> If you have a signal in your model:
>
> QSignal0 modelChanged
>
> and you connect
>
> modelChanged.connect(view, "update()");
>
> then when you do:
>
> // update model in model thread
> modelChanged.emit();
>
> update() will be called on the view in the GUI thread. This is a queued
> connection. Actually it is an auto connection which decides to do the
> operation queued at run time because the emitter and receiver are in
> different threads.
>
> The connect statement can be performed from any thread.
OK, I think I understand this now. :) But it's still not working.
Firstly, is there a QSignal0 class? I've used QSignalEmitter.Signal0. Is
that what you meant, or have I missed something else? :)
More importantly, I've used something similar to the code above, and
still get an exception. Inside my view class, which extends QWidget,
I've created a field called updateSignal which is a Signal0. From inside
my view class constructor I make the queued connection explicitly like this:
this.updateSignal.connect(this, "update()",
Qt.ConnectionType.QueuedConnection);
and from the thread started by the model's Timer, I do the emit like this:
this.updateSignal.emit();
When I do the emit, I get a QThreadAffinityException with a message of
"QObject used from outside its own thread". The exception message says
the view object is the offender. I guess I can only get the view to do
the emit from within the main thread?
I've yet to try the other methods. Do you have any further advice on how
I can fix this one first?
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