Is there a better way to find midnight tomorrow?

Question

Is there a better way to do this?

-(NSDate *)getMidnightTommorow {
    NSCalendarDate *now = [NSCalendarDate date];
    NSCalendarDate *tomorrow = [now dateByAddingYears:0 months:0 days:1 hours:0 minutes:0 seconds:0];
    return [NSCalendarDate dateWithYear:[tomorrow yearOfCommonEra]
                                  month:[tomorrow monthOfYear]
                                    day:[tomorrow dayOfMonth]
                                   hour:0
                                 minute:0
                                 second:0
                               timeZone:[tomorrow timeZone]];
}

Note that I always want the next midnight, even if it happens to be midnight when I make that call, however if it happens to be 23:59:59, I of course want the midnight that is coming in one second.

The natural language functions seem flaky, and I'm not sure what Cocoa would do if I pass 32 in the "day" field. (If that'd work I could drop the [now dateByAddingYears:...] call)

Answer 1

From the documentation:

Use of NSCalendarDate strongly discouraged. It is not deprecated yet, however it may be in the next major OS release after Mac OS X v10.5. For calendrical calculations, you should use suitable combinations of NSCalendar, NSDate, and NSDateComponents, as described in Calendars in Dates and Times Programming Topics for Cocoa.

Following that advice:

NSDate *today = [NSDate date];

NSCalendar *gregorian = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];

NSDateComponents *components = [[NSDateComponents alloc] init];
components.day = 1;
NSDate *tomorrow = [gregorian dateByAddingComponents:components toDate:today options:0];
[components release];

NSUInteger unitFlags = NSYearCalendarUnit | NSMonthCalendarUnit |  NSDayCalendarUnit;
components = [gregorian components:unitFlags fromDate:tomorrow];
components.hour = 0;
components.minute = 0;

NSDate *tomorrowMidnight = [gregorian dateFromComponents:components];

[gregorian release];

(I'm not sure offhand if this is the most efficient implementation, but it should serve as a pointer in the right direction.)

Note: In theory you can reduce the amount of code here by allowing a date components object with values greater than the range of normal values for the component (e.g. simply adding 1 to the day component, which might result in its having a value of 32). However, although dateFromComponents: may tolerate out-of-bounds values, it's not guaranteed to. You're strongly encouraged not to rely on it.

Answer 2

Convert your current date and time to a Unix date (seconds since 1970) or DOS style (since 1980), then add 24 hours and convert it back. Then reset the hours, minutes and seconds to zero to get to midnight.

Answer 3

Nope - it'll be the same way you use to find midnight today.

Answer 4

sounds like a good title for a pop song:

"Is there a better way to find midnight tomorrow?"

Answer 5

[NSDate dateWithNaturalLanguageString:@"midnight tomorrow"];

Answer 6

NSCalendar *gregorian = [[[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar] autorelease];

NSDate *tomorrow = [NSDate dateWithTimeIntervalSinceNow:(24 * 60 * 60)];

NSDateComponents *components = [gregorian components:(NSYearCalendarUnit | NSMonthCalendarUnit | NSDayCalendarUnit) fromDate:tomorrow];

NSDate *midnight = [gregorian dateFromComponents:components];