具有未知列表模板参数的QVariant到QList(QVariant to QList with unkown list template parameter)

我有几个QObject类，它们可能有QList属性，它们存储QObject派生类的列表：

class Object : public QObject
{
    Q_OBJECT
    Q_PROPERTY(QList<AnotherQObject*> otherObjects READ otherObjects)
public:
    QList<AnotherQObject*> otherObjects() ct;
}
 
 QList<AnotherObject*>当然已在Q_DECLARE_METATYPE和qRegisterMetaType()注册。 
 
 现在我想用Qt的元对象系统读取这个属性。 在我想要读取列表的那一点上，我只关心（并且知道）列表包含指向对象的指针这一事实，这些对象是从QObject派生的，但不是精确的类型。 
 
 当我将属性的QVariant转换为QList<QObject *> ，列表为空。 
 
QObject *object = getObjectFromSomeWhere();
ct char* propertyame = getPropertyameFromSomeWhere(); // "otherObjects"

QVariant list = object->property(name);
QList<QObject *> objectList = list.value<QList<QObject *> >(); // empty

QList<AnotherQObject*> anotherObjectList = 
    static_cast<Object*>(object)->otherObjects() // not empty
 
 是否可以访问此列表？ 你有其他建议吗？ 
 
 编辑：虽然我发布了我的解决方案，我仍然喜欢评论:-) 
I have several QObject classes, which may have QList properties, which store lists of QObject derived classes:
 
class Object : public QObject
{
    Q_OBJECT
    Q_PROPERTY(QList<AnotherQObject*> otherObjects READ otherObjects)
public:
    QList<AnotherQObject*> otherObjects() ct;
}
 
QList<AnotherObject*> has of course been registered with Q_DECLARE_METATYPE and qRegisterMetaType().
 
ow I want to read this property with Qt's meta object system. At the point where I want to read the list, I only care about (and know) the fact that the list contains pointers to objects, which have been derived from QObject, but not the exact type.
 
When I convert the property's QVariant to QList<QObject *>the list is empty.
 
QObject *object = getObjectFromSomeWhere();
ct char* propertyame = getPropertyameFromSomeWhere(); // "otherObjects"

QVariant list = object->property(name);
QList<QObject *> objectList = list.value<QList<QObject *> >(); // empty

QList<AnotherQObject*> anotherObjectList = 
    static_cast<Object*>(object)->otherObjects() // not empty
 
Is it possible to access this list? Do you have other suggesti?
 
EDIT: Although I posted my solution, I would still love comments :-)
                
最满意答案
                
                    
                        
 如果您使用的是Qt 5.2或更高版本，则可以使用QSequentialIterable类，而无需额外模板，修改的宏和其他注册的开销。 
 
 http:///doc/qt-5/qsequentialiterable.html 
 
 如果QList<T*>存储在QVariant ，其中T是从QObject派生的，则可以从变量中获取QSequentialIterable实例，这允许您将列表中的每个项目作为QVariant进行迭代。 然后你可以简单地将每个项的值作为QObject* ，然后添加到QList<QObject*> （确保你测试了类型是从QObject派生的假设）： 
 
QList<QObject*> qobjectList;
if ((QMetaType::QVariantList)) {
    QSequentialIterable iterable = variant.value<QSequentialIterable>();
    foreach (ct QVariant& item, iterable) {
        QObject* object = item.value<QObject*>();
        if (object) { qobjectList.append(object); }
    }
}
If you're using Qt 5.2 or higher there is a QSequentialIterable class that allows you to do this without the overhead of extra templates, modified macros, and additional registrati.
 
http:///doc/qt-5/qsequentialiterable.html
 
If you have a QList<T*> stored in a QVariant, where T is derived from QObject, you can get a QSequentialIterable instance from the variant, which allows you to iterate over each item in the list as a QVariant. Then you can simply get the value of each item as a QObject*, adding to a QList<QObject*> as you go (make sure you test the assumption that the type is derived from QObject, though):
 
QList<QObject*> qobjectList;
if ((QMetaType::QVariantList)) {
    QSequentialIterable iterable = variant.value<QSequentialIterable>();
    foreach (ct QVariant& item, iterable) {
        QObject* object = item.value<QObject*>();
        if (object) { qobjectList.append(object); }
    }
}