派生类和基类都需要虚拟析构函数吗？

发布于 2025-01-09 18:47:48 字数 817 浏览 0 评论 0原文

假设我们有以下内容：

#include <iostream>

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

struct B : A
{
    // no need to be virtual?
    ~B() { std::cout << "destr B\n"; }
};

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

现在，我创建一个 C 实例并将其分配给其基类 B 的指针。

int main()
{
    B* b = new C{};
    delete b;
    return 0;
}

输出是：

destr C
destr B
destr A

让我有点惊讶的是对象被正确销毁（所有三个析构函数都被调用）。根据输出，我会说 ~B() 是一个虚拟析构函数：~B() 分派到 ~C()，然后~B() 完成，最后调用 ~A()。这个说法正确吗？

我知道，如果基类的某些函数是虚拟的，则派生类中重写基类中的函数的函数也是虚拟的。我不知道析构函数也是如此。因为我使用 virtual 函数说明符声明了 ~A()，所以 ~B() 是虚拟的吗？

原文

Say we have the following:

#include <iostream>

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

struct B : A
{
    // no need to be virtual?
    ~B() { std::cout << "destr B\n"; }
};

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

Now, I create an instance of C and assign it to a pointer of its base class B.

int main()
{
    B* b = new C{};
    delete b;
    return 0;
}

The output is:

destr C
destr B
destr A

What surprised me a little is that the object gets destroyed correctly (all three destructors are called). According to the output, I would say ~B() is a virtual destructor: ~B() dispatches to ~C(), then ~B() finishes and finally, ~A() is called. Is this statement correct?

I know that, if some function of a base class is virtual, the function in the derived class which overrides the one in the base class is virtual as well. I did not know that was true for destructors as well. Is ~B() virtual because I declared ~A() with the virtual function specifier?

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