如何访问多个级别的覆盖元素？

发布于 2024-09-14 00:17:15 字数 827 浏览 6 评论 0原文

如何访问基类的基类的重写成员？

#include <iostream>
using namespace std;

class A {public: char x; A(){x='A';};};
class B1 : public A {public: char x; B1(){x='B';};};
class B2 : public A {public: char x; B2(){x='B';};};
class C : public B1, public B2 {public: char x; C(){x='C';};};

int main(){
    C c;
    cout << c.x << endl; // prints C

    cout << c.B1::x << endl; // prints B
    cout << ((B1&) c).x << endl; // prints B

    // cout << c.A::x << endl; // normally prints A but doesn't work here
    cout << ((A&) c).x << endl; // prints A
    return 0;
}

引用（或指针）方式是唯一的可能性吗？我尝试了 A::B::x 来链接作用域运算符，但这不起作用。假设我想保留“双A”，即不使继承虚拟。

((B&) c).x 似乎是 cB::x 的一个很好的“解决方法”，但它们在虚函数的情况下并不相等，是吗？？

原文

How do I access overridden members of base classes of base classes?

#include <iostream>
using namespace std;

class A {public: char x; A(){x='A';};};
class B1 : public A {public: char x; B1(){x='B';};};
class B2 : public A {public: char x; B2(){x='B';};};
class C : public B1, public B2 {public: char x; C(){x='C';};};

int main(){
    C c;
    cout << c.x << endl; // prints C

    cout << c.B1::x << endl; // prints B
    cout << ((B1&) c).x << endl; // prints B

    // cout << c.A::x << endl; // normally prints A but doesn't work here
    cout << ((A&) c).x << endl; // prints A
    return 0;
}

Is the reference (or pointer) way the only possibility?
I tried A::B::x in order to chain scope operators, but that doesn't work.
Suppose i want to keep the "double A", i.e. not make the inheritance virtual.

((B&) c).x seems to be a good "workaround" to c.B::x but they aren't equal in case of virtual functions, are they?

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凡间太子 2024-09-21 00:17:15

您可以通过引用强制转换加上作用域运算符来访问 A 的两个版本，如下所示： ((B1&) c).A::x 和 ((B2&) c).A ::x。

cout << ((A&)c).x << endl; 在我的编译器上失败，因为编译器不知道您要操作 A 数据的哪个副本。

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单身情人 2024-09-21 00:17:15

c.A::x

c.A::x

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伏妖词 2024-09-21 00:17:15

它不会工作，编译器应该告诉你原因。至少我的:-)：“C 的基础不明确”。答案是虚拟继承

#include <iostream>
using namespace std;

class A {public: char x; A(){x='A';};};
class B1 : public virtual A {public: char x; B1(){x='B';};};
class B2 : public virtual A {public: char x; B2(){x='B';};};
class C : public B1, public B2 {public: char x; C(){x='C';};};

int main(){
    C c;
    cout << c.x << endl; // prints C

    cout << c.B1::x << endl; // prints B
    cout << ((B1&) c).x << endl; // prints B

    cout << c.A::x << endl; // normally prints A but doesn't work here
    //cout << ((A&) c).x << endl; // prints A
    return 0;
}

It won't work and the compiler should tell you why. At least mine does :-) : "A ambigous base for C". The answer is virtual inheritance

#include <iostream>
using namespace std;

class A {public: char x; A(){x='A';};};
class B1 : public virtual A {public: char x; B1(){x='B';};};
class B2 : public virtual A {public: char x; B2(){x='B';};};
class C : public B1, public B2 {public: char x; C(){x='C';};};

int main(){
    C c;
    cout << c.x << endl; // prints C

    cout << c.B1::x << endl; // prints B
    cout << ((B1&) c).x << endl; // prints B

    cout << c.A::x << endl; // normally prints A but doesn't work here
    //cout << ((A&) c).x << endl; // prints A
    return 0;
}

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