am using the following piece of code to ensure that my application only runs once, however I have a few questions about it.
static Mutex m_Mutex; << in c# I assume that when the methods are static, so are the private members.
public static void Run(Form mainForm)
{
if(IsFirstInstance())
{
Application.ApplicationExi
Application.Run(mainForm);
}
}
The following is causing me some confusion, when the application starts it creates a newly named Mutex. It then calls WaitOne and determines whether it receives any messages, if it does it returns true otherwise false. My questions are this, when the next application comes to create a new mutex with the same name, what exactly happens? Does it deny it therefore this is why the WaitOne does not receive any messages, also the timespan is set to zero, how can the mutex determine if any messages have been recieved in zero time?
static bool IsFirstInstance()
{
m_Mutex = new Mutex(false,"SingletonApp Mutext");
bool owned = false;
owned = m_Mutex.WaitOne(TimeSpan.Z
return owned ;
}
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I don't understand your question. The mutex has to remain allocated for the duration of the application, otherwise a second instance would not see the mutex as being in use. When the process is closed, the mutex will be released automatically. Therefore, while it may be good practice to release the mutex upon application exit, you must no do this earlier or the second instance detection will fail.
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by: DespPosted on 2004-10-14 at 01:25:54ID: 12305979
I have given a solution to a similar problem here:
http://www.experts-exchang
unfortunately i cant test your code as i dont have VS.Net right now, you can find TAD's comment very useful for using the singleton way to avoid the multiple instanced of the same application and yes private members can also be static