TopMost form in a thread?

问题:

I am using the following code to open a form in a new thread:

private void button1_Click(object sender, EventArgs e)
{

    Thread thread = new Thread(ThreadProc);
    thread.Start();
}


public void ThreadProc()
{

    Form form = new Form();
    form.TopMost = true;
    form.ShowDialog();
}

But the newly created form isn't TopMost even though I set it to true.

How can I make a form in a thread TopMost ?

回答1:

Usually you don't need another thread, you open the form as usual in modal or non modal mode, if the form needs to do a heavy process then you do the process inside a thread.

Specific to your question one option is to run the form from an Application.Run as described here.

public partial class Form1 : Form
{
    public Form1()
    {
        InitializeComponent();
    }

    private void button1_Click(object sender, EventArgs e)
    {
        Thread thread = new Thread(ThreadProc);
        thread.Start();
    }


    public void ThreadProc()
    {
        using (Form1 _form = new Form1())
        {
            _form.TopMost = true;
            Application.Run(_form);
        }
    }
}

That will launch a new thread with its own message pump and will keep it as a TopMost form.

回答2:

Just ran into this problem myself. It seems that if the form has an Owner, then TopMost works as expected. If the owning form was created on another thread, though, it's a little tricky to set. Here's what I used:

var form = new Form();

form.Shown += (sender, e) => {
    Control.CheckForIllegalCrossThreadCalls = false;
    form.Owner = /* Owning form here */;
    form.CenterToParent();      // Not necessary
    Control.CheckForIllegalCrossThreadCalls = true;

    form.TopMost = true;        // Works now!
};

Application.Run(form);

回答3:

Instead of calling ShowDialog directly, try using this.Invoke to gain ownership of the form.

private void button1_Click(object sender, EventArgs e)
{

    Thread thread = new Thread(ThreadProc);
    thread.Start();
}


public void ThreadProc()
{

    Form form = new Form();
    form.TopMost = true;
    this.Invoke((Action)delegate() { form.ShowDialog(); });
}