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