[FIXED] Changed Properties do not trigger signal

Issue

For a subclass of QObject I'd like to define a property. On change, a signal should be emitted. According to the documentation, something like this: p = Property(int, _get_p, _set_p, notify=p_changed) should work, but the signal is not emitted by the change. Full example here:

from PySide2.QtCore import QObject, Property, Slot, Signal


class A(QObject):
    p_changed = Signal()

    def __init__(self):
        super(A, self).__init__()
        self._p = 0

    def _get_p(self):
        print(f"Getting p (val: {self._p})")
        return self._p

    def _set_p(self, v):
        print(f"Setting new p: {v}")
        self._p = v
    p = Property(int, _get_p, _set_p, notify=p_changed)


if __name__ == "__main__":
    import sys
    from PySide2.QtWidgets import QApplication, QWidget

    app = QApplication([])


    class W(QWidget):
        def __init__(self, *args):
            super().__init__(*args)
            self.a = A()
            self.a.p_changed.connect(self.test_slot)

            print(self.a.p)
            self.a.p = 42
            # Expecting "Got notified"!
            print(self.a.p)

            # This would print "Got notified":
            # self.a.p_changed.emit()

        @Slot()
        def test_slot(self):
            print("Got notified")


    w = W()
    w.show()
    sys.exit(app.exec_())

Solution

That you associate a signal to a QProperty does not imply that it will be emitted automatically but that you have to emit it explicitly.

def _set_p(self, v):
    if self._p != v:
        print(f"Setting new p: {v}")
        self._p = v
        self.p_changed.emit()

For more information read The Property System.

Answered By - eyllanesc