iphone - How to check if MKCoordinateRegion contains CLLocationCoordinate2D without using MKMapView?

Question

I need to check if user location belongs to the MKCoordinateRegion. I was surprised not to find simple function for this, something like: CGRectContainsCGPoint(rect, point).

I found following piece of code:

CLLocationCoordinate2D topLeftCoordinate = 
    CLLocationCoordinate2DMake(region.center.latitude 
                               + (region.span.latitudeDelta/2.0), 
                               region.center.longitude 
                               - (region.span.longitudeDelta/2.0));


    CLLocationCoordinate2D bottomRightCoordinate = 
    CLLocationCoordinate2DMake(region.center.latitude 
                               - (region.span.latitudeDelta/2.0), 
                               region.center.longitude 
                               + (region.span.longitudeDelta/2.0));

        if (location.latitude < topLeftCoordinate.latitude || location.latitude > bottomRightCoordinate.latitude || location.longitude < bottomRightCoordinate.longitude || location.longitude > bottomRightCoordinate.longitude) {

    // Coordinate fits into the region

    }

But, I am not sure if it is accurate as documentation does not specify exactly how the region rectangle is calculated.

There must be simpler way to do it. Have I overlooked some function in the MapKit framework documentation?

Answer 1

I'm posting this answer as the accepted solution is not valid in my opinion. This answer is also not perfect but it handles the case when coordinates wrap around 360 degrees boundaries, which is enough to be suitable in my situation.

+ (BOOL)coordinate:(CLLocationCoordinate2D)coord inRegion:(MKCoordinateRegion)region
{
    CLLocationCoordinate2D center = region.center;
    MKCoordinateSpan span = region.span;

    BOOL result = YES;
    result &= cos((center.latitude - coord.latitude)*M_PI/180.0) > cos(span.latitudeDelta/2.0*M_PI/180.0);
    result &= cos((center.longitude - coord.longitude)*M_PI/180.0) > cos(span.longitudeDelta/2.0*M_PI/180.0);
    return result;
}