QList in QVariant, QVariant::type() returns weird

2020-07-22 16:21发布

I'm trying to store QList<int> in QVariant and then do a typecheck to determine exact type of value stored in QVariant (for serialization). QVariant::type() works fine with scalar types like int or QString, but this is what I got with QList<int>:

QList<int> list{1,2,3};
QVariant v = QVariant::fromValue<QList<int>>(list);
qDebug() << v.type() << "|" << v.typeName() << "|" << v.userType() << "|" << QVariant::typeToName(v.type());

Produces:

QVariant::QWidget* | QList<int> | 1030 | QWidget*

Where this QVariant::QWidget* even came from? I can't find it in QVariant's enum Type, nor value 1030. Q_DECLARE_METATYPE(QList<int>); has no effect.

What's wrong here? Or should I use typeName() since it returns correct type name instead of type()?

Qt 5.13.1, clang 8.0.1, Linux 64-bit

标签: c++ qt qvariant
1条回答
趁早两清
2楼-- · 2020-07-22 17:03

QList<T> isn't really a pre-defined variant type (except QList<QVariant>) so it technically resolves to a QMetaType::UserType. It's a bit convoluted to follow, and definitely depends on which Qt modules have been included (the meta type system is designed to be expandable this way -- eg. w/out GUI module you wouldn't have any of those types available).

So... it's actually QMetaType that "knows" about the underlying custom types. From given example QVariant::typeToName(v.userType()) prints QList<int> correctly. I've found that QVariant::userType() always returns the actual meta type, whereas just type() seems fairly useless (even the docs are convoluted trying to explain how it works).

To add to the confusion, QVariant::Type seems to be deprecated... if you click the "QVariant::Type" link in the docs for type() it goes to the "obsolete members" section. Where that enum isn't even listed.

Actually QMetaType doesn't "know" the QList<int> type either until it is used (or registered). Example:

    qDebug() << QMetaType::type("QList<int>");  // prints 0
    QList<int> list{1,2,3};
    QVariant v = QVariant::fromValue(list);
    qDebug() << QMetaType::type("QList<int>");  // prints 1028 on my test project, yours may vary

Or, if registered specifically:

    qDebug() << qRegisterMetaType<QList<int> >("QList<int>");  // prints 1028
    qDebug() << QMetaType::type("QList<int>");  // prints 1028
    QList<int> list{1,2,3};
    QVariant v = QVariant::fromValue(list);
    qDebug() << QMetaType::type("QList<int>");  // still 1028
查看更多
登录 后发表回答