虚拟继承中谁调用构造函数？
#include<iostream>
class base{
public:
base(){std::cout<<"In base";}
};
class dv1:virtual private base {
public:
dv1(){std::cout<<"In DV1";}
};
class dv2:virtual private base {
public:
dv2(){std::cout<<"In DV2";}
};
class drv : public dv1, public dv2 {
public:
drv() {std::cout<<"Why is this working";}
};
int main() {
drv obj;
return 0;
}
在虚拟继承的情况下，调用构造函数不是大多数派生类的责任吗？ 注意：这里的base是虚拟的、私有的继承。
#include<iostream>
class base{
public:
base(){std::cout<<"In base";}
};
class dv1:virtual private base {
public:
dv1(){std::cout<<"In DV1";}
};
class dv2:virtual private base {
public:
dv2(){std::cout<<"In DV2";}
};
class drv : public dv1, public dv2 {
public:
drv() {std::cout<<"Why is this working";}
};
int main() {
drv obj;
return 0;
}
Isn't in case of virtual inheritance, it is the responsibility of most derive class to call the constructor?
Note: Here base is inherited virtually and privately.
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drv的构造函数没有显式调用其基类的构造函数，因此编译器将生成对基类的无参数构造函数的调用Your constructor of
drvdidn't explicitly call the constructors of its base class(es), so the compiler will generate a call to the parameterless constructor of the base class