为什么这会调用复制构造函数，而不是移动构造函数？

Question

I have a class, PlayerInputComponent :

.h:

class PlayerInputComponent
{
public:
    PlayerInputComponent(PlayerMoveComponent& parentMoveComponent_, std::unique_ptr<IRawInputConverter> inputConverter_);
    PlayerInputComponent(PlayerInputComponent&& moveFrom);
    void update();

private:
    std::unique_ptr<IRawInputConverter> inputConverter;
    PlayerMoveComponent& parentMoveComponent;
};
}

.cpp:

PlayerInputComponent::PlayerInputComponent(PlayerMoveComponent& parentMoveComponent_, std::unique_ptr<IRawInputConverter> inputConverter_) :
    parentMoveComponent(parentMoveComponent_),
    inputConverter(std::move(inputConverter_))
{
}

PlayerInputComponent::PlayerInputComponent(PlayerInputComponent&& moveFrom) :
    parentMoveComponent(moveFrom.parentMoveComponent),
    inputConverter(moveFrom.inputConverter.release())
{
}

and a class, PlayerMoveComponen t, that contains a PlayerInputComponent member and initializes it using a std::unique_ptr passed as a parameter. Its constructor:

PlayerMoveComponent::PlayerMoveComponent(/* other parameters */ std::unique_ptr<IRawInputConverter> inputConverter) :
    //other initializations
    inputComponent(PlayerInputComponent(*this, std::move(inputConverter)))
{
}

I defined my own move constructor for the PlayerInputComponent class since my understanding is that a default move constructor won't be constructed for a class which contains a reference member. In this case though I know that the reference will remain in scope for duration of the PlayerInputComponent object's lifetime.

Since I'm initializing the PlayerMoveComponent 's inputComponent variable from a temporary, I believe one of the following two things is supposed to happen:

1. PlayerInputComponent 's move constructor is used to initialize the playerInputComponent member variable.
2. The move is elided by the compiler.

However, Visual Studio 2012 spits this out:

error C2248: 'std::unique_ptr<_Ty>::unique_ptr' : cannot access private member declared in class 'std::unique_ptr<_Ty>'
1>          with
1>          [
1>              _Ty=SDLGame::IRawInputConverter
1>          ]
1>          c:\program files\microsoft visual studio 11.0\vc\include\memory(1447) : see declaration of 'std::unique_ptr<_Ty>::unique_ptr'
1>          with
1>          [
1>              _Ty=SDLGame::IRawInputConverter
1>          ]
1>          This diagnostic occurred in the compiler generated function 'PlayerInputComponent::PlayerInputComponent(const PlayerInputComponent &)'

Why is the copy constructor being called here? Making the PlayerInputComponent class's parentMoveComponent member a regular ParentMoveComponent instance, rather than a reference, gets rid of the error, but I don't understand why - I've tested and verified that move constructing objects with reference members works so long as you provide your own move constructor, so what's the deal?

Answer 1

I'm sorry in advance if this doesn't really answer your question, I just want to react on the apparent complexity of your problem. If I may, wouldn't this be a thousand times simpler:

/********************     **********     ********************/

class C {};
class B;



class A
{
public:

    A(): _b(nullptr), _c(nullptr) {}
    A( B *b, C *c ): _b(b), _c(c) {}
    A( A&& a ): _b(a._b), _c(a._c) {}

private:

    C *_c;
    B *_b;
};



class B
{
public:

    B( /* other parameters */ C *c ): _a( A(this,c) ) {}

private:

    A _a;
};


    /********************     **********     ********************/


int main()
{
    C c;
    B b(&c);
}

and yet achieve the same thing? I have nothing against using the new features in c++11, like std::unique_ptr , but IMHO, ensuring that a pointer can never be dereferenced from two places should not be a matter of run-time checking (except maybe in very rare cases), but a matter of design.. shouldn't it?

Answer 2

If you initialize a new Object using = , the copy constructor will be triggered by default. To trigger the move constructor, you need to alter the behavior of operator= You can find an example here Hope I helped you.