问题描述
问题解答
回答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
