我应该如何确保对移动构造函数的调用? (移动语义和右值引用)

How should I ensure the call to the move constructor? (move semantics and rvalue reference)

我有这段代码

#include <iostream>
#include <vector>

using namespace std;

class ArrayWrapper {
    private:
        int *_p_vals;
        int _size;

    public:
        //default constructor
        ArrayWrapper () : _p_vals( new int[ 64 ] ) , _size( 64 ) {
            cout << "default constructor called\n";
        }

        //constructor 2
        ArrayWrapper (int n) : _p_vals( new int[ n ] ) , _size( n ) {
            cout << "constructor 2 called\n";
        }

        //move constructor
        ArrayWrapper (ArrayWrapper&& other) : _p_vals(other._p_vals), _size(other._size) {  
            cout << "move constructor called\n";
            other._p_vals = nullptr;
            other._size = 0;
        }

        // copy constructor
        ArrayWrapper (const ArrayWrapper& other) : _p_vals( new int[ other._size  ] ) , _size( other._size ) {
            cout << "copy constructor called\n";
            for ( int i = 0; i < _size; ++i )
                _p_vals[i] = other._p_vals[i];
        }

        ~ArrayWrapper () {
            delete [] _p_vals;
        }

        void set_val (std::initializer_list <int> vals) {
            int i = 0;
            for (auto val : vals) {
                _p_vals[i] = val;
                i++;
            }
        }
        void print_val () const {
            for (auto i = 0; i < _size ; i++){
                cout <<_p_vals[i] << ":" ;
            }
            cout << endl;
        }
};

ArrayWrapper getArrayWrapper () {
    ArrayWrapper A(5);
    A.set_val({1,2,3,4,5});
    return A;
}

int main () {
    ArrayWrapper arr {getArrayWrapper()};
    arr.print_val();
    return 0;
}

我正在尝试确保正在调用移动构造函数。但是一些默认的复制构造函数正在被使用,因为输出只是 this

constructor 2 called
1:2:3:4:5:

我正在使用以下 g++ 版本

g++ (Ubuntu 6.3.0-12ubuntu2) 6.3.0 20170406

我是否应该进行所有构造函数调用explicit。我想知道这会有什么帮助。有没有办法强制程序调用我上面写的移动构造函数和复制构造函数?

感谢评论者,我发现编译器正在进行 return 值优化 (RVO),这会阻止对上面编写的构造函数的调用。我找到了一种禁用 RVO 的方法。 g++ 中的 -fno-elide-constructors 开关有效。