In QT & C++, Covert QMap<QString, QMap<QString, int> > to a single QVariant type.

Question

Apparently QVariant (needed for QSettings class) supports creation from QMap<QString, QVariant>

But trying to initialise something like this:

QMap<QString, QVariant(QMap<QString, QVariant>)> i;

Returns

instantiated from here. <-- points to code above.

function returning a function.

So then i tried the QMap<QString, QVariant> overload for QVariant() and got

error: no matching function for call to 'QVariant::QVariant(QMap<QString, QMap<QString, int> >&)'

Now i tried a typecast QMap<QString, (QVariant)QMap<QString, QVariant> > i;

and got template argument 2 is invalid and invalid type in declaration before ';' token

So whats the required voodoo to convert nested QMaps to a QVariant object?

Answer 1

The error being reported is that QVariant(...) is not a type, but a function (c-tor).

You should have just used: Map<QString, QVariant> i; and used QVariant(QMap<QString, QVariant>) only when assigning values to the map. The point is QVariant is anything really. So a map of QVariants, can have an int in one position (contained in the QVariant) and a QDate in another. So when declaring the type, you can't specify which types you want QVariant to hold.

Answer 2

1. In QMap<QString, QVariant(QMap<QString, QVariant>)>, you have defined a map from a string to a function type. What you really want is a QMap<QString, QVariant>.

2. You don't want a QMap<QString,(QVariant)QMap<QString, QVariant> > because that's just syntactically incorrect. Both template parameters need to be type names, and typecast can't be part of at type name.

3. Putting a QMap<QString, int> (or almost any other type of QMap) into a QVariant won't work. The only QMap type that can be converted into a QVariant is a QMap<QString,QVariant>.

   There's a typedef for this type that may be useful: QVariantMap. If you stick to using QVariantMap for this situation, then things will work properly for you.