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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