Question

在以下代码中，使用Multi Path Inheritance解析了Virtual Class 构造函数是如何工作的？无法继承构造函数或虚拟或静态。

/*Multi Path Inheritance*/

class A{

public:
    int a;
    A(){
        a=50;
    }
};


class B:virtual public A{

public:
    /*B(){
        a = 40;
    }*/

};

class C:virtual public A{

public:
    /*C(){
        a = 30;
    }*/

};

class E:virtual public A{

public:
    E(){
        a = 40;
    }

};

class D : public B, public C, public E{

public:
    D(){
        cout<<"The value of a is : "<<a<<endl;  
    }

};

int main(int argc, char *argv[]){

    D d;
    return 0;
}

Answer 1

基于标准12.6.2 / 10的以下配额，因此将在以下orde中调用构造函数体：A-> B-＆gt; C-＆gt; D，因此a的最终值将为40。

在非委托构造函数中，初始化继续进行以下顺序：

 — First, and only for the constructor of the most
   derived class (1.8), virtual base classes are initialized in the order
   they appear on a depth-first left-to-right traversal of the directed
   acyclic graph of base classes, where “left-to-right” is the order of
   appearance of the base classes in the derived class
   base-specifier-list. 

 — Then, direct base classes are initialized in
   declaration order as they appear in the base-specifier-list
   (regardless of the order of the mem-initializers).

Answer 2

你可以找到很多关于虚拟继承的信息和例子here（是的，它实际上是在msdn上，有多奇怪:)）

对于构造函数，构造函数在您指定它们时被调用。如果没有为virtua-base类构造函数指定调用，

类中任何位置的虚拟基类的构造函数继承层次结构由“最派生”类调用构造

（请阅读here）。

多路径继承和

2 个答案: