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 dateByAdd         


        

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];
[components 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.