Question

Here is the code

NSDate* d = [NSDate dateWithTimeIntervalSince1970:32.4560];
double ti = [d timeIntervalSince1970];
NSLog(@"Interval: %f %f %f %f",ti,32.4560,ti*1000.0,32.4560*1000.0);

the output is

Interval: 32.456000 32.456000 32455.999970 32456.000000

Why NSDate return the value which lose some precisions?

Answer 1

That’s not the problem of NSDate itself. It’s in the nature of the floating point numbers themselves. I believe NSDate keeps its date from the OS X epoch(2001), not the UNIX epoch (1970). Let the difference in the two epochs be x.

Then what happens is this:

NSDate* d = [NSDate dateWithTimeIntervalSince1970:32.4560];
// at this point, d keeps 32.4560 + x
double ti = [d timeIntervalSince1970];
// ti is then (32.4560+x)-x

However, the floating point doesn’t have infinite precision. So, +x and then -x can introduce slight error in the calculation.

For more, read e.g. this Wikipedia article.

If you use the OS X epoch, you get what you naively expect:

NSDate* d = [NSDate dateWithTimeIntervalSinceReferenceDate:32.4560];
// at this point, d keeps 32.4560 + 0
double ti = [d timeIntervalSinceReferenceDate];
// ti is then (32.4560+0)-0, which is 32.4560 even in the floating point world.