如何將NSNumber與十進制值進行比較？

Question

我最近從iOS 5升級到iOS 6，發現此單元測試失敗

- (void)testCalculatesDistanceBetweenTwoPoints
{
    self.sut = [[DistanceCalculator alloc] init];
    CLLocationCoordinate2D start = {.latitude = 34.32, .longitude = 99.13};
    CLLocationCoordinate2D finish = {.latitude = 105.94, .longitude = 27.73};
    NSNumber *distance = [self.sut kilometresBetweenPlace1:start andPlace2:finish];
    NSNumber *expected = [NSNumber numberWithDouble:3822.23073702318];
    STAssertEqualObjects(distance, expected, @"");
}

失敗的斷言

'3822.23073702318'應該等於'3822.23073702318'

當我打印每個數字的原始值時，我得到的看起來像是相同的值

2012-10-23 20:01:42.970 HelloWorld[1573:c07] 1 3822.23073702318
2012-10-23 20:01:42.970 HelloWorld[1573:c07] 2 3822.23073702318

當我打印每個數字的類型時，我得到的看起來像是一樣的（我做錯了嗎？）

2012-10-23 20:06:37.309 HelloWorld[1611:c07] 1 __NSCFNumber 
2012-10-23 20:06:37.309 HelloWorld[1611:c07] 2 __NSCFNumber

如果有幫助，這里是完整的實現

- (NSNumber *)kilometresBetweenPlace1:(CLLocationCoordinate2D)place1 andPlace2:(CLLocationCoordinate2D)place2
{
    MKMapPoint start = MKMapPointForCoordinate(place1);
    MKMapPoint finish = MKMapPointForCoordinate(place2);

    double distance = MKMetersBetweenMapPoints(start, finish) / 1600;

    return [NSNumber numberWithDouble:distance];
}

Answer 1

對於任何可能關注此討論的人-這就是我最終得到的結果

- (void)testCalculatesDistanceBetweenTwoPoints
{
    double accuracy = 0.00000000001;
    self.sut = [[DistanceCalculator alloc] init];
    CLLocationCoordinate2D start = {.latitude = 34.32, .longitude = 99.13};
    CLLocationCoordinate2D finish = {.latitude = 105.94, .longitude = 27.73};
    NSNumber *distance = [self.sut kilometresBetweenPlace1:start andPlace2:finish];
    NSNumber *expected = [NSNumber numberWithDouble:3822.23073702318];
    STAssertEqualsWithAccuracy([distance doubleValue], [expected doubleValue], accuracy, @"");
}