怎么做：应用程序的第一个实例将执行一个进程，第二个实例将在以后终止该进程

Question

I want to create a console application that behaves as follows:

1. The first instance of the app will execute a process.
2. The second instance executed later will kill the process.

Is there a simple way to do so?

EDIT: The second instance also terminates the first instance and itself.

EDIT 2:

More details scenario is as follows:

Assume there is no instance of my application running. If I execute my application, the new instance will run. The application will create a process that execute Adobe Acrobat Reader X (for example).

Later, I execute my application again just to kill the running Adobe Acrobat Reader X and of course its host (the first instance of my application).

Answer 1

You need to implement a mutex to do this.

     private static Mutex mutex = null;

     private void CheckIfRunning() {

       string strId = "291B62B2-812A-4a13-A657-BA672DD0C93B";

        bool bCreated;

        try
        {
            mutex = new Mutex(false, strId, out bCreated);
        }
        catch (Exception)
        {
            bCreated = false;
            //Todo: Kill your process
        }

        if (!bCreated)
        {
            MessageBox.Show(Resources.lbliLinkAlreadyOpen, Resources.lblError,         MessageBoxButtons.OK, MessageBoxIcon.Warning);
            return;
        }
      }

Answer 2

You can create a process with a known name. Then when the application starts you could get a list with all the processes that are running. If the process is not there you can start it, if it's already there you can kill the process and exit.

A more elegant solution would be as Max suggested to use a Mutex to communicate between the processes. For example to be sure that you don't kill another process with the same name.