递归中的异常，无法退出，即使是抛出异常也不能退出！

Question

你这样递归，这不是try.catch里面套try.catch,然后里面该套try.catch，套无数次。。？？

Answer 1

我用TRY。catch实现抛出异常，好退出，不用也是一样看看下面的。它就是不退出！用BREAK,和return都不行！即使用application.exit(0),都不行，我很郁闷。。
 private void findbottom(Point pt)
        {
            //只找上边界
                  for (int i = 0; i < 8; i++)
                    {
                        currentp.X = pt.X + direct[i].X;
                        currentp.Y = pt.Y + direct[i].Y;
                        nextp.X = pt.X + direct[i + 1].X;
                        nextp.Y = pt.Y + direct[i + 1].Y;                        if (source[nextp.X, nextp.Y] == 0 && source[currentp.X, currentp.Y] == 255)
                        {
                            //线段中断
                            if (i == predot || pt.X >= fourpoint[3].X)
                            {
                               // break;
                                return;
                            }
                            bottomxarray[bottomtotal] = currentp.X;
                            bottomyarray[bottomtotal] = currentp.Y;
                            //找到的个数
                            bottomtotal++;
                            //不能这个方向找，线中断了
                            predot = (i + 4) % 8;
                            findbottom(currentp);                        }
                    }
               
        }

Answer 2

龙工已帮忙解决，谢谢！ private bool  findbottom(Point pt)
        {
            //只找上边界 
            if (pt.X >= fourpoint[3].X)
            {
                return true;
            }
            for (int i = 0; i < 8; i++)
            {
                currentp.X = pt.X + direct[i].X;
                currentp.Y = pt.Y + direct[i].Y;
                nextp.X = pt.X + direct[i + 1].X;
                nextp.Y = pt.Y + direct[i + 1].Y;
               
                if (source[nextp.X, nextp.Y] == 0 && source[currentp.X, currentp.Y] == 255)
                {
                    //线段中断 
                    if (i == predot)
                        return true;
                    bottomxarray[bottomtotal] = currentp.X;
                    bottomyarray[bottomtotal] = currentp.Y;
                    //找到的个数 
                    bottomtotal++;
                    //不能这个方向找，线中断了 
                     predot = (i + 4) % 8;                     if (findbottom(currentp))
                    {
                        return true;
                    }                }
            }            return false;
        } 

Answer 3

恩，还是你的代码问题。
我觉的递归完全不用try catch.
if (findbottom(currentp))
                    {
                        return true;
                    } 呵呵