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Math.DivRem Method (Int64, Int64, Int64)

Calculates the quotient of two 64-bit signed integers and also returns the remainder in an output parameter.

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

public static long DivRem (
	long a,
	long b,
	out long result
)
public static long DivRem (
	long a, 
	long b, 
	/** @attribute OutAttribute() */ /** @ref */ long result
)
JScript does not support passing value-type arguments by reference.

Parameters

a

The System.Int64 that contains the dividend.

b

The System.Int64 that contains the divisor.

result

The System.Int64 that receives the remainder.

Return Value

The System.Int64 containing the quotient of the specified numbers.

Exception typeCondition

DivideByZeroException

b is zero.

The following code example demonstrates the DivRem method.

// This example demonstrates Math.DivRem()
//                           Math.IEEERemainder()
using System;

class Sample 
{
    public static void Main() 
    {
    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.NewLine;
//
    Console.WriteLine("{0}Calculate the quotient and remainder of two Int32 values:", nl);
    intResult = Math.DivRem(int1, 2, out 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, out 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.
*/

Windows 98, Windows 2000 SP4, Windows Millennium Edition, 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.

.NET Framework

Supported in: 2.0, 1.1

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