On Thu, Dec 8, 2022 at 11:32 AM Vladimir Grigoriev via Std-Discussion <
std-discussion_at_[hidden]> wrote:

> Well, and what about this declaration
>
> void h( auto x ) requires requires { sizeof( x ) == sizeof __func__; };
>
> Is it valid?
>

Still no. The term "function-local predefined variable" and the variable
__func__ are described in [dcl.fct.def.general] in the context of a
function definition. Since this is not a function definition, there are no
function-local predefined variables. No __func__ variable exists for this
code at all, so statements about its scope don't apply.


>
> As it is said in the Standard the function-local predefined variable has
> function parameter scope.
>
> With best regards
> (Vlad from Moscow)
> You can meet me at http://cpp.forum24.ru/ or www.stackoverflow.com or
> http://ru.stackoverflow.com
>
>
>
> Четверг, 8 декабря 2022, 15:24 +03:00 от language.lawyer_at_[hidden]:
>
> > According to the C++ 20 Standard (Reference: 6.4.4 Function parameter
> scope)
> > «1 A function parameter (including one appearing in a lambda-declarator)
> or function-local predefined variable (9.5) has function parameter scope….»
>
> The next sentence: «The potential scope of a parameter or function-local
> predefined variable begins at its point of declaration»
>
> [basic.scope.pdecl]/9: «The point of declaration for a function-local
> predefined variable is immediately before the function-body of a function
> definition.»
>
> > So should this function declaration be valid?
> >
> > void f( size_t n = sizeof __func__ );
>
> No.
>
> > Pay attention to that there is used a function declaration that is not
> the function definition.
>
> There is no point of declaration of __func__ here, then.
>
>
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