title: Why does PyQt crashes without information? (exit code 0xC0000409)
body: <p>I'm trying to develop a software with PyQt, but I often get stuck on software crashes without debug information (only the exit code 0xC0000409). I'm using QThread, and I wrote a system like this:</p>
  
  <pre><code>class serialThreadC(QThread):
      updateOutBox = QtCore.pyqtSignal(str)
      updateStatus = QtCore.pyqtSignal(int)
  
      def __init__(self):
          super(serialThreadC, self).__init__()
          self.ser = False
          self.state = 0
          self.serialEnabled = False
  
      def run(self):
          while True:
              if self.state == -3 or self.state == -2:
                  if self.SerialEnabled:
                      self.updatePB(20)
              elif self.state == 0:
                  if self.serialEnabled:
                      self.updatePB(20)
  
      def ConnDisconn(self):
          self.serialEnabled = not self.serialEnabled
  
      def updatePB(self, stat):
          self.state = stat
          self.updateStatus.emit(self.state)
  
  serialThread = serialThreadC()
  serialThread.start()
  
  ## sw is a QDialog already loaded
  serialThread.updateOutBox.connect(sw.updateOutBox)
  serialThread.updateStatus.connect(sw.updateStatus)
  
  sw.PB_ConnDisconn.clicked.connect(serialThread.ConnDisconn)
  </code></pre>
  
  <p>I have crashes when I read/write <code>serialEnabled</code> in <code>run()</code> or in <code>ConnDisconn()</code>. I know that PyQt is not thread-safe and that a wrong handling of variables gives crashes of my type, but I can't understand what is wrong with my code. My idea (maybe wrong) is that all <code>serialThread</code> methods are executed on the same thread, also if they are connected to a gui (main thread). Is that wrong? In the same way, I emit events from <code>serialThread</code> and I connected them to the GUI, but that never gave me problems.</p>
  
  <p>Can you see the mistake I made? Is there a way to debug the code if there is a crash without other infos? (I use PyCharm 2017.1.3).</p>