How does “std::cout << std::endl;” compile?

Question

Most IO stream manipulators are regular functions with the following signature:

std::ios_base& func( std::ios_base& str );

However

Answer 1

You need to put << before the endl. It is a member of ofstream:

namespace std {
template  >
class basic_ostream {
public:
    basic_ostream& operator<<(
      basic_ostream& (*pf)(basic_ostream&));
    // ...
};

}

endl works like a "/n" to skip the line, so you need a cout line to skip