有没有这种情况[向下转换]

Question

我也碰到这个棘手的问题来到这里，所以基本上我要求写应该返回一个指向一个数字如果点我点击就在于人物和空值的函数没有更好的替代方案如果这一点不在任何数字。

CFigure *ApplicationManager::GetFigure(int x, int y) const 
{ 
    //If a figure is found return a pointer to it. 
    //if this point (x,y) does not belong to any figure return NULL 
    int c = 0; 
    for (size_t i = 0; i < FigCount; i++) 
    { 
     if (dynamic_cast<CRectangle*> (FigList[i])) 
     { 
      CFigure* basepointer = FigList[i]; 
      Point A = static_cast<CRectangle*>(basepointer)->GetCorner1(); 
      Point B = static_cast<CRectangle*>(basepointer)->GetCorner2(); 

      if ((x>=A.x && x<=B.x) || (x<=A.x && x>=B.x)) 
      { 
       if ((y >= A.y && x <= B.y) || (y <= A.y && x >= B.y)) 
       { 
        c++; 
       } 
      } 
     } 
     else if (dynamic_cast<CCircle*> (FigList[i])) 
     { 
      CFigure* basepointer = FigList[i]; 
      Point A = static_cast<CCircle*>(basepointer)->getCntr(); 
      int B = static_cast<CCircle*>(basepointer)->GetRadius(); 

      double distance = sqrt(pow((x - A.x), 2) + pow((y - A.y), 2)); 
      if (distance<=(double)B) 
      { 
       c++; 
      } 
     } 
     else if (dynamic_cast<CLine*> (FigList[i])) 
     { 
      CFigure* basepointer = FigList[i]; 
      Point A = static_cast<CLine*>(basepointer)->getPoint1(); 
      Point B = static_cast<CLine*>(basepointer)->getpoint2(); 
      double distance1 = sqrt(pow((x - A.x), 2) + pow((y - A.y), 2)); //Distance from point to P1 
      double distance2 = sqrt(pow((x - B.x), 2) + pow((y - B.y), 2)); //Distance from Point to P2 
      double distance3 = sqrt(pow((B.x - A.x), 2) + pow((B.y - A.y), 2)); //Distance from P1 to P2 
      if (distance1+distance2==distance3) 
      { 
       c++; 
      } 
     } 
     else 
     { 
      CFigure* basepointer = FigList[i]; 
      Point p1 = static_cast<CTriangle*>(basepointer)->getp1(); 
      Point p2 = static_cast<CTriangle*>(basepointer)->getp2(); 
      Point p3 = static_cast<CTriangle*>(basepointer)->getp3(); 
      float alpha = (((float)p2.y - (float)p3.y)*((float)x - (float)p3.x) + ((float)p3.x - (float)p2.x)*((float)y - (float)p3.y))/
       (((float)p2.y - (float)p3.y)*((float)p1.x - (float)p3.x) + ((float)p3.x - (float)p2.x)*((float)p1.y - (float)p3.y)); 
      float beta = (((float)p3.y - (float)p1.y)*((float)x - (float)p3.x) + ((float)p1.x - (float)p3.x)*((float)y - (float)p3.y))/
       (((float)p2.y - (float)p3.y)*((float)p1.x - (float)p3.x) + ((float)p3.x - (float)p2.x)*((float)p1.y - (float)p3.y)); 
      float gamma = 1.0f - alpha - beta; 
      if (alpha>0 && beta>0 && gamma >0) 
      { 
       c++; 
      } 
     } 
    } 

    ///Add your code here to search for a figure given a point x,y 
    if (c==0) 
    { 
     return NULL; 
    } 

} 

你可以看到，我还没有决定什么就又回来，但我的问题是这里使用动态转换的最佳解决方案？

-CLine，CTriangle，CRectangle和CCircle都是从CFigure

Answer 1

在课堂CFigure加

virtual bool isclicked(int x, int y) = 0; 

这是一个纯虚拟函数。 CFigure的所有子类都必须实现它。子类的实现检查点击是否在其边界内，并相应地返回true或false。

的减少ApplicationManager::GetFigure喜欢的东西

CFigure *ApplicationManager::GetFigure(int x, int y) const 
{ 
    for (size_t i = 0; i < FigCount; i++) 
    { 
     if (FigList[i]->isclicked(x,y)) 
     { 
      return FigList[i]; 
     } 
    } 
    return nullptr; 
} 

通过虚函数和多态的神奇，程序会找出子类的isclicked功能需要您进行任何进一步的努力，被称为其。

Answer 2

可以使用虚函数的处理进入每一个派生类，而不是测试的类型来决定如何处理它们派生类。

见Virtual Functions

而不是做这个的：

struct B {}; 
struct D1: B {}; 
struct D2: B {}; 

// ... 

void func(B* b) 
{ 
    int c = 0; 

    if(dynamic_cast<D1*>(b)) 
    { 
     // do D1 stuff 
     c = ... 
    } 
    else if(dynamic_cast<D2*>(b)) 
    { 
     // do D2 stuff 
     c = ... 
    } 
} 

你的目标应该是让每个子类知道如何计算自己：

struct B { virtual int calc() = 0; }; // virtual function calls derived type 
struct D1: B { int calc() override { int c = 0; /* D1 calculation */ return c; } }; 
struct D2: B { int calc() override { int c = 0; /* D2 calculation */ return c; } }; 

// ... 

void func(B* b) 
{ 
    int c = b->calc(); // virtual means the correct type's function is used 
}