C# mutex in release mode behaves different than in debug mode

Question

So I have the following code:

...
private static void Main(string[] args)
{
    string file=DateTime.Now.ToFileTime().ToString();
    File.AppendAllText(file, "Mutex\r\n");

    bool CreatedNew;
    Mutex mutex=new Mutex(true, AppDomain.CurrentDomain.FriendlyName, out CreatedNew);

    if(CreatedNew)
    {
        #if DEBUG
        File.AppendAllText(file, "Launching in DEBUG mode\r\n");
        #else
        File.AppendAllText(file, "Launching in RELEASE mode\r\n");
        #endif

        //Program.Launch();
        Program.ProcessArgsAndLaunch(args);
    }
    else
    {
        File.AppendAllText(file, "Handling dupe\r\n");
        Program.HandleDuplicate();
    }
}
...

I've checked innumerable articles here and other sites with no luck.

Basically, the code checks for a running instance of the app and if there is one, it switches to the main window of the running one. If not, it launches the app.

In Debug mode it all works as expected, the problem starts when I switch my configuration to Release: the app always starts (with Mutex seemingly doing nothing).

I've added conditionally compiled dumps that show in which mode the app is starting and the output changes based on the configuration, but sadly, so does the behaviour of the app.

This may be a race condition but I am not sure.

More code will be posted if needed.

Thanks.

Jon Skeet · Accepted Answer · 2014-05-07 18:32:53Z

Along with Juan's answer, there's a difference in terms of garbage collection between launching inside and outside a debugger. That's not quite the same as a Debug configuration vs a Release configuration, but it's something you should be aware of anyway.

In the debugger, a local variable will act as a GC root for its entire scope - but when you're not debugging, your mutex variable isn't a GC root at all, because you don't use it after initializing it. That means your Mutex can be garbage collected (and thus the native mutex will be released) pretty much immediately.

You should use a using statement to explicitly dispose the Mutex at the right time:

// Note that you don't need a variable here... you can have one if you
// want though
using (new Mutex(...))
{
    // Code here to be executed while holding the mutex
}

Thanks a lot Jon, this was the issue. As soon as I placed my code into a using statement, it all worked. And you're right, I didn't need the variable...

Juan Lopes · Accepted Answer · 2014-05-07 18:29:06Z

8

If you have one instance running as Debug and another as Release, they will not share the same mutex, because AppDomain.CurrentDomain.FriendlyName is different dependening on whether the hosting process is active or not.

As seen here: http://msdn.microsoft.com/en-us/library/ms242202.aspx

answered May 7, 2014 at 18:29

Juan Lopes

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1 Comment

nurchi Over a year ago

Sorry if this was unclear, I would compile under Debug or Release, then launch the newly compiled app from within Explorer. Then I would close (all) instance(s), clean solution, change configuration, then recompile, then launch (again, from Windows Explorer)...

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