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

Because basic_ostream has a templated overload of operator<< that expects just such a function pointer:

basic_ostream& operator<<(basic_ios& (*pf)(basic_ios&));