Question

鉴于代码：

#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;
}

有两个钻石继承事件。我希望在所有基础上访问指定的数据成员i。如何获得访问权限？示例中的代码产生错误：

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.

Live example

编译器是 clang 3.7.0 。

Answer 1

这是一个相当混乱的类层次结构。我希望你不打算在真实的应用程序中使用它。

这是绕过障碍的一种方法：

// 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);

数据成员访问歧义和钻石继承

1 个答案: