Does member function like operator<< , operator* need ADL to work?

February 20, 2022

I have a code snippet (Hypothetically):

#include <iostream>

struct Pirate {
 void song_name() {
  std::cout << "Bink's Sake\n";
 }

 Pirate& operator*(Pirate const& other) {
  // do something
  return *this;
 }
};

int main() {
 Pirate p1{} p2{};
 p1.song_name(); // does this use qualified or unqualifed name lookup?
 
 p1 * p2;

 std::cout << 5;
 std::cout << 'a';
}

Does p1 * p2 use qualified name lookup or unqualified name lookup or ADL?

std::cout << 5 transforms into std::cout.operator<<(5);
std::cout << 'a' transforms into std::operator<<(std::cout, 'a');

Does member functions require ADL to work?
Does the above two statments use qualified or unqualifed name lookup or ADL?

Thanks

>Solution :

A name is a qualified name if the scope to which it belongs is explicitly denoted using a scope-resolution operator (::) or a member access operator (. or ->).

Case 1

Thus, when you wrote:

p1.song_name(); //here p1.song_name is a qualified name

In the above statement, p1.song_name is a qualified name and so here qualified lookup takes place.

Case 2

Next, when your wrote:

p1 * p2;

The above statement is equivalent to:

p1.operator*(p2);

Since your class Pirate have an overloaded member function operator*, the above statement will use that member function. The above statement uses qualified lookup as well because we have used the member access operator ..