Re: [boost] [type traits extension] test for const volatile& as return type

Subject: Re: [boost] [type traits extension] test for const volatile& as return type
From: Stewart, Robert (Robert.Stewart_at_[hidden])
Date: 2011-07-18 14:28:55

Frédéric Bron wrote:
>
> Here are other parts of the standard:
> § 13.3.2 Viable functions:
> "for F to be a viable function, there shall exist for each
> argument an implicit conversion sequence
> (13.3.3.1) that converts that argument to the corresponding
> parameter of F. If the parameter has reference
> type, the implicit conversion sequence includes the operation
> of binding the reference, and the fact that a
> reference to non-const cannot be bound to an rvalue can affect
> the viability of the function (see 13.3.3.1.4)."
>
> § 13.3.3.1.4/3:
> "A standard conversion sequence cannot be formed if it requires
> binding a reference to non-const to an
> rvalue (except when binding an implicit object parameter; see
> the special rules for that case in 13.3.1).
> [Note: this means, for example, that a candidate function
> cannot be a viable function if it has a non-const
> reference parameter (other than the implicit object parameter)
> and the corresponding argument is a temporary
> or would require one to be created to initialize the reference
> (see 8.5.3). ]"
>
> So the question is: is "const volatile &" a "reference to non-
> const"?

Were you awake into the wee hours again?  "Reference to non-const" means T & or T volatile &.  "const" can't be in the type.

> Or does "reference to non-const" mean exaclty T const &
> (excluding additionnal volatile qualifier) or not?

T const & is a "reference to const" so it cannot be a "reference to non-const."

> If T const volatile & is considered as "reference to non-const"
> then Intel is wrong.

It is not reference to non-const.

> If not, Intel is right.

OK, assuming there's nothing else that bears on the situation.  (I haven't studied your problem, so I can't say.)

> §8.5.3/5 you quoted earlier suggests that "reference to non-
> const" is only T const & which would mean that Intel is wrong:

Did you mean to write "reference to const?"

_____
Rob Stewart                           robert.stewart_at_[hidden]
Software Engineer                     using std::disclaimer;
Dev Tools & Components
Susquehanna International Group, LLP  http://www.sig.com

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