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Function `std.datetime.date.DateTime.add`

Adds the given number of years or months to this DateTime, mutating it. A negative number will subtract.


						
				ref DateTime add(string units)
				(
				

				  long value,
				

				  AllowDayOverflow allowOverflow = AllowDayOverflow.yes
				

				) pure nothrow @nogc @safe
				

				if (units == "years" || units == "months");

Note that if day overflow is allowed, and the date with the adjusted year/month overflows the number of days in the new month, then the month will be incremented by one, and the day set to the number of days overflowed. (e.g. if the day were 31 and the new month were June, then the month would be incremented to July, and the new day would be 1). If day overflow is not allowed, then the day will be set to the last valid day in the month (e.g. June 31st would become June 30th).

Parameters

Name	Description
units	The type of units to add ("years" or "months").
value	The number of months or years to add to this `DateTime`.
allowOverflow	Whether the days should be allowed to overflow, causing the month to increment.

Returns

A reference to the DateTime (this).

Example

auto dt1 = DateTime(2010, 1, 1, 12, 30, 33);
dt1.add!"months"(11);
writeln(dt1); // DateTime(2010, 12, 1, 12, 30, 33)

auto dt2 = DateTime(2010, 1, 1, 12, 30, 33);
dt2.add!"months"(-11);
writeln(dt2); // DateTime(2009, 2, 1, 12, 30, 33)

auto dt3 = DateTime(2000, 2, 29, 12, 30, 33);
dt3.add!"years"(1);
writeln(dt3); // DateTime(2001, 3, 1, 12, 30, 33)

auto dt4 = DateTime(2000, 2, 29, 12, 30, 33);
dt4.add!"years"(1, AllowDayOverflow.no);
writeln(dt4); // DateTime(2001, 2, 28, 12, 30, 33)

Authors

Jonathan M Davis

License

Boost License 1.0.