虚拟继承和参数化的构造函数

Question

可能重复：
Default constructor and virtual inheritance

class Base 
{ 
private: 
    int number; 
protected: 
    Base(int n) : number(n) {} 
public: 
    virtual void write() {cout << number;}  
}; 

class Derived1 : virtual public Base 
{ 
private: 
    int number; 
protected: 
    Derived1(int n, int n2) : Base(n), number(n2) {} 
public: 
    virtual void write() {Base::write(); cout << number;} 
}; 

class Derived2 : virtual public Base 
{ 
private: 
    int number; 
protected: 
    Derived2(int n, int n2) : Base(n), number(n2) {} 
public: 
    virtual void write() {Base::write(); cout << number;} 
}; 

class Problematic : public Derived1, public Derived2 
{ 
private: 
    int number; 
public: 
    Problematic(int n, int n2, int n3, int n4) : Derived1(n, n2), Derived2(n, n3), number(n4) {} 
    virtual void write() {Derived1::write(); Derived2::write(); cout << number;} 
}; 

int main() 
{ 
    Base* obj = new Problematic(1, 2, 3, 4); 
    obj->write(); 
} 

换句话说：

Base 
| \ 
| \ 
| \ 
| \ 
D1 D2 
| /
| /
|/
|/
Problematic 

我试图得到“1 2 1 3 4“的输出。但是，编译器不断抱怨我需要Base中的无参数构造函数，但是当我添加一个构造函数时，“1”变成垃圾。关于如何处理它的任何想法？使用参数化构造函数可以解决菱形图案吗？

Answer 1

看看Calling a virtual base class's overloaded constructor，看起来如果继承是虚拟的，最派生类必须调用基类构造函数。

Problematic(int n, int n2, int n3, int n4) : Derived1(n, n2), Derived2(n, n3), Base(n), number(n4) {}