How Function Overloading is working with default parameter

Question

Consider the following functions definition

void fn(int a, int b)
{
  //...
  //...
}

void fn(int a, int b, int c = 0)
{
  //...
  //...
}

In the main function I am calling a fn with 2 arguments:

int main()
{
  fn(10, 15);
  return 0;
}

So would like to know how compiler handles this situation.

Answer 1

The compiler can't know, and thus throws an error:

prog.cpp: In function ‘int main()’: prog.cpp:15:12: error: call of
  overloaded ‘fn(int, int)’ is ambiguous prog.cpp:15:12: note:
  candidates are: prog.cpp:1:6: note: void fn(int, int) prog.cpp:7:6:
  note: void fn(int, int, int)

The error doesn't happen on declaration, but on call, when it's actually resolved.

Answer 2

In addition to other answers, I guess, that the following rules from C++ standard applies:

13.3 Overload resolution

Each of these contexts defines the set of candidate functions and the list of arguments in its own unique way. But, once the candidate functions and argument lists have been identified, the selection of the best function is the same in all cases:

• First, a subset of the candidate functions (those that have the proper number of arguments and meet certain other conditions) is selected to form a set of viable functions (13.3.2).
• Then the best viable function is selected based on the implicit conversion sequences (13.3.3.1) needed to match each argument to the corresponding parameter of each viable function.

3. If a best viable function exists and is unique, overload resolution succeeds and produces it as the result. Otherwise overload resolution fails and the invocation is ill-formed. When overload resolution succeeds, and the best viable function is not accessible (Clause 11) in the context in which it is used, the program is ill-formed.

By the way, there actually is a way to call specific overload:

#include <stdio.h>

int f(int a, int b)
{
    printf("f - version 1\n");
    return 0;
}

int f(int a, int b, int c = 10)
{
    printf("f - version 2\n");
    return 0;
}

int main(int argc, char * argv[])
{
    int (* fn1)(int, int);
    fn1 = f;

    fn1(5, 10);

    return 0;
}