线程中的TopMost表单？

Question

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 ?

Answer 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.

Answer 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);

Answer 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(); });
}