c++ - 编译器真的强制执行纯虚拟析构函数吗？

Question

为了验证声明“编译器和链接器强制存在纯虚拟析构函数的函数体。”来自 geeksforgeeks article ，我编译了这段代码:

class Base
{
public:
    virtual ~Base()=0; // Pure virtual destructor
};

class Derived : public Base
{
public:
    ~Derived()
    {
        std::cout << "~Derived() is executed";
    }
};

int main()
{
    //Derived d;   <<<
    return 0;
}

编译没有任何错误。那么为什么在这种情况下编译器没有选择强制函数体的存在呢？

Answer 1

因为如果您执行 ODR¹ 违规，编译器(实际上是整个翻译过程)没有强制执行任何操作。根据 [basic.def.odr/4] 处的 C++ 标准:

Every program shall contain exactly one definition of every non-inline function or variable that is odr-used in that program; no diagnostic required. The definition can appear explicitly in the program, it can be found in the standard or a user-defined library, or (when appropriate) it is implicitly defined (see [class.ctor], [class.dtor] and [class.copy]). An inline function shall be defined in every translation unit in which it is odr-used.

编译器完全有权判断您的程序实际上没有使用² Derived 的析构函数(因此是 Base 的析构函数)，只是懒得通知你。

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<子>¹ 一个一个定义定义规则规则
² What does it mean to “ODR-use” something?