有时遇到一种情况，.ShowDialog()不显示。也不报错。例如以下：

<span style="font-size:14px;"> private void button1_Click(object sender, EventArgs e)

        {

            Thread thread = new Thread(show);

            thread.Start();

        }

       void show()

       {

           Control.CheckForIllegalCrossThreadCalls = false;

           //this.Invoke(new Action(() =>

           //{

               if (saveFileDialog1.ShowDialog() == DialogResult.OK)

               { }

           //}));

       }</span>

原因分析：这属于线程间操作的一种异常。界面呈现和新创建的thread分开在两个线程中。在thread线程中

不可以进行界面呈现，即显示.ShowDialog()；

解决方法：1：加入參数this。

.ShowDialog(IWin32Window owner);    //owner:表示将拥有模式对话框的*窗体

<span style="font-size:14px;"> private void button1_Click(object sender, EventArgs e)

        {

            Thread thread = new Thread(show);

            thread.Start();

        }

       void show()

       {

           Control.CheckForIllegalCrossThreadCalls = false;

           //this.Invoke(new Action(() =>

           //{

               if (saveFileDialog1.ShowDialog(this) == DialogResult.OK)

               { }

           //}));

       }</span>

2:使用Invoke

        private void button1_Click(object sender, EventArgs e)

        {

            Thread thread = new Thread(show);

            thread.Start();

        }

       void show()

       {

           // Control.CheckForIllegalCrossThreadCalls = false;

           this.Invoke(new Action(() =>

           {

               if (saveFileDialog1.ShowDialog() == DialogResult.OK)

               { }

           }));

       }