C# 使用 terminal.gui 时无法找到对象引用

Question

我一直在玩弄一些 server/client 的东西，虽然我尝试使用 terminal.gui，但出于某种原因我不明白它的表现。我有 2 个脚本；一个处理后端事务，一个处理用户 interface/interraction.

脚本 1:

using UI;

namespace ServerTest
{
    public class Server
    {
        static void Main() 
        {
            UI.UI.main();
            StartServ();
            RecieveConn.Start();
            MngSockets.Start();
        }
    }
}

脚本 2：

using ServerTest;
using Terminal.Gui;

namespace UI
{
    public class UI
    {
        static Window win = new Window(new Rect(0, 0, Application.Top.Frame.Width, Application.Top.Frame.Height), "MyApp");
        public static void main()
        {
            Application.Init();
            win.Add(new Label(0, 0, "asd"));

            Application.Run(win);
        }
    }
}

这给了我错误：

System.TypeInitializationException
  HResult=0x80131534
  Message=The type initializer for 'UI.UI' threw an exception.
  Source=Server
  StackTrace:
   at UI.UI.main() in C:\Users\ThisPc\Desktop\ConsoleApp2\ConsoleApp2\ui.cs:line 29
   at ServerTest.Server.Main() in C:\Users\ThisPc\Desktop\ConsoleApp2\ConsoleApp2\Program.cs:line 23

Inner Exception 1:
NullReferenceException: Object reference not set to an instance of an object.

不过，如果我这样做，它会起作用：

using ServerTest;
using Terminal.Gui;

namespace UI
{
    public class UI
    {
        public static void main()
        {
            Window win = new Window(new Rect(0, 0, Application.Top.Frame.Width, Application.Top.Frame.Height), "MyApp");

            Application.Init();
            win.Add(new Label(0, 0, "asd"));

            Application.Run(win);
        }
    }
}

但是我不能在其他函数中使用 win 变量... 提前致谢 =)

Answer 1

感谢 mjwills 的回答...

就这样

using ServerTest;
using Terminal.Gui;

namespace UI
{
    public class UI
    {
        static Window win = null;
        public static void main()
        {
            win = new Window(new Rect(0, 0, Application.Top.Frame.Width, Application.Top.Frame.Height), "MyApp");

            Application.Init();
            win.Add(new Label(0, 0, "asd"));

            Application.Run(win);
        }
    }
}