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Home/ Questions/Q 926093
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T19:37:40+00:00 2026-05-15T19:37:40+00:00

QVariant (needed for QSettings class) supports creation from QMap<QString, QVariant> But trying to initialise

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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;

Gives the error:

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
invalid type in declaration before ‘;‘ token

So what’s the required voodoo to convert a nested QMap to a QVariant object?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-15T19:37:41+00:00Added an answer on May 15, 2026 at 7:37 pm

    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.

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