From C++ FAQ
"How can you reseat a reference to make it refer to a different object?"
But when I do this it compiles and executes fine.
int f = 5;
int y =4;
int& u = f;
u = y;
B& bRef = B();
bRef = B();
This code is inside my main() function.
Link of C++ FAQ https://isocpp.org/wiki/faq/references#reseating-refs
You don't reseat the reference, you simply assign to the referred object.
#include <iostream>
struct X{
void operator=(X const&){
std::cout << "Woops, assignment!\n";
}
};
int main(){
X x, y;
X& rx = x;
rx = y;
}
By saying int &u=f; and then u=y; you are assigning the value of y to f as f is referred to by the reference u. Hence you are not reseating the reference but simply changing the value of f.
You are not changing the reference while saying u = y;. You are just assigning value of y to variable f which still referred by u.
Please check the value of f to see the effect.
when you call 'u = y;',you just assign value of y to the variable that u reference to(its f). Basically,reference is implemented as a 'pointer' under varies compiler, the operation 'u = y;', just set the memory to value of y.
The value of a reference cannot be changed after it has been initialized; it always refers to the same object it was initialized to denote..
