Question

我在这里碰到了这种棘手的情况所以基本上我被要求编写一个函数，如果我点击的点位于图中，则返回指向图的指针，如果该点不在任何位置，则返回null图。

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;
    }

}

正如您所看到的，我还没有决定要返回什么，但我的问题是在这里使用动态转换最佳解决方案？

-CLine，CTriangle，CRectangle和CCircle都是来自CFigure

的派生类

Answer 1

在课程CFigure中添加

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

这是一个纯虚函数。 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。

将ApplicationManager::GetFigure缩减为

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魔力，程序将确定需要调用哪个子类的isclicked函数，而不需要您付出更多努力。

Answer 2

您可以使用虚函数将处理移动到每个派生类型，而不是测试类型以决定如何处理它们。

参见 Virtual Functions

而不是这样做：

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 = ...
    }
}

您的目标应该是让每个子类型都知道如何计算自己：

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
}

有没有更好的替代方案[转发]

2 个答案: