Finding the line and file in which a function was called

In a function, is there a way to tell where it was called from (namely, the line and file) without passing parameters? It would be my guess that __LINE__ and __FILE__ contain the line and file that the function is on, not where it was called from.

My current solution would be to pass them as defaulted parameters:

Code:

void myFunc( int = __LINE__, char* = __FILE__ );

but this seems like a bad way, since the user could specify those parameters and override the wanted behavior. Is there a better way? TIA

Don´t think there is. A debugger should solve your problem.

Would an inline wrapper function solve the problem too? Like the wrapper, only a full-fledged C++ function:

Code:

void func( char *p, int line, char *file ) {
  std::cout << p << " at line " << line << " in file " << file << std::endl;
}

inline void inline_wrap( char* a ) {
  func( a, __LINE__, __FILE__ );
}

int main( ) {
  inline_wrap( "hehe" );
  return 0;
}

I'm hoping that since it's inline, __LINE__ and __FILE__ will refer to the point where it's being called in the file, not where I defined it. The obvious answer is I'm going to go check...

EDIT: I just checked, no go. Still reports the line on which func() is called from inline_wrap. The macro solution will be somewhat difficult because this function is a class member template function (i.e., it looks like "some_object_name.alloc<some_type>( params )" ). Thanks for the help though.