How to tell if .net code is being run by Visual Studio designer

Question

Hello,

I am getting some errors thrown in my code when I open a win form in visual studio's designer. I would like to branch in my code and perform a different initialization if the form is being opened by designer than if it is being run for real.

How can I determine at run-time if the code is being executed as part of designer opening the form?

Thanks

Answer 1

System.Diagnostics.Debugger.IsAttached

Answer 2

I'm not sure if running in debug mode counts as real, but an easy way is to include an if statement in your code that checkes for System.Diagnostics.Debugger.IsAttached.

Answer 3

It's hack-ish, but if you're using vb.net and when you're running from within visual studio My.Application.Deployment.CurrentDeployment will be Nothing, because you haven't deployed it yet. I'm not sure how to check the equivalent value in C#.

Answer 4

You check the DesignMode property of your control: if (!DesginMode) { //Do production runtime stuff }

Note that this won't work in your constructor because the components haven't been initialized yet.

Answer 5

If you are in a form or control you can use the DesignMode property:

if (DesignMode)
{
        DesignMode Only stuff
}

Answer 6

To find out if you're in "design mode":

• Windows Forms components (and controls) have a DesignMode property.
• Windows Presentation Foundation controls should use the IsInDesignMode attached property.

Answer 7

System.ComponentModel.Component.DesignMode == true

Answer 8

The Control.DesignMode property is probably what you're looking for. It tells you if the control's parent is open in the designer.

In most cases it works great, but there are instances where it doesn't work as expected. First, it doesn't work in the controls constructor. Second, DesignMode is false for "grandchild" controls. For example, DesignMode on controls hosted in a UserControl will return false when the UserControl is hosted in a parent.

There is a pretty easy workaround. It goes something like this:

public bool HostedDesignMode
{
  get 
  {
     Control parent = Parent;
     while (parent!=null)
     {
        if(parent.DesignMode) return true;
        parent = parent.Parent;
     }
     return DesignMode;
  }
}

I haven't tested that code, but it should work.

Answer 9

The most reliable approach is:

public bool isInDesignMode
        {
            get
            {
                return (System.Diagnostics.Process.GetCurrentProcess().ProcessName == "devenv");
            }
        }

Answer 10

We use the following code in UserControls and it does the work. Using only DesignMode will not work in your app that uses your custom user controls as pointed out by other members.

    public bool IsDesignerHosted
    {
        get { return IsControlDesignerHosted(this); }
    }

    public bool IsControlDesignerHosted(System.Windows.Forms.Control ctrl)
    {
        if (ctrl != null)
        {
            if (ctrl.Site != null)
            {
                if (ctrl.Site.DesignMode == true)
                    return true;
                else
                {
                    if (IsControlDesignerHosted(ctrl.Parent))
                        return true;
                    else
                        return false;
                }
            }
            else
            {
                if (IsControlDesignerHosted(ctrl.Parent))
                    return true;
                else
                    return false;
            }
        }
        else
            return false;
    }

Answer 11

I found the DesignMode property to be buggy, at least in previous versions of Visual Studio. Hence, I made my own using the following logic:

Process.GetCurrentProcess().ProcessName.ToLower().Trim() == "devenv";

Kind of a hack, I know, but it works well.

Answer 12

Process.GetCurrentProcess().ProcessName.ToLower().Trim() == "devenv";

causes a memory leak. You must dispose the process object.