48

Is there a variable or a preprocessor constant that allows to know that the code is executed within the context of Visual Studio?

Peter Mortensen
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user310291
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  • Do you want this information to know if your code it being debugged? If yes there is a macro which tells you if a debugger is attached to your application refer: http://stackoverflow.com/questions/2188201/is-there-a-way-to-detect-if-a-debugger-is-attached-to-a-process-from-c – Ganesh R. Mar 11 '10 at 17:59
  • In C# you can use: http://stackoverflow.com/questions/361077/c-clr-remote-debugger-how-to-wait-until-attached – Ganesh R. Mar 11 '10 at 18:06

9 Answers9

67

Try Debugger.IsAttached or DesignMode property or get ProcessName or a combination, as appropriate

Debugger.IsAttached // or                                       
LicenseUsageMode.Designtime // or 
System.Diagnostics.Process.GetCurrentProcess().ProcessName

Here is a sample

public static class DesignTimeHelper {
    public static bool IsInDesignMode {
        get {
            bool isInDesignMode = LicenseManager.UsageMode == LicenseUsageMode.Designtime || Debugger.IsAttached == true;

            if (!isInDesignMode) {
                using (var process = Process.GetCurrentProcess()) {
                    return process.ProcessName.ToLowerInvariant().Contains("devenv");
                }
            }

            return isInDesignMode;
        }
    }
}
Fábio Augusto Pandolfo
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Asad
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    dont forget the crazy memory leak on Process.GetCurrentProcess() remember to dispose – Neil Feb 24 '11 at 13:34
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    In my code, I've added a static bool variable to know when IsInDesignMode was called and another one to hold the result of the function. At the beginning of the function, I return the result of the first call. With this, the inner of the function is just call once in design time and once at runtime. I don't see any conflict where the design time is merge with the runtime and this is much faster when you have a huge number of calls to this function. – Samuel Sep 23 '16 at 20:32
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    This does not differentiate between running the VS Designer/preview and running the app with the debugger attached. – Maslow Jun 24 '19 at 16:59
18

The DesignMode property isn't always accurate. We have had use this method so that it works consistently:

    protected new bool DesignMode
    {
        get
        {
            if (base.DesignMode)
                return true;

            return LicenseManager.UsageMode == LicenseUsageMode.Designtime;
        }
    }

The context of your call is important. We've had DesignMode return false in the IDE if running in an event under certain circumstances.

Digicoder
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6

There is the DesignMode property for Components. It is handy when you use the Design Viewer of VS.

But when you talk about debugging in Visual Studio you need to use the Debugger.IsAttached property. Then, you can use

#if DEBUG
#endif

too

tanascius
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4

I think the simplest and most reliable way to determine if your extension is executed in the WinForms designer is to check the current process. 

public static bool InVisualStudio() {
  return StringComparer.OrdinalIgnoreCase.Equals(
    "devenv", 
    Process.CurrentProcess.ProcessName);
}
JaredPar
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4

I use this extension method:

internal static class ControlExtension
{
    public static bool IsInDesignMode(this Control control)
    {
        while (control != null)
        {
            if (control.Site != null && control.Site.DesignMode)
                return true;
            control = control.Parent;
        }
        return false;
    }
}
Knyaz
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2

You can use this:

protected static bool IsInDesigner
{
    get { return (Assembly.GetEntryAssembly() == null); }
}
Pedro77
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2

There is a DesignMode property that you can check but in my experience it's not always accurate. You could also check to see if the executable is DevEnv.exe

Take a look here. Might make this question a dup but it all depends on what you're trying to accomplish.

Community
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Paul Sasik
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1

I use this code to distinguish whether it's running in Visual Studio or if it's deployed to customers.

if (ApplicationDeployment.IsNetworkDeployed) {
    // do stuff 
} else {
   // do stuff (within Visual Studio)
}

Works fine for me for ages. I skip some logic when inside Visual Studio (such as logging in to application etc).

MadBoy
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  • Doesn't that presume [ClickOnce](http://en.wikipedia.org/wiki/ClickOnce) is used (disguised in Visual Studio as "Publish")? Would it work if a setup project is used to produce an ordinary [MSI](http://en.wikipedia.org/wiki/Windows_Installer) installer? – Peter Mortensen Feb 20 '14 at 23:06
0

I want to add to this that in Visual Studio 2022 with .Net 6, the process that actually opens winforms designers is called DesignToolsServer.

DesignMode works outside of the constructor and checking for ProcessName = "DesignToolsServer" works within the constructor.

Branston
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  • As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Jan 21 '22 at 19:24