Returning const value from arithmetic operator overload with move assignment

Question

Let's say I have the following minimal example class:

#include <iostream>

class Foo {
public:
    Foo() = default;

    Foo(const Foo&) = default;
    Foo(Foo&&) noexcept = default;

    Foo& operator=(const Foo& rhs) {
        std::cout << "copy\n";

        return *this;
    }

    Foo& operator=(Foo&& rhs) noexcept {
        std::cout << "move\n";

        return *this;
    }

    Foo operator+(const Foo& rhs) const {
        Foo x; // with some calculation

        return x;
    }
};

int main() {
    Foo a, b, c;

    a = b + c;
}

This prints move as expected. Now according to Effective C++ Item 3, I should return const Foo from operator+ to avoid construct like a + b = c , ie:

// To avoid a + b = c
const Foo operator+(const Foo& rhs) const {}

Unfortunately, this suddenly starts calling copy assignment instead of move assignment operator. [I'm using gcc 4.8.4 on Ubuntu, but it is probably nothing related to compiler]

How can I ensure that a + b = c fails to compile and in the same time move assignment is called for a = b + c ? Or with the introduction of move semantics, is there no way to achieve both of them in the same time?

Answer 1

I have ended up using lvalue reference qualifier as pointed by Caninonos in comment and by max66 in now deleted answer (but 10k users can see it).

Foo& operator=(const Foo& rhs) & {}
Foo& operator=(Foo&& rhs) & noexcept {}

It is simple to implement and it provides a better interface design since assignment to anything other that lvalue doesn't sound meaningful and a possible source of bug.

However, it should be noted that the possibility of writing a + b = c by mistake is very low. Also compiler generated assignment operators are not lvalue reference qualified and we can write a + b = c with standard types, eg with std::string or with std::complex .