PyQT4 signal.connect是否保持对象存活？ [英] Does PyQT4 signal.connect keep objects live?

问题描述

如果我有一个信号，并且我注册一个对象函数到信号中，这会使对象保持活动并停止对象的垃圾回收？

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?

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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()

测试对象仍会得到信号吗？

Will the Test objecct still get the signal?

推荐答案

不，它不会让物体保持活力。试试看：

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()

输出：

first
Done
second

此行为仅适用于将信号连接到绑定方法。例如，如果您使用：

This behaviour only applies when connecting a signal to a bound method. As an example, if you use:

s.connect(lambda: self.done())

来代替，那么它将起作用。如果库不会在这里保留一个强有力的参考，那么你永远不能将一个匿名函数连接到一个信号。所以在这种情况下，pyqt必须确保它保持对函数的引用，并且在lambda的闭包中保持引用对象（ self ）

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.

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