c++ - explicit copy constructor compile error

Question

I was checking operator overloading in C++ and came across something I did not expect and have some doubts about it.

My copy constructor is declared and implemented as as

explicit Vector(const Vector& v);

Vector::Vector(const Vector& v) :
_x(v._x), _y(v._y), _z(v._z) {}

then I am overloading the compound assignment operators

Vector Vector::operator+(const Vector& v) const
{
    Vector tmp(*this);
    tmp += v;
    return tmp;
}

Vector Vector::operator-(const Vector& v) const
{
    Vector tmp(*this);
    tmp -= v;
    return tmp;
}

however, in the return statements I got an error saying no matching constructor for initialization of 'Vector'.

Since the only thing I added to my constructor was the explicit keyword, I deleted it and the code compiles just fine, why?

I also was checking new stuff from C++11 and occurred that I can declare my constructor like a moving-constructor

explicit Vector(const Vector&& v);

and the code compiles just fine. If I do that, do I have to have both copy and move constructors?

explicit Vector(const Vector& v);
explicit Vector(const Vector&& v);

or just having the move constructor will work fine? If I want to stick to C++11, what is the correct approach to follow?

Answer 1

You defined an explicit copy constructor, but the functions

Vector Vector::operator+(const Vector& v) const

and

Vector Vector::operator-(const Vector& v) const

must return by value, and cannot anymore, due to explicit (in other words, they cannot copy tmp into the returned object).

I also was checking new stuff from C++11 and occurred that I can declare my constructor like a moving-constructor explicit Vector(const Vector&& v); and the code compiles just fine. If I do that, do I have to have both copy and move constructors?

Not sure I understand what you mean here. You will have the same issue if you only declare an explicit move constructor (which will prevent the compiler from generating a default copy constructor). I am not able to produce a "compilable" code.