How to set seconds to zero for NSDate

NSDate is immutable, so you cannot modify its time. But you can create a new date object that snaps to the nearest minute:

NSTimeInterval time = floor([date timeIntervalSinceReferenceDate] / 60.0) * 60.0;
NSDate *minute = [NSDate dateWithTimeIntervalSinceReferenceDate:time];

Edit to answer Uli's comment

The reference date for NSDate is January 1, 2001, 0:00 GMT. There have been two leap seconds added since then: 2005 and 2010, so the value returned by [NSDate timeIntervalSinceReferenceDate] should be off by two seconds.

This is not the case: timeIntervalSinceReferenceDate is exactly synchronous to the wall time.

When answering the question I did not make sure that this is actually true. I just assumed that Mac OS would behave as UNIX time (1970 epoch) does: POSIX guarantees that each day starts at a multiple of 86,400 seconds.

Looking at the values returned from NSDate this assumption seems to be correct but it sure would be nice to find a definite (documented) statement of that.

Although this is an old question and Uli has given the "correct" answer, the simplest solution IMHO is to just subtract the seconds from the date, as obtained from the calendar. Mind that this may still leave milliseconds in place.

NSDate *date = [NSDate date];
NSDateComponents *comp = [[NSCalendar currentCalendar] components:NSCalendarUnitSecond
                                                         fromDate:date];
date = [date dateByAddingTimeInterval:-comp.second];

Here is a Swift extension for anyone who is interested:

extension Date {
    public mutating func floorSeconds() {
        let calendar = Calendar.current
        let components = calendar.dateComponents([.year, .month, .day, .hour, .minute], from: self)
        self = calendar.date(from: components) ?? self // you can handle nil however you choose, probably safe to force unwrap in most cases anyway
    }
}

Example usage:

let date = Date()
date.floorSeconds()

Using DateComponents is much more robust than adding a time interval to a date.

You can't directly manipulate the NSTimeInterval since that is the distance in seconds since the reference date, which isn't guaranteed to be a 00-second-time when divided by 60. After all, leap seconds may have been inserted to adjust for differences between solar time and UTC. Each leap second would throw you off by 1. What I do to fix the seconds of my date to 0 is:

NSDate              *   startDateTime = [NSDate date];
NSDateComponents    *   startSeconds = [[NSCalendar currentCalendar] components: NSSecondCalendarUnit fromDate: startDateTime];

startDateTime = [NSDate dateWithTimeIntervalSinceReferenceDate: [startDateTime timeIntervalSinceReferenceDate] -[startSeconds second]];

This takes care of leap seconds. I guess an even cleaner way would be to use -dateByAddingComponents:

NSDate              *   startDateTime = [NSDate date];
NSDateComponents    *   startSeconds = [[NSCalendar currentCalendar] components: NSSecondCalendarUnit fromDate: startDateTime];
[startSeconds setSecond: -[startSeconds second]];

startDateTime = [[NSCalendar currentCalendar] dateByAddingComponents: startSeconds toDate: startDateTime options: 0];

That way you're guaranteed that whatever special things -dateByAddingComponents: takes care of is accounted for as well.