Stuck in understanding following macro definition - c++

Question

Here is a macro definition:
#define print_debug(fmt, ...) \
__print_debug__(__FILE__, __func__, __LINE__, fmt, ## __VA_ARGS__)
I cannot locate the expansion of __FILE__,__func__ ,__LINE__ and ## __VA_ARGS__ in visual studio(I can get __VA_ARGS__ that corresponds to a variable arguments). In particular what dose ## __VA_ARGS__ mean?

Answer 1

__print_debug__ is a function like macro which will print all arguments passed to it.
Here
__FILE__ :--> will pass current File name
__func__:--> will pass current function name.
__LINE__ :--> will pass current line number
You can have a look at This Link for more information on it.
Now fmt and ## __VA_ARGS__ :
Let me take below example.
print_debug(fmt,var1,var2,var3)
the macro expansion will be as following:
__print_debug__(__FILE__, __func__, __LINE__, fmt,var1,var2,var3)
This is because ## will concatenate all the the variable list of argument i.e. __VA_ARGS__
Save following code in macro.c file and compile it with
gcc -E macro.c -o macro.i
command and have a look at the macro.i file.
#define print_debug(fmt, ...) \
__print_debug__(__FILE__, __func__, __LINE__, fmt, ## __VA_ARGS__)
void call_Macro()
{
print_debug(2,3,4,5) //Here 2 is nothing but fmt
}
int main()
{
call_Macro();
return 0; #
}