[Interest] How to create QObject-derived class instance from class name

[Nye]

> For now there is a design choice which I feel weak myself, to do factory
or to use QMetaType for creating instances.

Depends on the use case I suppose. If you know the types at compile time,
as is usually the case I'd go with the simple solution to make a factory. I
recently, however, had a case where the deserialization routine couldn't
know at compile time the exact type (i.e. for a library), so I had to put a
(gadget) base class and use the runtime type ids.

On Wed, Mar 2, 2016 at 9:31 AM, Sina Dogru <sinadooru at gmail.com> wrote:

> I thank you for your answers and solutions.
>
> Before I wrote, I tried to use QMetaType::create function to instantiate
> an instance but since QObject have deleted copy constructor and copy
> assignment operator, I have ended up using QObject pointer with QMetaType..
> And when I see that QMetaType::create() function, it just create a pointer
> to QObject but not the QObject instance.. So I thought QMetaType is not for
> the QObject-derived classes..
>
> But I have not realised that QMetaType::metaObjectForType function returns
> QMetaObject instance for the class not for the pointer to class.
>
> For now there is a design choice which I feel weak myself, to do factory
> or to use QMetaType for creating instances.
>
> Thank you,
> Sina
>
> 2016-03-01 18:51 GMT+02:00 André Somers <andre at familiesomers.nl>:
>
>>
>>
>> Op 01/03/2016 om 17:21 schreef Thiago Macieira:
>>
>>> On terça-feira, 1 de março de 2016 17:06:49 PST André Somers wrote:
>>>
>>>> The meta *object* system has no registration.
>>>>>
>>>>> The meta *type* system requires that the registered type be default-
>>>>> constructible and copyable, but QObject is not copyable. Therefore,
>>>>> QObject- derived classes cannot be registered with the meta type
>>>>> system.
>>>>>
>>>> Am I completely misinterpretting the documentation then?
>>>> http://doc.qt.io/qt-5/qmetatype.html#metaObject
>>>>
>>>> If I read that correctly, you can register a
>>>> pointer-to-a-QObject-derived-class-instance and use that. So, indeed,
>>>> you do not register the type, but the type*. And that has no problems
>>>> with being default constructed or copied.
>>>>
>>> Correct. You can't register a QObject class with the meta type system,
>>> but you
>>> can register a pointer to a QObject class. The problem is that
>>> QMetaType::create() will then create a pointer, not the object.
>>>
>> Of course. But... Again, if I read it correctly, you *can* then get the
>> QMetaObject from QMetaType, and using that, you can create an actual
>> instance.
>>
>> Just tested this trivial example:
>>
>> //main.cpp
>> int main(int argc, char *argv[])
>> {
>>     qRegisterMetaType<TestClass*>(); //this could be elsewhere of course
>>
>>     QCoreApplication a(argc, argv);
>>
>>     auto tId = QMetaType::type("TestClass*"); //just using the class name
>> with an *
>>     auto metaObject = QMetaType::metaObjectForType(tId);
>>     QObject* instance = metaObject->newInstance();
>>
>>     return a.exec();
>>     delete instance;
>> }
>>
>> //testclass.h
>> class TestClass: public QObject
>> {
>>     Q_OBJECT
>>
>> public:
>>     Q_INVOKABLE TestClass();
>> };
>>
>> Q_DECLARE_METATYPE(TestClass*)
>>
>> //testclass.cpp
>> TestClass::TestClass()
>> {
>>     qDebug() << "TestClass instance created";
>> }
>>
>> Which prints out "TestClass instance created" on the console. My
>> conclusion is that it works, and that you _can_ create a QObject derived
>> instance with just the class name.
>>
>> Again: I am not claiming that abusing this to skip defining a factory is
>> a good idea. Just that it is possible.
>>
>>
>> André
>>
>>
>> _______________________________________________
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>>
>
>
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