如何使用 Struct 打印我的交换 function

Question

Could you help me, how can I print my swap3 function? I would be very thankful. I am beginner in programming

#include <iostream>
using namespace std;

struct Pair{
    int x;
    int y;
};

Pair* swap3(const Pair& p){
    Pair *t = new Pair();
    t->x = p.y;
    t->y = p.x; 
    return t;
}

int main(){

    int f = Pair(2,3);
    swap3(f);
    cout<<f<<endl;
    return 0;


}

Is my main function false?

Answer 1

you need to overload ostream operator:

 friend std::ostream &operator<<(std::ostream &os, const Pair& pair) {
    os << pair.x << " " << pair.y << '\n';
    return os;
 }

If you are not comfortable with operator oveloading, you can simply print elements individually:

cout<< f.x <<" "<< f.y <<'\n';

---

The way you construct f , is wrong too ( int and Pair are not the same type). You can try list initialization instead:

Pair f{2,3};
auto s = swap3(f);
cout<<f<<endl;
delete s;

Note that you have memory leak in your code, because your function returns a pointer, you don't store it and you never delete it.

I recommend using smart pointers to avoid memory leak you had:

std::unique_ptr<Pair> swap3(const Pair& p){
    auto t = make_unique<Pair>(Pair{});
    t->x = p.y;
    t->y = p.x; 
    return t;
}

Live on Godbolt

PS I am not sure what you want from swap, in the code you posted, you don't need a pointer at all. I think swap should be written like:

void swap3(Pair& p1, Pair& p2){
    Pair tmp{p1.x, p1.y};
    p1.x = p2.x;
    p1.y = p2.y;
    p2.x = tmp.x;
    p2.y = tmp.y;
}

or:

void swap3(Pair& p){
    Pair tmp{p.x, p.y};
    p.x = tmp.y;
    p.y = tmp.x;
}

Live on Godbolt