谁在虚拟继承中调用构造函数？

Question

#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;
}

不是在虚拟继承的情况下,调用构造函数是大多数派生类的责任吗？注意:这里基地是虚拟和私下继承的.

Answer 1

您的构造函数drv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