Question

I have this code:

#include <iostream>
using namespace std;

struct X {
    int a = 1;
};

struct Y {
    X &_x;
    Y(X &x) : _x(x) {}
};

// intentionally typoed version of Y, without the reference in the constructor
struct Z {
    X &_x;
    Z(X x) : _x(x) {}
};

int main() {
    X x;
    Y y(x);
    Z z(x);
    cout << "x:   " << &x << endl;
    cout << "y.x: " << &y._x << endl;
    cout << "z.x: " << &z._x << endl;
}

I keep finding myself forgetting the & in the constructor of my classes of this format.

This outputs the following:

x:   0xbfa195f8
y.x: 0xbfa195f8
z.x: 0xbfa195fc

Why is the behaviour different in the case of y and z?

And why is it not an error to initialize the X &_x member with an instance of type X in the constructor of Y?

Answer 1

A copy of the object is created when the constructor is called, hence the different output for z.x. The lifespan of that object is very limited – it exists only in the constructor. This is undefined behavior and the reference will be come invalid.

To prevent such behavior in your applications it is a good practice to mark the copy constructor and assignment operator private.