我在这里遇到过这种棘手的情况,因此基本上我被要求编写一个函数,如果我单击的点位于图形中,则该函数应返回指向图形的指针,如果该点不位于任何图形中,则返回null。
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;
}
}
如您所见,我还没有决定要返回什么,但是我的问题是在这里使用动态投射最佳解决方案?
-CLine,CTriangle,CRectangle和CCircle都是从CFigure派生的类
最佳答案
在CFigure
类中添加
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。子类的实现检查单击是否在其边界内,并相应地返回true或false。将
ApplicationManager::GetFigure
减少到类似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;
}
借助虚拟函数和多态性的魔力,该程序可以弄清楚需要调用哪个子类的
isclicked
函数,而无需您费力。关于c++ - 除了这种情况,还有其他更好的选择吗?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/43453438/