C#的寫法

public struct EarthPoint
? {
?? public const double Ea = 6378137; // 赤道半徑 WGS84標(biāo)準(zhǔn)參考橢球中的地球長半徑(單位:m)?
?? public const double Eb = 6356725; // 極半徑??
?? public readonly double Longitude,Latidute;
?? public readonly double Jd;
?? public readonly double Wd;
?? public readonly double Ec;
?? public readonly double Ed;
?? public EarthPoint(double _Longitude,double _Latidute)
?? {
??? Longitude = _Longitude;
??? Latidute = _Latidute;
??? Jd = Longitude * Math.PI / 180; //轉(zhuǎn)換成角度
??? Wd = Latidute * Math.PI /180; //轉(zhuǎn)換成角度
??? Ec = Eb + (Ea - Eb) * (90 - Latidute) / 90;
??? Ed = Ec * Math.Cos(Wd);
?? }
?? public double Distance(EarthPoint _Point)
?? {
??? double dx = (_Point.Jd - Jd) * Ed;
??? double dy = (_Point.Wd - Wd) * Ec;
??? return Math.Sqrt(dx * dx + dy *dy);
?? }
? }

public static Double GetDistance(double _Longitude1,
?? double _Latidute1,
?? double _Longitude2,
?? double _Latidute2)
? {
?? EarthPoint p1 = new EarthPoint(_Longitude1,_Latidute1);
?? EarthPoint p2 = new EarthPoint(_Longitude2,_Latidute2);
?? return p1.Distance(p2);
? }

C#的寫法，可惜不會用，所以將上面的代碼改造一下，形成C++的寫法

C++的寫法：

#include <math.h>
const double Ea = 6378137; // 赤道半徑 WGS84標(biāo)準(zhǔn)參考橢球中的地球長半徑(單位:m)?
const double Eb = 6356725; // 極半徑
#define PI 3.1416926;
class EarthPoint
{
public:?
??? double Longitude,Latidute;
??? double Jd;
??? double Wd;
??? double Ec;
??? double Ed;

??? EarthPoint(double _Longitude,double _Latidute)
??? {
??????? Longitude = _Longitude;
??????? Latidute = _Latidute;
??????? Jd = Longitude * 3.1415926 / 180; //轉(zhuǎn)換成角度
??????? Wd = Latidute * 3.1415926 / 180; //轉(zhuǎn)換成角度
??????? Ec = Eb + (Ea - Eb) * (90 - Latidute) / 90;
??????? Ed = Ec * cos(Wd);
??? }

??? double Distance(EarthPoint * _Point)
??? {
??????? double dx = (_Point->Jd - Jd) * Ed;
??????? double dy = (_Point->Wd - Wd) * Ec;
??????? return sqrt(dx * dx + dy *dy);
??? }
};

double GetDistance(double _Longitude1,
?????????????????? double _Latidute1,
?????????????????? double _Longitude2,
?????????????????? double _Latidute2)
{
??? EarthPoint *p1 = new EarthPoint(_Longitude1,_Latidute1);
??? EarthPoint *p2 = new EarthPoint(_Longitude2,_Latidute2);
??? double d=p1->Distance(p2);
??? return d;
}

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