问题
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;
}
回答1:
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.
回答2:
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.
来源:https://stackoverflow.com/questions/26818908/is-this-an-example-of-reference-reassignment-c11