How to Run Only One Instance of Application

Question

I have a application which uses socket connection to send and receive data from another application. While creating socket it uses the port 4998 .

That is where my problem lie. Once I start my application the socket starts using port 4998. So if I want to execute the application again then I get socket binding error.

So I want to limit my application instance to one. That means if the application is already running and some one tries to run the application again by clicking the exe or shortcut icon it shouldn't run the program, instead it should bring the existing application to the Top.

Answer 1

You may used named mutex.

Code sample from the article:

WINAPI WinMain(
  HINSTANCE, HINSTANCE, LPSTR, int)
{
  try {
    // Try to open the mutex.
    HANDLE hMutex = OpenMutex(
      MUTEX_ALL_ACCESS, 0, "MyApp1.0");

    if (!hMutex)
      // Mutex doesn’t exist. This is
      // the first instance so create
      // the mutex.
      hMutex = 
        CreateMutex(0, 0, "MyApp1.0");
    else
      // The mutex exists so this is the
      // the second instance so return.
      return 0;

    Application->Initialize();
    Application->CreateForm(
      __classid(TForm1), &Form1);
    Application->Run();

    // The app is closing so release
    // the mutex.
    ReleaseMutex(hMutex);
  }
  catch (Exception &exception) {
    Application->
      ShowException(&exception);
  }
  return 0;
}

Answer 2

Create named event on the start and check the result. Close application if the event is already exist.

BOOL CheckOneInstance()
{
    m_hStartEvent = CreateEventW( NULL, TRUE, FALSE, L"EVENT_NAME_HERE" );
    if ( GetLastError() == ERROR_ALREADY_EXISTS ) {
        CloseHandle( m_hStartEvent ); 
        m_hStartEvent = NULL;
        // already exist
        // send message from here to existing copy of the application
        return FALSE;
    }
    // the only instance, start in a usual way
    return TRUE;
}

Close m_hStartEvent on the app exit.

Answer 3

When your application initializes, create a mutex. If it already exists, find the existing application and bring it to the foreground. If the application has a fixed title for its main window, it is easy to find with FindWindow.

m_singleInstanceMutex = CreateMutex(NULL, TRUE, L"Some unique string for your app");
if (m_singleInstanceMutex == NULL || GetLastError() == ERROR_ALREADY_EXISTS) {
    HWND existingApp = FindWindow(0, L"Your app's window title");
    if (existingApp) SetForegroundWindow(existingApp);
    return FALSE; // Exit the app. For MFC, return false from InitInstance.
}

Answer 4

/* I have found the necessary editing to be done. Added some extra code and edits that are needed. The present one is working perfectly for me. Thank you, Kirill V. Lyadvinsky and Remy Lebeau for the help!!

*/

bool CheckOneInstance()
{

    HANDLE  m_hStartEvent = CreateEventW( NULL, FALSE, FALSE, L"Global\\CSAPP" );

    if(m_hStartEvent == NULL)
    {
    CloseHandle( m_hStartEvent ); 
        return false;
    }


    if ( GetLastError() == ERROR_ALREADY_EXISTS ) {

        CloseHandle( m_hStartEvent ); 
        m_hStartEvent = NULL;
        // already exist
        // send message from here to existing copy of the application
        return false;
    }
    // the only instance, start in a usual way
    return true;
}

/* The above code works even when one tries to open up second instance FROM A DIFFERENT LOGIN LEAVING THE FIRST LOGIN OPEN with ITS INSTANCE RUNNING. */