为什么在虚拟继承中调用默认构造函数?

问题描述

我不明白为什么在下面的代码中，当我实例化 daughter 类型的对象时，会调用默认的 grandmother() 构造函数?

I don't understand why in the following code, when I instanciate an object of type daughter, the default grandmother() constructor is called ?

我认为应该调用 grandmother(int) 构造函数(遵循我的 mother 类构造函数的规范)，或者这段代码不应该在都是因为虚拟继承.

I thought that either the grandmother(int) constructor should be called (to follow the specification of my mother class constructor), or this code shouldn't compile at all because of the virtual inheritance.

这里编译器在我的背后默默地调用了grandmother默认构造函数，而我从来没有要求它.

Here compiler silently calls grandmother default constructor in my back, whereas I never asked for it.

#include <iostream>

class grandmother {
public:
    grandmother() {
        std::cout << "grandmother (default)" << std::endl;
    }
    grandmother(int attr) {
        std::cout << "grandmother: " << attr << std::endl;
    }
};

class mother: virtual public grandmother {
public:
    mother(int attr) : grandmother(attr) {
        std::cout << "mother: " << attr << std::endl;
    }
};

class daughter: virtual public mother {
public:
    daughter(int attr) : mother(attr) {
        std::cout << "daughter: " << attr << std::endl;
    }
};

int main() {
  daughter x(0);
}

推荐答案

使用虚继承时，虚基类的构造函数直接被最派生类的构造函数调用.在这种情况下，daughter 构造函数直接调用了 grandmother 构造函数.

When using virtual inheritance, the virtual base class's constructor is called directly by the most derived class's constructor. In this case, the daughter constructor directly calls the grandmother constructor.

由于您没有在初始化列表中显式调用grandmother 构造函数，因此将调用默认构造函数.要调用正确的构造函数，请将其更改为:

Since you didn't explicitly call grandmother constructor in the initialization list, the default constructor will be called. To call the correct constructor, change it to:

daugther(int attr) : grandmother(attr), mother(attr) { ... }

另请参阅此常见问题解答.

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