C 棱镜三维空间中两个坐标之间的距离

我有一个家庭作业问题，要求用户提供三维空间中矩形棱镜的顶点。我需要计算棱镜的表面积和体积。我必须有一个计算棱镜上两点之间距离的函数，我可以要求用户从用户那里获得坐标的信息

double cx1, cy1, cz1, cx2, cy2, cz2, cx3, cy3, cz3, cx4, cy4, cz4, cx5, cy5, cz5, cx6, cy6, cz6, cx7, cy7, cz7, cx8, cy8, cz8;

int main() {
    printf("Enter the first coordinate in the form x y z: \n");
    scanf("%lf %lf %lf", &cx1, &cy1, &cz1);
    printf("Enter the second coordinate in the form x y z: \n");
    scanf("%lf %lf %lf", &cx2, &cy2, &cz2);
    printf("Enter the third coordinate in the form x y z: \n");
    scanf("%lf %lf %lf", &cx3, &cy3, &cz3);
    printf("Enter the fourth coordinate in the form x y z: \n");
    scanf("%lf %lf %lf", &cx4, &cy4, &cz4 );
    printf("Enter the fifth coordinate in the form x y z: \n");
    scanf("%lf %lf %lf", &cx5, & cy5, &cz5);
    printf("Enter the sixth coordinate in the form x y z: \n");
    scanf("%lf %lf %lf", &cx6, &cy6, &cz6);
    printf("Enter the seventh coordinate in the form x y z: \n");
    scanf("%lf %lf %lf", &cx7, &cy7, &cz7);
    printf("Enter the eighth coordinate in the form x y z: \n");
    scanf("%lf %lf %lf",&cx8, &cy8, &cz8);


return get_dist(cx1, cx2, cy1, cy2);

}

然后，我将每个x、y和z坐标指定给一个变量，这都在主函数中…然后我对一个点执行此操作：

double get_dist(cx1,cx2,cy1, cy2){
    double distance1_2;
    distance1_2 = sqrt(((cx2 - cx1)*(cx2 - cx1)) - ((cy2 - cy1)*(cy2 - cy1)) );

    printf("%lf",distance1_2);

return 0;
}

---

它给了我正确的值2，但是有没有比单独做每个坐标更简单/更快的方法呢

这就是你可以通过这么多变量来减少头痛的原因：

• 创建一个结构，将点的所有3个坐标分组
• 为点创建和设置阵列

像这样：

struct Point3 {
  double x, y, z;
}; // <-- the semocolon (;) is mandatory here

// hack so that you can use `Point3` instead of `struct Point3` when 
// referring to the structure type. This is completely unnecessary in C++
typedef struct Point3 Point3; 

int const g_num_points = 8; // this is actually an instance where
                            // a global variable is arguably not that bad

int main() {
  Point3 points[g_num_points]; // avoid global variables like plague
  int i;
  int ret;

  for (i = 0; i < g_num_points; ++i) {
    printf("Enter the %d'th coordinate in the form x y z: \n", i);
    ret = scanf("%lf %lf %lf", &(points[i].x), &(points[i].y), &(points[i].z));
    if (ret != 3) {
       // deal with invalid input. Since this is homework
       // I will leave it as an exercise for you
    }
  }
}

struct Point3{
双x，y，z；
}; // 顺便说一下，您的距离计算是错误的。
Point3
（或smthg）结构中的第3组坐标。做一个8个点的数组。这是怎么回事？我用C不是C++，上面的代码我缩小它来测试代码，当我继续工作的时候，我意识到我需要包含Z，所以我把它添加到像X和Y一样的上面，所以我用类似Distangs1S2 2= SqRT（（Cx2-Cx1）*（Cx2-Cx1））代替它们。（cy2-cy1）*（cy2-cy1）-（cz2-cz1）*（cz2-cz1））；虽然我想我最初的问题需要正确的措辞，但我想知道是否有一种更快的方法来获得距离，而不需要单独做每一件事，尽管单独做也能给我正确的答案