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From: Kaz Kylheku <643-408-1753@kylheku.com>
Newsgroups: comp.lang.c
Subject: Re: __func__ is not a keyword
Date: Sun, 16 Mar 2025 19:05:05 -0000 (UTC)
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On 2025-03-16, Keith Thompson <Keith.S.Thompson+u@gmail.com> wrote:
> source code and its behavior indicate that gcc treats __func__ as
> a keyword, which is inconsistent with the info page.  For example,
> one would expect this:
>
> int main(void) {
>     {
>         int __func__;
>     }
> }
>
> to be accepted, with the inner definition of __func__ hiding the
> implicit static declaration, but gcc reports a syntax error.
>
> It's not a conformance issue, since __func__ is a reserved identifier
> and any code that can tell whether it's a keyword has undefined
> behavior.

But __func__ is not a reserved identifier! Inside a function, it's a
documented identifier with specified properties, and those properties do
not support an interpretation that it may be a keyword.

We have to distinguish between specific, defined, standard identifiers
allocated from a reserved namespace, and the reserved namespace itself.

If gcc (in c99 mode or later) *allowed*  int _Bool = 42;   would you
call that conforming, because _B* is in the reserved namespace, so
any behavior is okay?

Since __func__ is not described as existing outside of a function,
there, it is just an identifier landing in the reserved namespace.

-- 
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