How to pass caller function name to callee function using default parameters

Consider the below code :

#include <iostream>

void test_function(const char* caller_name = __FUNCTION__) {
  std::cout << caller_name << " ";
}

void func1() {
  test_function();
}

int main() {
  test_function();
  func1();
}

The above code prints nothing. Is there any way that test_function prints the name of the function from which it is called ?
Eg : For this case, output should be : main func1

I understand that desired output can be achieved if we make the argument to test_function, non default, and pass __FUNCTION__ from the caller function.
This could be achieved by using macro also, but it is not what I am looking for.