Is it possible to bring global function into the overload resolution with member function?

Question

Here is the corresponding question, what I want to know is, Is it possible to bring global function into the overload resolution with member function?

I tried in 2 ways, but both don't work:

void foo(double val) { cout << "double\n";}

class obj {
public:
  using ::foo; // (1) compile error: using-declaration for non-member at class scope
  void callFoo() { 
    using ::foo; // (2）will cause the global version always be called
    foo(6.4); 
    foo(0); 
  }
private:
  void foo(int val) {cout << "class member foo\n"; }
};

Answer 1

I doubt you can make the compiler call one or the other based on the type. You can of course use a local wrapper function, something like this:

  void callFoo() { 
    foo(6.4); 
    foo(0); 
  }
private:
  void foo(double val) { ::foo(val); }

The wrapper function should nicely inline into nothing, so there is no actual overhead when compiled with optimisation.

Or don't call the member and the global function the same name, which makes life a whole lot easier!