超多重非虚继承中基类的作用域运算符 [英] Scope operator for base class in super multiple non-virtual inheritance

问题描述

考虑这个(完全没有意义，但完全有效)类继承:

struct Area { int size;};结构模式 { 整数大小；};结构 R :区域，模式{}；结构 C:区域，模式 {}；结构 X: R , C {};

让我们看看这个伟大层次结构的图表:

区域模式|\/||\/||/\ ||/\|电阻率\/\/X

现在，如果我没记错的话，X 应该有 4 个 size 成员.

如何使用作用域运算符来引用它们?

显而易见的解决方案不起作用:

X x;x.R::Area::size = 24;

clang 错误:

<块引用>

23 : :23:3: 错误:从派生类X"到基类Area"的不明确转换:结构 X ->结构 R ->结构区结构 X ->结构 C ->结构区x.R::Area::size = 8;^产生了 1 个错误.

gcc 错误:

<块引用>

:在函数auto test()"中:23 : <source>:23:14: 错误: 'Area' 是'X' 的不明确基数x.R::Area::size = 8;^~~~

<小时>

一些急需的澄清:

• 我只是在胡闹，这不是真正的设计

  • 所以请不要指出设计的问题
  • 请不要认为这是一个好的设计.这是...不是 - 至少可以说

• 这完全是关于解决歧义的 C++ 语法.

  • 请不要建议不要这样做
  • 请不要建议虚拟继承.

解决方案

类似于 static_cast(x).Area::size = 8;

它应该是丑陋的:)

为了阐明为什么原始代码不起作用，值得一提的是，合格的 id 具有以下形式(除其他外)type-name::id 所以 xR::Area::y 等价于 使用 T = R::Area;x.T::y; 就消除歧义而言，这显然无济于事.

Consider this (completely non-nonsensical, but perfectly valid) class inheritance:

struct Area { int size; };
struct Pattern { int size; };

struct R : Area, Pattern {};
struct C : Area, Pattern {};

struct X: R , C {};

Let's see a graph of this great hierarchy:

Area  Pattern
  |\  /|   
  | \/ |
  | /\ |  
  |/  \|
  R    C
   \  /
    \/
    X

Now, if I am not mistaken, X should have 4 size members.

How to refer to them using the scope operator?

The obvious solution doesn't work:

X x;
x.R::Area::size = 24;

clang error:

23 : <source>:23:3: error: ambiguous conversion from derived class 'X' to base class 'Area':
    struct X -> struct R -> struct Area
    struct X -> struct C -> struct Area
  x.R::Area::size = 8;
  ^
1 error generated.

gcc error:

<source>: In function 'auto test()':
23 : <source>:23:14: error: 'Area' is an ambiguous base of 'X'
   x.R::Area::size = 8;
              ^~~~

---

Some much needed clarification:

• I was just messing around, it is not a real design

  • so please don't point the problems with the design
  • and please don't think this is a good design. It is... not - to say the least

• This is strictly about the C++ syntax to resolve the ambiguity.

  • please don't suggest to not do it
  • please don't suggest virtual inheritance.

解决方案

something like static_cast<R&>(x).Area::size = 8;

which is as ugly as it should be :)

To clarify why the original code doesn't work, it's worth mentioning that a qualified id has the form (among others) type-name::id so x.R::Area::y is equivalent to using T = R::Area; x.T::y; that clearly does not help as far as disambiguation is concerned.

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