c++ - 我的功能不会交互。那或者它不会返回值

Question

所以我一直在努力工作，但运气不佳。第一个函数正在工作，因为我通过对每个变量使用 cout 来确认它并且方程是正确的，因为我在没有函数的情况下尝试了它。问题是 function1engineer 没有为 distance 赋值，所以它变成了默认的 double 值。如何让 function1engineer 接受来自第一个函数的输入？

这是我的代码。

#include <iostream>
#include <cmath>
#include <math.h>
#include <iomanip>
#define PI 3.14159265
using namespace std;
//angle*PI/180 to work in c++ for angles.
void function1customer(double& x1, double& x2, double& y1, double& y2);
double function1engineer(double distance, double x1, double x2, double y1, double y2);

int main()
{
    int exitLoop;
    do
    {


        cout << "Choose an option.\n1: Given two points, compute the distance between them."
            << "\n2: Given two points, compute the horizontal angle from \nthe first point to the second."
            << "\n3: Given the elevation angle and velocity, compute the \n(horizontal) distance an object travels."
            << "\n4: Given a starting point, a distance, and a horizontal angle, \ncompute the destination point."
            <<endl;
        int opselect;
        cin >> opselect;
        switch(opselect)
            {
                //compute distance between 2 points
        case 1:
        double distance;
        double x1, x2, y1,y2;
        function1customer(x1,x2,y1,y2);
        cout << x1 << endl << x2 << endl << y1<< endl << y2 << endl;
        function1engineer;
        cout << setprecision(2)<< distance << endl;
        break;
            }
    cout << "Would you like to perform another calculation?\n0=No\n1=Yes" <<endl;
    cin >> exitLoop;
    }   
    while(exitLoop != 0);
}

void function1customer(double& x1, double& x2, double& y1, double& y2)
{
    cout << "Enter X1 cordinate in feet" <<endl;
    cin >> x1;

    cout << "Enter X2 cordinate in feet" <<endl;
    cin >> x2;

    cout << "Enter Y1 cordinate in feet" <<endl;
    cin >> y1;

    cout << "Enter Y2 cordinate in feet" <<endl;
    cin >> y2;
}

double function1engineer(double distance, double x1, double x2, double y1, double y2)
{
    distance = 0;
    distance = pow(  ( pow((x2-x1),2.0) + pow((y2-y1),2.0)     ),(1.0/2));
    return distance;
}

Answer 1

您需要通过引用传递distance，并且不需要显式返回它。 (就像您对 function1customer 的参数所做的那样。使用这个:

void function1engineer(double& distance, double x1, double x2, double y1, double y2)

调用代码为:

function1engineer(distance,x1,x2,y1,y2);