how does cout << actually work?

Question

I was wondering how std::cout is able to use << as it does.

My main puzzlement is with whether std::cout as an instance

Answer 1

std::cout is an instance of std::ostream. std::cout << "something" calls one of the operator<< overloads as would be done for any instance of std::ostream.

It's "special" in that it references the console, but otherwise it behaves exactly as an ofstream or an ostringstream would.

EDIT: Chaining works just like it works for any other operators. Consider:

class MyType
{
    friend std::ostream& operator<<(std::ostream& target, const MyType& source);
    int val;
public:
    MyType()
        : val(0)
    { }
    MyType& Add(int toAdd)
    {
        val += toAdd;
        return *this;
    }
};

MyType& operator+(MyType& target, int toAdd)
{
    return target.Add(toAdd);
}

std::ostream& operator<<(std::ostream& target, const MyType& source)
{
    target << source.val;
    return target; //Make chaining work
}

int main()
{
    MyType value1;
    value1 + 2 + 3 + 4;
    std::cout << value1 << " and done!" << std::endl;
}

In this case, the chaining for +s on MyType work for the same reason the <<s work on top of std::ostream. Both + and << are left-associative, which means they're evaluated from left to right. In the case of overloaded operators, the operator is replaced with the equivalent function call.

EDIT2: In a bit more detail:

Let's say you're the compiler and you're parsing

std::cout << value1 << " and done!" << std::endl;

First, << is left associative, so you start on the left. You evaluate the first << and turn it into the function call:

operator<<(std::cout, value1) << " and done!" << std::endl;

You then see that you once again have a std::ostream (the result of the call to operator<<), and a char *, which you once again turn into the function call:

operator<<(operator<<(std::cout, value1)," and done!") << std::endl;

and so on until you've processed the entire statement.