Question

考虑代码

struct Base{};
struct Derived: public Base{};

struct A: public Base{};

struct B: public A, public Base{};

struct C: public A, public Derived{}; // why no ambiguity here?

int main() {}

编译器（g ++ 5.1）警告

警告：由于含糊不清'Base'，'B'无法访问基座struct B: public A, public Base{};

我理解这一点，Base在B中重复。

为什么C没有警告？ C和A都继承Derived并且都继承自Base？
为什么要添加virtual
```
struct Derived: virtual Base{};
```

> Wandbox

警告：由于含糊不清B，C无法访问基座'Base'

警告：由于含糊不清'B'，struct B: public A, public Base{};无法访问基座'Base'

Answer 1

在B中，无法直接引用Base子对象的成员。考虑：

struct Base {
    int x;
};

struct B: public A, public Base {
    void foo() {
        int& x1 = A::x; // OK
        int& x2 = x; // ambiguous
        // no way to refer to the x in the direct base
    }
};

在C这不是问题。 x＆gt;都可以使用限定名称来引用：

struct C: public A, public Derived {
    void foo() {
        int& x1 = A::x; // OK
        int& x2 = Derived::x; // OK
    }
};

所以你得到的警告只有当 direct 基地也通过另一条路径继承时才有意义。

对于你的第二个问题，我无法在Coliru上使用C使用g ++ - 5.1重现警告。

Answer 2

没有办法明确地访问“B”中的Base成员，而在“C”中则可以，如下面的代码所示：

#include <iostream>

using namespace std;

struct Base
{
    void print()
    {
        cout << "Base" << endl;
    }
};

struct Derived : public Base {};

struct A : public Base
{
    void print()
    {
        cout << "A" << endl;
    }
};

struct B : public A, public Base
{
    void print()
    {
        A::print();

        //error (ambiguous), no way to access to Base::print => warning
        //Base::print();
    }
};

struct C : public A, public Derived
{
    void print()
    {
        A::print();
        Derived::print(); // Not Ambiguous, it's the Base inherited by 'Derived' which is used.
        // Still an error but you can access print indirectly through "Derived" => no warning needed
        //Base::print();
    }
};

int main() 
{
    B b;
    b.print();

    C c;
    c.print();

    return 0; 
}

多继承模糊基类

2 个答案: