Subtract(TdIntervalHour,Int32) Method

The specified number of hours is subtracted from the TdIntervalHour.

Syntax

'Declaration
 
Public Overloads Shared Function Subtract( _
   ByVal interval As TdIntervalHour, _
   ByVal hours As Integer _
) As TdIntervalHour

'Usage
 
Dim interval As TdIntervalHour
Dim hours As Integer
Dim value As TdIntervalHour
 
value = TdIntervalHour.Subtract(interval, hours)

public static TdIntervalHour Subtract( 
   TdIntervalHour interval,
   int hours
)

public:
static TdIntervalHour Subtract( 
   TdIntervalHour interval,
   int hours
)

A new TdIntervalHour as the result of substracting an Int32 from TdIntervalHour.

Exceptions

Exception	Description
System.OverflowException	The resultant operation returns a value greater than the TdIntervalHour.MaxValue or less than the TdIntervalHour.MinValue.

Remarks

If interval is TdIntervalHour.Null a Null will be returned.

The returned TdIntervalHour will contain a precision set to the TdIntervalHour.MaxPrecision.

Requirements

Target Platforms: Windows 8.1, Windows 10, Windows Server 2012 R2, Windows Server 2016, Windows Server 2019