大家讲道理2017-04-17 13:48:26
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