Finally, why this code only runs on the second line?
Explanation
This is a 32-bit Windows process. It has been multi-threaded and therefore executes instructions sequentially. The best we can do is to start the process on the first line and when it is running, the second thread is created on the second line. Therefore, the answer is that the first thread runs until the process is running on the second thread (i.e. the first thread is waiting for the second thread to run), and then they both exit.
As you can see, the first thread has a running status until the process is running on the second thread. Hence, it is necessary to block the thread from starting until the process is running.
As you can see, my code is very simple. I used process creation flags so that the thread will be created in the system after the process is running (on the second line).
@Justin Hewitt asks in the comments below: "How would the thread get "started" if the process has not been created yet?"
I'm sorry, I'm not sure I understand what you mean by that question. However, if it's a question about the thread status from the point of view of the main thread, I do not need to answer it because the main thread is blocked from starting until the process is running on the second thread.
A:
I got the same issue recently while using getWindow(). I didn't use any flag to create the thread so it was difficult to guess how the GetWindow() method works. But when I used this code instead of GetWindow() then it worked as expected,
Process[] processList = Process.GetProcessesByName("processName");
if (processList.Length > 1)
{
// Here you can specify process id to find the specific process
// that created the window.
Windows.Forms.Process p = processList[0];
Windows.Forms.Window window = Windows.Forms.MessageBox.Show(
"Do you really want to exit?", "Message",
MessageBoxButtons.YesNoCancel, MessageBoxIcon.Warning 01e38acffe
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