[Qt-interest] QMetaMethod invocation without compile-time knowledge of the return type

[Thiago Macieira]

Em Quinta-feira 24. Setembro 2009, às 00.15.40, André Fillipe de Oliveira e 
Silva escreveu:
> Introspector intro = getAnIntrospectorForMyObjectGraph();
> Author *author =
> intro.readProperty("rootObject.anotherObject.myObject.author[2]")
>     .value<Author*>();
> 
> and it will navigate from rootObject to myObject and return it's second
> author. All the path until "myObject" can be resolved using
> QObject::property(), but to resolve "myObject.author[2]" I will need to
> call, say, Author* MyObject::authorAt(int) using a QMetaMethod, to obtain
> the i-th author in the list of authors. The introspector, however, knows
> nothing about the Author class. It deals only with QObjects (Author, of
> course, extends QObject and Author* was registered with the Qt Metatype
> System).

Hi André

Actually, your introspector knows about QVariants too. That's your clue.

> My problem is: QMetaMethod::invoke has a QGenericReturnArgument parameter.
> QGenericReturnArgument is a template class, and one of it's parameters is
> the type of the return value. Therefore, I must know the type of the return
> value at compile time to properly construct a QGenericReturnArgument.

Hmm... you're confusing things.

QReturnArgument is a template class. QGenericReturnArgument isn't.

class Q_CORE_EXPORT QGenericReturnArgument: public QGenericArgument
{
public:
    inline QGenericReturnArgument(const char *aName = 0, void *aData = 0)
        : QGenericArgument(aName, aData)
        {}
};

So you can do something like:

QVariant aux(qMetaTypeId<Author *>(), 0);
QGenericReturnArgument returnArg("Author*", aux.data());

QMetaObject::invokeMethod(obj, "authorAt", returnArg, Q_ARG(int, 2));

> But that only works for a direct access. One might want to chain indexed
> accesses (e.g. "myObject.author[2].book[3]") and the return types of the
> intermediate acessess must somehow be specified. An option:

Chained access will be very difficult. Once you make a call, you get an opaque 
QVariant.

If all your return types are pointers to classes deriving primarily and non-
virtually from QObject, you can try something like this then:

QObject *pointer;
QGenericReturnArgument returnArg("Author*", &pointer);
QMetaObject::invokeMethod(obj, "authorAt", returnArg, Q_ARG(int, 2));

This is *technically* equivalent to the C++ code:

union { QObject *obj; Author *auth; } ptr;
ptr.auth = obj->authorAt(2);
use(ptr.obj);

It's not technically equivalent to:
QObject *obj2 = obj->authorAt(2);
use(obj2);

Please note the slight but important difference. When you assign an Author* to 
a QObject*, the compiler will generate the necessary pointer adjustments so 
that the QObject pointer points to the QObject sub-object inside the Author 
object. If you use the union, then the pointers will have the same value, but 
will not necessarily be correct.

I'm pointing this out to note that it isn't a generic solution. However, in 
the case of classes that have a Q_OBJECT tag, moc will complain if the QObject 
sub-object is not at offset zero. In other words, if you use Q_OBJECT, then the 
trick above is safe.

> Why is it that QMetaMethod::invoke cannot accept a (non-const reference to
> a) QVariant and just proceed?

It can :-)

> Why is it necessary to provide the type name at compile time?

It's not.

It's not even necessary to provide the type name at all, if you call 
qt_metacall directly. But please don't do that because in Qt 4.6 the semantics 
of the meta call have changed slightly:

QVariant outArg; // initialised to QVariant(2) somewhere else
QObject *returnArg;
void *array[] = { &returnArg, const_cast<void*>(outArg.constData()) };

obj->qt_metacall(QMetaObject::InvokeMetaMethod, methodId, array);
use(returnArg);

See also:
http://qt.gitorious.org/qt/qt/blobs/master/src/dbus/qdbusintegrator.cpp#line808
and lines 869, 890

-- 
Thiago Macieira - thiago (AT) macieira.info - thiago (AT) kde.org
  Senior Product Manager - Nokia, Qt Development Frameworks
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