c++ - Interesting behavior of compiler with namespaces

Question

Assume the following code:

#include <iostream>
using namespace std;

namespace X
{
  class A{};

  void f(A a){}

  void g(int a){}
}

int main()
{
  X::A a;
  f(a);
  g(5);
}

When I compile the code, the following compile error occurs:

main.cpp: In function 'int main()':
main.cpp: error: 'g' was not declared in this scope

So the function f is compiled perfectly, but g isn't. How? Both of them belong to the same namespace. Does the compiler deduce that function f belongs to the X namespace from the argument of type X::A? How does compiler behave in such cases?

Answer 1

X::A a;
f(a);

works because of Argument-Dependent Lookup (Also known as Koenig Lookup). a is an object of class A inside namespace X, when compiler searches a match-able function f, it will look into namespace X in this case. See Argument Dependent Lookup for more information.