使用inhertitance在C ++中销毁析构函数和成员变量的顺序是什么？

问题描述

非常类似的问题，除了不完全相同：

Very similar question as these, except not exactly: What is the order in which the destructors and the constructors are called in C++Order of member constructor and destructor calls

我想知道：是成员吗？在调用基类的析构函数之前或之后销毁的派生类的变量？

I want to know: are the member variables of the derived class destroyed before or after the destructor of the base class is called?

这是使用Visual Studio 2008的C ++。谢谢。

This is in C++ using Visual Studio 2008. Thanks.

推荐答案

构造函数：一垒，然后派生

constructor: first base, then derived

销毁：

• 〜衍生


• 〜会员派生


• ~base


• 〜会员基础

• ~derived
• ~member derived
• ~base
• ~member base

代码：

class member {
    string s;

public:
    member(string s) {
        this-> s = s;
    }

    ~member() {
        cout << "~member " << s << endl;
    }
};

class base {
    member m;
public:
    base() : m("base"){
    }

    ~base() {
        cout << "~base" << endl;
    }
};

class derived : base{
     member m2;
public:

    derived() :m2("derived") {    }

    ~derived() {
        cout << "~derived" << endl;
    }
};

int main(int argc, const char * argv[]) {
    derived s;

    return 0;
}

参考文献&虚拟析构函数

当您计划动态分配时（即当您使用关键字 new & 删除）派生对象，然后始终具有虚拟或受保护基础上的析构函数。在下面的示例中，动态删除基类引用上的对象会导致内存泄漏：

References & virtual destructor

When you plan to dynamically allocate (i.e. when you use the keywords new & delete) a derived object, then always have a virtual or a protected destructor on your base. Dynamically deleting the object on the base class reference would otherwise lead to memory leaks in the example below:

class base {
    member m;
public:
    base() : m("base"){
    }

    /* correct behaviour is when you add **virtual** in front of the signature */
    ~base() {
        cout << "~base" << endl;
    }
};

class derived : public base{
     member m2;
    char* longArray;
public:

    derived() :m2("derived") {
        longArray = new char[1000];
    }


    ~derived() {
        delete[] longArray; // never called
        cout << "~derived" << endl;
    }
};

int main(int argc, const char * argv[]) {
    base *s = new derived; // mind the downcast to **base**

    delete s; /* only the non-virtual destructor on the base and its members is called.
               No destructor on derived or its members is called.
               What happens to the memory allocated by derived?
               **longArray** is leaked forever.
               Even without **longArray**, it most probably **leaks** memory, as C++ doesn't define its behaviour
               */
    return 0;
}

输出：

• ~base


• 〜会员基础

仅基础数据已清理， longArray 泄露。

Only base data is cleaned up, and longArray leaks.

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