Form.ShowDialog(this) - c++编程基础

有时遇到一种情况，.ShowDialog()不显示，也不报错；如下：

 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)
               { }
           //}));
       }

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

不能够进行界面呈现，即显示.ShowDialog()；

解决方法：1：添加参数this。

.ShowDialog(IWin32Window owner); //owner:表示将拥有模式对话框的顶级窗口

 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)
               { }
           //}));
       }

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)
               { }
           }));
       }