为什么不调用派生的 class 析构函数？

Question

我正在使用指向派生类的指针进行一些练习，当我运行下面提供的代码时，我得到的输出是

Constructor A
Constructor B
Destructor A

有人能告诉我为什么 B::~B() 在这里没有被调用吗？

class A {
 public:
  A() { std::cout << "Constructor A\n"; }
  ~A() { std::cout << "Destructor A\n"; }
};

class B : public A {
 public:
  B() { std::cout << "Constructor B\n"; }
  ~B() { std::cout << "Destructor B\n"; }
};

int main() {
  A* a = new B;
  delete a;
}

Answer 1

因为覆盖的方法需要是虚拟的

class A{
public:
    A()
    {
        std::cout<<"Constructor A\n";
    }
    virtual ~A()
    {
        std::cout<<"Destructor A\n";
    }
};
class B : public A{
public:
    B()
    {
        std::cout<<"Constructor B\n";
    }
    virtual ~B()
    {
        std::cout<<"Destructor B\n";
    }
};

Answer 2

指针a的静态类型是A *。

A* a = new B;

因此在class A.

中搜索所有使用该指针的被调用成员函数

要调用指针的动态类型的析构函数，即class B，需要在class中将析构函数声明为virtual A。例如：

#include <iostream>

class A {
 public:
  A() { std::cout << "Constructor A\n"; }
  virtual ~A() { std::cout << "Destructor A\n"; }
};

class B : public A {
 public:
  B() { std::cout << "Constructor B\n"; }
  ~B() override { std::cout << "Destructor B\n"; }
};

int main() {
  A* a = new B;
  delete a;
}

为什么不调用派生的 class 析构函数？

Why isn't the derived class destructor being called?

c++

inheritance

destructor

class

declaration