confusing output

Question

Below given program prints x = 10 y = 0

#include<iostream>
using namespace std;

class Test
{
private:
  int x;
  int y;
public:
  Test (int x = 0, int y = 0) { this->x = x; this->y = y; }
  Test setX(int a) { x = a; return *this; }
  Test setY(int b) { y = b; return *this; }
  void print() { cout << "x = " << x << " y = " << y << endl; }
};

int main()
{
  Test obj1;
  obj1.setX(10).setY(20);
  obj1.print();
  return 0;
}

but if we replace

Test setX(int a) { x = a; return *this; }
Test setY(int b) { y = b; return *this; }

with

Test &setX(int a) { x = a; return *this; }
Test &setY(int b) { y = b; return *this; }

output is x = 10 y = 20 Can anyone explain why this is happening?

Answer 1

When you return a plain object rather than a reference, you're returning a copy of the object on which you called the method. setY acts on that copy (which is a temporary), then the copy is destructed.

When you return a reference, all the calls operate on the same object, so you get the behavior you're looking for.