[PySide] QThread crash before emitting finished signal
Sebastian Elsner
sebastian at risefx.com
Fri Jun 8 08:13:29 CEST 2012
Strange, actually you would just need to remember the the photoworker,
because it automatically keeps a reference to the thred, which is
available via the QObject.thread() member. If this does not work for you
just make a list of tuples [(worker1,thread1),(worker2,thread2,...]. Ive
used this multiple times before and I had no problems. you could also
google for Qhtread queue, which sould give you advice on how to build a
real queue of threads (I assume you would only have like 10 workers
running, else it could blow your system with load)
You dont have to call quit(). Its more for killing a thread forcibly.
You should only do this in special cases. connect to the finished signal
to see when its done and create a member variable like self.cancel which
you can set to true from outside the thread and have the thread check
regularily in the workerloop if the user wants to cancel.
Am 08.06.2012 02:28, schrieb William Dias:
> Thanks, Sebastian.
>
> I did what you've said. The only thing is that I must keep the
> references to both photoWorker and photoThread, otherwise it won't
> work. I thought about keeping the reference to the thread object
> inside the photoWorker and keep a list of photoWorkers in the main
> thread. Is it a good approach?
>
> Should I call thread.quit() to finish the thread execution?
>
> Regards,
>
> 2012/6/7 Sebastian Elsner <sebastian at risefx.com
> <mailto:sebastian at risefx.com>>
>
> Hi,
>
> 1) I'd add an __init__ like this.
>
> def __init__(self,parent=None):
> QObject.__init__(self,parent)
>
> 2) I'd move the work method stuff to the init. It looks like you
> are doing the setup there. init is the place to be for this.
>
> 3) QMetaObject.invokeMethod(photoWorker, "work",
> Qt.QueuedConnection) looks very strange
> use:
> photoThread.started.connect(photoWorker.doWork)
> instead. place before photoThread.start().
>
>>
>> Apart from that, my threads are being destroyed as photoWorker
>> and photoThread objects run out of scope. How can I prevent
>> python from doing this? Making them instance objects is not an
>> option since every time my program reenters the method where they
>> are declared it spawns new threads.
>
> Make a list and add the photoWorker instances to the list. Clean
> the list up now and then.
>
>
>>
>> Thanks,
>>
>>
>> 2012/6/7 William Dias <william.dias at gmail.com
>> <mailto:william.dias at gmail.com>>
>>
>> Thank you, Sebastian. I followed the procedures described in
>> the document and it worked. But I have another question.
>> Take a look at the code bellow:
>>
>> class PhotoWorker(QObject):
>>
>> photoStatus = Signal(str, str)
>>
>> @Slot()
>> def work(self):
>> self.db = sqlite_database.SQLiteDatabase("db1")
>> self.timer = QTimer()
>> self.timer.timeout.connect(self.__resendPhoto)
>> logger.info <http://logger.info>("photo worker started")
>> self.timer.start(10000)
>>
>> @Slot()
>> def __resendPhoto(self):
>> data = self.db.selectPhoto()
>> if data != None:
>> self.filename = data[0]
>> self.tags = data[1]
>> logger.info <http://logger.info>("trying to resend photo " +
>> self.filename)
>> status = sendPhoto(self.tags, self.filename)
>> self.photoStatus.emit(self.filename, status)
>>
>> And in a GUI class, I do the following:
>>
>> photoWorker = http_connection.PhotoWorker(self)
>> photoWorker.photoStatus.connect(self.photoStatus)
>> photoThread = QThread()
>> photoWorker.moveToThread(photoThread)
>> photoThread.start()
>> QMetaObject.invokeMethod(photoWorker, "work",
>> Qt.QueuedConnection)
>>
>> I realized that when I create the photoWorker object, if I
>> don't set its parent to self, my thread dies. Is this because
>> my photoWorker doesn't have its own event loop? I am bit
>> confused here. How should I proceed?
>>
>> Thank you.
>>
>> 2012/6/7 Sebastian Elner <sebastian at risefx.com
>> <mailto:sebastian at risefx.com>>
>>
>> Although it is in the documentation, QThread is not meant
>> to be used this way (subclassing). Have a look at this:
>> http://labs.qt.nokia.com/2010/06/17/youre-doing-it-wrong/
>> It helped me a lot. Also a complete stripped down example
>> would greatly help in a case of a segfault.
>>
>>
>>
>> On 06/07/2012 09:59 AM, William Dias wrote:
>>> Hi,
>>>
>>> I have a subclass of QThread which uploads pictures to a
>>> webserver. For each new picture created in the main
>>> thread, the program spawns a new thread to handle the
>>> job. I've override the __init__ method to be able to
>>> pass some parameters to the thread object, as you can
>>> see bellow.
>>>
>>> class PhotoConnection(QThread):
>>> photoStatus = Signal(str, str)
>>>
>>> def __init__(self, tags, filename, parent=None):
>>> QThread.__init__(self, parent)
>>> self.tags = tags
>>> self.filename = filename
>>>
>>> def run(self):
>>> status = sendPhoto(self.tags, self.filename)
>>> self.photoStatus.emit(self.filename, status)
>>>
>>>
>>> In the GUI thread, the method responsible dor creating
>>> the threads is the following:
>>>
>>> def sendPhoto(self):
>>> photoConnection =
>>> PhotoConnection(self.tags_to_string, self.filename,
>>> self)
>>> photoConnection.photoStatus.connect(self.photoStatus)
>>> photoConnection.finished.connect(self.threadFinished)
>>> photoConnection.start()
>>>
>>>
>>> In the PhotoConnection, if I don't set the parent of my
>>> Qthread object to self (for example, if I set to None)
>>> or if I don't override the __init__ method and create
>>> another method to pass the arguments, when the thread
>>> finishes the execution I got a segfault.
>>> Why is this happening? Can you help me?
>>>
>>> Thanks,
>>>
>>>
>>> _______________________________________________
>>> PySide mailing list
>>> PySide at qt-project.org <mailto:PySide at qt-project.org>
>>> http://lists.qt-project.org/mailman/listinfo/pyside
>>
>>
>> --
>> Sebastian Elsner - pipeline td - r i s e | fx
>>
>> t:+49 30 20180300 <tel:%2B49%2030%2020180300> sebastian at risefx.com <mailto:sebastian at risefx.com>
>> www.risefx.com <http://www.risefx.com>
>>
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>>
>>
>> --
>> William Marques Dias
>>
>>
>>
>>
>> --
>> William Dias
>
>
>
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> --
> William Dias
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