Strange program output with post-increment

Question

So this question is basically guess the output program. I will be specific.

What I understood is value of x is passed by reference to y, which means whatever changes we do on x is reflected on y and vice versa. In line 6 we pass the value of x to y by pass by reference.

In line 7, x is incremented, so x becomes 11 and so does y. Now I expected the output to be 11 11, because we are printing x first, so x which was 11 will be printed , and as of y, it is post increment so 11 will be printed first and then only y is incremented. But I was wrong(obvious) the answer was 12 11.

I tried debugging and it didn't explain why it happened either. After some googling, I found that there is something called cout rule where execution of cout statement happens from right to left, which means in line 8, due to post increment y is printed first, and then incremented so x gets changed and becomes 12. But again the output in that case is 11 12 and not 12 11. Please can you explain if the cout rule is really true and if yes can you explain how is it really done or there is something else to this? Thanks.

#include <iostream>
using namespace std;
int main()
{
    int x = 10;
    int &y = x;             // line 6
    x++;                    // line 7
    cout<< x << " " << y++; // line 8
    return 0; 
}

Answer 1

Some compilers will print 11 11

while others will print 12 11

The standard doesn't define the order in such as cases.

I'm still don't understand why you used the reference? it doesn't change anything, it just like the following:

#include<iostream> 
using namespace std;
int main()
{
  int x = 10;       
  x++;       
  cout<< x << " " << x++;         
  return 0; 
}