pass std::endl to std::operator <<

Question

In this Stack Overflow answer it says that std::cout << "Hello World!" << std::endl; is the same as

std::operator<<(std::operator<<(std::cout, "Hello World!"), std::endl);

But when I compile the above line code, it doesn't compile! Then after trying something else I found that the reason it doesn't compile is because of std::endl, if I replace std::endl by "\n" then it works. But why you can not pass std::endl to std::operator<<?

Or more simply, isn't std::cout<<std::endl; the same as std::operator<<(std::cout, std::endl);?

EDIT

When compile with icpc test.cpp, the error message is error: no instance of overloaded function "std::operator<<" matches the argument list argument types are: (std::ostream, <unknown-type>) std::operator<<(std::cout, std::endl);

and g++ test.cpp gives much much longer error message.

Answer 1

It's because the answer there is a bit wrong. std::endl is a manipulator function, there is no overload for them in definitions of standalone operator<< of ostream. It is a member function of basic_ostream.

In other word, the presented invocation is wrong. It should be one of the following:

#include <iostream>
int main() {
    std::endl(std::operator<<(std::cout, "Hello World!"));
    std::operator<<(std::cout, "Hello World!").operator<<(std::endl);

    //of course if you pass new line as a supported type it works
    std::operator<<(std::operator<<(std::cout, "Hello World!"), '\n');
    std::operator<<(std::operator<<(std::cout, "Hello World!"), "\n");
    std::operator<<(std::operator<<(std::cout, "Hello World!"), string("\n"));
    return 0;
}

Live examples.

Well, some people do say that stream library does not have the prettiest design in the standard.

Answer 2

I dont know about this topic, but i think these 2 questions and answers are somewhat related to your question and might help you figure out a solution

operator << must take exactly one argument

Does std::cout have a return value?