__func__ not replaced in preprocessed output

Question

I was reading about __FUNCTION__ / __func__ in C/C++ (they are used to print the name of the function in which they are used). Every place I read they said that these are macros and are replaced at preprocessing time. So, I investigated this by seeing the preprocessed output using command gcc -E prog.c. But I saw that neither __func__ nor __FUNCTION__ were replaced by the preprocessor by the function name.

So, is it a macro? If not, what is it and how is it implemented?

EDIT

Even tried cpp prog.c. But still not replaced.

Also __FILE__, __LINE__, and __FUNCTION__ usage in C++ this post says that it never affects performance. Please clarify.

Answer 1

They are implemented as (and are) "magic variables". The manual says:

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.

They typically cannot be implemented as preprocessor macros; the preprocessor doesn't parse function scopes. Of course a preprocessor could be made to understand enough of the syntax to know where functions begin and end, but typically they're not operating at that level.

Answer 2

__func__ is essentially a pre-defined variable, not a macro. Where __FUNCTION__ will be expanded, __func__ will not, and will be used as if you defined the following in your function:

static const char __func__[] = "function-name";

or

static const char __func__[] = __FUNCTION__;

Please refer to:

The intent of __func__ is similar to __FUNCTION__, but the delivery is different.

http://www.open-std.org/jtc1/sc22/WG21/docs/papers/2004/n1642.html