Does PyQT4 signal.connect keep objects live?

问题

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?

回答1:

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.

来源：https://stackoverflow.com/questions/17211797/does-pyqt4-signal-connect-keep-objects-live