# Math.IEEERemainder Method

.NET Framework 2.0

Returns the remainder resulting from the division of a specified number by another specified number.

Namespace: System
Assembly: mscorlib (in mscorlib.dll)

## Syntax

```public:
static double IEEERemainder (
double x,
double y
)
```
```public static double IEEERemainder (
double x,
double y
)
```
```public static function IEEERemainder (
x : double,
y : double
) : double
```

x

A dividend.

y

A divisor.

#### Return Value

A number equal to x - (y Q), where Q is the quotient of x / y rounded to the nearest integer (if x / y falls halfway between two integers, the even integer is returned). If x - (y Q) is zero, the value +0 is returned if x is positive, or -0 if x is negative. If y = 0, NaN (Not-A-Number) is returned.

## Remarks

This operation complies with the remainder operation defined in Section 5.1 of ANSI/IEEE Std 754-1985; IEEE Standard for Binary Floating-Point Arithmetic; Institute of Electrical and Electronics Engineers, Inc; 1985.

## Example

```// This example demonstrates Math.DivRem()
//                           Math.IEEERemainder()
using namespace System;
int main()
{
int int1 = Int32::MaxValue;
int int2 = Int32::MaxValue;
int intResult;
Int64 long1 = Int64::MaxValue;
Int64 long2 = Int64::MaxValue;
Int64 longResult;
double doubleResult;
double divisor;
String^ nl = Environment::NewLine;

//
Console::WriteLine( "{0}Calculate the quotient and remainder of two Int32 values:", nl );
intResult = Math::DivRem( int1, 2, int2 );
Console::WriteLine( "{0}/{1} = {2}, with a remainder of {3}.", int1, 2, intResult, int2 );

//
Console::WriteLine( "{0}Calculate the quotient and remainder of two Int64 values:", nl );
longResult = Math::DivRem( long1, 4, long2 );
Console::WriteLine( "{0}/{1} = {2}, with a remainder of {3}.", long1, 4, longResult, long2 );

//
String^ str1 = "The IEEE remainder of {0:e}/{1:f} is {2:e}";
divisor = 2.0;
Console::WriteLine( "{0}Divide two double-precision floating-point values:", nl );
doubleResult = Math::IEEERemainder( Double::MaxValue, divisor );
Console::Write( "1) " );
Console::WriteLine( str1, Double::MaxValue, divisor, doubleResult );
divisor = 3.0;
doubleResult = Math::IEEERemainder( Double::MaxValue, divisor );
Console::Write( "2) " );
Console::WriteLine( str1, Double::MaxValue, divisor, doubleResult );
Console::WriteLine( "Note that two positive numbers can yield a negative remainder." );
}

/*
This example produces the following results:

Calculate the quotient and remainder of two Int32 values:
2147483647/2 = 1073741823, with a remainder of 1.

Calculate the quotient and remainder of two Int64 values:
9223372036854775807/4 = 2305843009213693951, with a remainder of 3.

Divide two double-precision floating-point values:
1) The IEEE remainder of 1.797693e+308/2.00 is 0.000000e+000
2) The IEEE remainder of 1.797693e+308/3.00 is -1.000000e+000
Note that two positive numbers can yield a negative remainder.
*/

```
```// This example demonstrates Math.DivRem()
// Math.IEEERemainder()
import System.*;

class Sample
{
public static void main(String[] args)
{
int int1 = Int32.MaxValue;
int int2 = Int32.MaxValue;
int intResult;
long long1 = Int64.MaxValue;
long long2 = Int64.MaxValue;
long longResult;
double doubleResult;
double divisor;
String nl = Environment.get_NewLine();
//
Console.WriteLine("{0}Calculate the quotient and "
+ "remainder of two Int32 values:", nl);
intResult = System.Math.DivRem(int1, 2, int2);
Console.WriteLine("{0}/{1} = {2}, with a remainder of {3}.",
new Object[] { (Int32)int1, (Int32)2,
(Int32)intResult, (Int32)int2 });
//
Console.WriteLine("{0}Calculate the quotient and remainder"
+ " of two Int64 values:", nl);
longResult = System.Math.DivRem(long1, 4, long2);
Console.WriteLine("{0}/{1} = {2}, with a remainder of {3}.",
new Object[] { (Int64)long1, (Int32)4,
(Int64)longResult, (Int64)long2    });
//
String str1 = "The IEEE remainder of {0:e}/{1:f} is {2:e}";
divisor = 2.0;
Console.WriteLine("{0}Divide two double-precision"
+ " floating-point values:", nl);
doubleResult = System.Math.IEEERemainder(System.Double.MaxValue, divisor);
Console.Write("1) ");
Console.WriteLine(str1,
((System.Double)System.Double.MaxValue).ToString("e"),
((System.Double)divisor).ToString("f"),
((System.Double)doubleResult).ToString("e"));

divisor = 3.0;
doubleResult = System.Math.IEEERemainder(System.Double.MaxValue, divisor);
Console.Write("2) ");
Console.WriteLine(str1,
((System.Double)System.Double.MaxValue).ToString("e"),
((System.Double)divisor).ToString("f"),
((System.Double)doubleResult).ToString("e"));
Console.WriteLine("Note that two positive numbers can"
+ " yield a negative remainder.");
} //main
} //Sample

/*
This example produces the following results:

Calculate the quotient and remainder of two Int32 values:
2147483647/2 = 1073741823, with a remainder of 1.

Calculate the quotient and remainder of two Int64 values:
9223372036854775807/4 = 2305843009213693951, with a remainder of 3.

Divide two double-precision floating-point values:
1) The IEEE remainder of 1.797693e+308/2.00 is 0.000000e+000
2) The IEEE remainder of 1.797693e+308/3.00 is -1.000000e+000
Note that two positive numbers can yield a negative remainder.
*/

```

## Platforms

Windows 98, Windows 2000 SP4, Windows CE, Windows Millennium Edition, Windows Mobile for Pocket PC, Windows Mobile for Smartphone, Windows Server 2003, Windows XP Media Center Edition, Windows XP Professional x64 Edition, Windows XP SP2, Windows XP Starter Edition

The .NET Framework does not support all versions of every platform. For a list of the supported versions, see System Requirements.

## Version Information

#### .NET Framework

Supported in: 2.0, 1.1, 1.0

#### .NET Compact Framework

Supported in: 2.0, 1.0