关于运算符重载的一个问题
#include <bits/stdc++.h>#define rep(i,a,n) for(int i=a;i<n;++i)
#define pb push_back
#define close() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define INF 0x3f3f3f3f
using namespace std;
typedef pair<int,int>P;
class Node{
private:
double x,y;
public:
Node(){x=0,y=0;}
Node(double x,double y){this->x=x,this->y=y;}
Node operator+(const Node &p){
Node sum;
sum.x=x+p.x;
sum.y=y+p.y;
return sum;
}
Node operator-(const Node &p){
Node sum;
sum.x=x-p.x;
sum.y=y-p.y;
return sum;
}
void Show(){
cout<<x<<" "<<y<<endl;
}
Node operator*(double n){
Node sum;
sum.x=x*n;
sum.y=y*n;
return sum;
}
/*
friend Node operator*(double n,Node p){
Node sum;
sum.x=p.x*n;
sum.y=p.y*n;
return sum;
}
*/
Node operator*(double n,const Node& p){
return p*n;
}
};
int main()
{
Node p1;
Node p2(3,4);
Node p3(4,5);
p2=p2+p3;
p2=5*p2;
p2.Show();
return 0;
}
为什么注释下面的return p*n会报错呢？C++ primer plus上说可以把友元函数改成return p*n 这种，为啥不对呢？ 这好像变成了自己调用自己的无限递归了 本帖最后由 qq1242009750 于 2018-9-10 11:52 编辑
Node operator*(double n,const Node& p){
return p*n; }
参数太多了，这个最多也就跟一个参数 自己 + 1参数 == 二元运算符
#include <iostream>
#define rep(i,a,n) for(int i=a;i<n;++i)
#define pb push_back
#define close() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define INF 0x3f3f3f3f
using namespace std;
typedef pair<int, int>P;
class Node {
private:
double x, y;
public:
Node(int i) { x = i; y = i; }
Node() { x = 0, y = 0; }
Node(double x, double y) { this->x = x, this->y = y; }
Node operator+(const Node &p) {
Node sum;
sum.x = x + p.x;
sum.y = y + p.y;
return sum;
}
Node operator-(const Node &p) {
Node sum;
sum.x = x - p.x;
sum.y = y - p.y;
return sum;
}
void Show() {
cout << x << " " << y << endl;
}
Node operator*(double n) {
Node sum;
sum.x = x * n;
sum.y = y * n;
return sum;
}
// friend Node operator*(double n,Node p){
// Node sum;
// sum.x=p.x*n;
// sum.y=p.y*n;
// return sum;
// }
Node operator*(Node &n) {
Node tmp;
tmp.x = x * n.x;
tmp.y = y * n.y;
return tmp;
}
};
int main()
{
Node p1;
Node p2(3, 4);
Node p3(4, 5);
p2 = p2 + p3;
p2 = (Node)5 * p2; //把5强制转换为Node类型但前提是Node必须有一个int类型的构造函数不用此方法请使用友元函数
p2.Show();
return 0;
}
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