[PySide] QObject.destroyed() is not emitted
Christian Tismer
tismer at stackless.com
Thu Nov 29 19:57:02 CET 2012
Hi,
On 11/29/12 6:37 PM, Stephan Deibel wrote:
> Alexey Vihorev wrote:
>> Thanks, nice find, but... I hit the next hurdle trying to go this
>> way. The signal QObject.destroyed(obj) is passing no arguments
>> (probably because obj is already destroyed), so my static method has
>> nothing to work on. Which kind of devaluates the whole idea, IMHO.
>> And in PyQt4 it*does* pass the object. Even more: in PyQt4 there is
>> no need for static method approach, as it works perfectly with
>> instance methods:
>
> Yea, you would have to bind the necessary data to the callback like this:
>
> def on_destroy(val1=self.whatever, val2=self.something):
> print 'destroyed'
> self.destroyed.connect(on_destroy)
Here is a not really nice hack, but it works, based on Stefan's
bug tracker code:
The idea is to simply capture the __dict__ of the object and to abuse it
as the dict of a helper class. This way, although the object goes away,
it __dict__ survives, and we can work on it after deletion.
from PySide.QtCore import QAbstractTableModel
class MyModel(QAbstractTableModel):
def __init__(self, *args):
super(MyModel, self).__init__(*args)
# This one does not work because the instance is destroyed already
# so the method call does not happen
self.destroyed.connect(self.onDestroy)
# This does work because the instance doesn't have to exist anymore
def on_destroy():
print "destroyed signal: function"
self.destroyed.connect(on_destroy)
self.important = 42
def onDestroy(self):
print('destroyed signal: method')
def do_something(self):
self.important += 1
m = MyModel()
class Helper(object):
def __init__(self, inst):
self.__dict__ = inst.__dict__
h = Helper(m)
m.do_something()
m.important
del m
print(h.important)
print("Done")
----------------------- output:
>>> m.do_something()
>>> m.important
43
>>>
>>> del m
destroyed signal: function
>>> print(h.important)
43
>>>
>>> print("Done")
Done
--
Christian Tismer :^) <mailto:tismer at stackless.com>
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