Re: [boost] [type_traits][function_types] Discard param const qualification, bug or feature?

Subject: Re: [boost] [type_traits][function_types] Discard param const qualification, bug or feature?
From: Andrey Semashev (andrey.semashev_at_[hidden])
Date: 2013-09-25 05:09:19

On Wed, Sep 25, 2013 at 12:48 PM, Rob Stewart <robertstewart_at_[hidden]>wrote:

> On Sep 25, 2013, at 2:57 AM, Krzysztof Czainski <1czajnik_at_[hidden]>
> wrote:
>
> > 2013/9/25 Mostafa <mostafa_working_away_at_[hidden]>
> >
> >> Both the type_traits and function_types library discard parameter
> const-qualifiers when decomposing function types. Is this a bug or a
> >> feature? If a feature, then it's surprising enough behaviour that I
> think it warrants some documentation in both libraries.
>
> [snip]
>
> >> typedef void (foo)(int const);
> >
> > The two function declarations:
> > void f(int);
> > void f(int const);
> > declare the same function, not an overload.
>
> Right. Top level const is not part of the function's type.
>
> > Const added to an argument type passed by value doesn't change the
> function signature. It is because the parameter will be copied anyway, and
> the const refers to how the argument
> > will be treated inside the function's implementation.
>
> Exactly.
>

<complain_mode>

I have to say that while these rules are logical and understandable when
explained, types of function arguments are a constant source of confusion.
The described above quirk with const arguments can be considered quite rare
as people usually don't declare arguments that way. But C++11 brought us
rvalue references, and the following:

  foo(int&& n)
  {
    // n is _not_ rvalue reference here
  }

I understand the rationale for this, and it seems the right thing. But once
in a while, when yet another fellow developer asks me why is that so, I
wonder if it would be better if the standard was more straight-forward in
this part.

</complain_mode>