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Monday, June 25th 2012, 7:58am

QSharedPointer with signals and slots

I have recently started using QSharedPointer for a particular type of object which is shared between many classes. Previously i had just used a raw pointer and managed the creation and deletion of the objects in one place. But now i realise it will be much easier (and actually necessary) to do reference counting.

For some reason, there are very few examples out there on how to use QSharedPointer, so i find myself posting here.

One problem i have ran into is using signals and slots with the objects that are shared-pointed-to. Previously i had done this:

Source code

MyObject* object;   // Subclass of QObject.
AnotherObject* something;
connect(object, SIGNAL(updated()), something, SLOT(update()));  // Notify when my object has changed

But now that i have changed MyObject* to QSharedObject<MyObject> (typedef'd as MyObjectPtr), the compiler complains:


error C2664: 'bool QObject::connect(const QObject *,const char *,const QObject *,const char *,Qt::ConnectionType)' : cannot convert parameter 1 from 'MyObjectPtr' to 'const QObject *'
No user-defined-conversion operator available that can perform this conversion, or the operator cannot be called

I read that i need to use Q_DECLARE_METATYPE and qRegisterMetaType, but i'm not sure about that.
QObject::connect takes a pointer, but here i am passing a QSharedPointer object. Would doing the above allow that type to be passed? Do i still need to do that if the MyObject class is already a subclass of QObject?
And is it safe to just pass the raw pointer obtained from


Monday, June 25th 2012, 8:13am

why dont you wait for a reply first?…gnals-and-slots :rolleyes:
If you have a problem, CUT and PASTE your code. Do not retype or simplify it. Give a COMPLETE and COMPILABLE example of your problem. Otherwise we are all guessing the problem from a fabrication where relevant details are often missing.


Tuesday, June 26th 2012, 12:10am

Sorry, i didn't know which forum would get answered first, if at all.
I have replied to the other one.