Question

我有一个问题，我需要帮助解决。我应该创建一个计算两个位置之间距离的函数。位置存储在两个数组中，我可以使用我需要的多个参数进行此计算

static double[] Latitudes = new double[] { 
  59.3261917, 57.7010496, 59.8939529, 65.5867395, 60.11021, 52.5069312, 48.859 
};

static double[] Longitudes = new double[] { 
  17.7018773, 11.6136602, 10.6450348, 22.0422998, 24.7385057, 13.1445521, 2.2069765 
};

我得到了一个可以帮助我计算距离的等式

distance = Math.sqrt( (x1 - x2)2 + (y1 - y2)2  )

我的问题是我无法将数组中的元素转换为函数内的变量

Answer 1

首先，您需要确定要比较的位置。这将通过索引完成。让我们说你想比较索引0的位置和索引2的位置。然后获取正确变量的代码是：

double x1 = Latitudes[0];
double y1 = Longitudes[0];
double x2 = Latitudes[2];
double y2 = Longitudes[2];

然后，您可以将这些值提供给您的函数。您的功能代码错误，无法编译。正确调用该函数将是：

double distance = Math.Sqrt(Math.Pow(x1 - x2, 2) + Math.Pow(y1 - y2, 2));

为了为您的班级提供更完整的功能，并记住您的阵列是静态的，这将允许您根据索引获得任意两个给定点的距离。另外，我希望这是一个家庭作业，所以我倾向于你要求创建一个类似于此的功能：

double CalculateDistance(int index1, int index2)
{
    double x1 = Latitudes[index1];
    double y1 = Longitudes[index1];
    double x2 = Latitudes[index2];
    double y2 = Longitudes[index2];

    return Math.Sqrt(Math.Pow(x1 - x2, 2) + Math.Pow(y1 - y2, 2));
}

然后您可以按如下方式调用此函数：

double distance = CalculateDistance(0, 2);

Answer 2

提取方法，将问题分成次要：

 // Initial step:
 // Distance between points
 private static double Distance(double x1, double y1, double x2, double y2) {
   return Math.Sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2)); 
 }

 // Next step: 
 // Distance between points given as arrays' items indexes
 private static double Distance(double[] xs, double[] ys, int indexFrom, indexTo) {
   return Distance(xs[indexFrom], ys[indexFrom], xs[indexTo], ys[indexTo]);
 }

然后使用

 // What is the distance between 0-th and 2-nd point?
 double result = Distance(Latitudes, Longitudes, 0, 2);

 Console.WriteLine(result);

 // What is the distance between all the points?
 for (int from = 0; from < Math.Min(Latitudes.Length, Longitudes.Length); ++from) 
   for (int to = from + 1; to < Math.Min(Latitudes.Length, Longitudes.Length); ++to) {
     Console.WriteLine($"Distance from item #{from} to item #{to} is {Distance(Latitudes, Longitudes, from, to)}");
   }

Answer 3

显示所有对的小说

if (Latitudes.Length == Longitudes.Length)
{
    for (int i = 0; i < Latitudes.Length - 1; i = i + 2)
    {
        double x1 = Longitudes[i];
        double x2 = Longitudes[i + 1];
        double y1 = Latitudes[i];
        double y2 = Latitudes[i + 1];
        double distance = Math.Sqrt(Math.Pow(x1 - x2, 2) + Math.Pow(y1 - y2, 2));

        Console.WriteLine($"x1 = {x1}; x2 = {x2}; y1 = {y1}; y2 = {y2}; distance {distance}");
    }
}

从数组计算不同位置的函数c＃

3 个答案: