c++ - 使用继承在 C++ 中调用析构函数和销毁成员变量的顺序是什么？

Question

与这些问题非常相似，但不完全相同: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:派生类的成员变量是在调用基类的析构函数之前还是之后销毁的？

这是在 C++ 中使用 Visual Studio 2008。谢谢。

Answer 1

构造函数:先基后导出

破坏:

• ~派生
• ~成员派生
• ~基地
• ~成员(member)群

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;
}

引用和虚析构函数

当您计划动态分配(即当您使用关键字 new 和 delete)派生对象时，总是有一个 virtual 或者你的基础上的 protected 析构函数。动态删除基类引用上的对象否则会导致下面示例中的内存泄漏:

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:

• ~基地
• ~成员(member)群

只有基础数据被清理，longArray泄漏。