)
'
``````

## Round(Double, Int32, MidpointRounding)

Rounds a double-precision floating-point value to a specified number of fractional digits, and uses the specified rounding convention for midpoint values.

``````public:
static double Round(double value, int digits, MidpointRounding mode);``````
``public static double Round (double value, int digits, MidpointRounding mode);``
``static member Round : double * int * MidpointRounding -> double``
``Public Shared Function Round (value As Double, digits As Integer, mode As MidpointRounding) As Double``

#### 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

Specification for how to round `value` if it is midway between two other numbers.

#### Returns

Double

The number nearest to `value` that has a number of fractional digits equal to `digits`. If `value` has fewer fractional digits than `digits`, `value` is returned unchanged.

#### Exceptions

`digits` is less than 0 or greater than 15.

`mode` is not a valid value of MidpointRounding.

### Remarks

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

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.

`````` double posValue =  3.45;
double negValue = -3.45;

// Round a positive and a negative value using the default.
double 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)\n", result, negValue);

// Round a positive value using a MidpointRounding value.
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)\n",
result, posValue);

// Round a negative value using a MidpointRounding value.
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)\n",
result, negValue);
// 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(-3.45, 1, MidpointRounding.ToEven)
//       -3.5 = Math.Round(-3.45, 1, MidpointRounding.AwayFromZero)
``````
``````Module Example
Public Sub Main()
Dim posValue As Double = 3.45
Dim negValue As Double = -3.45

' Round a positive and a negative value using the default.
Dim result As Double = 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 using a MidpointRounding value.
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 positive value using a MidpointRounding value.
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()
End Sub
End Module
' 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(-3.45, 1, MidpointRounding.ToEven)
'       -3.5 = Math.Round(-3.45, 1, MidpointRounding.AwayFromZero)
``````

### 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.

``````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
``````
``````Module Example
Public Sub Main()
Dim values() As Double = { 2.125, 2.135, 2.145, 3.125, 3.135, 3.145 }
For Each value As Double In values
Console.WriteLine("{0} --> {1}", value,
Math.Round(value, 2, MidpointRounding.AwayFromZero))
Next
End Sub
End Module
' 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
``````

## Round(Double, MidpointRounding)

Rounds a double-precision floating-point value to the nearest integer, and uses the specified rounding convention for midpoint values.

``````public:
static double Round(double value, MidpointRounding mode);``````
``public static double Round (double value, MidpointRounding mode);``
``static member Round : double * MidpointRounding -> double``
``Public Shared Function Round (value As Double, mode As MidpointRounding) As Double``

#### Parameters

value
Double

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

mode
MidpointRounding

Specification for how to round `value` if it is midway between two other numbers.

#### Returns

Double

The 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. Note that this method returns a Double instead of an integral type.

#### Exceptions

`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) method, the Round(Double, MidpointRounding) method with a `mode` argument of ToEven, and the Round(Double, MidpointRounding) method with a `mode` argument of AwayFromZero.

``````using System;

public class Example
{
public static void Main()
{
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("{0,-10} {1,-10} {2,-10} {3,-15}", "Value", "Default",
"ToEven", "AwayFromZero");
foreach (var value in values)
Console.WriteLine("{0,-10:R} {1,-10} {2,-10} {3,-15}",
value, Math.Round(value),
Math.Round(value, MidpointRounding.ToEven),
Math.Round(value, MidpointRounding.AwayFromZero));
}
}
// The example displays the following output:
//       Value      Default    ToEven     AwayFromZero
//       12         12         12         12
//       12.1       12         12         12
//       12.2       12         12         12
//       12.3       12         12         12
//       12.4       12         12         12
//       12.5       12         12         13
//       12.6       13         13         13
//       12.7       13         13         13
//       12.8       13         13         13
//       12.9       13         13         13
//       13         13         13         13
``````
``````Module Example
Public Sub Main()
Dim values() As Double = { 12.0, 12.1, 12.2, 12.3, 12.4, 12.5, 12.6,
12.7, 12.8, 12.9, 13.0 }
Console.WriteLine("{0,-10} {1,-10} {2,-10} {3,-15}", "Value", "Default",
"ToEven", "AwayFromZero")
For Each value In values
Console.WriteLine("{0,-10} {1,-10} {2,-10} {3,-15}",
value, Math.Round(value),
Math.Round(value, MidpointRounding.ToEven),
Math.Round(value, MidpointRounding.AwayFromZero))
Next
End Sub
End Module
' The example displays the following output:
'       Value      Default    ToEven     AwayFromZero
'       12         12         12         12
'       12.1       12         12         12
'       12.2       12         12         12
'       12.3       12         12         12
'       12.4       12         12         12
'       12.5       12         12         13
'       12.6       13         13         13
'       12.7       13         13         13
'       12.8       13         13         13
'       12.9       13         13         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.

``````using System;

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

Console.WriteLine();

value = 11.5;
}

{
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
``````
``````Module Example
Public Sub Main()
Dim value As Double = 11.1
For ctr As Integer = 0 To 5
Next
Console.WriteLine()

value = 11.5
End Sub

Private Function RoundValueAndAdd(value As Double) As Double
Console.WriteLine("{0} --> {1}", value, Math.Round(value,
MidpointRounding.AwayFromZero))
Return value + .1
End Function
End Module
' 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
``````

## 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.

``````public:
static double Round(double value, int digits);``````
``public static double Round (double value, int digits);``
``static member Round : double * int -> double``
``Public Shared Function Round (value As Double, digits As Integer) As Double``

#### 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

`digits` is less than 0 or greater than 15.

### Remarks

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

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.

``````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
``````
``````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
``````
``````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 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
``````
``````Module Example
Public Sub Main()
Dim values() As Double = { 2.125, 2.135, 2.145, 3.125, 3.135, 3.145 }
For Each value As Double In values
Console.WriteLine("{0} --> {1}", value, Math.Round(value, 2))
Next
End Sub
End Module
' 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
``````

## Round(Double)

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

``````public:
static double Round(double a);``````
``public static double Round (double a);``
``static member Round : double -> double``
``Public Shared Function Round (a As Double) As Double``

#### 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. 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 `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:

``````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.

``````using namespace System;

void main()
{
Console::WriteLine("Classic Math.Round in CPP");
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
}
``````
``````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
``````
``````Module Module1

Sub Main()
Console.WriteLine("Classic Math.Round in Visual Basic")
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
End Sub

End Module
``````

### 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.

``````using System;

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

Console.WriteLine();

value = 11.5;
}

{
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
``````
``````Module Example
Public Sub Main()
Dim value As Double = 11.1
For ctr As Integer = 0 To 5
Next
Console.WriteLine()

value = 11.5
End Sub

Private Function RoundValueAndAdd(value As Double) As Double
Console.WriteLine("{0} --> {1}", value, Math.Round(value))
Return value + .1
End Function
End Module
' 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
``````

## Round(Decimal, Int32)

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

``````public:
static System::Decimal Round(System::Decimal d, int decimals);``````
``public static decimal Round (decimal d, int decimals);``
``static member Round : decimal * int -> decimal``
``Public Shared Function Round (d As Decimal, decimals As Integer) As Decimal``

#### 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

`decimals` is less than 0 or greater than 28.

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. 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".

### Example

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

``````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
``````
``````Public Module Example
Sub Main()
Console.WriteLine(Math.Round(3.44, 1))
Console.WriteLine(Math.Round(3.45, 1))
Console.WriteLine(Math.Round(3.46, 1))
Console.WriteLine()

Console.WriteLine(Math.Round(4.34, 1))
Console.WriteLine(Math.Round(4.35, 1))
Console.WriteLine(Math.Round(4.36, 1))
End Sub
End Module
' The example displays the following output:
'       3.4
'       3.4
'       3.5
'
'       4.3
'       4.4
'       4.4
``````

## Round(Decimal)

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

``````public:
static System::Decimal Round(System::Decimal d);``````
``public static decimal Round (decimal d);``
``static member Round : decimal -> decimal``
``Public Shared Function Round (d As Decimal) As Decimal``

#### 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

The result is outside the range of a Decimal.

### Examples

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

``````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
``````
``````Module Example
Public Sub Main()
For value As Decimal = 4.2d To 4.8d Step .1d
Console.WriteLine("{0} --> {1}", value, Math.Round(value))
Next
End Sub
End Module
' 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. 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".

## Round(Decimal, MidpointRounding)

Rounds a decimal value to the nearest integer, and uses the specified rounding convention for midpoint values.

``````public:
static System::Decimal Round(System::Decimal d, MidpointRounding mode);``````
``public static decimal Round (decimal d, MidpointRounding mode);``
``static member Round : decimal * MidpointRounding -> decimal``
``Public Shared Function Round (d As Decimal, mode As MidpointRounding) As Decimal``

#### Parameters

d
Decimal

A decimal number to be rounded.

mode
MidpointRounding

Specification for how to round `d` if it is midway between two other numbers.

#### Returns

Decimal

The integer nearest `d`. If `d` is halfway between two numbers, one of which is even and the other odd, then `mode` determines which of the two is returned. Note that this method returns a Decimal instead of an integral type.

#### Exceptions

`mode` is not a valid value of MidpointRounding.

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".

### Example

The following example displays values returned by the Round(Decimal) method, the Round(Decimal, MidpointRounding) method with a `mode` argument of ToEven, and the Round(Decimal, MidpointRounding) method with a `mode` argument of AwayFromZero.

``````Console.WriteLine("{0,-10} {1,-10} {2,-10} {3,-15}", "Value", "Default",
"ToEven", "AwayFromZero");
for (decimal value = 12.0m; value <= 13.0m; value += 0.1m)
Console.WriteLine("{0,-10} {1,-10} {2,-10} {3,-15}",
value, Math.Round(value),
Math.Round(value, MidpointRounding.ToEven),
Math.Round(value, MidpointRounding.AwayFromZero));
// The example displays the following output:
//       Value      Default    ToEven     AwayFromZero
//       12         12         12         12
//       12.1       12         12         12
//       12.2       12         12         12
//       12.3       12         12         12
//       12.4       12         12         12
//       12.5       12         12         13
//       12.6       13         13         13
//       12.7       13         13         13
//       12.8       13         13         13
//       12.9       13         13         13
//       13         13         13         13
``````
``````Module Example
Public Sub Main()
Console.WriteLine("{0,-10} {1,-10} {2,-10} {3,-15}", "Value", "Default",
"ToEven", "AwayFromZero")
For value As Decimal = 12.0d To 13.0d Step .1d
Console.WriteLine("{0,-10} {1,-10} {2,-10} {3,-15}",
value, Math.Round(value),
Math.Round(value, MidpointRounding.ToEven),
Math.Round(value, MidpointRounding.AwayFromZero))
Next
End Sub
End Module
' The example displays the following output:
'       Value      Default    ToEven     AwayFromZero
'       12         12         12         12
'       12.1       12         12         12
'       12.2       12         12         12
'       12.3       12         12         12
'       12.4       12         12         12
'       12.5       12         12         13
'       12.6       13         13         13
'       12.7       13         13         13
'       12.8       13         13         13
'       12.9       13         13         13
'       13         13         13         13
``````