无法访问的类型由于私有继承 [英] Inaccessible type due to private inheritance

问题描述

g ++ 是拒绝我访问类型，只是因为它恰好是一个私人的父亲。这是否有意义？

g++ is denying me access to a type, just because it happens to be a private grand-father. Does this make sense?

struct A {};

struct B : private A {};

struct C : B {
  void foo(A const& a) {}
};

编译这会产生：

1:10: error: ‘struct A A::A’ is inaccessible
6:12: error: within this context

我的观点是：我从来不想访问 A 作为祖先。事实上，如果 A 是 B 的私有祖先，则不应该对任何人完全不可见， c $ c> B （即 C ）？

My point is: I never wanted to access A as an ancestor. In fact, if A is a private ancestor of B, shouldn't this be completely invisible to anybody but B (i.e. C)?

使用 protected 继承，但在我的例子中它没有什么意义。

Of course, I could use protected inheritance but in my case it doesn't really make sense.

推荐答案

这是因为注入的类名隐藏中的全局 A C 。虽然 A 是可见的，它是不可访问的（因为它作为私有导入），因此错误。您可以通过在全局命名空间中查找 A 来访问

This is due to the injected class name from A hiding the global A inside C. Although A is visible, it is not accessible (since it is imported as private), hence the error. You can access A by looking it up in the global namespace:

void foo(::A const& a) {}

这篇关于无法访问的类型由于私有继承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！