const reference to non-const object

Question

In the following, would there be a temporary object created before const reference is used to a non-const object?

const int y = 2000;
const int &s = y // ok, const reference to const object.

int x = 1000;
const int &r = x; // any temporary copy here?

If no then, how does this work?

   const int z = 3000;
   int &t = z // ok, why can't you do this?

Answer 1

No.

A reference is simply an alias for an existing object. const is enforced by the compiler; it simply checks that you don't attempt to modify the object through the reference r.^* This doesn't require a copy to be created.

Given that const is merely an instruction to the compiler to enforce "read-only", then it should be immediately obvious why your final example doesn't compile. const would be pointless if you could trivially circumvent it by taking a non-const ref to a const object.

_{* Of course, you are still free to modify the object via x. Any changes will also be visible via r, because they refer to the same object.}

Answer 2

In

int x = 1000;
const int &r = x;

the right-hand side is an lvalue and the type of x is the same as the type of the reference (ignoring cv-qualifications). Under these circumstances the reference is attached directly to x, no temporary is created.

As for "how does this work"... I don't understand what prompted your question. It just works in the most straighforward way: the reference is attached directly to x. Nothing more to it.

You can't do

const int z = 3000;
int &t = z;

because it immediately violates the rules of const-correctness.