继承中调用构造函数/析构函数的顺序

Order of calling constructors/destructors in inheritance(继承中调用构造函数/析构函数的顺序)

问题描述

关于创建对象的一个​​小问题.假设我有这两个类:

A little question about creating objects. Say I have these two classes:

struct A{
    A(){cout << "A() C-tor" << endl;}
    ~A(){cout << "~A() D-tor" << endl;}
};

struct B : public A{
    B(){cout << "B() C-tor" << endl;}
    ~B(){cout << "~B() D-tor" << endl;}

    A a;
};

并且在 main 我创建了一个 B 的实例:

and in main I create an instance of B:

int main(){
    B b;
}

注意 B 派生自 A 并且还有一个 A 类型的字段.

Note that B derives from A and also has a field of type A.

我想弄清楚规则.我知道构造对象时首先调用其父构造函数，析构时反之亦然.

I am trying to figure out the rules. I know that when constructing an object first calls its parent constructor, and vice versa when destructing.

字段(A a; 在这种情况下)呢?当B被创建时，它什么时候会调用A的构造函数?我还没有定义初始化列表，是否有某种默认列表?如果没有默认列表?还有关于破坏的同样问题.

What about fields (A a; in this case)? When B is created, when will it call A's constructor? I haven't defined an initialization list, is there some kind of a default list? And if there's no default list? And the same question about destructing.

推荐答案

• 构造总是从基础 class 开始.如果有多个基class，那么从最左边的基开始构造.(旁注:如果存在虚拟继承，则优先级更高).
• 然后构造成员字段.它们在声明它们的顺序
• 最后，class 本身被构造
• 析构函数的顺序正好相反
• Construction always starts with the base class. If there are multiple base classes then, construction starts with the left most base. (side note: If there is a virtual inheritance then it's given higher preference).
• Then the member fields are constructed. They are initialized in the order they are declared
• Finally, the class itself is constructed
• The order of the destructor is exactly the reverse

不管初始化列表如何，调用顺序都是这样的:

Irrespective of the initializer list, the call order will be like this:

1. Base class A 的构造函数
将构造
2. class B 的名为a(类型class A)的字段
3. 派生class B的构造函数
1. Base class A's constructor
2. class B's field named a (of type class A) will be constructed
3. Derived class B's constructor

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