数据成员访问歧义和菱形继承

Data member access ambiguity and diamond inheritance

本文关键字：继承 访问 歧义 数据成员      更新时间：2023-10-16

给定代码:

#include <cassert>
#include <cstdlib>
int
main()
{
    struct A { int i; A(int j) : i(j) { ; } };
    struct E { int i = 3; };
    struct B : A, E { using A::A; };
    struct C : A, E { using A::A; };
    struct D : B, C { D(int i, int j) : B{i}, C{j} { ; } };
    D d{1, 2};
    assert(d.B::A::i == 1);
    assert(d.C::A::i == 2);
    assert(d.B::E::i == 3);
    assert(d.C::E::i == 3);
    return EXIT_SUCCESS;
}

有两个菱形继承发生。我想访问所有基地的特定数据成员i。如何获得访问权限?示例中的代码产生一个错误:

main.cpp:13:12: error: ambiguous conversion from derived class 'D' to base class 'A':
    struct D -> struct B -> struct A
    struct D -> struct C -> struct A
    assert(d.B::A::i == 1);
           ^
/usr/include/assert.h:92:5: note: expanded from macro 'assert'
  ((expr)                                                               
    ^
main.cpp:14:12: error: ambiguous conversion from derived class 'D' to base class 'A':
    struct D -> struct B -> struct A
    struct D -> struct C -> struct A
    assert(d.C::A::i == 2);
           ^
/usr/include/assert.h:92:5: note: expanded from macro 'assert'
  ((expr)                                                               
    ^
main.cpp:15:12: error: ambiguous conversion from derived class 'D' to base class 'E':
    struct D -> struct B -> struct E
    struct D -> struct C -> struct E
    assert(d.B::E::i == 3);
           ^
/usr/include/assert.h:92:5: note: expanded from macro 'assert'
  ((expr)                                                               
    ^
main.cpp:16:12: error: ambiguous conversion from derived class 'D' to base class 'E':
    struct D -> struct B -> struct E
    struct D -> struct C -> struct E
    assert(d.C::E::i == 3);
           ^
/usr/include/assert.h:92:5: note: expanded from macro 'assert'
  ((expr)                                                               
    ^
4 errors generated.
<<p> 生活例子/kbd>

编译器是clang 3.7.0.

这是一个相当混乱的类层次结构。我希望您不打算在实际应用程序中使用它。

这里有一个绕过障碍的方法:

// Get references to the B and C parts of D.
B& b = d;
C& c = d;
// Now you can get the A::i and the E::i.
assert(b.A::i == 1);
assert(c.A::i == 2);
assert(b.E::i == 3);
assert(c.E::i == 3);
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