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是否可以为复制基的虚拟函数提供不同的定义?

virtual-functions overriding multiple-inheritance c++

0

pasha · 技术社区 · 10 年前

我知道该声明, “复制基类的虚拟函数可以被派生类中的(单个)函数重写。” 我很想知道是否可以提供一个以上的定义,以防正确定义和调用?

为了更清楚, 我的类和类层次结构如下

     A  A
     |  |
     B  C
     \ /
      D

//i.e., I am not deriving B and C from a virtual base A.

class A {
    virtual void f(){
         cout<<"belongs to A"<<endl;
    }
}
class B: public A {
    void f(){
         cout<<"belongs to B"<<endl;
    }
}
class C: public A {
    void f(){
         cout<<"belongs to C"<<endl;
    }
}
class D: public B, public C {
    void f(){  //This overrides definitions B::f() and C::f()
         cout<<"belongs to D"<<endl;
    }
}
B* b;
C* c;
D d;
b = &d;
c = &d;
b->f();//This would output "belongs to D"
c->f();//This also would output "belongs to D"

//Now I want to redefine f() in D like this
class D: public B, public C {
    void B::f() {
         cout<<"belongs to D::B::f()"<<endl;
    }
    void C::f() {
         cout<<"belongs to D::C::f()"<<endl;
    }
}
//So that 
B* b;
C* c;
D d;
b = &d;
c = &d;
b->f();//This would output "belongs to D::B::f()"
c->f();//This also would output "belongs to D::C::f()"

3 回复 | 直到 10 年前

1

2

Guillaume Gris 10 年前

在某些时候,您需要能够告诉编译器您要使用哪个实现。

这样做的一种方法是:

  A
 / \
B   C
 \ /
  D

struct A
{
  virtual void f();
};

struct B : public A
{
  void f() override;
};

struct C : public A
{
  void f() override;
};

struct D : public B,C
{
};

int test(int num) {

  D d;
  d.f();    // undefined
  d.A::f(); // Impossible because conversion from D to A is ambiguous
  d.B::f(); // calls B implementation
  d.C::f(); // calls C implementation
}

2

0

aliasm2k 10 年前

下面是一个虚拟函数的基本示例

class Person
{
    int age;
    public:
        Person(int=0);
        virtual void describe() = 0;
        int getAge();
};

Person :: Person(int age)
{
    this->age = age;
}

int Person :: getAge()
{
    return this->age;
}

class Student : public Person
{
    public:
        Student(int=0);
        void describe();
};

Student :: Student(int age) : Person(age) {}
void Student :: describe() 
{
    std::cout << "Hi, I am a student, ";
    std::cout << this->getAge() << " years old!";
}

class Employee : public Student
{
    public:
        Employee(int=0);
        void describe();
};

Employee :: Employee(int age) : Person(age) {}
void Employee :: describe()
{
    std::cout << "Hi, I am an employee, ";
    std::cout << this->getAge() << " years old!";
}

3

0

pasha 10 年前

简短的答案是不 .好奇的家伙

答案由@curiousguy在评论中提供。