具有混合虚拟和非虚拟基础的 C++11 类格？

Question

在 C++11 (N3485) 10.1.4 [class.mi] 中它说:

For each distinct occurence of a non-virtual base class in the class lattice of the most derived class, the most derived object shall contain a corresponding distinct base class subobject of that type.

For each distinct base class that is specified virtual, the most derived class shall contain a single base class object of that type.

考虑以下 C++11 代码:

struct B {};

struct BV : virtual B {};
struct BN : B {};

struct C1 : BV, BN {};
struct C2 : BV, BN {};

struct D : C1, C2 {};

首先，为了清楚起见，D的类格有多少个顶点？

其次，标准要求D类型的最派生对象有多少个不同的B类型子对象？

更新:

下面哪个是类格？

(1)

    B     B     B    B
    ^     ^     ^    ^
    |     |     |    |
    BV    BN    BV   BN
    ^     ^     ^    ^
    |     |     |    |
     \   /       \  /
       C1         C2
         \        /
          \      /
           -  D -

(2)

    B<---------
    ^          \
    |           |
    |     B     |    B
    |     ^     |    ^
    |     |     |    |
    BV    BN    BV   BN
    ^     ^     ^    ^
    |     |     |    |
     \   /       \  /
       C1         C2
         \        /
          \      /
           -  D -

(3)

       B   
      /  \     
     /    \ 
    BV    BN
    | \  / |
    |  \/  |
    |  / \ |
    | /   \|
    C1     C2
     \    /
      \  /
       D

如果意图是 (1) 那么是否不可能有任何不是树的 DAG？ (即钻石是不可能的)如果是这样，将它称为类树不是更好吗？

如果是 (2) 是否足以说“对于类格中基类的每次出现，都有一个相应的基类子对象”？也就是说，如果格子的构建已经依赖于虚拟和非虚拟基类关系来选择边和顶点？

如果是 (3) 那么标准中的语言是不是不正确，因为类格中的类只能出现一次？

Answer 1

Which of the following is the class lattice?

2

演示:

#include <iostream>

struct B {};

struct BV : virtual B {};
struct BN : B {};

struct C1 : BV, BN {};
struct C2 : BV, BN {};

struct D : C1, C2 {};

int
main()
{
    D d;
    C1* c1 = static_cast<C1*>(&d);
    BV* bv1 = static_cast<BV*>(c1);
    BN* bn1 = static_cast<BN*>(c1);
    B* b1 = static_cast<B*>(bv1);
    B* b2 = static_cast<B*>(bn1);
    C2* c2 = static_cast<C2*>(&d);
    BV* bv2 = static_cast<BV*>(c2);
    BN* bn2 = static_cast<BN*>(c2);
    B* b3 = static_cast<B*>(bv2);
    B* b4 = static_cast<B*>(bn2);
    std::cout << "d = " << &d << '\n';
    std::cout << "c1 = " << c1 << '\n';
    std::cout << "c2 = " << c2 << '\n';
    std::cout << "bv1 = " << bv1 << '\n';
    std::cout << "bv2 = " << bv2 << '\n';
    std::cout << "bn1 = " << bn1 << '\n';
    std::cout << "bn2 = " << bn2 << '\n';
    std::cout << "b1 = " << b1 << '\n';
    std::cout << "b2 = " << b2 << '\n';
    std::cout << "b3 = " << b3 << '\n';
    std::cout << "b4 = " << b4 << '\n';
}

我的输出:

d = 0x7fff5ca18998
c1 = 0x7fff5ca18998
c2 = 0x7fff5ca189a0
bv1 = 0x7fff5ca18998
bv2 = 0x7fff5ca189a0
bn1 = 0x7fff5ca18998
bn2 = 0x7fff5ca189a0
b1 = 0x7fff5ca189a8
b2 = 0x7fff5ca18998
b3 = 0x7fff5ca189a8
b4 = 0x7fff5ca189a0

If it is (2) wouldn't it be sufficient to say "for each occurence of a base class in the class lattice there is a corresponding base class subobject" ?. That is, if the construction of the lattice already depends on virtual and non-virtual base class relationships to select edges and vertices?

合并您的建议...

A base class specifier that contains the keyword virtual, specifies a virtual base class. For each distinct occurrence of a non-virtual base class in the class lattice of the most derived class, the most derived object (1.8) shall contain there is a corresponding distinct base class subobject of that type. For each distinct base class that is specified virtual, the most derived object shall contain a single base class subobject of that type.

我不是标准语言一半的专家。但是，当我阅读您修改后的规范时，我看不出如何:

class V { /∗...∗/ };
class A : virtual public V { /∗ ... ∗/ };
class B : virtual public V { /∗ ... ∗/ };
class C : public A, public B { /∗...∗/ };

图 4 中的结果:

   V
  / \
 /   \
A     B
 \   /
  \ /
   C

我没有在标准中看到另一个地方指定虽然 V 在 C 下面的类层次结构中出现了两次，但只有一个 V 类型的子对象 实际存在是因为使用了 virtual 关键字。