If I have a signal and I register an objects function to the signal will this keep the object live and stop the garbage collection of that object?
E.g.
class Signals():
signal = Qt.pyqtSignal()
def __init__(self):
QObject.__init__(self)
class Test();
def __init__(self, s):
s.connect(self.done)
def done(self):
print("Done")
s = Signals()
t = Test(s.signal)
t = None
s.signal.emit()
Will the Test objecct still get the signal?
No, it won't, it's not enough to keep the object alive. Just try it:
from PyQt4.QtCore import *
app = QCoreApplication([])
class Signals(QObject):
signal = pyqtSignal()
def __init__(self):
QObject.__init__(self)
class Test():
def __init__(self, s):
s.connect(self.done)
def done(self):
print("Done")
s = Signals()
t = Test(s.signal)
print("first")
s.signal.emit()
app.processEvents()
t = None
print("second")
s.signal.emit()
app.processEvents()
Output:
first
Done
second
This behaviour only applies when connecting a signal to a bound method. As an example, if you use:
s.connect(lambda: self.done())
instead, then it will work. If the library wouldn't keep a strong reference here, then you could never connect an anonymous function to a signal. So in this case pyqt has to ensure that it keeps a reference to the function, and the object (self) keeps to be referenced in the closure of the lambda.
