When auto can and can not deduce const

[Paul]

Below is quoted from a C++ book. I don't understand why the cases auto& g = ci; and auto &h = 42; are different. auto& h = 42; is an error because we can't bind a plain reference to a literal. So why is auto& g = ci; allowed? If auto& h = 42 binds a plain reference to a literal then why doesn't auto& g = ci; bind a plain reference to a const int?

Many thanks for your help,
Paul



BEGIN QUOTE
int i = 0, &r = i;
auto a = r; // a is an int (r is an alias for i, which has type int)

Second, auto ordinarily ignores top-level consts. As usual in initializations, low-level consts, such as when an initializer is a pointer to const, are kept

const int ci = i, &cr = ci; auto b = ci; // b is an int (top-level const in ci is dropped)

auto c = cr; // c is an int (cr is an alias for ci whose const is top-level) auto d = &i; // d is an int*(& of an int object is int*)
auto e = &ci; // e is const int*(& of a const object is low-level const) If we want the deduced type to have a top-level const, we must say so explicitly
const auto f = ci; // deduced type of ci is int; f has type const int We can also specify that we want a reference to the auto-deduced type. Normal initialization rules still apply
auto &g = ci; // g is a const int& that is bound to ci
auto &h = 42; // error: we can't bind a plain reference to a literal
const auto &j = 42; // ok: we can bind a const reference to a literal
END QUOTE

[Paul]

On Saturday, June 13, 2015 at 1:56:22 PM UTC+1, Stefan Ram wrote:
> Paul <peps...@gmail.com> quotes:

> >const auto &j = 42; // ok: we can bind a const reference to a literal
>

> The reference is /not/ bound to the literal, but to the
> /value/ of the literal!
>
> Justification:
>
> A literal is a part of the source-code model, a value is a
> part of the run-time model. A reference is a part of the
> run-time model; it can only bind to other run-time entities.
>
> Example:
>
> const auto &i = 042;
> const auto &j = 0042;
>
> These are two different literals, but they have the same value.
> Where the references bound to the literals, it should be possible
> two see the difference between the literals using the references.
> But it is not. The reference are bound to the /values/ of the literals,
> and the values are the same, so the value of the references also
> is the same.

Thanks. The error is in the text I'm quoting from not mine (not that anyone said otherwise).

Paul

[Paul]

On Saturday, June 13, 2015 at 1:44:42 PM UTC+1, Stefan Ram wrote:
> Paul <peps...@gmail.com> writes:
> >const int ci = i
> ...

> >why is auto& g = ci; allowed
>

> »A const object is an object« 3.9.3p1.1

Sorry, I still don't get it. The examples below are totally unclear to me. Why is auto &h = 42; wrong? If the answer is because an auto should refer to an object, not a value then why is const auto& j = 42; ok?

Is the rule that auto can't refer to a value but const auto can?

Paul

auto &g = ci; // g is a const int& that is bound to ci

auto &h = 42; // error: we can't bind a plain reference to a value

[Richard Damon]

The issue isn't auto, but what a reference can bind to.

int &h = 42; // error
int const&j = 42; // ok

const references can bind to a temporary, regular references can't.

42 has type "int", if it had type "const int" then auto would work.
(Maybe that is a 'defect' in the standard??)

[Paul]

Thanks, I understand now.

Paul

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