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

Question

I am getting some errors thrown in my code when I open a Windows Forms 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?

Illfavored

Answer 1

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.

Suet

DesignMode is not 100% reliable. – Choi

...see JohnV's answer for discussion of (and solutions to) the DesignMode problems. – Suet

See NET3's answer for a working solution. No disrespect to JohnV intended. – Wellrounded

I'm checking DesignMode and LicenseUsage and it is reporting wrong – Patency

Answer 2

if (System.ComponentModel.LicenseManager.UsageMode == System.ComponentModel.LicenseUsageMode.Designtime)
{
  // Design time logic
}

Shrike

+1. Alternately, LicenseUsageMode.Runtime, for anything that should run only at runtime. – Wellrounded

+1 This worked a treat for me. Had been using DesignMode which it turns out is completely undependable. Switched all usages to this "UsageMode" alternative. – Savaii

The LicenseManager solution has worked for me in suppressing code that is executed during .cctor module loading and was causing bad behavior in the WPF designer. – Diatom

This one worked for me, .DesignMode property didn't work for me. – Cristiano

In Visual Studio 2017, this works in a control's constructor, but not in the control's load event handler; there, it is always set to runtime <sigh> Not hard to work around of course (just set a member variable in constructor), but just... weird. – Am

Doesn't work for Dotnet Core EF migrations, unfortunately. – Pennebaker

Answer 3

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.

Miniver

it doesnt work DesignMode is protected. But the problem you mention is exactly what i am getting. But I also get it if put code in OnLoad() instead of .cor() – Glottis

Answer 4

The most reliable approach is:

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

Soteriology

This can also be used to determine whether you are running within IDE debugger, if you use 'process.ProcessName.Contains("vshost")' as criteria. – Xylophone

Amendment für VS2013: As XAML editor finally got its own process, you need to check for "XDesProc" – Catlett

While this is a perfectly reasonable solution and likely very reliable, there's something about this answer that sends shivers down my spine. – Inbound

...unless your own application's process name is called "devenv" (and I guess some more exceptions might apply). I wouldn't call "most reliable approach" the one, that have obvious flaw. Duh, user may change the filename to devenv.exe and your application will suddenly stop working properly. – Bulgarian

JetBrains's Rider team doesn;t like this answer :) – Antipodal

FWIW: in Net6, the designer exe is no longer devenv.exe but rather designtoolsserver.exe. Then this tip also works there – Belief

Answer 5

The most reliable way to do this is to ignore the DesignMode property and use your own flag that gets set on application startup.

Class:

public static class Foo
{
    public static bool IsApplicationRunning { get; set; }
}

Program.cs:

[STAThread]
static void Main()
{
     Foo.IsApplicationRunning = true;
     // ... code goes here ...
}

Then just check the flag whever you need it.

if(Foo.IsApplicationRunning)
{
    // Do runtime stuff
}
else
{
    // Do design time stuff
}

Bobsled

I spent a lot of time on trying different solutions, but eventually I also came up with the idea of just setting a flag. It's the only way that works 100%. – Choi

This doesn't work for code libraries though, and this would then have to be included in the call to all methods that would need it. – Hermann

Use a global variable and set it once. – Choi

I don't really see the point of this one. Unless I'm missing something, it doesn't answer the question of how to tell if running in the debugger - it requires you to manually sets a flag to indicate that. – Micronutrient

Answer 6

I had the same problem in Visual Studio Express 2013. I tried many of the solutions suggested here but the one that worked for me was an answer to a different thread, which I will repeat here in case the link is ever broken:

protected static bool IsInDesigner
{
    get { return (Assembly.GetEntryAssembly() == null); }
}

Xenolith

Answer 7

The devenv approach stopped working in VS2012 as the designer now has its own process. Here is the solution I am currently using (the 'devenv' part is left there for legacy, but without VS2010 I am not able to test that though).

private static readonly string[] _designerProcessNames = new[] { "xdesproc", "devenv" };

private static bool? _runningFromVisualStudioDesigner = null;
public static bool RunningFromVisualStudioDesigner
{
  get
  {
    if (!_runningFromVisualStudioDesigner.HasValue)
    {
      using (System.Diagnostics.Process currentProcess = System.Diagnostics.Process.GetCurrentProcess())
      {
        _runningFromVisualStudioDesigner = _designerProcessNames.Contains(currentProcess.ProcessName.ToLower().Trim());
      }
    }

    return _runningFromVisualStudioDesigner.Value;
  }
}

Hermann

How did you find this out? – Zacynthus

Too long ago for me to remember that I'm afraid. I think I just noticed the odd processes in Task Manager and looked them up, but I can't say for sure though. – Hermann

Because I'm really grateful for this remark, I was already wondering why some of my Components where causing crashes in the new Visual Studio, sadly uncaught Exceptions can cause a crash of the whole Visual Studio (i.e. an error in an unmanaged Dispose method) – Zacynthus

Glad you found it useful. :) Sorry I can't be of more help as to how I found the error in the first place. – Hermann

Answer 8

System.Diagnostics.Debugger.IsAttached

Affenpinscher

Answer 9

/// <summary>
/// Are we in design mode?
/// </summary>
/// <returns>True if in design mode</returns>
private bool IsDesignMode() {
    // Ugly hack, but it works in every version
    return 0 == String.CompareOrdinal(
        "devenv.exe", 0,
        Application.ExecutablePath, Application.ExecutablePath.Length - 10, 10);
}

Cloddish

Answer 10

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#.

Maffick

Answer 11

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;
    }

Basically the logic above boils down to:

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

Stringer

Answer 12

using (System.Diagnostics.Process process = System.Diagnostics.Process.GetCurrentProcess())
{
    bool inDesigner = process.ProcessName.ToLower().Trim() == "devenv";
    return inDesigner;
}

I tried the above code (added a using statement) and this would fail on some occasions for me. Testing in the constructor of a usercontrol placed directly in a form with the designer loading at startup. But would work in other places.

What worked for me, in all locations is:

private bool isDesignMode()
{
    bool bProcCheck = false;
    using (System.Diagnostics.Process process = System.Diagnostics.Process.GetCurrentProcess())
    {
        bProcCheck = process.ProcessName.ToLower().Trim() == "devenv";
    }

    bool bModeCheck = (System.ComponentModel.LicenseManager.UsageMode == System.ComponentModel.LicenseUsageMode.Designtime);

    return bProcCheck || DesignMode || bModeCheck;
}

Maybe a bit overkill, but it works, so is good enough for me.

The success in the example noted above is the bModeCheck, so probably the DesignMode is surplus.

Fertility

I think this is the best solution as it covers all situations. Worked perfectly for me. – Therontheropod

Answer 13

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.

Fanelli

Answer 14

You check the DesignMode property of your control:

if (!DesignMode)
{
//Do production runtime stuff
}

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

Trough

Answer 15

When running a project, its name is appended with ".vshost".

So, I use this:

    public bool IsInDesignMode
    {
        get
        {
            Process p = Process.GetCurrentProcess();
            bool result = false;

            if (p.ProcessName.ToLower().Trim().IndexOf("vshost") != -1)
                result = true;
            p.Dispose();

            return result;
        }
    }

It works for me.

Basaltware

This lets you know if a program is being run from VS, not if methods are invoked form the VS designer, which is what Zvi is trying to achieve. – Hermann

Answer 16

If you created a property that you don't need at all at design time, you can use the DesignerSerializationVisibility attribute and set it to Hidden. For example:

protected virtual DataGridView GetGrid()
{
    throw new NotImplementedException("frmBase.GetGrid()");
}

[DesignerSerializationVisibility(DesignerSerializationVisibility.Hidden)]
public int ColumnCount { get { return GetGrid().Columns.Count; } set { /*Some code*/ } }

It stopped my Visual Studio crashing every time I made a change to the form with NotImplementedException() and tried to save. Instead, Visual Studio knows that I don't want to serialize this property, so it can skip it. It only displays some weird string in the properties box of the form, but it seems to be safe to ignore.

Please note that this change does not take effect until you rebuild.

Disagreeable

Answer 17

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

if (DesignMode)
{
        DesignMode Only stuff
}

Felodese

This doesn't seem to work in controls - always returns false :( – Hearttoheart

Answer 18

System.ComponentModel.Component.DesignMode == true

Laaland

It says it's not accessible because it's Protected – Forelli

Answer 19

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.

Secondclass

That won't work if the designer is hosted somewhere other than Visual Studio. For instance, the web page designer is also used in SharePoint Designer and Expression Web. The workflow designer can be hosted anywhere. – Fleabitten

Process.GetCurrentProcess().ProcessName.ToLower().Trim() == "devenv"; causes a memory leak. You must dispose the process object. – Richman

This is the single cause of a MASSIVE memory leak on my application. – Catechism

This is KEY! This line does cause a massive memory leak. Use dispose! – Catechism

To clarify for everyone, your code should look something like this: System.Diagnostics.Process process = System.Diagnostics.Process.GetCurrentProcess(); bool inDesigner = process.ProcessName.ToLower().Trim() == "devenv"; process.Dispose(); if (isDesigner) ... – Denver

@Catechism remember to dispose -- using (var p = Process.GetCurrentProcess()) { ... } – Womankind

Out of curiosity, why isn't it garbage collected by default? – Biometrics

Answer 20

To solve the problem, you can also code as below:

private bool IsUnderDevelopment
{
    get
    {
        System.Diagnostics.Process process = System.Diagnostics.Process.GetCurrentProcess();
        if (process.ProcessName.EndsWith(".vshost")) return true;
        else return false;
    }

}

Neveda

that checks if run in the debugger, but we talk about the VS Designer – Zacynthus

Answer 21

Here's another one:

        //Caters only to thing done while only in design mode
        if (App.Current.MainWindow == null){ // in design mode  }

        //Avoids design mode problems
        if (App.Current.MainWindow != null) { //applicaiton is running }

Turnery

Answer 22

After testing most of the answers here, unfortunately nothing worked for me (VS2015). So I added a little twist to JohnV's answer, which didn't work out of the box, since DesignMode is a protected Property in the Control class.

First I made an extension method which returns the DesignMode's Property value via Reflection:

public static Boolean GetDesignMode(this Control control)
{
    BindingFlags bindFlags = BindingFlags.Instance | BindingFlags.NonPublic | BindingFlags.Static;
    PropertyInfo prop = control.GetType().GetProperty("DesignMode", bindFlags);
    return (Boolean)prop.GetValue(control, null);
}

and then I made a function like JohnV:

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

This is the only method that worked for me, avoiding all the ProcessName mess, and while reflection should not be used lightly, in this case it did all the difference! ;)

EDIT:

You can also make the second function an extension method like this:

public static Boolean IsInDesignMode(this Control control)
{
    Control parent = control.Parent;
    while (parent != null)
    {
        if (parent.GetDesignMode())
        {
            return true;
        }
        parent = parent.Parent;
    }
    return control.GetDesignMode();
}

Progenitor

How slow is calling that function at runtime? – Tirewoman

Just use control.Site.DesignMode instead of reflection. Its the same thing. Do ensure to null check – Antipodal

Answer 23

For WPF (hopefully this is useful for those WPF people stumbling upon this question):

if (System.ComponentModel.DesignerProperties.GetIsInDesignMode(new DependencyObject()))
{
}

GetIsInDesignMode requires a DependencyObject. If you don't have one, just create one.

Extravert

Answer 24

    /// <summary>
    ///  Whether or not we are being run from the Visual Studio IDE
    /// </summary>
    public bool InIDE
    {
        get
        {
            return Process.GetCurrentProcess().ProcessName.ToLower().Trim().EndsWith("vshost");
        }
    }

Munich

Answer 25

Here's a flexible way that is adaptable to where you compile from as well as whether or not you care which mode you're in.

string testString1 = "\\bin\\";
//string testString = "\\bin\\Debug\\";
//string testString = "\\bin\\Release\\";

if (AppDomain.CurrentDomain.BaseDirectory.Contains(testString))
{
    //Your code here
}

Zellazelle

This does not seem to be anywhat near reliable? What if the user changes the directory where he runs from? – Clustered