不，你不需要调用基析构函数，基析构函数总是由派生析构函数调用。请参阅我的相关回答在这里的破坏顺序。

要理解为什么要在基类中使用虚析构函数，请参阅下面的代码:

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


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

当你这样做时:

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

然后，如果在B中没有虚析构函数，则只会调用~B()。但是由于您有一个虚析构函数，首先将调用~D()，然后是~B()。

不。与其他虚方法不同的是，在其他虚方法中，您需要显式地从Derived调用Base方法来“链接”调用，编译器会生成代码，以调用析构函数的构造函数的相反顺序调用析构函数。

不，你永远不会调用基类析构函数，它总是自动调用，就像其他人指出的那样，但这里是概念的证明和结果:

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;
}

输出结果为:

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函数，那么情况3的结果将与情况1和2相同。

其他人说了什么，但还要注意，您不必在派生类中声明析构函数虚值。一旦将析构函数声明为虚函数，就像在基类中所做的那样，所有派生析构函数都将是虚函数，无论是否这样声明。换句话说:

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

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

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

不，析构函数是按照构造的相反顺序自动调用的。(基类最后)。不要调用基类析构函数。

不。它会被自动调用。