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 some manipulators (including the most frequently used ones - std::endl and std::flush) are templates of the following form:

template< class CharT, class Traits >
std::basic_ostream<CharT, Traits>& func(std::basic_ostream<CharT, Traits>& os);

Then, how does the compilation of std::cout << std::endl; succeed given that the following example fails:

$ cat main.cpp 
#include <iostream>

int main()
{
    auto myendl = std::endl;
    std::cout << myendl;
}

$ g++ -std=c++11    main.cpp   -o main
main.cpp: In function ‘int main()’:
main.cpp:5:24: error: unable to deduce ‘auto’ from ‘std::endl’
     auto myendl = std::endl;
                        ^

It is clear that the context (in std::cout << std::endl;) helps the compiler to disambiguate the reference to std::endl. But what are the rules that govern that procedure? It looks like a real challenge for overloading resolution, which has to answer two questions at once:

1. Which specialization of std::endl<CharT, Traits>() does std::endl refer to?
2. Which function does the operator<< refer to?

Template argument deduction (1) should happen before overload resolution (2), but it seems that (at least some part of) (2) is required to be performed in order for (1) to succeed.

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Somewhat related but no-way duplicate questions are:

• Does std::endl work with both cout and wcout?
• Why does endl(std::cout) compile
• How does std::flush work?

None of those questions and neither answers to them address the workings of template argument deduction that should precede overload resolution but must be helped by the latter.

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Follow-up question: How does overload resolution work when an argument is an overloaded function?

Answer 1

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

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

}

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

Answer 2

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

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

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<char, std::char_traits<char>>, aka std::ostream. So the member function of cout looks like

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

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<char, std::char_traits<char>>& (*pf)
    (std::basic_ostream<char, std::char_traits<char>>&) = std::endl;

Answer 3

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

basic_ostream<charT, traits>& operator<<(basic_ios<charT, traits>& (*pf)(basic_ios<charT, traits>&));

Answer 4

Given a statement expression of the form

 std::cout << std::endl;

The compiler has information about the type of std::cout - which is a specialisation of the templated std::basic_ostream which looks something like (omitting the containing namespace std).

template <class charT, class traits = char_traits<charT> >
    class basic_ostream
{
    public:
        basic_ostream<charT,traits>& operator<<(
            basic_ostream<charT,traits>& (*pf)(basic_ostream<charT,traits>&));
};

Since the compiler has information about the type of std::cout it knows what charT and traits are to specialise the preceeding template.

The above causes std::endl in the expression std::cout << std::endl to match to the specific std::basic_ostream<charT, traits>& endl( std::basic_ostream<charT, traits>&).

The reason type deduction doesn't work in

 auto myendl = std::endl;

is because std::endl is a templated function, and this declaration provides no information to specialise that template (i.e. picking what charT or traits are). If it can't specialise the templated std::endl, it can't infer that function's return type, so the type deduction fails.