c – 在对象被激活之后,使用隐式转换而不是赋值运算符调用构造函数的要点是什么？

#include <iostream>
using std::cout;
using std::endl;

class CBox
{
public:
    CBox(double lv = 1.0,double wv = 1.0,double hv = 1.0) :
        m_Length {lv},m_Width {wv},m_Height {hv}
    {
        cout << "Constructor called" << endl;   
    }

    //CBox& operator=(doublE)
    //{
    //  cout << "Assignment operator called" << endl;
    //  return *this;
    //}

    double volume() const { return m_Length* m_Width* m_Height; }

private:
    double m_Length;
    double m_Width;
    double m_Height;
};

int main()
{
    CBox Box {1.0,1.0,1.0}; // no need for intializer list,but put it anyway
    Box = 2.0; // why is this calling constructor again ?!
    cout << Box.volume() << endl; // prints 2
}

请注意,我有意地注释掉了重载的赋值运算符.

执行此程序会产生以下输出：

Constructor called
Constructor called
2

我注意到,即使Box对象已被初始化,下一个语句也是故意地调用构造函数.这是什么意思？

我知道这可以通过explicit关键字来防止,所以构造函数将是：

explicit CBox(double lv = 1.0,m_Height {hv}
    {
        cout << "Constructor called" << endl;   
    }

但我宁愿期望影响如下结构：

CBox Box = 2.0;

当然,当我取消注释重载的赋值运算符时,它优先于构造函数.