Is there a way to swap 2 structs?

Question

In c++ you can swap 2 numbers without 3rd variable:

int x = 10;
int y = 5;
x ^= y; 
y ^= x; 
x ^= y; //x = 5, y = 10

So I wonder if there is a similar way to swap 2 structs without 3rd variable:

struct Position{int x, y;};
int main()
{
   Position firstPose{x = 10, y = 5;};
   Position secPose{x = 5, y = 10;};
   //Now I want to swap firstPose with secPose.
}

Is this possible? if so; how to do it

Answer 1

There is no a standard way to swap two structure without an intermediate copy. Arguably, one of the main benefit of swap "is" the intermediate copy, this wonderful article explain how swap is a crucial part to achieve "strong exception guarantee". https://www.stroustrup.com/except.pdf

Furthermore, if the goal is to not make a copy of the struct (because is resource intensive) you can design your class using the pimpl idiom and swap just the pointer (you will still have a third variable but it will be just the raw pointer to the structure).

If you want to use C++ effectively make yourself familiar with exception safety, it is truly of of the area where the language gives its best

A bit old but still good article: http://www.gotw.ca/gotw/008.htm

At the end, the final solution is create a custom swap function:

#include <iostream>
#include <string>

struct Position{int x, y;};

void swap(Position& a, Position& b)
{
    a.x ^= b.x; 
    b.x ^= a.x; 
    a.x ^= b.x;

    a.y ^= b.y; 
    b.y ^= a.y; 
    a.y ^= b.y; 
}

int main()
{
    Position a = { 10, 100};
    Position b = { 20, 200};
  
    swap(a, b);
    std::cout << "a:" << a.x << "," << a.y << std::endl;
    std::cout << "b:" << b.x << "," << b.y << std::endl;
}

IMHO, the last option is more for personal amusement than real production code.