Question

我正在尝试使用城市的经度和纬度制作城市距离计算器。我遇到的问题是把城市变成一个阵列。我的数组尝试在main方法中被注释掉了。请让我知道我做错了什么。

#include <iostream>
#include <math.h>
#define pi 3.14159265358979323846

using namespace std;

double distance(double lat1, double lon1, double lat2, double lon2, char unit);
double deg2rad(double deg);
double rad2deg(double rad);

class City
{
public:
    //declare variables
    string name;
    double latitude;
    double longitude;

    City(string, double, double); //constructor
    double distance(string, string);
};

double cityDistance(City a, City b);

int main() {

    //construct cities
    City Providence("Providence", 41.8239890, -71.4128340);
    City Cranston("Cranston", 41.7798230, -71.4372800);
    City NewYork("NewYork", 40.7143530, -74.0059730);
    City Boston("Boston", 42.3584310, -71.0597730);
    City Killington("Killington", 43.6775680, -72.7798250);
    City Springfield("Springfield", 42.1014830, -72.5898110);
    City Bridgeport("Bridgeport", 41.1865480, -73.1951770);
    City Cambridge("Cambridge", 42.3736160, -71.1097340);
    City Norwalk("Norwalk", 41.1175970, -73.4078970);
    City Quincy("Quincy", 42.2528770, -71.0022710);

    //put city into array
//  City cityArray[] = { };
//  cityArray[0] = {"Providence", 41.8239890, -71.4128340};
//  cityArray[1] = {"Cranston", 41.7798230, -71.4372800};
//  cityArray[2] = {"NewYork", 40.7143530, -74.0059730};
//  cityArray[3] = {"Boston", 42.3584310, -71.0597730};
//  cityArray[4] = {"Killington", 43.6775680, -72.7798250};
//  cityArray[5] = {"Springfield", 42.1014830, -72.5898110};
//  cityArray[6] = {"Bridgeport", 41.1865480, -73.1951770};
//  cityArray[7] = {"Cambridge", 42.3736160, -71.1097340};
//  cityArray[8] = {"Norwalk", 41.1175970, -73.4078970};
//  cityArray[9] = {"Quincy", 42.2528770, -71.0022710};

    //calculate distance
    cout << "Distance between Providence and Cranston: " << distance(41.8239890, -71.4128340, 41.7798230, -71.4372800, 'M') << endl;
    cout << "Distance between NewYork and Boston: " << distance(40.7143530, -74.0059730, 42.3584310, -71.0597730, 'M') << endl;
    cout << "Distance between Killington and Norwalk: " << distance(43.6775680, -72.7798250, 41.1175970, -73.4078970, 'M') << endl;

    //print distance using city name
    cout << endl;
    cout << "Distance between Providence and Cranston: " << cityDistance(Providence, Cranston) << endl;
    cout << "Distance between NewYork and Boston: " << cityDistance(NewYork, Boston) << endl;
    cout << "Distance between Killington and Norwalk: " << cityDistance(Killington, Norwalk) << endl;

    //end
    return 0;
}

City::City(string n, double lt, double lg)
{
    name = n;
    latitude = lt;
    longitude = lg;
}

double deg2rad(double deg)
{
    return (deg * pi / 180);
}

double rad2deg(double rad)
{
      return (rad * 180 / pi);
}

double distance(double lat1, double lon1, double lat2, double lon2, char unit)
{
    double theta, dist;
    theta = lon1 - lon2;
    dist = sin(deg2rad(lat1)) * sin(deg2rad(lat2)) + cos(deg2rad(lat1)) * cos(deg2rad(lat2)) * cos(deg2rad(theta));
    dist = acos(dist);
    dist = rad2deg(dist);
    dist = dist * 60 * 1.1515;
    switch(unit) {
        case 'M':
            break;
        case 'K':
            dist = dist * 1.609344;
            break;
        case 'N':
            dist = dist * 0.8684;
            break;
      }
      return(dist);
}


double cityDistance(City a, City b)
{
    return distance(a.latitude, a.longitude, b.latitude, b.longitude, 'M');
}

Answer 1

你现在的做法是创建一个大小为0的数组。然后你尝试将你的城市放在不存在的单元格0到9中。

解决该问题的一种方法是声明一个大小为10的数组，而不在声明点指定内容。

City cityArray[10];

然后像您一样单独分配每个单元格。然而，这不是最佳解决方案，它甚至无法工作，因为您尚未定义默认的无参数构造函数。

您可以改为创建一个您确切希望数组保持的城市数组。请参阅下面的代码。

如果你想使用类似结构的初始化，就像你一样，代码将是这样的：

//put city into array
City cityArray[] = {
  {"Providence", 41.8239890, -71.4128340},
  {"Cranston", 41.7798230, -71.4372800},
  {"NewYork", 40.7143530, -74.0059730},
  {"Boston", 42.3584310, -71.0597730},
  {"Killington", 43.6775680, -72.7798250},
  {"Springfield", 42.1014830, -72.5898110},
  {"Bridgeport", 41.1865480, -73.1951770},
  {"Cambridge", 42.3736160, -71.1097340},
  {"Norwalk", 41.1175970, -73.4078970},
  {"Quincy", 42.2528770, -71.0022710}
};

但是，您已经为City类定义了一个构造函数，因此，这将无法编译。你可以这样做：

City cityArray[] = {
  City("Providence", 41.8239890, -71.4128340),
  City("Cranston", 41.7798230, -71.4372800),
  City("NewYork", 40.7143530, -74.0059730),
  City("Boston", 42.3584310, -71.0597730),
  City("Killington", 43.6775680, -72.7798250),
  City("Springfield", 42.1014830, -72.5898110),
  City("Bridgeport", 41.1865480, -73.1951770),
  City("Cambridge", 42.3736160, -71.1097340),
  City("Norwalk", 41.1175970, -73.4078970),
  City("Quincy", 42.2528770, -71.0022710)
};

您还可以重复使用刚刚创建的City对象：

City cityArray[] = {
  Providence,
  Cranston,
  NewYork,
  Boston,
  Killington,
  Springfield,
  Bridgeport,
  Cambridge,
  Norwalk,
  Quincy
};

但是在上面的代码中，您必须知道对象Providence，Cranston等将被复制到数组中。因此，您将拥有相同数据的两个独立副本，例如一个在Providence变量中，一个在cityArray[0]中。改变一个不会影响另一个。

无论你选择什么情况，这都会创建一个固定大小的数组（因为c ++中的所有数组都有固定的大小）。您无法再添加任何城市。

如果您希望阵列是“可扩展的” - 您希望将来添加新城市，那么您应该使用其中一个STL集合而不是数组。例如vector<City>或map<City>。如果您打算搜索特定城市的集合，例如地图特别有用，例如以它的名字命名。

Answer 2

这是何时使用std::vector或std::map的一个很好的例子。在这种特殊情况下，我认为std::map将是更好的选择（因为您不必迭代整个列表来查找您正在寻找的2个城市）：

std::map<std::string, City> cities;
cities["Providence"] = City("Providence", 41.8239890, -71.4128340);
// ...

Example

Answer 3

在C ++中，数组不是动态的。您需要使用您将需要的确切大小声明您的数组。您也不能使用该表示法填充类，它不是有效的C ++构造，因为您没有声明默认构造函数。尝试这样的事情：

//put city into array
City cityArray[9];
cityArray[0].name = "Providence";
cityArray[0].latitude = 41.8239890;
cityArray[0].longitude = -71.4128340;
cityArray[1] = ...

当然，您可以使用类的构造函数和类似vector的STL容器，使其更简洁，并使其更具动态性。

explicit vector<City*> cityArray(9);
cityArray.push_back(new City("Providence", 41.8239890, -71.4128340));
...

向量的构造函数中的参数（大小）是可选的，但如果明确告诉它为您要存储的最可能数量的城市保留内存，则会获得更好的性能。如果超过该限制，矢量将自动增长。

Answer 4

使用此

City cityArray[10] = { 
     City ("Providence", 41.8239890, -71.4128340),
    City ("Cranston", 41.7798230, -71.4372800),
    City ("NewYork", 40.7143530, -74.0059730),
    City ("Boston", 42.3584310, -71.0597730),
    City ("Killington", 43.6775680, -72.7798250),
    City ("Springfield", 42.1014830, -72.5898110),
    City ("Bridgeport", 41.1865480, -73.1951770),
    City ("Cambridge", 42.3736160, -71.1097340),
    City ("Norwalk", 41.1175970, -73.4078970),
    City ("Quincy", 42.2528770, -71.0022710)
 };

c ++将对象放入数组中

4 个答案: