Re: __func__ is not a keyword

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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