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

Question

这是一个三角类Triangle，重载运算符“+”来实现两个三角形对象的面积之和
如何改进成计算任意多个三角形的面积之和？
我这样重载“+”符合实现对两个三角形对象的面积之和吗？

#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 Triangle operator+(Triangle b1,Triangle b2)
    {
        Triangle b;
        b.x=b1.x+b2.x;
        b.y=b1.y+b2.y;
        b.z=b1.z+b2.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;
    a.input();
    b.input();
    (a+b).output();
    (a+b).area();
    return 0;
}

Answer 1

一般不会(a + b).area(); 这样调用，用起来不自然。
而是会实现类似下面的一个重载函数

    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;
    }
    

这样，打印的时候只要 cout << a + b << endl;即可
代码稍微改了下，具体的算法没看，

#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:
我更新了下，这里其实没必要用友元函数，直接如下用就行

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;

}

你用friend来处理的话，返回值也是一个Triangle,可以递归的再去加另外一个Triangle,就实现多个Triangle连加的形式