DateTime.Subtraction 演算子

定義

指定した DateTime から指定した DateTime または TimeSpan を減算します。Subtracts a specified DateTime orTimeSpan from a specified DateTime.

オーバーロード

Subtraction(DateTime, TimeSpan)

指定した日付と時刻から指定した時間間隔を減算して、新しい日付と時刻を返します。Subtracts a specified time interval from a specified date and time and returns a new date and time.

Subtraction(DateTime, DateTime)

指定した日付と時刻から指定したもう 1 つの日付と時刻を減算して、時間間隔を返します。Subtracts a specified date and time from another specified date and time and returns a time interval.

Subtraction(DateTime, TimeSpan)

指定した日付と時刻から指定した時間間隔を減算して、新しい日付と時刻を返します。Subtracts a specified time interval from a specified date and time and returns a new date and time.

public:
 static DateTime operator -(DateTime d, TimeSpan t);
public static DateTime operator - (DateTime d, TimeSpan t);
static member ( - ) : DateTime * TimeSpan -> DateTime
Public Shared Operator - (d As DateTime, t As TimeSpan) As DateTime

パラメーター

d
DateTime

減算される日時の値。The date and time value to subtract from.

t
TimeSpan

減算する時間間隔。The time interval to subtract.

戻り値

d の値から t の値を減算した値を保持するオブジェクト。An object whose value is the value of d minus the value of t.

例外

結果として返された DateTimeMinValue より小さいか、MaxValue より大きいです。The resulting DateTime is less than MinValue or greater than MaxValue.

次の例は、Subtract メソッドと減算演算子を示しています。The following example demonstrates the Subtract method and the subtraction operator.

System::DateTime date1 = System::DateTime( 1996, 6, 3, 22, 15, 0 );
System::DateTime date2 = System::DateTime( 1996, 12, 6, 13, 2, 0 );
System::DateTime date3 = System::DateTime( 1996, 10, 12, 8, 42, 0 );

// diff1 gets 185 days, 14 hours, and 47 minutes.
System::TimeSpan diff1 = date2.Subtract( date1 );

// date4 gets 4/9/1996 5:55:00 PM.
System::DateTime date4 = date3.Subtract( diff1 );

// diff2 gets 55 days 4 hours and 20 minutes.
System::TimeSpan diff2 = date2 - date3;

// date5 gets 4/9/1996 5:55:00 PM.
System::DateTime date5 = date1 - diff2;
System.DateTime date1 = new System.DateTime(1996, 6, 3, 22, 15, 0);
System.DateTime date2 = new System.DateTime(1996, 12, 6, 13, 2, 0);
System.DateTime date3 = new System.DateTime(1996, 10, 12, 8, 42, 0);

// diff1 gets 185 days, 14 hours, and 47 minutes.
System.TimeSpan diff1 = date2.Subtract(date1);

// date4 gets 4/9/1996 5:55:00 PM.
System.DateTime date4 = date3.Subtract(diff1);

// diff2 gets 55 days 4 hours and 20 minutes.
System.TimeSpan diff2 = date2 - date3;

// date5 gets 4/9/1996 5:55:00 PM.
System.DateTime date5 = date1 - diff2;
Dim date1 As New System.DateTime(1996, 6, 3, 22, 15, 0)
Dim date2 As New System.DateTime(1996, 12, 6, 13, 2, 0)
Dim date3 As New System.DateTime(1996, 10, 12, 8, 42, 0)

Dim diff1 As System.TimeSpan
' diff1 gets 185 days, 14 hours, and 47 minutes.
diff1 = date2.Subtract(date1)

Dim date4 As System.DateTime
' date4 gets 4/9/1996 5:55:00 PM.
date4 = date3.Subtract(diff1)

Dim diff2 As System.TimeSpan
' diff2 gets 55 days 4 hours and 20 minutes.
diff2 = System.DateTime.op_Subtraction(date2, date3)

Dim date5 As System.DateTime
' date5 gets 4/9/1996 5:55:00 PM.
date5 = System.DateTime.op_Subtraction(date1, diff2)

注釈

このメソッドは、dのティック値から t のティック値を減算します。This method subtracts the ticks value of t from the ticks value of d.

この演算子の同等のメソッドは DateTime.Subtract(DateTime)The equivalent method for this operator is DateTime.Subtract(DateTime)

こちらもご覧ください

Subtraction(DateTime, DateTime)

指定した日付と時刻から指定したもう 1 つの日付と時刻を減算して、時間間隔を返します。Subtracts a specified date and time from another specified date and time and returns a time interval.

public:
 static TimeSpan operator -(DateTime d1, DateTime d2);
public static TimeSpan operator - (DateTime d1, DateTime d2);
static member ( - ) : DateTime * DateTime -> TimeSpan
Public Shared Operator - (d1 As DateTime, d2 As DateTime) As TimeSpan

パラメーター

d1
DateTime

減算対象の日付および時刻の値 (被減数)。The date and time value to subtract from (the minuend).

d2
DateTime

減算する日付および時刻の値 (減数)。The date and time value to subtract (the subtrahend).

戻り値

d1d2 の間の時間間隔、つまり d1 から d2 を引いた値です。The time interval between d1 and d2; that is, d1 minus d2.

次の例は、Subtract メソッドと減算演算子を示しています。The following example demonstrates the Subtract method and the subtraction operator.

System::DateTime date1 = System::DateTime( 1996, 6, 3, 22, 15, 0 );
System::DateTime date2 = System::DateTime( 1996, 12, 6, 13, 2, 0 );
System::DateTime date3 = System::DateTime( 1996, 10, 12, 8, 42, 0 );

// diff1 gets 185 days, 14 hours, and 47 minutes.
System::TimeSpan diff1 = date2.Subtract( date1 );

// date4 gets 4/9/1996 5:55:00 PM.
System::DateTime date4 = date3.Subtract( diff1 );

// diff2 gets 55 days 4 hours and 20 minutes.
System::TimeSpan diff2 = date2 - date3;

// date5 gets 4/9/1996 5:55:00 PM.
System::DateTime date5 = date1 - diff2;
System.DateTime date1 = new System.DateTime(1996, 6, 3, 22, 15, 0);
System.DateTime date2 = new System.DateTime(1996, 12, 6, 13, 2, 0);
System.DateTime date3 = new System.DateTime(1996, 10, 12, 8, 42, 0);

// diff1 gets 185 days, 14 hours, and 47 minutes.
System.TimeSpan diff1 = date2.Subtract(date1);

// date4 gets 4/9/1996 5:55:00 PM.
System.DateTime date4 = date3.Subtract(diff1);

// diff2 gets 55 days 4 hours and 20 minutes.
System.TimeSpan diff2 = date2 - date3;

// date5 gets 4/9/1996 5:55:00 PM.
System.DateTime date5 = date1 - diff2;
Dim date1 As New System.DateTime(1996, 6, 3, 22, 15, 0)
Dim date2 As New System.DateTime(1996, 12, 6, 13, 2, 0)
Dim date3 As New System.DateTime(1996, 10, 12, 8, 42, 0)

Dim diff1 As System.TimeSpan
' diff1 gets 185 days, 14 hours, and 47 minutes.
diff1 = date2.Subtract(date1)

Dim date4 As System.DateTime
' date4 gets 4/9/1996 5:55:00 PM.
date4 = date3.Subtract(diff1)

Dim diff2 As System.TimeSpan
' diff2 gets 55 days 4 hours and 20 minutes.
diff2 = System.DateTime.op_Subtraction(date2, date3)

Dim date5 As System.DateTime
' date5 gets 4/9/1996 5:55:00 PM.
date5 = System.DateTime.op_Subtraction(date1, diff2)

注釈

Subtraction(DateTime, DateTime) メソッドは、減算を実行するときに、2つの DateTime 値の Kind プロパティの値を考慮しません。The Subtraction(DateTime, DateTime) method does not consider the value of the Kind property of the two DateTime values when performing the subtraction. DateTime オブジェクトを差し引く前に、オブジェクトが同じタイムゾーンの時刻を表していることを確認します。Before subtracting DateTime objects, ensure that the objects represent times in the same time zone. それ以外の場合、結果にはタイムゾーン間の違いが含まれます。Otherwise, the result will include the difference between time zones.

注意

DateTimeOffset.Subtraction(DateTimeOffset, DateTimeOffset) メソッドは、減算を実行するときのタイムゾーン間の違いを考慮します。The DateTimeOffset.Subtraction(DateTimeOffset, DateTimeOffset) method does consider the difference between time zones when performing the subtraction.

この演算子の同等のメソッドは DateTime.Subtract(DateTime)The equivalent method for this operator is DateTime.Subtract(DateTime)

こちらもご覧ください

適用対象