Example of where using pyqtSignal.connect as a decorator will NOT work

signal_connection_decorator_fail.py

from PyQt4.QtCore import pyqtSignal, QObject
from PyQt4.QtGui import QApplication, QFrame, QGridLayout, QPushButton


class View(QFrame):
    openStuff = pyqtSignal()

    def __init__(self, parent=None):
        super(View, self).__init__(parent)

        self._button = QPushButton('Open')

        # Notify outside world (controller)
        self._button.clicked.connect(self.openStuff.emit)

        layout = QGridLayout()
        layout.addWidget(self._button)
        self.setLayout(layout)


class Controller(QObject):
    def __init__(self, view):
        super(Controller, self).__init__()

        self._view = view
        self._view.openStuff.connect(self.clicked)

    # FIXME: Cannot write a decorator like this, no way to have a reference to
    # `self` because when this is evaluated `self` doesn't fully exist yet.
    # Remember, decorators are evaluated at definition time!
    #@self._view.openStuff.connect
    def clicked(self):
        print 'Button clicked'


if __name__ == "__main__":
    import sys
    app = QApplication(sys.argv)

    view = View()
    controller = Controller(view)

    view.show()
    sys.exit(app.exec_())