为什么使用此全局`operator <<`无法编译？

Question

Here is the interface in a nutshell

#include <iostream>

struct Graph
{
    typedef int Parameter;

    struct Render
    {
        void to (std :: ostream &) {}
    };

    Render render (Parameter) {}
};

std :: ostream &
operator << (std :: ostream &, Graph :: Render &);

int main ()
{
    Graph () .render (0) .to (std :: cout);

    // std :: cout << Graph () .render (0);
}

The above will compile without complaint, unless you uncomment the last line.

Why doesn't the global operator<< compile?

Answer 1

You have overloaded your operator only for mutable lvalues, so the prvalue (the temporary value that is Graph().render(0) ) does not bind to it.

You can either change your operator overload to use a const reference (which will accept lvalues and rvalues):

std::ostream & operator<<(std::ostream &, const Graph::Render &);
//                                        ^^^^^^^^^^^^^^^^^^^^^

Or you can add an additional rvalue overload:

std::ostream & operator<<(std::ostream &, Graph::Render &&);
//                                        ^^^^^^^^^^^^^^^^

If in doubt, go with the first solution. It would be extremely surprising if printing a string representation of the object would require its mutation.

(It is also kind of weird that Graph::render should return a new Render object by value, but that's your decision.)