From Newsgroup: comp.lang.c

__func__ is just an string that has name of the function it is in,
so for example:

#include <stdio.h>
int
main(int argc, char** argv) {
printf("we are on: %s\n", __func__);
return 0;
}

Will print:

we are on: main

My understand is that __func__ is not a keyword and that is something defined inside the functions.. so I don’t know why gcc and clang
complains in the file scope.

Yes, it is a sort of atomic identified reserved by the compiler.
You can't redefine it, just like any normal identifier, if that
makes sense. The compiler errors are from C not understanding
the code.

"The identifier __func__ shall be implicitly declared by the translator
as if, immediately following
the opening brace of each function definition, the declaration" ...

What it refers to is:

type
myfunction(...) { // <--- from here on, __func__ is available.
...
}
--
"I know thy works, that thou art neither cold nor hot: I would thou wert
cold or hot. So then because thou art lukewarm, and neither cold nor
hot, I will spue thee out of my mouth." (Rev. 3:15-16)
--- Synchronet 3.21f-Linux NewsLink 1.2

From Newsgroup: comp.lang.c

You're replying to an article posted by Thiago Adams more than a
year ago.

You deleted the attribution line. Please don't do that.

Mario Rosell <usenet@mariorosell.es> writes:

__func__ is just an string that has name of the function it is in,
so for example:

#include <stdio.h>
int
main(int argc, char** argv) {
printf("we are on: %s\n", __func__);
return 0;
}

Will print:

we are on: main

It's not *just* a string. It's an implicitly defined static array of
char.

My understand is that __func__ is not a keyword and that is something
defined inside the functions.. so I don’t know why gcc and clang
complains in the file scope.

Yes, it is a sort of atomic identified reserved by the compiler.
You can't redefine it, just like any normal identifier, if that
makes sense. The compiler errors are from C not understanding
the code.

"The identifier __func__ shall be implicitly declared by the translator
as if, immediately following
the opening brace of each function definition, the declaration" ...

What it refers to is:

type
myfunction(...) { // <--- from here on, __func__ is available.
...
}

This was all discussed at the time, and more accurately than in your
response.

The code in the original message was:

int __func__ i =1;

which is a syntax error even if a non-reserved identifier is used.
It's likely that Thiago meant to write:

int __func__ = 1;

But that has undefined behavior, because __func__ is a reserved
identifier due to the double underscore.

There are some oddities in the way gcc and clang treat __func__.
If you're curious, you can look up the thread from last year.
(gcc seems to treat __func__ as a keyword, which it isn't, but
that doesn't seem to cause any conformance issues, just some odd
wording in some diagnostics for bad code.)
--
Keith Thompson (The_Other_Keith) Keith.S.Thompson+u@gmail.com
void Void(void) { Void(); } /* The recursive call of the void */
--- Synchronet 3.21f-Linux NewsLink 1.2