Detecting design mode from a Control's constructor

Question

Following-on from this question, is it possible to detect whether one is in design or runtime mode from within an object's constructor?

I realise that this may not be possible, and that I'll have to change what I want, but for now I'm interested in this specific question.

Chloe

Answer 1

You can use the LicenceUsageMode enumeration in the System.ComponentModel namespace:

bool designMode = (LicenseManager.UsageMode == LicenseUsageMode.Designtime);

Torse

Elegant solution, it works better than C# functionality ISite.DesignMode. – Hayrick

@Filip Kunc: if this doesn't work in OnPaint, you may check this condition in constructor and store it in a class field. – Euphorbia

This also doesn't work when overriding WndProc in a user control. Has to use @Euphorbia suggestion – Hulett

putting it in the construction is a nice idea IMil, it worked for me.. i tried to puting it on a static classes field, but (i think) static classes fields initialized when you first called them, so not a safe solution.. – Grovel

This does NOT work if LicenseManager.UsageMode is used in a UserControl A, and this UC A is used in another UserControl or Form B in another Assembly. In this case, I still get Runtime instead of DesignTime. – Drue

Found a solution that works also inside multi-level-nested user controls, see below: stackoverflow.com/a/65849349 – Drue

@TobiasKnauss you are right, UsageMode alone wont cover all cases but a combination of UsageMode and DesignMode covers all cases. For e.g. this answer is quite complete as far as I know: stackoverflow.com/a/2693338 (your answer as well) – Pottle

This only works for WinForms on .NET Framework. Microsoft is not supporting LicenseManager.UsageMode on .NET Core. It will always return Runtime even within the designer. – Woodham

Doesn't work in all cases. – Pertinent

Answer 2

Are you looking for something like this:

public static bool IsInDesignMode()
{
    if (Application.ExecutablePath.IndexOf("devenv.exe", StringComparison.OrdinalIgnoreCase) > -1)
    {
        return true;
    }
    return false;
}

You can also do it by checking process name:

if (System.Diagnostics.Process.GetCurrentProcess().ProcessName == "devenv")
   return true;

Daudet

Works in OnPaint, derived classes, constructors, etc. Best solution yet. – Shoulder

IMHO, this looks like an ugly solution. – Proceeding

Attention possible memory leak here. Process must be disposed. – Womanhater

While I'm sure this will work fine in most use cases, this solution has one principal flaw: Visual Studio is (at least in theory) not the only designer host. Therefore, this solution will only work if your designer is hosted by an application called devenv. – Housman

Why not just return Application.ProductName.Contains("Visual Studio"); if you know your application name does not meet the criteria? – Depressive

Works on VS2013, unlike the currently accepted answer. – Incus

@MobyDisk LicenseManager.UsageMode is only valid during control's the constructor. You need to persist its value in the constructor if you want to reference it later. It is generally the best solution to this problem. – Observer

Works in 2015 but I set a bool in the constructor to avoid perf loss during run time (if called frequently) – Phocaea

This approach is useful, LicenseManager.UsageMode seems to be broken by the new Xaml designer in VS2022. This approach allows you to detect if the process is "WpfSurface" which means the code is running in the Xaml designer. Good point by @Womanhater - best to wrap the call in a using statement: using (var process = System.Diagnostics.Process.GetCurrentProcess()){ return process.ProcessName; } – Perfecto

NOTE: in VS-2019, the process name reported for devenv when using NET-5 Winforms is "DesignToolsServer" and not "devenv". – Dangelo

Answer 3

Component ... as far as I know does not have the DesignMode property. This property is provided by Control. But the problem is when CustomControl is located in a Form in the designer, this CustomControl is running in runtime mode.

I have experienced that the DesignMode property works correct only in Form.

Leacock

Thanks for the tip! I never realized that before, but it makes perfect sense. Using the LicenseManager method provided by adrianbanks works perfect in these cases, where the control is embedded in another control/form. +1 for each! – Lacework

+1 You are absolutely right, this has been my experience as well. When you place a user control on a form, if there are any mouseenter or load events DesignMode will still come up as false because you are not in designmode for this control. In my experience it causes visual studio to crash pretty hard. – Derwin

Answer 4

Controls(Forms, UserControls etc.) inherit Component class which has bool property DesignMode so:

if(DesignMode)
{
  //If in design mode
}

Sharpeared

Which isn't set when the constructor runs, aka the initial issue of the OP. The first moment you can use it is in OnHandleCreated. – Dubbing

Downvoted the wrong answer to the question – Pottle

Answer 5

IMPORTANT

There is a difference of using Windows Forms or WPF!!

They have different designers and and need different checks. Additionally it's tricky when you mix Forms and WPF controls. (e.g. WPF controls inside of a Forms window)

If you have Windows Forms only, use this:

Boolean isInWpfDesignerMode   = (LicenseManager.UsageMode == LicenseUsageMode.Designtime);

If you have WPF only, use this check:

Boolean isInFormsDesignerMode = (System.Diagnostics.Process.GetCurrentProcess().ProcessName == "devenv");

If you have mixed usage of Forms and WPF, use a check like this:

Boolean isInWpfDesignerMode   = (LicenseManager.UsageMode == LicenseUsageMode.Designtime);
Boolean isInFormsDesignerMode = (System.Diagnostics.Process.GetCurrentProcess().ProcessName == "devenv");

if (isInWpfDesignerMode || isInFormsDesignerMode)
{
    // is in any designer mode
}
else
{
    // not in designer mode
}

To see the current mode you can show a MessageBox for debugging:

// show current mode
MessageBox.Show(String.Format("DESIGNER CHECK:  WPF = {0}   Forms = {1}", isInWpfDesignerMode, isInFormsDesignerMode));

Remark:

You need to add the namespaces System.ComponentModel and System.Diagnostics.

Anchie

I think your naming is misleading. When using for WinForms the naming is 'isInWpfDesignerMode' and for WPF it's 'isInFormsDesignerMode' – Bodrogi

Answer 6

You should use Component.DesignMode property. As far as I know, this shouldn't be used from a constructor.

Finial

This doesn't work when your control is inside of another control or form that is being designed. – Observer

Actually it works pretty good in my components. I always had to add if (!DesignMode) to OnPaint methods to make sure it doesn't spam the design time. – Oxyhydrogen

Answer 7

You can use this

if (DesignerProperties.GetIsInDesignMode(this))
{
...
}

Sweyn

This answer is for WPF, the question is about WinForms. – Triptych

Answer 8

Like many others, I have had this problem several times already when designing Windows Forms UserControls.
But today, I had a situation where none of the mentioned solutions worked for me.
The problem is, that LicenseManager.UsageMode only works reliably in the constructor, while DesignMode only works outside the constructor and not always. This is my experience, and this is what is said in a discussion on GitHub.
And another problem comes with inheritance, and embedding user controls in another user controls in another user controls. At the latest in the 2nd level of embedding a user controls, both ways fail!

This can be shown in the UserControls that I created for this test. Every UC has 3 labels:

1. its (project name) and type name

2. The values of

   • DesignMode (true: "DM=1"),
   • LicenseManager.UsageMode == LicenseUsageMode.Designtime, queried locally, (true: "local_LM-DT=1")
   • LicenseManager.UsageMode == LicenseUsageMode.Designtime, queried from a private field that was written in the constructor (true: "ctor_LM-DT=1")

   all taken inside the constructor ("CTOR") and inside a method that was called from the constructor ("CFCtor")

3. The same values as in 2)
   all taken inside the Load event ("Load()") and inside a method that was called from the Load event ("CFLoad")

The UserControls and the Form that I created are (all screenshots shown them in the WinForms Designer):

• UserControl1:

  • contains 3 labels

  
  The Designer does not execute the construtor or events, therefore the labels are not filled.

• UserControl1a:

  • inherits from UserControl1
  • contains 2 more labels

  
  The Designer executes the construtor and events of the parent UserControl.

• UserControl2: contains

  • contains 3 labels
  • contains 1 UserControl1
  • contains 1 UserControl1a

  
  The Designer executes the construtors and events of the embedded UserControls.
  Only 1 level of embedding.

• UserControl3:

  • contains 3 labels
  • contains 1 UserControl2

  
  The Designer executes the construtors and events of the embedded UserControls.
  2 level of embedding: The values inside the UserControl at 2nd embedding level are wrong.

• Form1:

  • contains 3 labels
  • contains 1 UserControl1
  • contains 1 UserControl1a
  • contains 1 UserControl2
  • contains 1 UserControl3.

  The Designer executes the construtors and events of the embedded UserControls.
  3 level of embedding: The values inside the UserControl at 2nd and 3rd embedding level are wrong.

As you can see from the screenshots, "ctor_LM-DT" is always 1.
This means, that storing the value from the LicenseManager in a member field is necessary to get a valid status of the Designer usage:

private LicenseUsageMode m_ctorLMUsageMode = LicenseManager.UsageMode;

For the sake of completeness, here's some of my code that can be used to reproduce the test:

public static string CreateText(bool i_isInDesignMode, LicenseUsageMode i_localLicenseUsageMode, LicenseUsageMode i_ctorLicenseUsageMode)
{
  return $"DM={(i_isInDesignMode ? 1 : 0)} local_LM-DT={(i_localLicenseUsageMode == LicenseUsageMode.Designtime ? 1 : 0)} ctor_LM-DT={(i_ctorLicenseUsageMode == LicenseUsageMode.Designtime ? 1 : 0)}";
}

The other UserControls are identical or similar:

public partial class UserControl1 : UserControl
{
  private LicenseUsageMode m_ctorLMUsageMode = LicenseManager.UsageMode;

  public UserControl1()
  {
    InitializeComponent();

    label2.Text = $"CTOR: {CInitTester.CreateText (DesignMode, LicenseManager.UsageMode, m_ctorLMUsageMode)}";
    CalledFromCtor();
  }

  private void UserControl1_Load(object sender, EventArgs e)
  {
    label3.Text = $"Load(): {CInitTester.CreateText(DesignMode, LicenseManager.UsageMode, m_ctorLMUsageMode)}";
    CalledFromLoad();
  }

  private void CalledFromCtor()
  {
    label2.Text += $"\r\nCFCtor: {CInitTester.CreateText (DesignMode, LicenseManager.UsageMode, m_ctorLMUsageMode)}";
  }

  private void CalledFromLoad()
  {
    label3.Text += $"\r\nCFLoad: {CInitTester.CreateText (DesignMode, LicenseManager.UsageMode, m_ctorLMUsageMode)}";
  }
}

Drue

Thanks a lot for your answer, it is quite thorough.. Just wanted to point out that this answer also covers all your cases: stackoverflow.com/a/2693338. While your answer is certainly simpler, as in this, the other answer is easier, as in more convenient. +1 and cheers – Pottle

One advantage of your approach is that it works without a control instance. – Pottle

Answer 9

With cooperation of the designer... It can be used in Controls, Components, in all places

    private bool getDesignMode()
    {
        IDesignerHost host;
        if (Site != null)
        {
            host = Site.GetService(typeof(IDesignerHost)) as IDesignerHost;
            if (host != null)
            {
                if (host.RootComponent.Site.DesignMode) MessageBox.Show("Design Mode");
                else MessageBox.Show("Runtime Mode");
                return host.RootComponent.Site.DesignMode;
            }
        }
        MessageBox.Show("Runtime Mode");
        return false;
    }

MessageBox.Show( lines should be removed. It only makes me sure it works correctly.

Leacock

Answer 10

I wasn't able to get any of these solutions to work for me in Visual Studio 2019 when creating a WinForms app on .NET Core 3.1.

Both Appllication.ProcessName and Process.ProcessName are returning "DesignToolsServer" for me and LicenseManager.UsageMode returns LicenseUsageMode.Runtime when the Control is in another control or just on a form itself.

I did get it to work using Application.ProcessName == "DesignToolsServer".

Gonophore

Came here to share that WinForms on .NET Core had an issue raised about always reporting Runtime usage mode. As of this writing, Microsoft has decided NOT to support LicenseManager.UsageMode in .NET Core. – Woodham

Answer 11

Another interesting method is described on that blog: http://www.undermyhat.org/blog/2009/07/in-depth-a-definitive-guide-to-net-user-controls-usage-mode-designmode-or-usermode/

Basically, it tests for the executing assembly being statically referenced from the entry assembly. It circumvents the need to track assembly names ('devenv.exe', 'monodevelop.exe'..).

However, it does not work in all other scenarios, where the assembly is dynamically loaded (VSTO being one example).

Mckenzie

The link is (effectively) broken. It now redirects to the latest blog post (currently 2016-03) instead. – Contrariety

And now its just a spam site. – Foothill

It's archived here: web.archive.org/web/20111015201417/http://www.undermyhat.org/… – Wickman

Answer 12

This is the method I used in my project:

//use a Property or Field for keeping the info to avoid runtime computation
public static bool NotInDesignMode { get; } = IsNotInDesignMode();
private static bool IsNotInDesignMode()
{
    /*
    File.WriteAllLines(@"D:\1.log", new[]
    {
        LicenseManager.UsageMode.ToString(), //not always reliable, e.g. WPF app in Blend this will return RunTime
        Process.GetCurrentProcess().ProcessName, //filename without extension
        Process.GetCurrentProcess().MainModule.FileName, //full path
        Process.GetCurrentProcess().MainModule.ModuleName, //filename
        Assembly.GetEntryAssembly()?.Location, //null for WinForms app in VS IDE
        Assembly.GetEntryAssembly()?.ToString(), //null for WinForms app in VS IDE
        Assembly.GetExecutingAssembly().Location, //always return your project's output assembly info
        Assembly.GetExecutingAssembly().ToString(), //always return your project's output assembly info
    });
    //*/

    //LicenseManager.UsageMode will return RunTime if LicenseManager.context is not present.
    //So you can not return true by judging it's value is RunTime.
    if (LicenseUsageMode.Designtime == LicenseManager.UsageMode) return false;
    var procName = Process.GetCurrentProcess().ProcessName.ToLower();
    return "devenv" != procName //WinForms app in VS IDE
        && "xdesproc" != procName //WPF app in VS IDE/Blend
        && "blend" != procName //WinForms app in Blend
        //other IDE's process name if you detected by log from above
        ;
}

Attention!!!: The code returned bool is indicating NOT in design mode!

Cooksey

Answer 13

Yes, you can check for whether you're in "design mode" from within an object's constructor. But using the WinForms DesignMode property doesn't always work as expected. An alternative:

This is my technique to check DesignMode in C# using Visual Studio and it does work in constructors.

// add this class...
public static class Globals
{
    static Globals() => DesignMode = true;
    public static bool DesignMode { get; set; }
}

// and modify your existing class...
public static class Program
{
    public static void Main()
    {
        Globals.DesignMode = false;
        // ...
        // ... and then the rest of your program
        //
        //  in any of your code you can check Globals.DesignMode for
        //  the information you want.
    }
}

This solution is lightweight and simple. The downside is that you have to remember to clear the flag in your Main code.

When checking for "design mode," we're essentially checking for whether our code is being executed because our whole program is being run or because parts of our code are being executed by the VS designer. With this solution, the flag only gets set to false when the whole program is being run.

Backing

Answer 14

The LicenseManager solution does not work inside OnPaint, neither does this.DesignMode. I resorted to the same solution as @Jarek.

Here's the cached version:

    private static bool? isDesignMode;
    private static bool IsDesignMode()
    {
        if (isDesignMode == null)
            isDesignMode = (Process.GetCurrentProcess().ProcessName.ToLower().Contains("devenv"));

        return isDesignMode.Value;
    }

Be aware this will fail if you're using any third party IDE or if Microsoft (or your end-user) decide to change the name of the VS executable to something other than 'devenv'. The failure rate will be very low, just make sure you deal with any resulting errors that might occur in the code that fails as a result of this and you'll be fine.

Electrokinetic

Answer 15

If you want to run some lines when it is running but not in the Visual Studio designer, you should implement the DesignMode property as follows:

// this code is in the Load of my UserControl
if (this.DesignMode == false)
{
    // This will only run in run time, not in the designer.
    this.getUserTypes();
    this.getWarehouses();
    this.getCompanies();
}

Albeit

Answer 16

    private void CtrlSearcher_Load(object sender, EventArgs e)
    {
           if(!this.DesignMode) InitCombos();
    }

Kaiulani

While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value. – Equipoise

Answer 17

Timers that are enabled by default can cause crash when using custom/user controls. Disable them by default, and enable only after design mode check

   public chartAdapter()
    {
        try
        {

            //Initialize components come here
            InitializeComponent();

            //Design mode check
            bool designMode = (LicenseManager.UsageMode == LicenseUsageMode.Designtime);
            if (designMode)
                return;

            //Enable timers ONLY after designmode check, or else crash
            timerAutoConnect.Enabled = timerDraw.Enabled = true;

Penthouse

Answer 18

As of .net 6+ you have an additional property IsAncestorSiteInDesignMode.

We have it like this for WinForms .net 6+:

        public static bool IsRealDesignerMode(this Control c)
        {
            return System.ComponentModel.LicenseManager.UsageMode == System.ComponentModel.LicenseUsageMode.Designtime
                || c.IsAncestorSiteInDesignMode
                || System.Diagnostics.Process.GetCurrentProcess().ProcessName == "devenv";
        }

Malinin

Answer 19

The only way I could get this to work properly was by creating a static var "static bool IsRuntime;". This will have a default "false" value when in design mode. Then in the "main" function I just set it to true. Simple solution that just works.

function MyEvent()
{
  if (MyClass.IsRuntime == false)
  {
     // Design mode!
  }
  else
  {
     // Runtime mode!
  }
}

Main:

[STAThread]
static void Main()
{
  MyClass.IsRuntime = true;
  (...)
}

Pillow