C++： Questions about pass by reference

Question

Why the output is 57? Can someone pls explain to me...int &y = x will pass the x value to y by reference right?Thats mean at that point, y=0. And then y = 5. And x still equals to 0. Am i right?
#include <iostream>
using namespace std;

int main()
{
int x=0;int &y=x;y=5;
while(x<=5){
cout<<y++;
x++;
}
cout<<x;
return 0;
}

Answer 1

汝風留名 when x and y are referring to the same memory then if value at y is changed, it means that even value at x will change.
So 5<=5 is true and y++ = 5 is printed.
But now y is 6 and x++ is done.
So when while loop checks for condition again now x is 7.
So condition fails and 7 is printed.
Thus your output 57.

Answer 2

The last run of the loop will be when x (which is also y) is 5.

After it is couted, it will be upped two times (y++, x++), so it ends up at 7.

Answer 3

Thx～

Answer 4

Kit D. Cat Oh, that’s cool. Does “int &y = x;” work in C as well? Does it work for function calls? If so, that could be useful so I don’t have to pass a pointer and dereference it whenever I need to use it.

Answer 5

Kit D. Cat ok