多线程调用同一控件

Question

下面代码  2个线程  一个加点  一个刷新  调用同一控件 报错 "集合已修改；可能无法执行枚举操作。"  
请问应该怎么修改?  放在同一线程 会造成加点太少  
另外问下 windows最快刷新频率大约多少毫秒?
public Form1()
        {
            InitializeComponent();            chart1.Series["Series1"].Color = Color.Lime;
            chart1.Series["Series1"].Points.AddXY(0, 0);            Thread t1 = new Thread(DrawQuxian);
            t1.Start();            Thread t2 = new Thread(FlashQuxian);
            t2.Start();
        }        int i = 0;
        private void DrawQuxian()
        {
            while (true)
            {
                //拿点要求10毫秒以内
                this.chart1.Series["Series1"].Points.AddXY(i, i%10);
                i++;
                Thread.Sleep(10);
            }
        }        private void FlashQuxian()
        {
            while (true)
            {
                try
                {
                    if (this.chart1.InvokeRequired)
                    {
                        this.Invoke((EventHandler)(delegate
                        {
                            //刷新要求1000毫秒以内
                            this.chart1.Invalidate();
                        }));
                    }
                }
                catch
                {                }                Thread.Sleep(1000);
            }    
        }

Answer 1

        bool flag;
        private void button1_Click(object sender, EventArgs e)
        {
            flag = false;
            Thread thread = new Thread(new ThreadStart(target));
            thread.IsBackground = true;
            thread.Start();
            progressBar1.Maximum = 100;
            Thread threada = new Thread(new ThreadStart(target));
            threada.IsBackground = true;
            threada.Start();
            progressBar1.Maximum = 100;
        }        public void target()
        {
            MethodInvoker mi = new MethodInvoker(countPro);
            while (!flag)
            {
                this.Invoke(mi);
                
                Thread.Sleep(100);
            }        }

Answer 2

        public void countPro()
        {
            int min;
            double numerator, denominator, completed;
            min = this.progressBar1.Minimum;
            if (this.progressBar1.Value == this.progressBar1.Maximum)
            {
                this.progressBar1.Value = this.progressBar1.Minimum;
            }
            else
            {
                this.progressBar1.PerformStep();
            }
            //显示进度
            this.textBox1.Text = this.progressBar1.Value.ToString();
            numerator = this.progressBar1.Value - min;
            denominator = this.progressBar1.Maximum - this.progressBar1.Minimum;
            completed = (numerator / denominator) * 100.0;
            this.label2.Text = Math.Round(completed) + "%";
        }

Answer 3

1楼的方法可行。lz到MSDN查查，有这方面的例子

Answer 4

 this.chart1.Series["Series1"].Points.AddXY(i, i%10);
放到Invoke里去做
操作Series的代码全部回主线程操作即可。

Answer 5

可以锁定集合的SyncRoot，不仅要在DrawQuxian内添加，在绘图函数内亦要将对集合的整个枚举过程包起来

Answer 6

一个是要对共享数据进行同步，另外最好不要用Thread.Sleep()方法来控制刷新频率，使用个定时器