Re: __func__ is not a keyword

scott@slp53.sl.home (Scott Lurndal) writes:

Keith Thompson <Keith.S.Thompson+u@gmail.com> writes:

Thiago Adams <thiago.adams@gmail.com> writes:

This program does not compile..in gcc and clang
>
int __func__  i =1;

>
I'm mildly curious how gcc and clang treat "__func__" internally
that leads to this odd behavior.

>
From the gcc info page:
>
   GCC provides three magic variables that hold the name of the current
   function, as a string.  The first of these is '__func__', which is part
   of the C99 standard:
>
    The identifier '__func__' is implicitly declared by the translator as
   if, immediately following the opening brace of each function definition,
   the declaration
>
     static const char __func__[] = "function-name";
>
   appeared, where function-name is the name of the lexically-enclosing
   function.  This name is the unadorned name of the function.

But that doesn't tell us how it's treated internally and it's
inconsistent with gcc's actual behavior.  Both a look at gcc's
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.

--
Keith Thompson (The_Other_Keith) Keith.S.Thompson+u@gmail.com
void Void(void) { Void(); } /* The recursive call of the void */