C++虚拟基类:不会调用父级的复制构造函数

Question

我在下面的代码中看到了三个类.请注意我是如何编写复制构造函数的.

#include <iostream>

class Abstract
{
public:
    Abstract(){};
    Abstract( const Abstract& other ): mA(other.mA){};
    virtual ~Abstract(){};

    void setA(double inA){mA = inA;};
    double getA(){return mA;};

    virtual void isAbstract() = 0;
protected:
    double mA;
};

class Parent : public virtual Abstract
{
public:
    Parent(){};
    Parent( const Parent& other ): Abstract(other){};
    virtual ~Parent(){};

};


class Child : public virtual Parent
{
public:
    Child(){};
    Child( const Child& other ): Parent(other){};
    virtual ~Child(){};

    void isAbstract(){};
};


int main()
{
    Child child1;
    child1.setA(5);

    Child childCopy(child1);
    std::cout << childCopy.getA() << std::endl;
    return 0;
}

现在为什么Abstract()在构造Abstract( const Abstract& other )时调用而不是复制构造childCopy函数？

不应该Child(other)打电话Parent(other)？而且不应该Parent(other)反过来打电话Abstract(other)？

Answer 1

虚拟基类只能由派生程度最高的类初始化.从非最派生类调用虚拟基础的构造函数将被忽略,并替换为默认构造函数调用.这是为了确保虚拟基础子对象仅初始化一次:

正确的代码应该将构造函数调用放在最派生的类' ctor-initializer中:

Child(Child const& other)
    : Abstract(other) // indirect virtual bases are
                      // initialized first
    , Parent(other) // followed by direct bases
{ }