Is it possible to make macros that are default arguments expand at call site?

#include <stdio.h>

void print(int a = __LINE__){printf("hello %d\n", a);}

void main(){
  print();
  print();
  print();
  print();
}

The __LINE__ macro in this case expands to 3, so the print function is called 4 times with the same value. Is there a way to convince the compiler to expand this macro at the callsite, so that the print function would be called with 6,7,8,9 instead of 3,3,3,3 with features that exist in C++11?

My usecase:

In my application, I provide multiple functions that take a unique ID. The ID should be unique per callsite/location (so if the function is called two times throught the same statement, it should receive the same id). Currently, the user always has to manually type a LOCATION macro at the callsite, like this:

#define S1(x) #x //voodoo to concat __FILE__ and __LINE__
#define S2(x) S1(x)
#define LOCATION __FILE__ S2(__LINE__)

do_stuff1(arguments, LOCATION)
do_stuff2(arguments, LOCATION)

It would be more convenient if I could save them the typing, without creating a macro for each function like this:

#define do_stuff1(do_stuff1_imp(arguments, LOCATION))
#define do_stuff2(do_stuff2_imp(arguments, LOCATION))

Hence I thought a default argument would do the trick. Is there any way to achieve this?

The __LINE__ macro in this case expands to 3

Sure because your function declaration is done at line 3 and the prepreprocessor expands it from there.

Another macro instead of a function declaration would do the trick well (taken from @Ejay's comment):

#define print() printf("hello %d\n", __LINE__)

See a working example here.