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QMetaType :: Float not in QVariant :: Type

I have an application that is working fine, but it was not compiled with warnings enabled. I'm trying to get it back and sort it but out of ideas on how to fix it. I have:

QVariant someVar     
QVariant::Type variantType = someVar.type();
switch (variantType) {
    case QMetaType::QString:
        doSomething1();
        break;
    case QMetaType::Float:
        doSomething2();
        break;
}

and got this warning / error:

error: case value ‘135’ not in enumerated type ‘QVariant::Type’ [-Werror=switch]

in the line QMetaType :: Float. I checked the QT docs and listed QMetaType :: Float as value 38. What might be causing this?

The closest thing I can find is at https://github.com/qbittorrent/qBittorrent/issues/2510 which has the same error. Has anyone encountered this before?

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3 answers


Qt played some dirty tricks with the two enumerations ( QMetaType::Type

and QVariant::Type

). Citing 4.8.4 docs on QVariant::type()

:

Returns the storage type of the value stored in the variant. Although declared as a return function QVariant::Type

, the return value must be interpreted as QMetaType::Type

. In particular, QVariant::UserType

it is only returned here if the value is equal or greater QMetaType::User

.

Note that the returned values ​​in the QVariant::Char

thru QVariant::RegExp

and QVariant::Font

thru QVariant::Transform

ranges correspond to the QMetaType::QChar

thru QMetaType::QRegExp

and QMetaType::QFont

thru ranges QMetaType::QQuaternion

.

Also note that the types void*

, long

, short

, unsigned long

, unsigned short

, unsigned char

, float

, QObject*

and QWidget*

are given in QMetaType::Type

but not in QVariant::Type

, and they can be returned by this function. However, they are considered user-defined types when tested against QVariant::Type

.

In other words, the function QVariant::type()

returns the values QMetaType::Type

entered as QVariant::Type

, and the two enums share many (but not all) of their counters. This makes working with them in a strict type system tricky - they're basically wibbly-wobbly typey-wypey stuff.



In your case, note that the enumerator QMetaType::Float

is one that does not have a direct equivalent in QVariant::Type

.

I would say that the best way to silence the warning would be to change variantType

to QMetaType::Type

, perhaps on initialization on include and / or Qt-specific comments, if necessary.

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QVariant::Type

does not contain named entry Float

(see qvariant.h )



+1


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Just turn it on QMetaType::Type

. This is indeed the case because the value QVariant::type()

matches the value QMetaType::Type

, even though the return type matters QVariant::Type

. This is the quirk / bug API you have to work with:

QVariant variant = ...;
switch (static_cast<QMetaType::Type>(variant.type())) {
  ...
}
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