Ensuring only one application instance [duplicate]

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What is the correct way to create a single instance application?

I have a Winforms app, which launches a splash screen via the following code:

Hide(); bool done = false; // Below is a closure which will work with outer variables. ThreadPool.QueueUserWorkItem(x => { using (var splashForm = new SplashScreen()) { splashForm.Show(); while (!done) Application.DoEvents(); splashForm.Close(); } }); Thread.Sleep(3000); done = true;

The above is in the main form's codebehind and called from the load event handler.

However, how can I ensure that only one instance of the application will load at a time? In the load event handler of the main form, I could check if the process list is on the system (via GetProcessesByName(...)), but is there a better method?

Using .NET 3.5.

You should call Application.Run(splashForm) instead of a DoEvents() loop. — SLaks
– SLaks, Commented Jun 26, 2011 at 19:34

michalczerwinski · Accepted Answer · 2011-06-26 19:54:44Z

67

GetProcessesByName is slow way of checking if another instance is running. The fastest and elegant method is using mutex:

[STAThread] static void Main() { bool result; var mutex = new System.Threading.Mutex(true, "UniqueAppId", out result); if (!result) { MessageBox.Show("Another instance is already running."); return; } Application.Run(new Form1()); GC.KeepAlive(mutex); // mutex shouldn't be released - important line }

Please also bear in mind that the code you presented is not the best approach. As it was advised in one of comments calling DoEvents() in a loop is not the best idea.

answered Jun 26, 2011 at 19:54

michalczerwinski

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5 Comments

monstr Over a year ago

GC.KeepAlive(mutex); - doesn't work for me by some reasons. I was forsed to use private static mutex;

Alex B. Over a year ago

I may be wrong, but shouldn't 'GC.KeepAlive(mutex);' line be before 'Application.Run(new Form1());'? 'Application.Run()' method starts the program's message loop and does not return until 'Form1' is closed.

michalczerwinski Over a year ago

@AlexB.: no, this is on purpose. The goal is to prevent mutex from being released, which can occured when a new form is opened and Form1 is closed. Before Form1 is closed there is no risk of mutex being released

bh_earth0 Over a year ago

in if block, putting "application.exit()" instead of "return ", works form me, but i put the mutex check in the form1's form_load .

qakmak Over a year ago

Why some people recommend use mutex.ReleaseMutex(), what's the difference between GC.KeepAlive(mutex) ?

Denis Biondic · Accepted Answer · 2011-06-26 21:19:53Z

static class Program { // Mutex can be made static so that GC doesn't recycle // same effect with GC.KeepAlive(mutex) at the end of main static Mutex mutex = new Mutex(false, "some-unique-id"); [STAThread] static void Main() { // if you like to wait a few seconds in case that the instance is just // shutting down if (!mutex.WaitOne(TimeSpan.FromSeconds(2), false)) { MessageBox.Show("Application already started!", "", MessageBoxButtons.OK); return; } try { Application.EnableVisualStyles(); Application.SetCompatibleTextRenderingDefault(false); Application.Run(new Form1()); } finally { mutex.ReleaseMutex(); } // I find this more explicit } }

One note about the some-unique-id -> this should be unique on the machine, so use something like your company name / application name.

Edit:

http://sanity-free.org/143/csharp_dotnet_single_instance_application.html

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