抽象类、继承和虚拟析构函数

Question

即使B的析构函数不是virtual，我仍然可以通过B指针调用C的析构函数。

这是否意味着只有最外层的抽象class需要有一个virtual析构函数？

如果是这样，为什么它会这样工作？

是因为B继承了A的析构函数吗？

#include <iostream>

struct A {
    virtual ~A() {
        std::cout << "~A\n";
    }

    virtual void
    function_a() = 0;
};

struct B : A {
    /*
    virtual ~B() {
        std::cout << "~B\n";
    }
    */

    virtual void
    function_b() = 0;
};

struct C : B {
    ~C() override {
        std::cout << "~C\n";
    }

    void
    function_a() override {
        std::cout << "function_a\n";
    }

    void
    function_b() override {
        std::cout << "function_b\n";
    }
};

int
main() {
    B * b = new C();

    b->function_a();
    b->function_b();

    delete b;
}

Answer 1

B和C的析构函数也是virtual。析构函数不会被继承，但是如果基础class的析构函数是virtual，派生的析构函数会覆盖它并且也是virtual；尽管 virtual 是否明确指定。

Even though destructors are not inherited, if a base class declares its destructor virtual, the derived destructor always overrides it.

和

Then this function in the class Derived is also virtual (whether or not the keyword virtual is used in its declaration) and overrides Base::vf (whether or not the word override is used in its declaration).

抽象类、继承和虚拟析构函数

Abstract classes, inheritance and virtual destructors

c++

inheritance

overriding

abstract-class

virtual