Converting QList to QVariant

Question

The class contains this:

Q_PROPERTY(QList<double> switch1 READ switch1 WRITE setSwitch1 NOTIFY switch1Changed)

void setSwitch2(QList<double> arg)
{
    if (m_switch2 != arg)
    {
        m_switch2 = arg;
        emit switch2Changed(arg);
    }
}

The below works:

setSwitch2(QList<double>::fromVector(QVector<double>::fromStdVector(data->switch2)));

but now my datatype is QVariantList instead of QList<double>.

How should I replace QList with QVariant now?

This doesn't work:

setSwitch1(QVariantList::fromVector(QVector<QVariant>::fromStdVector(data->switch1)));

The body of you question does not make sense after reading the title. It does not even have a QVariant anywhere. — László Papp, Jan 06 '14 at 06:19
It does have a QVariantList. Actually I want the C++ array to be accessible from QML, hence I wanted to replace QList with QVariant or QVariantList (whichever of them works) @LaszloPapp — Aquarius_Girl, Jan 06 '14 at 06:22

score 2 · Accepted Answer · edited Jun 20 '20 at 09:12

2

Just use this constructor:

QVariant::QVariant(const QList & val)

Constructs a new variant with a list value, val.

I.e. when storing a QList as a QVariant, the template type of the QList has to be a type that is OK for QVariant. There is no other constructor or conversion method for this.

You should be writing this:

QVariant variant(
    QList<double>::fromVector(QVector<double>::fromStdVector(data->switch2))
);

setSwitchVariant(variant);

edited Jun 20 '20 at 09:12

Community

1
1

answered Jan 06 '14 at 06:07

László Papp

51,870
39
111
135

@user462608: your code does not show the whole context, but this is the only way of the conversion. I am not sure what you mean by "what should I replace with what?". – László Papp Jan 06 '14 at 06:10
1

Due to QList being a class template it is just not possible to have a generic constructor (as proposed in the accepted answer) for QVariant from a QList. Constructing such QVariant can be achieved with QVariant::fromValue(). Although the result will be a QVariant containing a QList. – StefanQ Apr 06 '18 at 11:23
In addition to what StefanQ wrote, there is in fact no `QVariant(const QList &)` constructor (that's not even valid syntax since QList needs a type). There is one specifically for a `QList` but that is not the same thing. Qt cannot determine that the `double` in this case can be converted to QVariant. So the suggested code throws a compile error (at least with Qt 5.9+): "cannot convert argument 1 from 'QList' to 'QVariant::Type'". – Maxim Paperno Oct 01 '19 at 03:49
PS. Further details about this/related issue: https://stackoverflow.com/questions/58158034/qlist-in-qvariant-qvarianttype-returns-weird-type/58178040#58178040 – Maxim Paperno Oct 01 '19 at 04:52

score 1 · Answer 2 · answered Sep 19 '14 at 13:33

1

If no need constructor:

QList<QUrl> params;
QVariant varParams;
varParams.setValue<QList<QUrl>>( params );

answered Sep 19 '14 at 13:33

ZolotovPavel

138
6

Serge · Answer 3 · 2019-09-02T07:08:41.200

1

I understand that it is not relevant for the author. But maybe someone will come in handy.

qRegisterMetaTypeStreamOperators<QList<double>>("Stuff");
QList<double> lst;
// convert
QVariant varLst = QVariant::fromValue(lst);
// back
lst = varLst.value<QList<double>>();

edited Sep 02 '19 at 07:08

answered Sep 02 '19 at 05:45

Serge

41
4

Andreas Hagmüller · Answer 4 · 2020-11-16T22:44:46.370

0

There is the QVariant constructor for lists: QVariant(const QList &list) and the typedef QVariantList. Both togheter in a nice template:

template <class T> static QVariant toVariant(const QList<T> &list)
{
    QVariantList variantList;
    variantList.reserve(list.size());
    for (const auto& v : list)
    {
        variantList.append(v);
    }
    return variantList;
}

edited Nov 16 '20 at 22:44

answered Apr 27 '20 at 22:30

Andreas Hagmüller

1
1

Converting QList to QVariant

4 Answers4