为什么在随后的私人派生类中简单提及基指针/引用被禁止？ [英] Why a simple mention of base pointer/reference in subsequent privately derived classes is forbidden?

问题描述

struct B {};
struct D : private B {
  B* fun () { return new D; }  // ok
}
struct DD : public D {
  B* foo () { return 0; } // error: ‘struct B B::B’ is inaccessible !
};

这个错误对我来说似乎不合理。如果我们可以在全局范围中使用简单 B * ，那么为什么不在其私有派生类？ g ++ demo 。

This error seems unreasonable to me. If we can use simple B* in global scope then why not in its privately derived classes? g++ demo.

我们不会尝试转换 DD * 到 B * ，这是语言规则禁止的（ ，此，此是相关问题）。

请注意，如果我将 B * foo（）更改为 int foo（）好的。

We are Not trying to convert DD* to B*, which is forbidden by the language rules (this, this, this are related questions).
Note that, if I change B* foo() to int foo(), things go fine.

推荐答案

显然，编译器认为 B B 的私人构造函数。

So apparently the compiler thinks B is referring to the private constructor of B rather than the type.

符合 B 显然修正了这个错误：

Qualifying B apparently fixes that error:

class B* foo () { return 0; }

或此：

::B* foo () { return 0; }

我不知道为什么会发生，但也许这会有帮助。

I don't know why that's happening, but maybe this will help.

更新：也许它与11.2.4的标准相关？唯一的问题是我的标准版不够好，不能完全理解。

Update: maybe it's related to 11.2.4 of standard? The only problem is that my standardese isn't good enough to fully understand it.

（对不起图片，复制/粘贴不适用于我）

(sorry for the image, copy/pasting isn't working for me)

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