c++ - 析构函数可以是最终的吗？

Question

C++ 标准是否允许将析构函数声明为 final？像这样:

 class Derived: public Base
 {
      ...
      virtual ~Derived() final;
 }

如果是这样，那是否会阻止派生类的声明:

 class FurtherDerived: public Derived {// allowed?
 }

如果允许，编译器是否可能发出警告？将析构函数声明为 final 是否是一个可行的习惯用法，用于指示一个类不打算用作基类？

(有no point in doing this in a ultimate base class，只有派生类。)

Answer 1

May a C++ destructor be declared as final?

是的。

And if so, does that prevent declaration of a derived class:

是的，因为派生类必须声明一个析构函数(由您显式或由编译器隐式)，并且该析构函数将覆盖声明为 final 的函数，该函数格式错误.

规则是[class.virtual]/4 :

If a virtual function f in some class B is marked with the virt-specifier final and in a class D derived from B a function D​::​f overrides B​::​f, the program is ill-formed.

这是不正确的推导本身，它不必使用。

Is declaring a destructor to be final a workable idiom for indicating that a class is not intended to be used as a base class?

有效，但您应该只标记类 final。它更明确一些。