std::tuple member by member comparison fails

Question

I wanted to test this very interesting answer and came out with this minimal implementation:

class A
{
    enum M { a };
    std::tuple<int> members;

public:

    A() { std::get<M::a>(members) = 0; }
    A(int value) { std::get<M::a>(members) = value; }
    A(const A & other) { members = other.members; }

    int get() const { return std::get<M::a>(members); }

    bool operator==(A & other) { return members == other.members; }
};

and a simple test:

int main() {

    A x(42);
    A y(x);

    std::cout << (x==y) << std::endl;

    return 0;
}

Everything's fine, until I define a simple struct B {}; and try to add an instance of it as a member. As soon as I write

std::tuple<int, B> members;

the operator== isn't ok anymore and I get this message from the compiler (gcc 5.4.1):

error: no match for ‘operator==’ (operand types are ‘std::__tuple_element_t<1ul, std::tuple<int, B> > {aka const B}’ and ‘std::__tuple_element_t<1ul, std::tuple<int, B> > {aka const B}’)
  return bool(std::get<__i>(__t) == std::get<__i>(__u))
                                 ^

I tried providing an operator== to B :

struct B
{
    bool operator==(const B &){ return true; }
};

and had an extra from compiler:

candidate: bool B::operator==(const B&) <near match>
     bool operator==(const B &){ return true; }
          ^

Can anyone explain what's wrong with B struct? Is it lacking something, or else?

Answer 1

It's ultimately const correctness. You didn't const qualify B 's (or A 's, for that matter) comparison operator and its parameter consistently.

Since tuple's operator== accepts by a const reference, it cannot use your const-incorrect implementation. And overload resolution fails as a consequence.

Tidying up all of those const qualifiers resolves all the errors .