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?

Oliver Charlesworth · Accepted Answer · 2011-09-13 19:50:06Z

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.}

In particular, the value of r can change, even though it is declared const.
Usually, compilers implement the reference with a simple pointer.
@quant_dev Given OP's sample, the source code that mentions x and source code that mentions r are compiled to identical machine code. No pointers involved (on all compilers I've seen). Not that it's relevant anyway.
@Cubbi: this all depends on the context, whenever the compiler can just alias the names, it will do so, but if the reference is passed to a function or stored in a type, etc. the compiler must store a reference to the object, and that is usually done with pointers.

AnT stands with Russia · Accepted Answer · 2011-09-13 19:00:16Z

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.

so what you are saying is that there is no implicit conversion taking place on x "from int to const int " during reference binding is that right? If yes can you provide link to the standard of the statement which proves your point..
@sparsh goyal: Process of reference initialization is described in 9.4.4/5 (eel.is/c++draft/dcl.init.ref#5). This situation immediately falls under 5.1 and 5.1.1. No conversion "from int to const int" is involved there.

AlvaroAV · Accepted Answer · 2014-06-24 11:19:19Z

The understanding on the Reference (&) answers this question..

Reference is just an alias to the variable that it is assigned to it..

And const is a constraint imposed by the compiler to the variable that is declared as const

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

In this case, its a const reference to a non const variable. So you cannot change the data of x with reference variable r(just acts a read only).. yet you can still change the data x by modifying x

const int z = 3000; int &t = z

In this case, non const reference to const member which is meaningless. You are saying reference can allow you to edit a const member(which is never possible)..

So if you want to create a reference for a const member, it has to be like the first case you mentioned

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

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const reference to non-const object

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