Welcome to ShenZhenJia Knowledge Sharing Community for programmer and developer-Open, Learning and Share
menu search
person
Welcome To Ask or Share your Answers For Others

Categories

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.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
thumb_up_alt 0 like thumb_down_alt 0 dislike
94 views
Welcome To Ask or Share your Answers For Others

1 Answer

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;
}

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
thumb_up_alt 0 like thumb_down_alt 0 dislike
Welcome to ShenZhenJia Knowledge Sharing Community for programmer and developer-Open, Learning and Share
...