我是否需要显式调用基本虚拟析构函数？

在C ++中（使用虚拟析构函数）覆盖类时，我在继承类上再次将析构函数实现为虚拟形式，但是我是否需要调用基本析构函数？

如果是这样的话，我想是这样的...

MyChildClass::~MyChildClass() // virtual in header
{
    // Call to base destructor...
    this->MyBaseClass::~MyBaseClass();

    // Some destructing specific to MyChildClass
}

我对吗？

不，析构函数将以相反的构造顺序自动调用。（最后是基类）。不要调用基类的析构函数。

不，您不需要调用基本析构函数，派生析构函数总是为您调用基本析构函数。 请查看我的相关答复，以了解销毁顺序。

要了解为什么要在基类中使用虚拟析构函数，请参见以下代码：

class B
{
public:
    virtual ~B()
    {
        cout<<"B destructor"<<endl;
    }
};


class D : public B
{
public:
    virtual ~D()
    {
        cout<<"D destructor"<<endl;
    }
};

当您这样做时：

B *pD = new D();
delete pD;

然后，如果您在B中没有虚拟析构函数，则仅会调用〜B（）。但是，由于您具有虚拟析构函数，因此先调用〜D（），然后调用〜B（）。

其他人说的话，但也请注意，您不必在派生类中将析构函数声明为虚拟。像在基类中一样声明了虚拟的析构函数之后，无论您是否声明它们，所有派生的析构函数都将是虚拟的。换一种说法：

struct A {
   virtual ~A() {}
};

struct B : public A {
   virtual ~B() {}   // this is virtual
};

struct C : public A {
   ~C() {}          // this is virtual too
};

不会。与其他虚拟方法不同，在其他虚拟方法中，您将从派生对象显式调用Base方法来“链接”调用，编译器生成代码以按与调用其构造函数相反的顺序来调用这些析构函数。

不，您永远不会调用基类析构函数，它总是像其他人指出的那样自动调用，但这是带有结果的概念证明：

class base {
public:
    base()  { cout << __FUNCTION__ << endl; }
    ~base() { cout << __FUNCTION__ << endl; }
};

class derived : public base {
public:
    derived() { cout << __FUNCTION__ << endl; }
    ~derived() { cout << __FUNCTION__ << endl; } // adding call to base::~base() here results in double call to base destructor
};


int main()
{
    cout << "case 1, declared as local variable on stack" << endl << endl;
    {
        derived d1;
    }

    cout << endl << endl;

    cout << "case 2, created using new, assigned to derive class" << endl << endl;
    derived * d2 = new derived;
    delete d2;

    cout << endl << endl;

    cout << "case 3, created with new, assigned to base class" << endl << endl;
    base * d3 = new derived;
    delete d3;

    cout << endl;

    return 0;
}

输出为：

case 1, declared as local variable on stack

base::base
derived::derived
derived::~derived
base::~base


case 2, created using new, assigned to derive class

base::base
derived::derived
derived::~derived
base::~base


case 3, created with new, assigned to base class

base::base
derived::derived
base::~base

Press any key to continue . . .

如果将基类析构函数设置为虚拟的，那么情况3的结果将与情况1和2相同。

不，它会自动调用。

仅当声明了基类析构函数时，C ++ 中的析构函数才会按其构造顺序（先导出，然后继承）自动调用。virtual

如果不是，则在删除对象时仅调用基类析构函数。

示例：没有虚拟析构函数

#include <iostream>

using namespace std;

class Base{
public:
  Base(){
    cout << "Base Constructor \n";
  }

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

};

class Derived: public Base{
public:
  int *n;
  Derived(){
    cout << "Derived Constructor \n";
    n = new int(10);
  }

  void display(){
    cout<< "Value: "<< *n << endl;
  }

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

int main() {

 Base *obj = new Derived();  //Derived object with base pointer
 delete(obj);   //Deleting object
 return 0;

}

输出量

Base Constructor
Derived Constructor
Base Destructor

示例：使用基本虚拟析构函数

#include <iostream>

using namespace std;

class Base{
public:
  Base(){
    cout << "Base Constructor \n";
  }

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

};

class Derived: public Base{
public:
  int *n;
  Derived(){
    cout << "Derived Constructor \n";
    n = new int(10);
  }

  void display(){
    cout<< "Value: "<< *n << endl;
  }

  ~Derived(){
    cout << "Derived Destructor \n";
    delete(n);  //deleting the memory used by pointer
  }
};

int main() {

 Base *obj = new Derived();  //Derived object with base pointer
 delete(obj);   //Deleting object
 return 0;

}

输出量

Base Constructor
Derived Constructor
Derived Destructor
Base Destructor

建议声明基类析构函数，virtual否则会导致未定义的行为。

参考：虚拟析构函数