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

The operator<< in question is a member of std::basic_ostream:

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

Since the call is to std::cout << std::endl, or equivalently std::cout.operator<<(std::endl), we already know the exact instantiation of basic_ostream: std::basic_ostream>, aka std::ostream. So the member function of cout looks like

std::ostream& operator<<(std::basic_ostream>& (*pf)
    (std::basic_ostream>&));

This member function is not a function template, just an ordinary member function. So the question remaining, is can it be called with the name std::endl as an argument? Yes, initializing the function argument is equivalent to a variable initialization, as though we had written

std::basic_ostream>& (*pf)
    (std::basic_ostream>&) = std::endl;