覆盖在非虚拟函数 [英] override on non-virtual functions

问题描述

在C ++ 11 FDIS它说：

  

如果一个虚函数标有的virt-符覆盖和不覆盖的成员函数
  一个基类，是病态的程序。 [示例：

 结构B {
    虚拟无效F（INT）;
};
结构D：B {
    无效F（长）覆盖; //错误：错误的签名重写B ::˚F
    无效F（INT）覆盖; // 好
};
 

如果 B ::˚F就不会被标记虚拟什么？是计划不周形成的，然后呢？或者是覆盖然后将ignored`。我找不到这种情况下，在性病的任何文本处理。

更新1/2 （合并）我转交了对C ++编辑的要求寻找到的东西。谢谢约翰内斯的要指出这一点给我。

• 无效F（长）覆盖不会覆盖功能，ESP。没有虚拟的，


• 因此不虚


• 所以文若一个虚函数标有......并不适用


• 因此，例如不匹配文本。

但是，意识到这一点，我发现，认为越权上下文关键字的意图无法得到满足：如果函数名称的拼写错误或错误的参数类型确实让本身的功能非虚，那么标准的文本永远不会应用 - 与覆盖，也不起作用。

最佳的解决方案可能是

• 将虚拟中的例子的功能前

解决方案

  

如果 B ::˚F不会被虚标什么？在程序不当形成的，然后呢？

是的，是的。因为为了 覆盖的的东西，这东西的具有的是虚的。否则，它不是 覆盖的，它的的隐藏的。所以，肯定的回答从你的问题如下报价

The C++11 FDIS it says

If a virtual function is marked with the virt-specifier override and does not override a member function of a base class, the program is ill-formed. [ Example:

struct B {
    virtual void f(int);
};
struct D : B {
    void f(long) override; // error: wrong signature overriding B::f
    void f(int) override; // OK
};

What if B::f would not have been marked virtual? Is the program ill-formed, then? Or is override then to be ignored`. I can not find any handling of this case in the std text.

Update 1/2 (merged) I forwarded a request to the C++ Editors to look into things. Thanks Johannes to pointing that out to me.

• "void f(long) override" does not override a function, esp. no virtual one,
• therefore it is not virtual
• therefore the text "If a virtual function is marked with..." does not apply
• therefore the example does not match the text.

But by realizing this I found, that the intention of the "override" contextual keyword can not be met: if a typo in the function name or the wrong argument type does make the function itself non-virtual, then the standard's text never applies -- and "override" is rendered useless.

The best possible solution may be

• putting "virtual" in front of the example's functions

解决方案

What if B::f would not have been marked virtual? Is the program ill-formed, then?

Yes, it is. Because in order to override something, that something has to be virtual. Otherwise it's not overriding, it's hiding. So, the positive answer follows from the quote in your question.

这篇关于覆盖在非虚拟函数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！