Reputation: 2363
As I understand it, one cannot change the reference variable once it has been initialized. See, for instance, this question. However, here is a minmal working example which sort of does reassign it. What am I misunderstanding? Why does the example print both 42 and 43?
#include <iostream>
class T {
int x;
public:
T(int xx) : x(xx) {}
friend std::ostream &operator<<(std::ostream &dst, T &t) {
dst << t.x;
return dst;
}
};
int main() {
auto t = T(42);
auto q = T(43);
auto &ref = t;
std::cerr << ref << std::endl;
ref = q;
std::cerr << ref << std::endl;
return 0;
}
Upvotes: 0
Views: 120
Reputation: 118340
You're not changing the reference here.
You are replacing the object the reference is referring to.
In other words: after the assignment, your t
is replaced by q
.
ref
is still a reference to t
.
Upvotes: 4
Reputation: 29983
That does not perform a reference reassignment. Instead, it copy assigns the object in variable q
into the object referenced by ref
(which is t
in your example).
This also justifies why you got 42 as output: the default copy assignment operator modified the first object.
Upvotes: 3