c++ - 友元函数运算符重载参数表中如何写多个参数一起进行操作

Answer 1

Normally (a + b).area(); is not called like this, it is unnatural to use.
Instead, an overloaded function similar to the following will be implemented

    friend ostream &operator<<(ostream &os, const Triangle &b) {
        float s, area;
        s = (b.x + b.y + b.z) / 2;
        area = sqrt(s * (s - b.x) * (s - b.y) * (s - b.z));
//        cout << "area is " << area << endl;
        os << "area is " << area <<endl;
        return os;
    }
    

In this way, when printing, just cout << a + b << endl;
The code has been slightly changed, but I haven’t seen the specific algorithm,

#include <iostream>
#include <math.h>

using namespace std;

class Triangle {
private:

    int x, y, z;
public:

    Triangle() { }

    void area() {
        float s, area;
        s = (x + y + z) / 2;
        area = sqrt(s * (s - x) * (s - y) * (s - z));
        cout << "area is " << area << endl;
    }

    friend ostream &operator<<(ostream &os, const Triangle &b) {
        float s, area;
        s = (b.x + b.y + b.z) / 2;
        area = sqrt(s * (s - b.x) * (s - b.y) * (s - b.z));
//        cout << "area is " << area << endl;
        os << "area is " << area <<endl;
        return os;
    }

    friend Triangle operator+(Triangle left, Triangle right) {
        Triangle b;
        b.x = left.x + right.x;
        b.y = left.y + right.y;
        b.z = left.z + right.z;
        return b;
    }

    void input() {
        cin >> x >> y >> z;
    }

    void output() {
        cout << "triangle three horizon length is " << x << "  " << y << "  " << z << endl;
    }
};

int main() {

    Triangle a, b, c;
    a.input();
    b.input();
    c.input();
    (a + b).output();
//    (a + b).area();
    cout << a + b + c <<endl;
    return 0;
}      
  
  

PS:
I updated it. There is actually no need to use friend functions here, just use them as follows

Triangle operator+(Triangle other)
{

Triangle ret;
ret.x = this->x + other.x;
ret.y = this->y + other.y;
ret.z = this->z + other.z;
return ret;

}

If you use friend to handle it, the return value will also be a Triangle. You can recursively add another Triangle to achieve the form of adding multiple Triangles together