Windows Forms帮助窗体调用Show()无法正常显示标签的问题

我用Windows Forms开发系统托盘应用，通过自定义的SystemTrayApplicationContext（继承自ApplicationContext）启动，代码如下：

SystemTrayApplicationContext appContext = new SystemTrayApplicationContext();
Application.ApplicationExit += appContext.HandleExit;
Application.Run(appContext);

仅保留系统托盘元素时运行正常，现在希望按下游戏手柄按键（使用Xinputium库，输入读取功能正常）时显示帮助界面。我在SystemTrayApplicationContext的构造函数中编程创建了包含Label的Form实例，同时设置了任务栏的ContextMenuStrip和NotifyIcon：

using System;
using System.Drawing;
using System.Windows.Forms;
using XInputium;
using XInputium.XInput;

public class SystemTrayApplicationContext : ApplicationContext
{
    private readonly XGamepad xinputDevice = new XGamepad();

    private Form helpForm;

    private NotifyIcon trayIcon;

    private ContextMenuStrip contextMenuStrip;

    private ToolStripMenuItem exitLabel;

    public SystemTrayApplicationContext()
    {
        // Setup help screen
        Label label = new Label();
        label.Text = "test";
        label.Size = new Size(300, 300);
        label.Anchor = AnchorStyles.Left;
        label.TextAlign = ContentAlignment.TopCenter;

        this.helpForm = new Form();
        this.helpForm.Size = label.Size;
        this.helpForm.ShowInTaskbar = false;
        this.helpForm.ShowIcon = false;
        this.helpForm.ControlBox = false;
        this.helpForm.FormBorderStyle = FormBorderStyle.None;
        this.helpForm.StartPosition = FormStartPosition.CenterScreen;
        this.helpForm.Controls.Add(label);

        // SetupContextMenu
        this.exitLabel = new ToolStripMenuItem("Exit");
        this.exitLabel.Anchor = AnchorStyles.Right;
        this.exitLabel.Click += this.HandleExit;

        this.contextMenuStrip = new ContextMenuStrip();
        this.contextMenuStrip.ShowImageMargin = false;
        this.contextMenuStrip.Items.Add(exitLabel);

        this.trayIcon = new NotifyIcon()
        {
            Icon = Properties.Resources.AppIcon,
            Visible = true,
            Text = "My System Tray App",
            ContextMenuStrip = this.contextMenuStrip,
        };

        // Setup Xinput events
        this.xinputDevice.ButtonPressed += this.HandleXinputButtonPressed;
        this.xinputDevice.ButtonReleased += this.HandleXinputButtonReleased;
    }

    public void HandleExit(object sender, EventArgs e)
    {
        trayIcon.Visible = false;
        Application.Exit();
    }

    private void HandleXinputButtonPressed(object sender, DigitalButtonEventArgs<XInputButton> e)
    {
        if (e.Button.Button == XButtons.A)
        {
            this.helpForm.Show();
            //this.helpForm.ShowDialog();
        }
    }

    private void HandleXinputButtonReleased(object? sender, DigitalButtonEventArgs<XInputButton> e)
    {
        if (e.Button.Button == XButtons.A)
        {
            this.helpForm.Hide();
        }
    }
}

问题：使用Show()显示帮助窗体时，窗体出现但Label不显示；使用ShowDialog()时渲染正常，但会阻塞应用。希望使用Show()，但不清楚渲染异常的原因，尝试过设置Label的Visible属性、窗体的AutoSize等均无效。

问题根源是XInputium的事件回调在非UI线程触发，而Windows Forms控件的所有操作必须在创建它的UI线程执行。直接在非UI线程调用helpForm.Show()会导致控件渲染异常，所以Label无法显示；ShowDialog()内部会自动切换到UI线程上下文，因此能正常渲染，但会阻塞线程。

解决方法是在事件回调中通过Invoke切换到UI线程再操作窗体：

修改按键按下和释放的处理方法：

private void HandleXinputButtonPressed(object sender, DigitalButtonEventArgs<XInputButton> e)
{
    if (e.Button.Button == XButtons.A)
    {
        this.helpForm.Invoke(() => this.helpForm.Show());
    }
}

private void HandleXinputButtonReleased(object? sender, DigitalButtonEventArgs<XInputButton> e)
{
    if (e.Button.Button == XButtons.A)
    {
        this.helpForm.Invoke(() => this.helpForm.Hide());
    }
}

这样就能保证窗体和控件的操作都在UI线程执行，Show()时Label也能正常显示，同时不会阻塞应用。

内容的提问来源于stack exchange，提问作者Anderson Urbano