Preventing PyQt to silence exceptions occurring in slots

Question

As far as I can see, if an exception occurs in a slot under PyQt, the exception is printed to screen, but not bubbled. This creates a problem in my testing strategy, because if an exception occurs in a slot, I will not see the test fail.

Here is an example:

import sys from PyQt4 import QtGui, QtCore  class Test(QtGui.QPushButton):     def __init__(self, parent=None):         QtGui.QWidget.__init__(self, parent)         self.setText("hello")         self.connect(self, QtCore.SIGNAL("clicked()"), self.buttonClicked)      def buttonClicked(self):         print "clicked"         raise Exception("wow")  app=QtGui.QApplication(sys.argv) t=Test() t.show() try:     app.exec_() except:     print "exiting" 

Note how the exception never quits the program.

Is there a way to work around this problem?

Answer 1

Can create a decorator that wraps PyQt' new signal/slot decorators and provides exception handling for all slots. Can also override QApplication::notify to catch uncaught C++ exceptions.

import sys import traceback import types from functools import wraps from PyQt4 import QtGui, QtCore  def MyPyQtSlot(*args):     if len(args) == 0 or isinstance(args[0], types.FunctionType):         args = []     @QtCore.pyqtSlot(*args)     def slotdecorator(func):         @wraps(func)         def wrapper(*args, **kwargs):             try:                 func(*args)             except:                 print "Uncaught Exception in slot"                 traceback.print_exc()         return wrapper      return slotdecorator  class Test(QtGui.QPushButton):     def __init__(self, parent=None):         QtGui.QWidget.__init__(self, parent)         self.setText("hello")         self.clicked.connect(self.buttonClicked)      @MyPyQtSlot("bool")     def buttonClicked(self, checked):         print "clicked"         raise Exception("wow")  class MyApp(QtGui.QApplication):     def notify(self, obj, event):         isex = False         try:             return QtGui.QApplication.notify(self, obj, event)         except Exception:             isex = True             print "Unexpected Error"             print traceback.format_exception(*sys.exc_info())             return False         finally:             if isex:                 self.quit()  app = MyApp(sys.argv)  t=Test() t.show() try:     app.exec_() except:     print "exiting" 

Answer 2

You could exit the application with a non-zero return code to indicate that an exception has occurred.
You can catch all exception by installing a global exception hook. I added an example below, but you probably will want to adjust it to your needs.

import sys from PyQt4 import QtGui, QtCore  class Test(QtGui.QPushButton):     def __init__(self, parent=None):         QtGui.QWidget.__init__(self, parent)         self.setText("hello")         self.connect(self, QtCore.SIGNAL("clicked()"), self.buttonClicked)      def buttonClicked(self):         print "clicked"         raise Exception("wow")  sys._excepthook = sys.excepthook def exception_hook(exctype, value, traceback):     sys._excepthook(exctype, value, traceback)     sys.exit(1) sys.excepthook = exception_hook  app=QtGui.QApplication(sys.argv) t=Test() t.show() try:     app.exec_() except:     print "exiting"