java - iOS 和 Android : Same function, 相同的输入，不同的十进制输出

Question

我们在 iOS 和 Android 中使用相同的函数，相同的输入数据，输出的是第二位/第三位小数。

我无法理解原因，我们确实需要完全相同的输出。

按照iOS功能:

func angleBoundSetFromPosition(userPosition: CLLocationCoordinate2D, distance: Double, bearing: Double) -> (CLLocationCoordinate2D) {
    let dist = Float(distance) / Float(6371)
    let brg = Float(bearing.degreesToRadians)
    let lat1 = Float(userPosition.latitude).degreesToRadians
    let lon1 = Float(userPosition.longitude).degreesToRadians
    let lat2 = asin(sin(lat1) * cos(dist) + cos(lat1) * sin(dist) * cos(brg))
    let lon2 = lon1 + atan2(sin(brg) * sin(dist) * cos(lat1), cos(dist) - sin(lat1) * sin(lat2))
    print("for: \(bearing), dist: \(dist), bearing: \(brg),  lat1: \(lat1), lon1: \(lon1), lat2: \(lat2), lon2: \(lon2)")
    return CLLocationCoordinate2D(latitude: Double(lat2.radiansToDegrees), longitude: Double(lon2.radiansToDegrees))
}

iOS 扩展:

extension FloatingPoint {
    var degreesToRadians: Self { return self / 180 * .pi}
    var radiansToDegrees: Self { return self * 180 / .pi }
}

这里是java:

public static LatLng getDestinationPoint(LatLng source, double brng, double dist) {
            dist = dist / 6371;
            brng = Math.toRadians(brng);

            double lat1 = Math.toRadians(source.latitude), lon1 = Math.toRadians(source.longitude);
            double lat2 = Math.asin(Math.sin(lat1) * Math.cos(dist) +
                    Math.cos(lat1) * Math.sin(dist) * Math.cos(brng));
            double lon2 = lon1 + Math.atan2(Math.sin(brng) * Math.sin(dist) *
                            Math.cos(lat1),
                    Math.cos(dist) - Math.sin(lat1) *
                            Math.sin(lat2));
            if (Double.isNaN(lat2) || Double.isNaN(lon2)) {
                return null;
            }
            return new LatLng(Math.toDegrees(lat2), Math.toDegrees(lon2));
    }

输入:

UserPosition(45.6091463, 8.8496677)
distance = 10
called once with bearing = -45, and once with bearing 135

输出:

安卓

{latitude 45.621861885764005, longitude 8.83148183051733}
{latitude 45.596425230386295, longitude 8.867843324603944}

iOS

{latitude 45.67270183849526, longitude 8.758660791031469}
{latitude 45.54551866514264, longitude 8.940468486905017}

Answer 1

将函数从一种语言转换为另一种语言时，应尽量使用相同的变量名。在这种情况下，对于 Java，您可能还需要 statically import all Math functions所以你没有一个实际上最终使比较变得不可能的函数。

这是 swift 函数:

let lat2 = asin(sin(lat1)) * cos(newDistance) + cos(lat1) * sin(newDistance) * cos(newBearing)
                         ^

这是美化后的 Java 函数:

double lat2 = asin(sin(lat1) * cos(newDistance) + cos(lat1) * sin(newDistance) * cos(newBearing));
                                                                                                ^

您的 Swift 函数在 sin 之后有一个括号，而 Java 公式在末尾有一个括号。

因此，分歧来自于这两个方程都不匹配的事实，而不是任何形式的错误双重数学。

double 漂移通常只发生在小范围内。如果您的测试数据显示出截然不同的值，则更有可能是实现不正确，而不是解释。

这是盒装测试代码。

public static void main(String[] args) {
    System.out.println(getDestinationPoint(new LatLng(45.6091463, 8.8496677), -45, 10));
    System.out.println(getDestinationPoint(new LatLng(45.6091463, 8.8496677), 135, 10));
}

public static LatLng getDestinationPoint(LatLng source, double bearing, double distance) {
    double newDistance = distance / 6371;
    double newBearing = toRadians(bearing);
    double lat1 = toRadians(source.latitude);
    double lon1 = toRadians(source.longitude);
    double lat2 = asin(sin(lat1) * cos(newDistance) + cos(lat1) * sin(newDistance) * cos(newBearing));
    double lon2 = lon1 + atan2(sin(newBearing) * sin(newDistance) * cos(lat1), cos(newDistance) - sin(lat1) * sin(lat2));
    if (Double.isNaN(lat2) || Double.isNaN(lon2)) {
        return null;
    }
    return new LatLng(toDegrees(lat2), toDegrees(lon2));
}

static class LatLng {
    double latitude;
    double longitude;

    public LatLng(double lat, double lon) {
        latitude = lat;
        longitude = lon;
    }

    public String toString() {
        return latitude + ", " + longitude;
    }
}