Question

给出以下代码：
（这主要是关于Function（）方法中发生的事情，其余只是设置/上下文。）

enum class class_type { a, b };

class Base {
   public:
    Base(class_type type) : type(type) {}

    class_type type;
};

class DerivedA : public Base {
   public:
    DerivedA() : Base(class_type::a) {}

    void FunctionA() {}
};

class DerivedB : public Base {
   public:
    DerivedB() : Base(class_type::b) {}

    void FunctionB() {}
};

void Function(Base& base) {
    switch (base.type) {
        case class_type::a: {
            DerivedA& temp = (DerivedA&)base; // Is this the best way?
            temp.FunctionA();
            break;
        }
        case class_type::b: {
            base.FunctionB(); // This obviously doesn't work.
            break;
        }
    }
}

int main() {
    DerivedA derived_class;
    Function(derived_class);
}

我在DerivedA上做这件事的方式是最好/最有效的方式吗？我觉得有更好的方法可以做到这一点，但我不知道怎么做。

Answer 1

答案是您不这样做，它完全由多态处理，请阅读以下代码：

并尝试将其映射到您的代码：

夏普是您的基础
矩形是您的 DerivedA
三角形是您的 dB

#include <iostream> 
using namespace std;

class Shape {
   protected:
      int width, height;

   public:
      Shape( int a = 0, int b = 0){
         width = a;
         height = b;
      }
      int area() {
         cout << "Parent class area :" <<endl;
         return 0;
      }
};
class Rectangle: public Shape {
   public:
      Rectangle( int a = 0, int b = 0):Shape(a, b) { }

      int area () { 
         cout << "Rectangle class area :" <<endl;
         return (width * height); 
      }
};

class Triangle: public Shape {
   public:
      Triangle( int a = 0, int b = 0):Shape(a, b) { }

      int area () { 
         cout << "Triangle class area :" <<endl;
         return (width * height / 2); 
      }
};

// Main function for the program
int main() {
   Shape *shape;
   Rectangle rec(10,7);
   Triangle  tri(10,5);

   // store the address of Rectangle
   shape = &rec;

   // call rectangle area.
   shape->area();

   // store the address of Triangle
   shape = &tri;

   // call triangle area.
   shape->area();

   return 0;
}

Answer 2

这足以满足您的需求：

DerivedA& temp = static_cast<DerivedA&>(base);
temp.FunctionA();

内幕下，这与您刚进行的C样式强制转换相同，但将其显式表示是一种很好的做法。

调用不属于基类的派生类的函数的最佳方法是什么？

2 个答案: