正确的方式继承从一个虚拟类与非虚拟父 [英] Correct way to inherit from a virtual class with non-virtual parent

问题描述

我写了这个测试代码，使用三种类型： struct One 是一个没有虚拟成员的普通类型， struct Two：One 有一个纯虚函数和一个虚析构函数， struct Three：Two implements 两个

I've written this test code that uses three types: struct One is a normal type with no virtual members, struct Two : One has a pure virtual function and a virtual destructor, and struct Three : Two implements Two's interface.

#include <iostream>

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

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

    virtual void test() = 0;
};

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

    virtual void test() {
        std::cout << "Three::test()\n";
    }
};

int main()
{
    Two* two = new Three;
    two->test();

    One* one = two;
    delete one;
}

不出所料，输出为

Three :: test（）

〜One（）

Three::test()
~One()

有没有办法解决这个问题，而不是使每个析构函数虚拟？或者应该程序员只是小心不要遇到这种情况？我发现很奇怪，编译这个时没有警告。

Is there any way to fix this other than making every destructor virtual? Or should programmers just be careful not to run into this situation? I find it odd that there's no warning when compiling this.

推荐答案

唯一的修复不是通过指向 c $ c>。

The only "fix" is not to delete the objects through a pointer to One.

如果这是一个常见的问题，或者不是，这取决于你的类是如何使用的。例如，标准库包含没有虚拟析构函数的 unary_function 结构，但我们几乎看不到这样的误用。

If this is a frequent problem, or not, depends on how your classes are used. For example, the standard library contains structs like unary_function without a virtual destructor, but we hardly ever see it misused like this.

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