c++ - 了解 N3690 草案中的 §11.2-4

Question

在 C++ N3690 草案中。

A base class B of N is accessible at R, if

• an invented public member of B would be a public member of N, or
• R occurs in a member or friend of class N, and an invented public member of B would be a private or protected member of N, or
• R occurs in a member or friend of a class P derived from N, and an invented public member of B would be a private or protected member of P, or
• there exists a class S such that B is a base class of S accessible at R and S is a base class of N accessible at R.

有人可以为我提供上述语句的代码吗？

我试过这样做。

class B {
public:
    int m;
};

class S: private B {
    N r;
};

class N: private S {
    void f() {
        B* p = this;
    }
};

int main()
{
    return 0;
}

Answer 1

您的N 不是S 类的成员或好友。你的 r 是，但它不是 N，它只有 N 类型。

使其成为成员:

struct B {
    int m;
};

struct S : private B {
    struct N;
};

struct S::N : private S {
    void f() {
        B* p = this;
    }
};

让它成为 friend :

struct B {
    int m;
};

struct S : private B {
    friend struct N;
};

struct N : private S {
    void f() {
        B* p = this;
    }
};