Way to determine whether executing in IDE or not? [duplicate]
Asked Answered
H

2

14

In C#/VB in Visual Studio 2010, is there way in the code to determine whether the program is currently running in the IDE or not?

eg. If ProgramRunningInIDE Then MessageBox.Show exc.Message
Heighttopaper answered 8/3, 2013 at 11:18 Comment(4)
Or just look at the Related list on the right side of this pageUpcoming
I believe this is a good use for #if DEBUG.Unobtrusive
@IAbstract, not so if you're running in release configuration.Hooge
@Morrison: once I followed the string of dupe questions I learned something new.Unobtrusive
H
36

You could check if the debugger is attached with:

System.Diagnostics.Debugger.IsAttached

This essentially does the same thing.

Hooge answered 8/3, 2013 at 11:22 Comment(2)
Does this return true if in the IDE but in release mode?Glossa
@SteveSmith technically it can return false even in debug mode, so long as a debugger is not attached. The debugger can be attached to release builds too, in which case this will return true.Hooge
S
-1

There is an IsInDesignMode property you can use. In some circumstances it isn't accurate, though, so you additionally may want to check the UsageMode.

public static bool IsRunningInIdeContext
{
    get {
        if (DesignerProperties.IsInDesignMode)
            return true;
        return LicenseManager.UsageMode == LicenseUsageMode.Designtime;
    }
}
Smutch answered 8/3, 2013 at 11:35 Comment(3)
IsInDesignMode can be used by a control (in a library) to understand if it's hosted in the designer(visual studio) or in a running application, but it have nothing to do about detecting if the application have been launghed by the IDE or not. IsInDesignMode will return true only for a control that have been istantiated BY the IDE, not if the app is launched by the IDEPutter
Then please clarify that in your question. This is pretty much what I understand by 'the program is currently running in the IDE'.Smutch
@Smutch During design-time your program isn't actually running, is it?Josephinajosephine

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