メモ : この名前空間、クラス、およびメンバは、.NET Framework Version 1.1 だけでサポートされています。
2 つの 32 ビット符号付き整数の商を返し、出力パラメータとして剰余を渡します。
Overloads Public Shared Function DivRem( _
ByVal a As Integer, _ ByVal b As Integer, _ <Out()> ByRef result As Integer _) As Integer
[C#]
public static int DivRem(inta,intb, out intresult);
[C++]
public: static int DivRem(inta,intb, [ Out] int* result);
[JScript]
public static function DivRem(
a : int,b : int,result : int) : int;
パラメータ
- a
被除数を格納している System.Int32 。 - b
除数を格納している System.Int32 。 - result
剰余を受け取る System.Int32 。
戻り値
指定した数値の商を格納している System.Int32 。
使用例
' This example demonstrates Math.DivRem()
' Math.IEEERemainder()
Imports System
Class Sample
Public Shared Sub Main()
Dim int1 As Integer = Int32.MaxValue
Dim int2 As Integer = Int32.MaxValue
Dim intResult As Integer
Dim long1 As Long = Int64.MaxValue
Dim long2 As Long = Int64.MaxValue
Dim longResult As Long
Dim doubleResult As Double
Dim divisor As Double
Dim nl As [String] = 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)
'
Dim str1 As [String] = "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.")
End Sub 'Main
End Class '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 -1.995840e+292
'Note that two positive numbers can yield a negative remainder.
'
[C#]
// 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 -1.995840e+292
Note that two positive numbers can yield a negative remainder.
*/
[C++]
// This example demonstrates Math.DivRem()
// Math.IEEERemainder()
#using <mscorlib.dll>
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(S"{0}Calculate the quotient and remainder of two Int32 values:", nl);
intResult = Math::DivRem(int1, 2, &int2);
Console::WriteLine(S"{0}/{1} = {2}, with a remainder of {3}.", __box(int1), __box(2), __box(intResult), __box(int2));
//
Console::WriteLine(S"{0}Calculate the quotient and remainder of two Int64 values:", nl);
longResult = Math::DivRem(long1, 4, &long2);
Console::WriteLine(S"{0}/{1} = {2}, with a remainder of {3}.", __box(long1), __box(4), __box(longResult), __box(long2));
//
String* str1 = S"The IEEE remainder of {0:e}/{1:f} is {2:e}";
divisor = 2.0;
Console::WriteLine(S"{0}Divide two double-precision floating-point values:", nl);
doubleResult = Math::IEEERemainder(Double::MaxValue, divisor);
Console::Write(S"1) ");
Console::WriteLine(str1, __box(Double::MaxValue), __box(divisor), __box(doubleResult));
divisor = 3.0;
doubleResult = Math::IEEERemainder(Double::MaxValue, divisor);
Console::Write(S"2) ");
Console::WriteLine(str1, __box(Double::MaxValue), __box(divisor), __box(doubleResult));
Console::WriteLine(S"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 -1.995840e+292
Note that two positive numbers can yield a negative remainder.
*/
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必要条件
プラットフォーム: Windows 98, Windows NT 4.0, Windows Millennium Edition, Windows 2000, Windows XP Home Edition, Windows XP Professional, Windows Server 2003 ファミリ