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Math.DivRem Method (Int32, Int32, Int32)

Calculates the quotient of two 32-bit signed integers and also returns the remainder in an output parameter.

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

public:
static int DivRem (
	int a, 
	int b, 
	[OutAttribute] int% result
)
public static int DivRem (
	int a, 
	int b, 
	/** @attribute OutAttribute() */ /** @ref */ int result
)
Not applicable.

Parameters

a

The System.Int32 that contains the dividend.

b

The System.Int32 that contains the divisor.

result

The System.Int32 that receives the remainder.

Return Value

The System.Int32 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 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.
*/

Windows 98, Windows Server 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 Microsoft .NET Framework 3.0 is supported on Windows Vista, Microsoft Windows XP SP2, and Windows Server 2003 SP1.

.NET Framework

Supported in: 3.0, 2.0, 1.1

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