c++ - 在 C++ 继承中，当指向基类的指针对象指向派生类时，不调用派生类析构函数

Question

我是新手，我知道这是一个非常基本的概念，也可能是重复的。 一旦调用构造函数，就必须调用其对应的析构函数，这不是真的吗？ [在 Dev C++ 上运行的代码]

class Base
    {
     public:
            Base() { cout<<"Base Constructor\n";}
            int b;
     ~Base() {cout << "Base Destructor\n"; }
    };

class Derived:public Base
{
 public:
        Derived() { cout<<"Derived Constructor\n";}
        int a;
 ~Derived() { cout<< "Derived Destructor\n"; }
}; 
int main () {
Base* b = new Derived;    
//Derived *b = new Derived;
 delete b;
    getch();
    return 0;
}

给出输出

Base Constructor
Derived Constructor
Base Destructor

Answer 1

您的代码有未定义的行为。基类的析构函数必须是 virtual 才能使以下具有定义的行为。

Base* b = new Derived;    
delete b;

来自 C++ 标准:

5.3.5 Delete

3 In the first alternative (delete object), if the static type of the operand is different from its dynamic type, the static type shall be a base class of the operand’s dynamic type and the static type shall have a virtual destructor or the behavior is undefined.

所以在你的例子中，静态类型是Base，动态类型是Derived。所以 Base 的析构函数应该是:

virtual ~Base() {cout << "Base Destructor\n"; }