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c++ - 数据成员访问歧义和菱形继承

转载 作者:搜寻专家 更新时间:2023-10-31 02:19:59 24 4
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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(钻石问题)事件。我想访问所有库中的指定数据成员 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

最佳答案

这是一个相当困惑的类层次结构。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);

关于c++ - 数据成员访问歧义和菱形继承,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33220947/

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