用于在多重继承中覆盖函数的语法

Question

class A
{
   protected:
    void func1() //DO I need to call this virtual?
};
class B
{
   protected:
    void func1() //and this one as well?
};
class Derived: public A, public B
{
    public:
    //here define func1 where in the code there is
    //if(statement) {B::func1()} else {A::func1()}
};

你如何覆盖 func1？或者你可以只定义它

class Derived: public A, public B
{
    public:
    void func1()
};

没有任何虚拟或覆盖？我不明白可访问性。谢谢。

Answer 1

Leonard Lie，

要

重写，您可以简单地声明具有相同名称的函数，要实现代码注释中的功能，您需要将变量传递给 派生的func1（）

例如：

#include <iostream>
using namespace std;
class A
{
   protected:
    void func1()    {   cout << "class An";    } //DO I need to call this virtual?
};
class B
{
   protected:
    void func1()    {   cout << "class Bn";    } //and this one as well?
};
class Derived: public A, public B
{
    public:
    //here define func1 where in the code there is
    //if(statement) {B::func1()} else {A::func1()}
    void func1(bool select = true)
    {
        if (select == true)
        {
            A::func1();
        }
        else
        {
            B::func1();
        }
    }
};
int main()
{
   Derived d;
   d.func1();          //returns default value based on select being true
   d.func1(true);      //returns value based on select being set to true
   d.func1(false);     // returns value base on select being set to false
   cout << "Hello World" << endl; 
   return 0;
}

这应该可以满足您的需求，我使用了布尔值，因为只有 2 个可能的版本，但您可以使用 enum 或 int 来适应具有更多选项的情况。